Talk:Monty Hall problem/Arguments/Archive 10

Continued from the talk page
Agreed. Now Martin says "it has to do with the stage before mathematics becomes involved, where the problem is put into mathematical language". If only that were the case, I would not bring it up this question here yet again. I think most people agree that it is useful to separate these two phases: translate from natural to mathematical language; then solve within a formal framework. The first phase depends on personal taste, culture, experience, tradition. It's an art. The second phase is a matter of logical deduction. However, the "solution" we are talking about is sloppy because it does part of the translation, gives the appearance that a shift has been made to a formal mathematical analysis, but then makes an intuitive jump which in fact depends on a silently made further translation step. If Monty Hall paradox is about being careful with probabilistic reasoning, then this particular solution is a bad solution since it is careless with probabilistic reasoning. It counts on the reader being convinced by the authority of the writer together with some intutive gut feelings, not by thought.

Martin also wrote "The step you call non-trivial is as obvious to most people as is ...". It's completely obvious to most people that the answer to the Monty Hall problem is don't switch because the chance is 50/50... Completely obvious and wrong. The history of statistics is full of terrible disasters because steps which statisticians called non-trivial were completely obvious to most people ... and wrong. Eg. Sally Clark, Lucia de Berk, ... The moral is: be careful with probabilistic reasoning. Sloppy reasoning costs lives. Richard Gill (talk) 07:20, 27 May 2013 (UTC)


 * Richard, I have nothing whatever against doing things properly, as you say, at can be a matter of life and death. So why not do just that, formulate the problem with no assumptions at the start and only add the assumptions in when we have the problem in mathematical form? To save us from infinite speculation let us only consider facts given in Whitaker's problem statement.  We can also assume the usual game rules as we are both agreed on the effects of changing them.


 * We start with three doors and the player chooses one of them. The player can choose any door.  We must therefore start with three options that the player chooses door1, door 2, or door 3.  Intuitively, by symmetry etc we can argue that it does not matter which door the player chooses, but we have agreed to do things properly, no assumptions until we have everything in good mathematical order.  It could make a difference if the car was not placed uniformly.  Do you agree? Martin Hogbin (talk) 08:50, 27 May 2013 (UTC)


 * I start with the car being hidden behind one of three doors, that was the first step. The second step is that the player chooses a door. The third step is that the host opens a door. The fourth and last step is that the player makes a choice.
 * I am following the careful, systematic, traditional, reliable way of solving problems like this, as explained by Ruma Falk (1992): (1) write down the historical sequence of actions/events, (2) figure out the probabilities of what happens at each step given the past, (3) figure out the conditional probability of what we want to know given the information we have obtained - both of these in the context of (or if you prefer, relative to) the sequence of events we are talking about. No need for any original flash of insight. Just follow the standard methodology and see where it takes us.
 * Since we are hoping to advise the player on what action to take, I propose to study mathematically how the player's knowledge about the situation he is in changes as the history develops, as he and others take actions and he observes what happens. "Probability" will be "the player's (subjective) probability".
 * We (you, me and the player) don't know anything about how the car was placed. Initially (end of step 1) the car is equally likely behind any of the three doors.
 * Step 2: player chooses a door. Since we have no further information it doesn't make a difference which door we choose.
 * Let's suppose we choose door 1. At the end of step 2, the car is still equally likely behind any of the three doors, and we have chosen door 1.
 * Step 3: host opens a door. He is constrained to open a different door from door 1, and to open a goat door (he knows where the goats are hidden so at least one of doors 2 and 3 hides a goat). We don't know how he chooses when he has a choice, so in that case either door would be for us equally likely. So either the car is behind door 1, and the host is equally likely to open door 2 or door 3, or the car is not behind door 1, and the host is equally likely to open door 2 or door 3, because the car is equally likely behind door 3 or door 2. Suppose he actually opens door 3.


 * Let's be accurate. If in step 2 the contestant has chosen door 1, i.e. before any host's action, the probability that the car is behind door 1 remained unchanged 1/3. Step 3: host opens a door. For the standard paradox, if the car is behind door 1, (although the host may actually be biased as much as he likes to be, but we – nor the contestant – nor the audience "know" about such given(?), not given(?) bias, nor its direction nor its extent, so no one can benefit from such ungiven knowledge) then for us, for the audience and for the contestant the host "seemed to be" equally likely to open door 2 or door 3. "For us" it was 50 % each. So for us it does not matter which door he actually indeed did open. But if the car is not behind door 1, then the probability that he opens his only goat-hiding door is 1, and in this case the probability that he shows the car is zero. (Btw, this fact, that in this case the host is slave to the act makes the "paradox" arise.) But neither the contestant, nor the audience nor "we" know about the actual location of the car, so at this stage of development "for all of us it seems / seemed" that the host is / was equally likely to open door 2 or door 3. "For us" it was 50 % each. Result: after the host has shown a goat, whether or not the car is behind door 1, "for all of us" it did not matter which door the host actually did open indeed, for us, for the audience and for the contestant, the host "seemed to be" equally likely to open door 2 or door 3. "For us" it was 50 % each. Which door he actually indeed opened did not matter, so the probability that the car is behind door 1 did not change, it is still 1/3. The host's knowledge of the location of the car, and of his (im)possible bias is a priori quite another story. Gerhardvalentin (talk) 22:16, 28 May 2013 (UTC)
 * Gerhard, you just wrote out Step 4 instead of reading it.
 * Richard, you are wrong again. Did you read what I tried to say, and did you read your steps? Step 4 seems correct, but is not accurate enough. The contestant's view is one thing. But even the contestant knows that the host in 2 out of 3 is unable to choose randomly between the two non selected doors, but in 2 out of 3 is subject to the condition that he has to show his only goat, because he will not show the car. There never is randomness in his opening of his "only goat door" in that 2 out of 3. Naively it may "seem" to be at random, but in 2/3 effectively it never is. This is known to the contestant, too. No wrong contraction, please. For the contestant it "seems" equally likely that the host will open and "has opened" either one of his two unselected doors. But in fact, in that 2 out of 3, it never really was. The fact that in 2 out of 3 there never is randomness in the host's action, lets the standard paradox arise. Fortunately the article finally says so explicitly, thanks Mlodinow. In the article, we have to underline that it just casually "seems to be equally likely" whereas intrinsically, in depth, it never is "at random". No OR. That's what I said for years, and the article should not obliterate and obnubilate, but should account for this fact. Gerhardvalentin (talk) 22:10, 30 May 2013 (UTC)
 * Gerhard, I stand by what I wrote in my Step 4. Nothing more needs to be said. There is no contradiction with your remarks, but they are superfluous.Richard Gill (talk) 14:10, 31 May 2013 (UTC)
 * Not superfluous.
 * Richard please could you just try to understand. It's not superfluous, because the article should not mislead. The reader should understand and "see" what the article says, at least where it is of importance. The most important aspect is that the host will always show a goat but never will show the car. It is finally this very condition that lets the paradox arise, see Mlodinow. "Twice as likely to open ...", as a statement to help the reader to grasp the paradox, IMO is never enough. Please help to breach the glass ceiling of obscurity of the article, to help the reader to face the paradox. "Twice as likely" is not enough, and a sloppy "if the car is not behind door 1, then the host is equally likely to open door 2 or door 3" is terribly misleading. Misleading despite the following "because the car...". Albeit "true" from your line of sight, but my warning is that it is terribly misleading for the reader. The article should become clearer, not even more cloudy. I would suggest:
 * If the car is not behind door 1, then the host never is equally likely to open door 2 or door 3, because he is bound to never show the car. It is of importance that in this case he is under constraint to show his only goat, be it behind door 2 or behind door 3. Finally this restriction is the reason for the emergence of the paradox. But, as the car is equally likely behind door 3 or door 2, for us it incurs that the host seems to be equally likely to open door 2 or door 3 though, in effect.
 * Can you see what I am after? Please help the article to become transparent. Make the article explicit, bring it into the open. No sloppy "if the car is not behind door 1, then the host is equally likely to open door 2 or door 3." So not "is superfluous" or "I don't understand what you say". Please try to help the article to wipe off the wall of cloudiness. Thanks and regards, Gerhardvalentin (talk) 21:58, 1 June 2013 (UTC)


 * Step 4: Whether or not the car is behind door 1, the chance that the host would opens door 3 rather than door 2 is 50%. So the fact of which door the host opens can't change our degree of belief that the car is behind door 1. The odds against door 1 used to be 2:1, so they remain 2:1. We recommend the player to switch. Richard Gill (talk) 16:54, 28 May 2013 (UTC)

Let us now take look at the non-mathematical steps that are necessary to reach that conclusion

1) 'We (you, me and the player) don't know anything about how the car was placed. Initially (end of step 1) the car is equally likely behind any of the three doors'.

2) 'player chooses a door. Since we have no further information it doesn't make a difference which door we choose'.

3) 'We don't know how he chooses when he has a choice, so in that case either door would be for us equally likely.'

I would argue that you have missed a step:

4) 'We see one of two possible goats revealed, since we have no further information it doesn't make a difference which goat is revealed.

Although I have no real problem with the above solution (apart from the goats, which is easily remedied) your solution does not match any of the solutions shown in the article. Most of these use more formal methods like setting up a sample space or using a standard notation. Martin Hogbin (talk) 21:56, 28 May 2013 (UTC)


 * Regarding your step 4), I understood that the questioner wants me to assume that the goats are not previously known to the player. The question does not name them by name or tell us that they have distinguishing features known to the player in advance. When the host opens a door and reveals a goat we certainly see and know which door but I have no reason to suppose that we know and recognise which goat. You said "We start with three doors and the player chooses one of them". I said similarly, but a little bit earlier, that we start with three doors and the quiz-show team hides a car behind one of them. Of course, the quiz-team presumably knows the goats personally and determines which goat goes behind which non-car-hiding door. I could add that to the story. But I have no reason at all to suppose that the player knows the goats personally so I don't do anything with this ingredient later! It's clearly the intention of the questioner that I do not complicate the already complicated problem in such a way. For the player it is just "a goat" which he gets to see. He wouldn't have known if had been "the other goat" instead because he did not know the two goats in advance and so can't recognise which one of the two he sees. But he certainly does know which door is opened and I already was forced to identify the doors from the start, whether or not the questioner intended me to do anything with this information.


 * I understand your point, but when you analyse it carefully, it disappears. I, obviously, agree with everything you say about why we should ignore which goat is revealed, that is why I say, 'We see one of two possible goats revealed, since we have no further information it doesn't make a difference which goat is revealed', but we still need to say it.  Even if the doors were not numbered or visible (the situation could be described verbally) we could still identify them as: the door originally chosen, the door opened by the host, and the third door.  In probability (as of course you know) we have to consider not just what did happen but what might have happened. Even if the doors were not numbered and we could not see them it could still be argued that we must ask the question as to what would happen if the host had chosen to open the other (than the one he actually did) door.  Do you agree?  A very good point that you made yourself sometime back was that one measure of how complete a solution is is how general it is, how it applies to variations of the problem.  The solution in which we consider which door the host opens is better than one which does not because it also covers the variant in which the host has a known door preference.  Even if the doors were not numbered or visible this information could me made available to us by saying, for example, 'the host opens a door to reveal a goat but he would have opened the other available door if possible under the rules'. Now the Morgan solution becomes essential. Martin Hogbin (talk) 08:17, 29 May 2013 (UTC)


 * I don't agree. I am not presenting my own research. I am trying to explain the standard approach to modelling and solution of probability problems. Falk (1992) explains exactly the same principles. Apply the standard routine to the three doors problem and you will automatically produce the standard conditional probability 2/3 solution (with three distinguished doors and a generic, not distinguished, goat). Apply the standard routine to the three shells problem and you will automatically produce the standard unconditional probability 2/3 "simple solution". The standard routine delivers the standard solutions to three boxes, three prisoners, Monty Fall, Monty Crawl, and hundreds of other similar problems.
 * But you can also be original and do it differently. Contrary to the obvious intention of the questioner, you can also make the goats individually known and recognisable to the player. Whether or not the questioner intended, following the standard probability approach forces you to distinguish between the doors. Richard Gill (talk) 07:16, 30 May 2013 (UTC)
 * I was asking about the two doors that the host has a choice between. You say above that we must make the statement 'We don't know how he chooses when he has a choice, so in that case either door would be for us equally likely'. If the doors are not numbered and not visible, do we still need to make that statement?  I believe that you have previously asserted that we do.Martin Hogbin (talk) 18:38, 30 May 2013 (UTC)
 * I have answered this question by giving you my solution to the three shells problem, which is different from my solution to three doors. To make it clearer still: suppose in the three doors problem, the player is blindfolded, can't see which door the host opens. He only knows a different door was opened revealing a goat and he's asked if he wants to switch to the remaining closed door, but is not told which door that is.
 * Or if you prefer, think of the three balls problem with a colour-blind player but not oolour-blind host. When figuring out how the player should update his personal probabilities, does the player have to think about how the host chooses between the two balls when they have the same colour? Richard Gill (talk) 13:57, 31 May 2013 (UTC)


 * Secondly, you did not recognise this solution from the article. Then you haven't read the article very well recently. This solution is the so-called conditional solution from the second part of the article. I deliberately did not use mathematical notation or other formal ingedients because I wanted to explain the standard solution in plain words and plain logical reasoning. The sequence of steps describes a branching tree of possibilities. The end points of the tree, the leaves, correspond to possible histories, and each possible history is one point of the sample space. I used Bayes's rule, odds form, to update probabilities in the light of the player's information in a dynamic way, i.e., step by step. I could also have just calculated the required conditional probability using Bayes's theorem, ie by the definition of a conditional probability, but I wanted to avoid formal calculations, I wanted to stick close to intuition and not intoduce formal notation. I was careful to motivate each new probability assignment in a consistent way, seeing probability as information (ie so-called Bayesian / subjective probability). The solution is the standard solution of Gillman (1991) and Falk (1992) (both prior to Morgan, and completely in the tradition of Bertrand's three boxes and Gardner's three prisoners) and Rosenthal (200?) and many many other authors, it's given in the article in various different forms.
 * Richard there is no need to dumb down anything for my benefit, I can always ask you for an explanation if I do not understand any notation. The critical thing that I wanted to do was to separate the mathematical calculations from the non-mathematical assumptions necessary to solve the problem.  You did this, but not in the way that I was expecting, or in any way that follows exactly what we have in the article.  Ultimately we are trying to improve the article.
 * Regarding Bayes' rule, odds form, has this method of solution been published anywhere?
 * Converting formal calculations written out at length in clumsy notation into a few sentences of plain English is not "dumbing down". It's translation from a specialist language and vocabulary into ordinary language. As I said, the steps in the argument correspond to the branches in a decision tree or the levels of splitting of the big table with the pictures. Rosenthal uses Bayes' rule. So does RD Gill. So do others. Bayes' rule is merely a convenient form of Bayes' theorem and both are just repackagings of the definition of conditional probability. Nothing new, standard approach going back more than 100 years.
 * This solution in this verbal form, and specifically using Bayes' rule (odds form), is already in the article, along with what is essentially the same solution just written out in more formal (specialistic, mathematical) languages and possibly using other forms of Bayes' theorem. And Bayes' rule is both fundamental and trivial: by the chain rule, twice, P(B) P(A | B) = P(A & B) = P(A) x P(B | A). Write out the same again with the same B, but with A' instead of A: P(B) P(A' | B) = P(A' & B) = P(A') x P(B | A'). Now divide one by the other: P(A|B)/P(A'|B) = (P(A)/P(A')) x (P(B|A/P(B'|A)). Posterior odds equals prior odds times likelihood ratio. Richard Gill (talk) 10:31, 30 May 2013 (UTC)
 * All I asked was whether this method of solution been published anywhere. Martin Hogbin (talk) 18:38, 30 May 2013 (UTC)
 * And I answered several times: Yes, all over the place (and I mentioned some references). And it's in the wikipedia article too. Richard Gill (talk) 13:43, 31 May 2013 (UTC)
 * Can you give me just one reference please, not WP. Martin Hogbin (talk)
 * Falk (1992). Who incidentally has no reference to Morgan et al. but does reference Gillman (1991) and Gillman (1992). Her paper was submitted April 1991, final version accepted December 1991. See her "stages (1), (2) and (3)", top of page 200.
 * Falk uses the standard formulation of Bayes' theorem to do the routine conditional probability calculation which follows the statistical modelling. Rosenthal (2005) "Monty Hall, Monty Fall, Monty Crawl" uses Bayes' rule (Bayes' theorem in odds form) which I think is more illuminating and which is how I did the final probability update in my step 4. Richard Gill (talk) 10:27, 1 June 2013 (UTC)
 * IMHO, the whole "conditional" part of the article could be replaced by what I have given you here, together with references to sources. One could add a discussion of how the solution would change (add one step) if the player is previously acquainted with the goats and can recognise which one he has seen when its door is opened, if Martin would be so kind as to publish this new variation first. (I am thinking of writing an article about three shells versus three doors and one could add the new variation, three doors and two personally known and easily distinguishable goats).. Richard Gill (talk) 05:05, 29 May 2013 (UTC)
 * That is a good idea but I see no need to use shells. In principle, shells are no different from doors. Martin Hogbin (talk) 08:45, 29 May 2013 (UTC)
 * Vos Savant and many of her readers thought that doors and shells are the same. I think they are not the same, but that doors can be reduced to shells by symmetry. The natural traditional probability approach to the three doors has four outcomes with probabilities 1/3, 1/3, 1/6, 1/6 and leads one to calculate a conditional probability (the player gets to see that two of the four outcomes have been excluded). The natural traditional probability approach to the three shells, indistinguishable to the player, has two outcomes of probabilities 2/3 and 1/3, no conditioning because neither outcome is excluded. Richard Gill (talk) 07:01, 30 May 2013 (UTC)
 * Neither doors nor shells are actually indistinguishable. We can define them to be so in the question if we like although this is really just an instruction to the person answering the question on how to treat the objects.  Three shells are the same as three doors and three indistinguishable shells are the same as three indistinguishable doors.  I know there are some conventions, such as those involving objects taken from urns, but it is always easier and less open to misunderstanding to say exactly what you mean. Martin Hogbin (talk) 18:38, 30 May 2013 (UTC)
 * I am not talking about conventions. The three shells appear indistinguishable to the player, though they are distinguishable to the host. That is the whole point of the three shell game. Probability is about information, and information depends on appearances. If we are talking about the subjective probabilities of the player, three shells are definitely not the same as three doors.
 * "Actually" is irrelevant. Appearances are what counts, here. Appearances, to whom. Richard Gill (talk) 11:50, 31 May 2013 (UTC)
 * If course I understand that probability is about information, but the distinction you seek to draw simply does not exist. Even if we cannot see the doors and they are not numbered, they are clearly identified in the problem.  The there is a door that the player initially chooses, we could ourselves chose to number it door 1.  There is the door that the host opens, which we could, if we wish, call door 3, and the remaining door we could call door 2.  These numbers identify individual and specific doors.  We could, for example, rerun the problem with the player initially opening the door that we numbered door 2.  Appearances are irrelevant. Martin Hogbin (talk) 08:47, 1 June 2013 (UTC)


 * If you want to make an argument based on information, it can only be on information that could possible affect the probability of interest. Deciding exactly what that is is not a matter of mathematics or even, the way that 'professionals in this field' do things but of personal understanding and choice, which ideally should be made clear. Martin Hogbin (talk) 08:47, 1 June 2013 (UTC)
 * Agreed. Exactly what I am trying to do. "Personal understanding and choice" has to be made explicit, brought out into the open, and discussed. The general procedure which I outlined gives us a framework in which to systematically explore these issues. Richard Gill (talk) 10:27, 1 June 2013 (UTC)
 * You agree? You agree that, if the doors are not numbered and not visible, we must still make the statement 'We don't know how he chooses when he has a choice [between doors to open], so in that case either door would be for us equally likely'?  Please to avoid any misunderstanding could you please include a simple 'yes I agree' or a 'no I do not agree' in your answer. Martin Hogbin (talk) 19:43, 1 June 2013 (UTC)

Would you mind giving me a simple answer
Richard, the question above is not intended to be a trick question and I am not trying to argue anything, I just want to know what your answer is to my question. If the intended question is not clear to you please ask me to clarify. Martin Hogbin (talk) 12:32, 3 June 2013 (UTC)

Dates of solutions
I am a little puzzled by the statement, '...standard solution of Gillman (1991) and Falk (1992) (both prior to Morgan...'. The only solution I can find by Gillman was published in 1992 and the Morgan paper was published in November 1991. Martin Hogbin (talk) 17:54, 29 May 2013 (UTC)
 * Gillman wrote an earlier paper in 1991. He refers to it in his '92 paper, not to Morgan et al. I deduce that he is not aware of Morgan et al. (or not interested). Falk's paper is primarily on the three prisoners problem. She refers to, and approves of, Mosteller's solution. She handles the Monty Hall problem too, and I think also Bertrand's box problem. No reference to Morgan et al. I should not have said "prior" to Morgan et al, but independently of Morgan et al. Richard Gill (talk) 06:33, 30 May 2013 (UTC)
 * That is just supposition on your part, it seems remarkable that Gillman used the same letter as a parameter.
 * The letter was p. Not a surprising symbol to use for an unspecified probability. I would be very surprised if Gillman failed to cite the publication from which he had got these ideas, if he had got them from someone else.
 * But we just need to dig up Gillman (1991) to verify these things. Richard Gill (talk) 11:45, 31 May 2013 (UTC)
 * So actually, just as Morgan et al claimed, there are no previously published versions of their solution. Martin Hogbin (talk) 18:38, 30 May 2013 (UTC)
 * Morgan et al. are the first to systematically study the biased host. That's not my point. You earlier seemed to suggest that Morgan et al. are the first to give a solution using conditional probability, and hence are responsible for all the discussions which go on here. You forget the vast literature on the three prisoners problem, which Falk identifies as the same problem. You forget Selvin's second solution. Morgan et al refer to Mosteller's solution ... Mosteller wrote on the three prisoners problem ... and gives the conditional probability solution. The point is, that it comes completely natural to the "professionals" in this field to solve MHP by a conditional probability calculation. That was not some weird aberation which was started by Morgan et al. Richard Gill (talk) 11:45, 31 May 2013 (UTC)
 * Your sloppy use of the word 'conditional' makes your response meaningless. Martin Hogbin (talk) 08:47, 1 June 2013 (UTC)
 * My careful use of a precise technical term apparently makes communication impossible. Richard Gill (talk) 10:20, 1 June 2013 (UTC)
 * There is nothing precise about your use of the technical term 'conditional'. Any solution to the MHP or the TPP obviously involves a conditional probability calculation strictly speaking.  If that is all you mean to say then there is indeed nothing more to say or discuss.
 * If, on the other hand, you are trying to say any more than that it would make discussion much easier if you were to clearly state exactly which event or events you are condition on. Martin Hogbin (talk) 19:38, 1 June 2013 (UTC)
 * Read Falk (1992). Top of page 200: "The expedient course in grappling with probability puzzles of this kind typically involves three major stages. (1) Uncover hidden assumptions and check they are warranted. (2) Explicate the random process that has generated the data. (3) Apply Bayes theorem. The first and second steps are obviously interwoven." The event which needs to be conditioned on in step (3), if any, follows from expressing the original question in terms of the mathematical model which results at the end of Falk's steps (1) and (2).
 * For example, in MHP, steps (1) and (2) might result in a three phase random process: (a) team hides car behind one of three doors, (b) player chooses door, (c) host opens different goat door. The randomness in this process is described by determining the conditional probabilities of what happens in each phase conditional on the result of the earlier phases. There seems to be very little disagreement in the literature, whether popular or academic, that (a) initially the car is behind each door with probability 1/3. Next (b) the player makes a choice of door independently of the location of the car, which means that conditional on the player's choice, the car is behind each door with conditional probability 1/3. Next (c) the host chooses a door to open conditional on the door hiding the car and the door chosen by the player. If he has a choice, his conditional probabilities of either choice is 1/2.
 * To carry out step (3) we ask ourself, in terms of the mathematical model which has now been established for the random process under consideration, what the player wants to know (where's the car?) and what he does know (his initial choice and the host's subsequent choice). And we calculate the conditional probability of each possible location of the car given the player's information. Richard Gill (talk) 07:14, 2 June 2013 (UTC)
 * You always seem to think that I am disagreeing with you or am trying to argue some point. All I am asking is that you make clear what events you have decided (using a procedure like that given above) to condition on.  For example rather than just say 'I recommend a conditional solution to he MHP', it would avoid unnecessary argument to say (and this is purely an example of what one might say not a claim to represent your opinion) 'I recommend a solution to the MHP in which we condition on, the door originally chosen by the player and the door opened by the host'.  Just saying we should use conditional probability means nothing.  We obviously always condition on the event that the host reveals a goat but this is normally taken as read.  It would just save pages of pointless discussion if you would make clear the events that you consider significant and on which we must therefore condition our probability.
 * Dear Martin, you are still not appreciating the technical meaning of the word "conditional" in this context. We do not calculate conditional probabilities given the event that the host reveals a goat. This event is certain, in the model which results from Falk's steps (1) and (2). Having completed steps (1) and (2) we calculate the probability of the possible locations of the car conditional on the information which the player has at their disposal, whatever that might be. Richard Gill (talk) 10:44, 3 June 2013 (UTC)
 * Yes that is what I said, we take it as read that a goat is always revealed so conditioning is trivially easy (do nothing) and we therefore do not need to do it. In Bayesian probability strictly speaking almost every probability is conditional. I understand and agree that in the MHP the event that the host reveals a goat is normally taken as certain but this is just the standard interpretation of the problem. In the non-standard interpretation in which the host chooses randomly the host might reveal the car, but that was not my point. I am trying to ask you as simple question, which you seem unwilling to answer. Here it is again.


 * In the (agreed conditional) solution of the standard MHP, what specific events is it necessary for us to condition on in order to have a solution that you would consider acceptable? Martin Hogbin (talk) 11:59, 3 June 2013 (UTC)


 * Short answer: what event do we condition on, in the probabilistic sense, depends on what we see as the random process describing the problem.
 * Long answer: once we have carried out steps (1) and (2) of Falk's programme, then Falk's step (3) "Apply Bayes theorem" can be performed automatically. She means: calculate P(A|B) = P(A&B)/P(B) where B represents what the player knows, and A stands for what they want to know. A, B and P are all parts of the mathematical model which came out of steps (1) and (2).
 * If P(B)=1, or if there are good reasons to argue that P(A&B)=P(A)P(B), we don't need a conditional probability, we don't need Bayes' theorem, because in both these cases we can argue directly that P(A|B)=P(A). We don't need a conditional probability. When mathematicians talk about "conditional solutions to MHP" they mean solutions involving a non-trivial calculation of P(A|B). Non-trivial because P(A|B) is not equal to P(A).
 * Suppose our mathematical model is (i) the car is hidden completely at random by the quiz team, (ii) the player chooses a door independently of the result of (i), (iii) the host opens a door different from the door hiding the car and door chosen by the player, choosing completely at random when he has a choice. The player knows which door he chose and which door the host opens. So B stands for the particular outcomes in the case at hand. The player wants to know what is the chance that the car is behind each of the three doors, given what he now knows. For instance: B=player chose door 1, host opened door 3; C= car is behind door 1. Because P(B)<1 and P(A|B) is different from P(A) (events A and B are not independent of one another) this is a non-trivial application of Bayes's theorem.
 * That is fine, I have no problem in conditioning on the two events that the player chooses door 1 and that the host opens door 3. That is the logical thing to do but that is not the same as conditioning on the one event, that the host opens door 3. Martin Hogbin (talk) 18:28, 4 June 2013 (UTC)
 * It is the same thing if the player's choice of door 1 is fixed, not random. Depends how you set up your probability model in Falk's phases (1) and (2). As I wanted to emphasize, what you should condition on - the player's information when they have to decide - follows logically from the probability model which you arrive at after steps (1) and (2). Richard Gill (talk) 05:10, 5 June 2013 (UTC)
 * We both know that it is the same thing if the player's choice of door 1 is fixed, but we also both know that it is the same thing if the host had opened door 2. To be consistent we should either explicitly condition on both events or provide rationales for not needing to condition on either.  That is my point.  There is no logical reason to treat the host's door choice differently from the player's door choice. Martin Hogbin (talk) 13:28, 5 June 2013 (UTC)
 * I treat them differently in my implementation of Falk's first two phases, because I am taking probability to mean the player's subjective probability. From the player's point of view, the location of the car is random but the door which is chosen is not necessarily random. But since they are independent and in phase 3 I calculate the player's conditional probability given what the player now knows, I'll get exactly the same result as if I assumed that the player picks door 1 with certainty. So I reiterate: different people might like to carry out phases 1 and 2 differently. What is conditioned on in phase 3 - the random events of phases 1 and 2 of which the player knows the outcome - is determined by the random process description which we settled on in the preliminary phases. Richard Gill (talk) 14:24, 5 June 2013 (UTC)
 * You cannot get away with that one. Your own choice does not count as random so we can treat it differently?  We have to consider the special case where we have chosen door 1 out of three possible doors.  The natural way for a mathematician to do that (so you tell me) is by conditioning on the specific choice.
 * Of course, you are free to treat the problem how you like but please do not try to claim that the 'proper' or 'logically' precise way to solve the problem is by conditioning on the host's choice but not your own. Be consistent, condition on both. Martin Hogbin (talk) 23:11, 5 June 2013 (UTC)
 * I agree, performing steps (1) and (2) is a partly subjective matter. Partly a matter of taste. If you do them differently then step (3) will be different. In particular, it might not require calculation of a conditional probability at all. I already argued that the three shells problem is different from the three doors problem in precisely this respect. I already wrote out what I think is a natural probability model for this problem, and it does not require calculation of a conditional probability at all. Whereas if you agree that the model with ingredients (i) to (iii) is somehow the right model for MHP, then it does require calculation of a conditional probability. Richard Gill (talk) 11:47, 4 June 2013 (UTC)


 * By the way I agree with Falk's advice, it is just a pity that no one seems to follow it. Martin Hogbin (talk) 11:16, 2 June 2013 (UTC)

Radio version of the show
This is my version of your shell problem, but it uses doors, cars, and goats. Suppose the show is on the radio. We, as listeners know the standard MHP rules and make the standard MHP assumptions.

We are told in order:


 * 1) The player has chosen a door.
 * 2) The host has opened a different door to reveal a goat.
 * 3) The player has the option of sticking with their originally chosen door or swapping to the unchosen, unopened one.

We know that there are three physical doors and that one was chosen by the player, one was opened by the host, and one neither. We are free to label those doors in our minds P,H, and N. The problem is to be solved from the state of knowledge of the listener.

The question is, 'Should the solution condition on the door opened by the host?'. Martin Hogbin (talk) 21:58, 4 June 2013 (UTC)
 * Do you want the listener to make a decision based on the listener's own knowledge? (That is different from advising the player what to do, based on the player's knowledge). The listener hasn't heard anything that wasn't certain in advance. There is nothing to condition on. This is indeed a version of the three shells problem. In the language of Falk's phase (3), we must calculate P(A), there is no information B. Richard Gill (talk) 05:26, 5 June 2013 (UTC)


 * But what if in step 2 the radio said:
 * "The host has opened a different door to reveal a goat, but he is well known to usually open the other one of his two doors, if ever possible."
 * Does such info matter now – regarding the listener's decision, based on the updated listener's knowledge?
 * IMO the crucial point is that – in the standard paradox – we never will have any of such additional info, neither. Gerhardvalentin (talk) 07:44, 5 June 2013 (UTC)
 * We are headed in the same direction Gerhard. I am trying to determine exactly what this extra 'information' is that we get in the real MHP.Martin Hogbin (talk) 07:57, 5 June 2013 (UTC)
 * The extra radio information would change the listener's subjective probability, though it wouldn't change his advice to the player, as we know from Morgan et al's biased host solution. But adding this radio info takes us away from standard MHP as Gerhard rightly remarks.Richard Gill (talk) 13:29, 5 June 2013 (UTC)

I said that the problem is to be solved from the state of knowledge of the listener. So you are saying that there is no information on B (the doors).


 * You told me that the listener doesn't know which doors were involved, so he does not compute a conditional probability. No possibilities have been excluded. He might as well not even have listened to the show at all. But It coincidentaly turned out that the listener's optimal decision rule "switch" coincides with the player's optimal decision rule, "switch, whichever door I chose and whichever door the host opened" Richard Gill (talk) 13:29, 5 June 2013 (UTC)

Suppose now the show went like this. We are told:


 * 1) The player has chosen door P.
 * 2) The host has opened door H to reveal a goat.
 * 3) The player has the option of sticking with their originally chosen door P or swapping to door N.

What extra information do we have? Martin Hogbin (talk) 07:23, 5 June 2013 (UTC)


 * I don't see the difference with the listener problem = three shell problem. Unless you are changing your notation. Are the letters P, H and N English language abbreviated names or are they mathematical variables? (We are approaching here the problems in philosophy and semantics concerning "rigid designators". Cf two envelope problem in the philosophical literature).
 * If P means "the door the player chooses" then the statement "player chooses door P" means "the player chooses the door he chooses" which is a tautology. If on the other hand P is a mathematical variable taking the values 1, 2, 3, then "the player chooses door P" is one of three different statements according to whether P = 1, 2, or 3. In mathematics and in logic we sometimes have to be more precise than we usually are in ordinary language. Richard Gill (talk) 14:03, 5 June 2013 (UTC)


 * The letters are just symbols, names given to the doors by the radio commentator. P is what the radio commentator has decided to call the door originally chosen by the player.  Does that make it a rigid designator?  What if the game is replayed and the commentator uses the same names for the same physical doors, for example he says, 'The player has chosen door N this time'? Are the letters rigid designators now?  What if the doors happen to have those letters written on them?


 * It is getting philosophical but that is not a bad thing. We may have got to the crux of the argument. Martin Hogbin (talk) 17:01, 5 June 2013 (UTC)


 * I agree. There is an ambiguity in the meaning of of indefinite article "a" which we run into when we try to translate "the player chooses a door" into the formal language of probability theory. And according to the philosophers, this ambiguity lies at the heart of the two envelopes problem. But I am not a philosopher or logician or linguist, I am a mathematician, so I find it hard to talk about this issue. I prefer to express myself in mathematics. Richard Gill (talk) 12:58, 6 June 2013 (UTC)
 * Sure but in the MHP we are faced with a natural language statement. All I am objecting to is your claim that a solution which conditions on the door number opened by the host is the way that professionals naturally solve the problem.  There is no one foolproof reason for my objection but there are three arguments which seriously weaken your assertion.  They are:


 * 1) If you explicitly condition on the host-door you should (for consistency) also explicitly condition on the player's initial choice of door.


 * 2) Depending on your exact interpretation of the problem, as you discuss above, if you explicitly condition on the host-door you should (for consistency) also condition on the identity of the goat revealed.


 * 3) It is not clear from the wording that we are intended to consider door identities at all, and we actually know that this was not the intention of Whitaker, vos Savant, or Selvin.


 * You have agreed points 1 and 3 above. All I am arguing against is your occasional assertion that the Morgan-style (explicitly conditioning on the host door only) is the 'logically precise', 'professional' or 'proper' way of solving the problem. It is just one way, better than some and less good than others. Martin Hogbin (talk) 08:58, 7 June 2013 (UTC)


 * Morgan et al. also condition on the door chosen by the player. It is not a question of being consistent, it is a question of being methodical (Falk's phases 1 to 3). The intention of vos Savant, Selvin, or Whitaker is irrelevant. I totally disagree that you should condition on the identity of the goat revealed. "Consistency" is irrelevant. You should be methodical and economical.
 * I agree that there is not just one way to solve MHP. You seem to be absolutely the first who wants to distinguish between the two goats and apparently believes that the player knows them personally, in advance. Richard Gill (talk) 10:54, 7 June 2013 (UTC)
 * Of course the intent of the questioner is relevant this is Seymann's illuminating contribution to the problem.


 * Where in Morgan's paper do they explicitly condition on the door chosen by the player? By 'explicitly condition', I mean either start with a sample space which includes the possibility that the player might initially choose any door and condition it in the actual choice of door 1; or apply Bayes' rule to the players initial door choice of 1.


 * What do you mean by 'There is an ambiguity in the meaning of of indefinite article "a" which we run into when we try to translate "the player chooses a door"? Do you mean that we might be referring a specific door?  What else can he choose? There are three real physical doors and he has chosen one of them.  At that point we could mark the door that he has chosen. Martin Hogbin (talk) 16:15, 7 June 2013 (UTC)

Conclusion

 * 1) Falk's 1992 conditional probabilty approach is sensible (model the random process carefully, and then use Bayes' theorem)
 * 2) It was not inspired by Morgan et al's fixation on the biased host but follows a long tradition in elementary probability theory going back 100 years, and as far as we can see, written completely independently of Morgan and co-authors. Richard Gill (talk) 14:31, 5 June 2013 (UTC)


 * Result as to the standard paradox:
 * Is it correct to resume that the conditional probability approach is (just only :-) a refined and proven method / technique to "calculate" the actual value / amount that has to be ascribed to the actual probability enquired, indifferent of the actual number of the door that coincidentally (for example) had been opened by the host, be it #2 or be it #3? If this is appropriate, then conditioning on the "number" of the door that had actually been opened indeed, is logical and correct.
 * Yes! Richard Gill (talk) 08:43, 6 June 2013 (UTC)


 * But: any upright person, especially a mathematician, at the latest after having seen the result true for the actually opened door #3, and basing on all available actual knowledge, "should" explicitly admit that the same (identical) result would be expectable when conditioning on the "other non selected door" (in this case #2) that by chance could have been opened also. Because a mathematician can spot that, basing on all actually available knowledge, the second non selected door evidently was equally likely to having been opened.
 * Yes, but not because the other door was equally likely to have been opened; it's because of deeper lying equalities. Yes, the perceptive mathematician notices that Prob(win by switching | Player opened door p, host opened dooor h) = 2/3 whatever the values of p and h, two different door names chosen from the set {"left", "middle", "right"} (six possible combinations). In terms of Martin's radio listener, the radio listener has the same probability of winning by switching as the player, even though the player knows more. Remember: the radio listener doesn't know anything which wasn't certain anyway, so the radio listener does not calculate a conditional probability. The radio listener is dealing with the much more simple three shells problem.
 * The perceptive mathematician realizes that the equality of the six different just mentioned player conditional probabilities is a direct logical consequence of the symmetries in the player's prior beliefs which led to the specification of equal probabilites 1/3 of location of car, and equal conditional probabilities 1/2 for door opened by host given door chosen by player and door hiding car, when the last two are the same. The perceptive mathematician then realizes that there are other smart ways to deduce the same answer, with even less calculations. Richard Gill (talk) 08:43, 6 June 2013 (UTC)


 * On the other hand, basing on "never available" knowledge can produce quite differing results, that forever will be regulated by dreamt-up (maybe completely wrong) additional assumptions, along with tacit acceptance of unfounded and thus wrong results. Wrong for the standard paradox. And that probability to win by switching must be between the range of 0 and 1 is trivial, and that it must be at least 1/2 to 1 is trivial also.


 * In controversial scenarios deviating from the standard paradox, any additional dreamt-up additional assumptions give "their resultant" results only and nothing more, correct only for their associated assumptions, but wrong outside their narrow associated assumptions. Nonsignificant and wrong for the standard paradox presented by MvS, but just only matter for training probability theory. Not tangent to the famous paradox, and not affecting it. Gerhardvalentin (talk) 17:49, 5 June 2013 (UTC)


 * Morgan et al were saying essentially the same thing Falk (and numerous others) have said. The biased host scenario is not a "fixation" but a device (used in exactly the same way by numerous others) to show exactly why the conditional probability approach is necessary - because even if you grant that the car is randomly (uniformly) distributed at the start, and the host must open a door showing a goat, and the host must make the offer to switch (all clarified by vos Savant in her subsequent columns) - the probability of winning by switching is still not necessarily 2/3.  If you want the answer to be 2/3 regardless of the specific doors involved, you must also constrain the host's choice to be random if the player initially chooses the car (a critical detail vos Savant never mentioned).


 * True, Morgan et al. later admitted that p=1/2 was the natural assumption to make so that their conditional probability answer 1/(1+p) becomes equal to 2/3. I agree that their main point - just like Gillman (1991: written simultaneously and independently of Morgan), and Falk (1992: written independently of Gillman and Falk, references added to Gillman during the refereeing process), Rosenthal (2005), and so many others - was that vos Savant's argument was wrong. Nobody disagrees with the numerical value. They disagreed with deduction process, more specificially, they disagreed with what was to be calculated. Richard Gill (talk) 08:54, 6 June 2013 (UTC)


 * If you don't constrain the host's choice (but do specify the car is initially hidden randomly, and the host must show a goat, and the host must make the offer to switch), the average across all players (which is what the "subjective" probability is measuring) is 2/3 - but this means something completely different from saying the probability is 2/3 in any specific case (such as the player picks door 1 and the host opens door 3). I would venture a guess that if you say "the probability" of winning by switching is 2/3 if you pick door 1 and the host opens door 3, most people would expect if you had a sample of 9000 shows then in the subset where the player picks door 1 and the host opens door 3 the car would be behind door 2 about 2/3 of the time.  The point here is that this won't be the case unless the host is picking randomly (uniformly) between door 2 and door 3 when the player picks door 1 and the car happens to be there.  Indeed - the car will be behind door 2 about 1/(1+p) and if you didn't know p beforehand you can determine it by examining this subset.  This possibility (figuring out p based on observation) is only possible if you know which door the host opens (which in any normal scenario you do know), and is why knowing this gives you additional information.


 * If you want the probability without vos Savant's clarifications, see the paper by Puza et al . The general answer (given only the usual statement of the problem) is the probability of winning by switching is between 0 and 1.  This paper shows you can make the answer pretty much anything you want by adding assumptions.  With the "usual" assumptions (specifically including the host chooses randomly if he has a choice) the answer is 2/3.  -- Rick Block (talk) 16:44, 5 June 2013 (UTC)


 * This is a wrong conclusion, see Falk. You do not "need to constrain" the host's choice to be at random if the player initially chooses the car. Quite the contrary: you only "may" take the host's choice to be NOT at random if you KNOW about his one-sided bias, its direction and its extent. Only then this will be admissible. Only then. Concerning the paradox presented by MvS, any such "nonproven assumption" necessarily will result in wrong values. Gerhardvalentin (talk) 18:10, 5 June 2013 (UTC)


 * What "observation"? The paradox results in the step sequence presented in condensed form of a "story" told by MvS. This "story" never did happen before on stage, nor will happen on stage, in reality. So on what "observation" did you base your nonproven assumptions? Gerhardvalentin (talk) 18:29, 5 June 2013 (UTC)


 * I think I said that some years ago. Even with the standard game rules, taking a frequentist approach without any assumptions about the distributions involved means that the problem is insoluble (the answer is from 0 to 1).  On the other hand taking a Bayesian approach, given only the standard rules, the answer is 2/3.  Martin Hogbin (talk) 19:34, 5 June 2013 (UTC)
 * You need to be a little more specific about what you mean by "the standard rules". In particular, the vos Savant/Whitaker statement of the problem is not enough to conclude the answer is 2/3 - even for a Bayesian.  And, if we're adding constraints to make the Bayesian answer 2/3 why not go just a little bit farther and include the constraints making the frequentist answer 2/3 as well?  -- Rick Block (talk) 22:04, 5 June 2013 (UTC)
 * The standard rules are that the host always opens an unchosen goat-hiding door and always offers the swap. That is sufficient for a Bayesian to say the answer is 2/3. I have no objection to adding the common assumptions about distributions (the producer's car and goat placement, the player's door choice, and the host's legal door choice all uniform at random) to the frequentist approach, it is hard to see what else they might reasonably be. If we make no assumptions about these distributions, even with the standard game rules, the problem is insoluble.  No doubt if we consider all things permitted by Whitaker's statement the problem becomes even more complicated, I must look at the paper.Martin Hogbin (talk) 23:05, 5 June 2013 (UTC)
 * A serious frequentist like me therefore comes up with the following (game theoretic) solution: advise the player to choose their door in advance uniformly at random, and thereafter to switch whatever happens. The player who follows this advice wins with objective (frequentist) (unconditional) probability 2/3. And there is no way to do better.
 * The problem with a subjective probability approach is "garbage in garbage out". We put subjective probabilities into the problem and get subjective probabilities out. It's soft. No objective guarantees. Richard Gill (talk) 09:22, 6 June 2013 (UTC)
 * Caution: the host's legal door choice may "seem to be at random", but the paradox arises because it is not completely uniform at random. In 2 out of 3, if the player selected one of the two goats, the host's legal door choice is / was extremely biased, because he never opens the unselected door hiding the car, as he has no choice but to open the other unselected door only – the door that hides the second goat, in that 2 out of 3.
 * Added: in this "wrong-guess-scenario" (2 in 3) the host is "slave to the act" and his action is not at random, but fixed. This fixation secures that, in the "wrong-guess-scenario" (2 in 3) switching to the still closed second host's door wins the car for sure. On the other hand, the fact that in this wrong-guess-scenario the car is equally likely to be behind both unselected host's doors, guarantees that the host will open any of them with probability 1/2 and because of this "seems" to be equally likely to open any of them.
 * As to the remaining 1 in 3 where the guest's first selection was the car by luck, the host just opens one of the two unselected doors, both hiding goats, and we have no information on how he will choose in that case. He "may" be biased as much as he likes, but that does not matter, we couldn't care less which one he opens then, because we have no evidence whatsoever that he indeed "is"  biased / or not. To get the correct answer for any actual game, we may never adhere to any not "given" one-sided bias, because consequently this would lead to a completely wrong result for any "actual" game. In making this mistake, though, one would have to "repair" this mistake by neutralization. As some do by neutralizing the mistake in taking q to necessarily be 1/2.
 * Result: because, for any actual game, to mull over any imaginable but unknown one-sided host's bias must necessarily give a wrong actual value of probability to win by switching in the actual game, the only correct prophylaxis / prevention is to declare the host to be "unbiased" and to be equally likely to open any of his two doors if both hide goats. To declare that the host, if he has a choice, then this choice is completely at random.
 * Please consider that this conclusion never is some "additional assumption", but that such "close of argument" (Krauss, Selvin) is just a necessary prevention to avoid false assumptions like considering any actual unknown host's one-sided bias "to be given", for the actual game presented in the form of a story where the famous paradox clearly arises.
 * The article should help the reader to spot the arising paradox that isn't so hard to detect, although not everyone is capable to detect that occurrence.
 * The article should, in the first line, report on the sudden arising of the clear paradox in its proper scenario, to help to understand where it arises, and in which other scenarios (in adding not given additional preconditions) it will not arise. Gerhardvalentin (talk) 08:13, 6 June 2013 (UTC)


 * @Rick Block, Richard Gill: The crucial difference between the rule that the host has to open a not chosen door with a goat and to offer a switch (which means that the contestant before his very choice has to determine two doors of which the host has to open one with a goat), and the other rules ("assumptions") is that the first is necessary to guarantee an "average chance" of 2/3, completely independent of other assumptions. The host may do what he wants: Always hide the car behind a special door, always open the "rightmost" door if he has the choice and so on; he cannot avoid this average 2/3-chance for the contestant: If by his behaviour he decreases this chance for one case, he increases this chance exactly weighted for the other case leaving the average chance constant. I think we could consider a simpler example to see what matters: The contestant has to guess whether the host has a coin in his hand or not. Surely if the host never has a coin in his hand (p = 0), and we ask: What is the probability for the contestant to win if he says "Yes", it is not very surprising that if we play (or think ...) this game 10000 times the contestant ever loses. But by randomly saying "Yes" or "No" the "average chance" for the contestant is 1/2. And in the same sense the average chance in MHP without any additional assumptions is 2/3: It is the contestant who can fix this chance by randomly choosing a door.


 * Shortly: Is the contestant wrong if he calculates a stake assuming exactly a 2/3 chance? - Is he more wrong if he calculates the stake just after the host has opened the door with a goat? - Is he more wrong if he calculates the stake after he has chosen door 1 and the host has opened door 3? - Is he more wrong than if he did the same for the numbers 1 to 4 of a dice?--Albtal (talk) 13:29, 6 June 2013 (UTC)


 * I understand your point and I'm confident Richard does as well. Do you understand ours?  In particular, if you had a sample of 9000 shows (run under the rules only that the host must open a not chosen door with a goat and make the offer to switch), and someone posed the vos Savant/Whitaker wording of the MHP to you - which uses as an example the case where the player chooses Door 1 and the host opens Door 3 - wouldn't your natural response be to look at those cases in your sample matching the cited example?  This will presumably be a fairly large number of samples and the fraction of times the car is behind Door 2 in this sample should (by the law of large numbers) fairly accurately reflect the probability of winning by switching in this specific case.  If your only rules are that the host must open a not chosen door with a goat and must make the offer to switch, what will this fraction be?  Is the question the MHP asks "what is the best strategy to follow if you're going to be on such a show", or is it "should you switch after having chosen, say, Door 1 and having seen the host open, say, Door 3"?  If the former, picking a random door and switching is minimax optimal across all host behaviors (as long as the host always reveals a goat and always offers the switch).  If the latter and you know nothing about the host behavior (except that he always reveals a goat and always offers the switch), your overall chance of winning by switching (across all door combinations) is 2/3, but your specific chance of winning by switching if you've picked door 1 and have seen the host open door 3 is between 0 and 1 (between 1/2 and 1, i.e. never worse than staying, if the initial car location is randomized).  If you want the answer to be 2/3 regardless of the door combination (which I think is clearly the intent - it was Selvin's intent since he as much as said so in his second letter about the problem) you have to randomize both the initial car location (as vos Savant does in her cup experiment) and the host's choice in the case where the player initially picks the car (and yes Gerhard, this happens only 1/3 of the time - in the other 2/3 of the time the host's action is fixed) - or you need to be talking about a "subjective" probability that effectively treats the doors as indistinguishable (meaning if you have a sample of 9000 shows your answer will be close to the fraction of times all 9000 players would win by switching, but not necessarily the fraction of times the car is behind Door 2 if the player initially picks Door 1 and the host opens Door 3).  Is there anything I've said here that is in any way controversial? -- Rick Block (talk) 00:25, 7 June 2013 (UTC)
 * To avoid arguing about things on which we agree (which seems to happen a lot here) let me make some statements which we probably all agree on.


 * 1)Assuming the standard game rules (that the host always opens an unchosen door to reveal a goat, and always offers the swap) regardless of the host's policy and the car placement, a player can guarantee a 2/3 chance of winning by swapping by choosing their initial door uniformly at random.


 * 2)Taking a Bayesian perspective but assuming the standard game rules the probability of winning by switching is 2/3.


 * 3)Taking a frequentist approach and assuming the standard game rules and applying the principle of indifference to the undefined distributions in the problem, the probability of winning by switching is 2/3.


 * 4))Taking a frequentist approach and assuming the standard game rules but taking the undefined distributions in the problem to be unknown, the problem is insoluble (the probability of winning by switching can be from 0 to 1).


 * 5)Without assuming any specific game rules the problem is insoluble (the probability of winning by switching can be from 0 to 1).


 * Does anyone disagree with any of these statements? Martin Hogbin (talk) 09:14, 7 June 2013 (UTC)


 * Let me comment in short. On 1): A guarantee of a 2/3 chance of winning by swapping, does not in itself guarantee a 2/3 chance of winning by swapping in the situation of the player. And then, the question is not to advice the player about their initial choice. 2+3): What do you mean by: the probability of winning by switching? The a-priori, the posteriori, a conditional probability? 4+5): The solution may be given in terms of these distributions. Nijdam (talk) 18:43, 7 June 2013 (UTC)
 * OK, for 1, I mean the probability prior to the opening of a door by the host choosing of a door by the player. I do think any other interpretation makes much sense since we do not know what events will follow this.  So, to put it another way, if the game is replayed and, each time the player chooses a door uniformly at random, the player will win 2/3 of the time.
 * For 2 and 3 I think all the probabilities you mention will be 2/3.
 * For 4 it is easily possible to envisage distributions for which the probability of winning by switching is 0 or 1.
 * For 5 also.
 * Do you agree? Martin Hogbin (talk) 19:40, 7 June 2013 (UTC)
 * For 1: As I said, the question is not to advice the player about their initial choice.
 * For 2+3: Just stating this does not contribute to the solution.
 * For 4+5: Agree, but such distributions are mainly rather artificial. Nijdam (talk) 10:24, 8 June 2013 (UTC)
 * 1 I understand what you mean, but if the player thought that the host was, for example, using their historical knowledge of typical player behaviour to reduce the probability that the player would win the prize by switching, the player would be best to choose a door uniformly at random and to tell the host that they had done so. The host would then realise the futility of trying to outwit the player.
 * 2+3 Shows that with reasonable assumptions the conditional solution gives the same answer as the unconditional one. I think that is a useful contribution to the subject.
 * 4+5 The only reasonable distributions are 2 and 3. Martin Hogbin (talk) 10:57, 8 June 2013 (UTC)
 * I venture to agree with 1). This, of course, means that the chance to win by switching is 2/3 in all cases: The contestant has chosen door 1, the host has opened door 3; the contestant has chosen door 2, the host has opened door 1 etc ... The crucial simple reasoning without any calculations is, that there isn't any reason to assume a difference between the two doors of which the host has to open one with a goat. By randomly choosing the door we protect us against any strategy of the host.
 * Surprisingly for me I read in Steinbach (p. 7) who wrote when referencing Gillman, that he (Steinbach) doesn't know a mathematical theory which approaches such problems. Before I believed, as I wrote here earlier, that this should be a simple case in game theory. Maybe too simple to arise there? This "new mathematics" seems to be nothing else than some sort of indifference principle: If there is no specific information, make no difference, or "take the (overall) average", or p = 1/2. So I think if we have clarified what "assumption" means here there should be no controversy about this point: Do we mean that "the task setting has to be completed by an additional assumption", or "to solve such (complete) problems we apply our indifference principle" by the means described?--Albtal (talk) 14:46, 7 June 2013 (UTC)
 * Sorry Albtal, I meant to say 'disagree'. Is there any statement above that you disagree with? Martin Hogbin (talk) 15:48, 7 June 2013 (UTC)
 * It functioned well with "agree", too. I think that 1) is the best. I cannot say that I disagree with the others, but I am not very interested in studying the "exact definition" of "frequentist" or "bayesian" and so on, what would be a requirement for profound answers. To answer the original problem we should use the simplest concept of probability. And I think that the discussion of door numbers and so on is simply a wrong track caused by a dubious source.--Albtal (talk) 16:31, 7 June 2013 (UTC)


 * And 1) functions well without your corrections above ...--Albtal (talk) 21:07, 7 June 2013 (UTC)


 * @Rick Block: I thought that I understood your arguments well, and I don't know why you just repeated them in your new comment. I also thought that I repeated your crucial argument by the example Have I a coin in my hand?. And what about the stake?--Albtal (talk) 11:31, 7 June 2013 (UTC)


 * @Rick Block: Yes, I think that Have I a coin in my hand? is a good simple example to demonstrate all crucial aspects of your arguments. But to eliminate the process of analogy formation I'll answer your original questions. And although it's hard to argue straightforward I try to do it. If I had a sample of 9000 shows and someone posed me the question in the version "door 1/door 3" I should say, ignoring the samples, that the chance is 2/3 for switching. And if I had to answer whether the case "door 1/door 3" should have a 2/3-distribution in the sample I should say "Absolutely no", because it could even be that the players always had chosen door 2 (or door 3); so we have no single example for "door 1/door 3" in the set. But if the players had chosen randomly? Then it could be, for example, that in the subset "door 1/door 3" switching has always won. So, if the players chose randomly, but the host didn't: Where can we find the 2/3-distribution? - Normally not for the "door 1/door 3" case, and if, it wouldn't tell us anything for the actual game; for the host could change his p from game to game, or sometimes not, and so on. - So the host can effectuate by his strategy that there is no 2/3-chance for the combination "door 1/door 3"? - Yes, he can. But only by changing the chance for "door 2/door 3", exactly weighted around 2/3 to the other direction. So, strictly spoken, we cannot answer the question "What is the chance for "door 1/door 3"?", but we can say: "Based on my knowledge, all combinations have the chance 2/3. And this is what matters. It may be that I just left the 2/3-path in one direction or another. But my centroid is 2/3." (From time to time we should remember which path we are following here ... BTW: What I really said in 1991, you can read in the two letters attached in Ziegenproblem. When I told a friend those days that p = 1/2 (for the host's strategy if the contestant has chosen the car) is only an assumption which could be "splitted" by the host even into 0 / 1 he said "oh, this would be rather hair-splitting", the same what Steinbach wrote later - and which has been my own opinion. But we may treat it - as a footnote.)--Albtal (talk) 20:43, 7 June 2013 (UTC)

Again the conjuring trick: Before and After

 * Perhaps I'm confused. Above you say Martin's #1 (pick a random door and switch ensures winning with probability 2/3) "of course, means that the chance to win by switching is 2/3 in all cases" - but just above you say if you had a sample of 9000 shows (with a meaningfully large subset where the player who picked door 1 and the host opened door 3) you would absolutely not expect the car to be behind door 2 2/3 of the time.  Isn't this precisely what it means for the chance to win by switching to be 2/3 in all cases, or do you not believe in the law of large numbers?We can answer the question "what is the chance for door 1/door 3" - and the answer is 1/(1+p) (nearly by definition).  Whether the host is consistent or not, over the long haul we could measure the average value for p (and there are 3 pairs of p/q values where in each pair p=1-q).  I'd venture a guess that if a real host does not use some randomization strategy (flipping a coin), he'd hardly be able to avoid expressing a preference.It is perhaps hair-splitting, because even knowing p you can't improve your chances as measured at the beginning of the game beyond 2/3 (unless the car is not hidden randomly - but this is a different issue) - and if the host must show a goat and must make the offer to switch you're never worse off switching.  However, if you're watching the show in Las Vegas and can place your bet after the host opens a door, you can do better than 2/3 if you know p.  -- Rick Block (talk) 00:59, 8 June 2013 (UTC)


 * We know that we don't know p. (And moreover we have to take into account that p may change "randomly" from game to game.) If we knew we could simply calculate the chance to win. (And this chance could also be less than 2/3, maybe 1/2.) The question here is: Is there a proper concept of probability which results "exactly" in a 2/3 chance instead of the interval [1/2, 1] (which in games with other rules could be for example [1/3, 1])? A player in Las Vegas could say: "If, according to the law of large numbers, you can guarantee that I'll win in the long run 2/3 of the time, averaged over all combinations, then I'll calculate my stake based exactly on a 2/3 chance in this actual situation where I have chosen door 1 and the host has opened door 3." - "But you don't know p." - "I'm not interested in p; I'm interested in getting rich." - "Warning! If one day you look back to all your MHP games it may even be that the winning rate for the combination "door 1/ door 3" is only 1/2." - "Then the winning rate for other combinations will be correspondingly greater. And as I don't know p, I cannot make any difference between the combinations. So based on my knowledge I have to guess a 2/3 chance for all combinations. Surely I chose my door randomly." - "But I am interested exactly in the answer to the question for the combination "door 1/ door 3". - "I gave you the answer: 2/3. And a warning to you: In your proposal [1/2, 1] the centroid 2/3 has disappeared completely. If, for example, someone would calculate a stake based on 3/4, he would be dead wrong."


 * Let us look at the game Have I a coin in my hand?. It seems to be an accepted way in game theory to guarantee a 1/2 chance by throwing a coin before the choice. But we could ask for the chance after the toss and before seeing the result. Suppose there are two players, one host. The first player guesses "Yes"; the second player tosses a coin and also says "Yes". Are their chances different? - No, they cannot be different. Strictly spoken, the chance is only "exactly" 1/2 before tossing a coin. And we cannot answer the question "What is the chance for a player who says "Yes"?", because we don't know p. In MHP we are in the situation after having tossed a coin. And guessing the chance 2/3 in MHP for "door 1/ door 3" is the same as guessing the chance 1/2 for "Yes" in the coin example.


 * If we really need new mathematics for this case I have a first proposal: Take the "average probability" over all combinations if you don't know p. - But the question discussed here is nothing else than the question "What is p?". - I don't know.--Albtal (talk) 12:14, 8 June 2013 (UTC)


 * Why do you insist we don't know p? The doors are numbered.  We can watch the show and keep a tally.  Even a few hundred would be enough to end up with a reasonably reliable estimate.  Do you have a coin in your hand?  I don't know, so my (random) yes/no guess has a 1/2 chance of being right.  And, if I do this repeatedly (guess randomly) over time I'll be right about 50% of the time regardless of who's asking.  Indeed, by this logic the MHP contestant has a 50% chance of winning by switching (and a contestant who picks randomly whether to switch indeed does have a 50% chance of winning).  But this is not the same as saying if you ask 100 people whether you have a coin in your hand that 50 times you will, or that out of 100 shows where the player picks door 1 and the host opens door 3 the car will be behind door 2 only 50 times.  To some extent this leads to a discussion about what is meant by "probability".  However, if I watch you and out of 100 times you've asked you only have a coin 10 times, haven't I discovered that "no" is a better guess the next time you ask (i.e. for whatever reason you do hold a coin only about 10% of the time)?  Similarly, by knowing the structural answer is 1/(1+p) if I see the car is behind door 2 90 times out of 100 instances where players picked door 1 and the host opened door 3 don't I know p is about 1/9 (i.e. for whatever reason, the host is choosing to open door 2 much more often when the car is behind door 1)?  If I'm watching this show with this host in Las Vegas and the player picks door 1 and the payoff reflects a 1/3 probability for door 1 and a 2/3 probability for the "switch" door, I'm betting on door 1 if the host opens door 2 and on door 2 if the host opens door 3.  Perhaps you're saying that unless we know why the host is not choosing randomly we don't truly know p (past performance is no guarantee of future results).  To some extent I agree.  But large numbers cannot lie.  -- Rick Block (talk) 17:32, 8 June 2013 (UTC)


 * We don't know p. And I think that the question here is nothing but "What is p?".


 * We have to solve the problem for a single situation. "Histories" and "Futures" are only experiments in mind for explanations. To estimate probabilities based on real samples is a new thread here in which I shall not participate.


 * The question "What is the chance in the door 1/door 3 situation" corresponds to the question "What is the chance for "Yes"" in the coin game. The answer 2/3 in the first corresponds to the answer 1/2 in the second game.


 * I dare to say that game theory reenforces that this chance for MHP without any additional rules is 2/3, which corresponds to the simple assumption that there is no difference for the contestant whether the host opens door 2 or door 3.


 * And I dare to say that wise guys who started their article with a fudge have stimulated other wise guys to vaste plenty of other peoples' time.--Albtal (talk) 10:54, 9 June 2013 (UTC)
 * I agree, it is a conjuring trick. Martin Hogbin (talk) 10:57, 9 June 2013 (UTC)

The host, if he has a choice, then this choice is completely at random.

Please consider that this conclusion never is some "additional assumption", but that such "close of argument" (Krauss, Selvin) is just a necessary prevention to avoid false assumptions like considering any actual unknown host's one-sided bias "to be given", for the actual game presented in the form of a story where the famous paradox clearly arises. MCDD obviously were not seeing the paradox arise. That it arises in the story of a one-time-show that never, in its clear scenario, was on stage. To dream about quite other stories and to dream of Las Vegas, and hoping to ... just shows that not everyone understand the paradox. But, in adding not given additional preconditions outside the original story, is sticking on methods to teach conditional probability theory. But this is long since ditched. Gerhardvalentin (talk) 11:32, 9 June 2013 (UTC)


 * If anyone is not willing to agree that saying "the probability of winning by switching if the player picks door 1 and the host opens door 3 is 2/3" means that over a large number of samples (where the player has picked door 1 and the host has opened door 3) the fraction of times the car is behind door 2 should approach 2/3, then what does this statement mean? -- Rick Block (talk) 18:24, 9 June 2013 (UTC)
 * That certainly is on of the most common definitions of probability (frequentist). I prefer the Bayesian approach where we talk in terms of degrees of belief and states of knowledge and use Bayes' rule but I have absolutely no objection to a pure frequentist approach to the problem. You need to be very careful though, as Morgan pointed out. — Preceding unsigned comment added by Martin Hogbin (talk • contribs)
 * What Morgan actually pointed out is that you need to be careful when designing an experiment about conditional probability. My question remains - if your notion of probability (whatever you want to call it) does not mean that the fraction of samples relative to a total approaches what you're calling the probability, what does it mean?   Side comment - I'm pretty sure Bayesians believe in the law of large numbers.  -- Rick Block (talk) 20:22, 9 June 2013 (UTC)
 * Morgan said, 'The modelling of conditional probabilities through repeated experimentation can be a difficult concept for the novice, for whom careful thinking through of the situation can be of considerable benefit'. What we are doing below is, 'modelling of conditional probabilities through repeated experimentation', and what I am suggesting that we do some, 'careful thinking through of the situation'. Martin Hogbin (talk) 20:50, 9 June 2013 (UTC)

Frequentists only
Let us consider how a frequentist, as defined above, might solve the MHP.

Although the actual question does not actually ask for it let us calculate, "the probability of winning by switching if the player picks door 1 and the host opens door 3"

This is the fraction of times the car is behind door 2 over a large number of samples (where the player has picked door 1 and the host has opened door 3). That sounds simple enough but we need to consider exactly what we repeat to get our large number of samples.

First, a stage hand places a car behind one of the three doors and a goat behind another door and another goat behind the remaining door. We first need to decide exactly what is repeated here. Do we imagine repeating the same door placements every time or should we take the three objects to be placed uniformly at random or should we envisage some other distribution? Maybe the stage hand has a preference for placing the car behind door 2 for example. Martin Hogbin (talk) 18:46, 9 June 2013 (UTC)
 * Haven't we have previously agreed, and didn't vos Savant (through the directions for her cup experiment) make it perfectly clear that the initial placement of the car is to be random (uniform)? Further, since in her cup experiment the non-winning cups are empty, isn't it also clear that the goats are not relevant (i.e. she considers replacing the goats by nothing to be equivalent)? -- Rick Block (talk) 20:11, 9 June 2013 (UTC)
 * So are we basing our method of repeating the game on what we know is vS's intended scenario. That would seem a reasonable option to me. Martin Hogbin (talk) 20:20, 9 June 2013 (UTC)
 * You know perfectly well what we're talking about - two scenarios where the difference is whether we constrain the host's pick between the two losing options in the case the player initially selects the car. In both, the "Bayesian probability" of winning by switching (for a player picking door 1 and seeing the host open door 3) is 2/3.  But in only one will the fraction of times the car is behind door 2 necessarily be 2/3 (in the case the player picks door 1 and the host opens door 3).  -- Rick Block (talk)
 * I do not understand what you are getting at. We have to decide exactly how the game is to be repeated. You mentioned vS's intended scenario as a basis for repeating the game which seems a perfectly reasonable option to me.  Are you objecting to that?  Would you prefer some other basis? Martin Hogbin (talk) 20:43, 9 June 2013 (UTC)
 * Again, you know perfectly well what we're talking about - i.e. the difference between vos Savant's cup experiment (from her 3rd column) where the method by which the host decides which of two losing cups to reveal is not specified, and a modified version where we add another randomization step in this case. -- Rick Block (talk) 21:40, 9 June 2013 (UTC)

I thought we were talking about how to model the MHP can we leave cups out of it please. We start at the beginning:

First, a stage hand places a car behind one of the three doors and a goat behind another door and another goat behind the remaining door. We first need to decide exactly what is repeated here. Do we imagine repeating the same door placements every time or should we take the three objects to be placed uniformly at random or should we envisage some other distribution? Maybe the stage hand has a preference for placing the car behind door 2 for example.

It is not a trick question, I simply ask what we do regarding the initial placement of the objects behind the doors when we repeat the game. What is your answer? Martin Hogbin (talk) 22:48, 9 June 2013 (UTC)
 * My answer is you already know my answer. Don't be a dick. -- Rick Block (talk) 22:56, 9 June 2013 (UTC)
 * I think we can go no further as you seem to be answering a simple, straightforward, and necessary question with a personal insult. Martin Hogbin (talk) 08:29, 10 June 2013 (UTC)
 * No, Martin. I already answered your question.  And you know I already answered it - you're pretending not to understand what I'm talking about when you obviously do.  In vos Savant's 3rd column she describes a cup experiment which she clearly intends to be equivalent to the MHP.  In this experiment the doors and cars/goats are replaced with cups and a penny representing the car (she completely omits the goats since they are nothing but "zonks", i.e. unwanted booby prizes - see discussion of this at Let's Make a Deal).  The cups are numbered 1,2 and 3.  The host rolls a die until it comes up 1,2, or 3 and hides the penny under the indicated cup (i.e. the car is to be hidden uniformly randomly behind one of the three doors, and the goats might as well be dispensed with).  The contestant rolls a die until it comes up 1,2, or 3 and initially chooses the indicated cup (i.e. the contestant picks a random door).  The host now "purposely" lifts up an empty cup (opens a losing door) from the unchosen two (how the host should choose between two losers is not specified).  She suggests repeating this 200 times with the contestant not switching, and then 200 times with the contestant switching.  Out of the 200 not switchers, you'll see about 67 winners and out of the 200 switchers, you'll see about 133 winners. My point, of which you are well aware, is that without specifying how the host chooses between two losers we have no particular guarantee the fraction of times #2 is the winner when the player initially picks #1 and the host reveals #3 (relative to the total times the player picks #1) approaches 2/3 over large numbers of trials.  You can do it vos Savant's way, but it will go faster and show the same result if the player always picks #1.  If you count how many times switching wins in the case the player picks #1 and the host reveals #3 the fraction relative to the total number of times the player picks #1 as the number of trials increases might be anything between 1/2 and 1 with these instructions.  For example, I'm right handed and with the cups laid out in front of me #1 to #3 right to left, when I'm playing the host I'll reveal #2 if possible (which will make the fraction 1).  However, when someone left handed does the same thing, they'll reveal #3 if possible (which will make the fraction approach 1/2).  If you want to know what the fraction will approach you have to specify how the host decides between two losers.  In particular, if you want the fraction to be 2/3, add a step where the host flips a coin to decide which loser to reveal (always flip a coin without showing the result to the contestant and use this coin flip to decide which of two losers to reveal - higher numbered [heads], or lower numbered [tails]).  So, you tell me.  What is the "probability" (frequentist if you must) of winning by switching to #2 if you initially pick #1 and the host reveals #3 given only vos Savant's instructions? -- Rick Block (talk) 15:24, 10 June 2013 (UTC)
 * I see no point whatsoever in discussing what you believe my opinion is of your opinion of one of vos Savant's explanatory simulations. If you want to discuss, "the probability of winning by switching if the player picks door 1 and the host opens door 3" in relation to the Whitaker/vS problem statement I am happy to do so.  You can tell me what you think is the right way to repeat the game, I can tell you what I think is the right way to do that, that makes life much simpler. Martin Hogbin (talk) 18:44, 10 June 2013 (UTC)
 * Didn't I just tell you, above (roll a die to decide where to put the car, etc)? What would be helpful here is if you would stop pretending you are not understanding what I'm talking about.  -- Rick Block (talk) 19:42, 10 June 2013 (UTC)
 * I really do not know exactly what you are getting at. But you seem to have moved on now, so, if we can get back to the game show involving one car two goats and three doors we can start to address your question.  You suggest that the stage hand is asked to roll a die to decide where to put the car and goats.  What is your logic for this?  Is it based on your understanding of what would be done in a real game show? Martin Hogbin (talk) 19:49, 10 June 2013 (UTC)
 * We pretty clearly want the car to be evenly distributed behind the doors over time, in a hard to guess way, i.e. with a uniform random distribution. Picking the car door based on the roll of a die is a reasonable way to ensure this.  That's why.  -- Rick Block (talk) 20:21, 10 June 2013 (UTC)
 * And, it's how vos Savant herself says to simulate it. -- Rick Block (talk) 20:25, 10 June 2013 (UTC)
 * I was not asking why you used a die just why the car and goats were to be placed randomly. There is no reason that the must be placed that way.  Maybe the stage hand was a bit lazy and tended to put the car behind the nearest door more often.  We could even have the car always put behind door 1.  Why do you want the items placed randomly behind the door?  Martin Hogbin (talk) 22:22, 10 June 2013 (UTC)
 * Random placement behind the doors is what vos Savant says to do. It's an explicit clarification in the same vein as "host must open a losing, unchosen door and must make the offer to switch". -- Rick Block (talk) 01:38, 11 June 2013 (UTC)
 * That is fine. The point is that the W/vS problem statement does not make this clear so we do have to make a decision of some kind.  We could for example, say that, as the information is not given as to how the car and goats were placed, we can only parameterise the problem and say that the stage hand places car behind door 1 a fraction s of the time and behind door 2 t of the time.  Of course, this would make the solution rather dull and generally incomprehensible (to the intended audience).
 * You have chosen to base your decision on what you believe to be the intention of vS when she worded the problem. I (and Seymann) agree with you.  To get an answer to any probability question we must consider exactly what scenario the questioner intended.  The context of the problem, a simple puzzle in a general interest magazine, in my opinion, confirms the view that we should take the car and goat placement to be uniform at random. — Preceding unsigned comment added by Martin Hogbin (talk • contribs)

Now the player chooses a door
So, the car and goats having been placed, what happens next? The player chooses a door. We assume, I guess, that the player does not always choose door 1 but that they can choose any door. Although it can be argued that it makes no difference, I believe that, continuing the logic started above, vS intended to indicate that the player might choose any door with equal probability. Of course, later in the calculation we will need to condition our probability in the door actually chosen by the player. — Preceding unsigned comment added by Martin Hogbin (talk • contribs)
 * Yes - the player is free to choose any door, and because the car was placed randomly the probability the player's chosen door hides the car is 1/3 (by anyone's definition of probability). If we repeat what we have so far 3000 times and the player always picks door 1, we'll see the player's initial pick hides the car about 1/3 of the time.  If we repeat what we have so far 3000 times and the player rolls a die to select the initial door (as per vos Savant), we'll see the player pick door 1 about 1000 times and of these the car will be behind door 1 about 1/3 of the time (ditto doors 2 and 3).
 * I just want to agree at this stage that the player can pick any door. Do you agree that we should assume that he picks uniformly at random.  I do not think that we should jump ahead with off-the-cuff calculations.  We are setting up a sample space that will cover all the possibilities.  later we can condition it according to what we might consider is described in the problem. Martin Hogbin (talk) 17:54, 11 June 2013 (UTC)
 * I think it doesn't matter how the player picks the initial door - the probability the car is behind this door is 1/3 regardless of how the player picks. This is what I said above.  If you prefer (and this matches vos Savant's instructions) we could force the player's pick to be random.  -- Rick Block (talk) 20:13, 11 June 2013 (UTC)

Moving along, now the host opens a door. As you already know, I want to contrast two different procedures. One (per vos Savant's explicit instructions from her cup simulation) is that we say only that the host "purposely" reveals an unchosen loser. This means the host must open a door and must not reveal the car, so if the car is behind one of the two unchosen doors the host must open the other door. However, if the car is behind neither unchosen door the host is free to open whichever one he wants. How he decides in this case is not specified. Why not? Because vos Savant did not specify. When we use this procedure, you (as the host) are free to use whatever algorithm you want (without telling anyone) to decide which of the two losers to reveal.The other procedure is the same, except we specify that the host must pick randomly between two losers (by flipping a coin, per what I already said above).The question is what is your expectation for the fraction of times the car will be behind door 2 if the player initially selects door 1 and the host opens door 3 following these two different procedures. Is it the same in both cases? -- Rick Block (talk) 15:36, 11 June 2013 (UTC)
 * I see what you were getting at before now.


 * So, the host has to open a door to reveal a goat. Sometimes he has two doors to choose from and has to decide which one to open.  The W/vS problem statement does not tell us how he decides, just as it does not tell us how the car and goat are placed and how the player chooses his door.  We have exactly the same options that we had before.  Last time you chose to assume that the choice was made uniformly at random because that is clearly how vS intended the problem to be understood.  lat me asl you in two stages what we should do now.


 * Do you agree that we should be consistent and base our decision on how vos Savant intended the problem to be understood?


 * Do you agree that she intended that the host should choose uniformly at random when he had a choice? Martin Hogbin (talk) 17:54, 11 June 2013 (UTC)


 * The car was hidden uniformly at random because this is explicitly how vos Savant says to simulate the problem. She said nothing whatsoever about how the host chooses between two losers, so we don't know her intent.  Given the precision of her instructions (and how she then proceeds to tally the results), it becomes clear she was not thinking about the conditional probability given the player picks door 1 and the host opens door 3 - but rather the overall chance of winning by switching (regardless of which door the player chooses and which door the host opens).  The former is the probability we're talking about, so we have two choices.  Proceed as per vos Savant's explicit instructions (how the host chooses between two losers remains unspecified) or constrain the host's choice in some manner.  I think I've made it abundantly clear I'd like to do both. -- Rick Block (talk) 20:13, 11 June 2013 (UTC)
 * Firstly, do you agree that we should be consistent and base our decision on how vos Savant intended the problem to be understood?


 * We know exactly how vos Savant intended the question to be understood because she tells us. She says in her response to Morgan, 'Pure probability is the paradigm, and we published no significant reason to view the host as anything more than an agent of chance who always opens a losing door and offers the contestant the opportunity to switch...'.


 * She also said, 'In the original column, no additional stated conditions appeared important to a general apprehension of the problem because circumstances in default are reasonably considered random'.


 * Her simulation suggestion is a different thing, she was was trying to convince people of the 2/3 answer and indeed may have overlooked some details. Are you talking about her shells suggestion or have I got that wrong or did she suggest another simulation with cups?  I do not think it is that important though.  You may think you can deduce how she expected the actual MHP question to be understood from her simulation but there is no need to do that; she tells us. Martin Hogbin (talk) 21:58, 11 June 2013 (UTC)


 * Is there some reason you're unwilling to follow through what happens in each case - one where we specify how the host chooses between two losers and one where we don't? The vos Savant simulation I'm talking about (like I've said already) is the one with cups that she describes in her 3rd column (all 4 columns are reproduced here).  Whatever her intent, her instructions are very clear.  And they don't specify how the host should make this choice.  You have said if we don't know how the host chooses in this case, the "subjectivist probability" of winning by switching is 2/3, even if we're considering the example case where the player picks door 1 and the host opens door 3.  I'm exploring precisely what this statement means.  In particular, I'm suggesting it doesn't mean that the fraction of times the car is behind door 2 when the player picks door 1 and the host opens door 3 will approach 2/3 in a large sample of shows - because if we don't know how the host chooses between two losers we can't know what value this fraction will approach (as I think you know, the value the fraction approaches can be anything between 1/2 and 1).  I will further assert that this fraction is what most people would consider to be "the probability" of winning in this case. -- Rick Block (talk) 06:32, 12 June 2013 (UTC)
 * I know exactly what happens in each case and do not dispute what you say but can we at least agree that we are discussing a difference in opinion in what vos Savant intended rather than some important issue of probability. In the two cases you describe above we get two different answers depending on how the host chooses which door to open. If we take a consistent view and follow exactly the same logic that we did with stage hand's choice of car placement and the player's initial door choice for the host's door choice than the answer is 2/3.  If, for whatever reason, we are inconsistent in our choice, and we take stage hand's choice of car placement and the player's initial door choice to be uniform at random but we take the host's door to be unknown then the answer is between 1 and 1/2. We can, of course, take all the unstated distributions to be unknown, this at least is a consistent approach to the problem.  Unfortunately it leads to the rather uninteresting result that the answer is from 0 to 1.


 * You have chosen to be inconsistent you choice of unknown distribution. What is your reason for this? Martin Hogbin (talk) 08:23, 12 June 2013 (UTC)


 * If someone says their subjective probability of some event is 2/3, it possibly means that they would be pretty indifferent between receiving 2 Euro's with certainty, or 3 Euro's if the event actually happened but 0 if not. Keynes famously said "in the long run we are all dead". Subjective probabilities are just a measure of subjective degrees of belief, which we can gauge by imagining whether or not we consider various bets fair or not. There is nothing at all in the Bayesian point of view to guarantee that if someone has a subjective probability of 2/3 on some event, that this event is going to happen 2/3 of the time in a long series of independent repetitions. There is no reason why subjective degrees of belief should match objective probabilities (Well, I must admit that I don't know what "most people" think; I am reporting here the official academic point of view on subjective probability). Richard Gill (talk) 08:01, 12 June 2013 (UTC)
 * That is an interesting but completely different subject that I would be happy to discuss with you elsewhere. Martin Hogbin (talk) 08:23, 12 June 2013 (UTC)
 * No, Martin, this is precisely the point. By using subjectivist probability you are moving to an abstract notion of probability that very, very, very few people (based on your arguments here over the last 4 years, I would conjecture not including you) actually understand.  In the context of the MHP, by using subjectivist probability you are denying the possibility that there can be any difference between the outcomes (after picking door 1) if the host opens door 2 or door 3.  Per the discussion with Albtal below, what this does is move the probabilities in these cases to what he calls the "centroid" - which is logically the same as moving the player's point of choice to before the host opens a door rather than after (which is how Gillman describes vos Savant's solution).  I don't have any particular problem with doing this as long you understand, and are willing to admit, what you're doing - and IMO (I've said this numerous times) this simply turns the problem into a mathematically obfuscated way (what Richard calls a joke) of asking whether you want the car if it is only behind, say, door 1, or behind either of the other two doors (i.e. completely avoids the paradoxical situation where you're looking at only two closed doors and an open door showing a goat, but yet the probability the car is behind either of the closed doors is not the same).  The difference here is precisely the point that Morgan says "confounds many".  Indeed it does. I thought I'd give this one more shot.  Based on your responses so far it's abundantly clear I'm wasting my time, so I'll stop.  -- Rick Block (talk) 16:09, 12 June 2013 (UTC)
 * Rick, by all means give up but please do not make such silly statements as, 'by using subjectivist probability you are denying the possibility that there can be any difference between the outcomes (after picking door 1) if the host opens door 2 or door 3'. Of course there can be a difference; that is what we are discussing above, and exactly what I have said.  If we take it that the stage hand places the car uniformly at random but that the host chooses a legal door non-randomly then there clearly is a difference between the 'conditional' and 'unconditional' answers.  My complain is that there is no justification for this assumption.  It was not the clearly-stated intention of either vos Savant or Selvin, it is not what we would expect from a real TV game show and it is not how a Bayesian would interpret the question.  So, yes, you are quite right, if you change the interpretation of the question to a perverse, unintended, and unrealistic one the answer might not be 2/3.  However, even Morgan now agree that the right answer is 2/3. Martin Hogbin (talk) 17:06, 12 June 2013 (UTC)

Discussion with Albtal

 * @frequentists: Look at MvS's cup example, take a variant with 1000000 cups (computers make it possible), and place the game in Las Vegas. All what you could say to the player, if he has chosen cup 1 and the host has lifted cups 2 to 777776 and cups 777778 to 1000000, is: "Your chance for switching now is between 1/2 and 1"; the same as for 3 cups. Here it is eminently eye-catching that the centroid, which is now 99,9999%, has completely vanished. And the poor player who would take your advice, would forfeit the chance of his life. But instead he would say: "Imagine that at this time the game took place a huge number of times in the world, and you could look at the results, then you would see - according to the law of "huge" numbers - that in "around" 99,9999% of the situations where the player has chosen cup 1 and the host did not lift cup 777777, switching has won." - One might say now, for example, that there could be hosts who never lift cup 777777 if possible. But others could prefer a fixed other cup if possible and so on, which leads to a 99,9999% proportion within the sample.--Albtal (talk) 22:18, 10 June 2013 (UTC)
 * You're simply avoiding the question - which is what conditions are necessary to guarantee the fraction of times switching wins in a large number of samples converges to the probability you're computing. With 1M cups (or doors) you'll need many, many millions of samples to show convergence to the probability.  My point is that unless you guarantee the host is picking which loser to reveal randomly (or picking a set of hosts with randomly distributed p's), you have no assurance this convergence will happen.  What this means (in frequentist terms) is that the probability is not what you're claiming it is.   It doesn't matter whether you're talking about a probability of 2/3 or a probability of 999999/1000000, whatever the probability is will be what the fraction approaches (given a sufficient number of samples), and, conversely, whatever value the fraction approaches must be what the probability is.  -- Rick Block (talk) 01:38, 11 June 2013 (UTC)


 * You answered your question, just as I did: The fraction approaches 999999/1000000 if we have 1000000 cups, 2/3 if we have 3 cups - if we pick a set of hosts with randomly distributed p's. But maybe now you have broken a law of frequentists.--Albtal (talk) 07:33, 11 June 2013 (UTC)


 * It sounds like we're agreeing here - i.e. by saying the probability is 2/3 (without knowing how the host chooses between two losers), the subjectivist is not saying this is the fraction of times the car will be behind door 2 given the player picks door 1 and the host opens door 3. Can we tease out a more concrete explanation of what the subjectivist actually is saying?  Over a large number of samples (where we don't know how the host is choosing between losers), is there anything that will necessarily be 2/3?  What is it?  -- Rick Block (talk) 15:53, 11 June 2013 (UTC)


 * We know for example, that with the strategy where the host always opens the rightmost door if possible (p=0), in 2/3 "of the cases" the chance for switching is 1/2, in "1/3 of the cases" 1. We can easily calculate the "weighted average chance": (2/3)*(1/2) + (1/3)*1 = 2/3. Instead of p=0 we can take an arbitrary value for p (0<=p<=1), and say: In 1/3 + (1/3)*p of the cases the chance for switching is 1/(1+p), in 1/3 + (1/3)*(1-p) of the cases this chance is 1/(2-p); and the ""weighted average chance", which we might call the Expected Value for the probability to win by switching, is, maybe trivially, independent of p:


 * ((1/3)*(1+p))/(1+p) + ((1/3)(2-p))/(2-p) = 2/3


 * The corresponding value for the 100 doors variant is 99/100.


 * The Expected Value also has much to do with the law of large numbers, the long-run average of the results of many independent repetitions of an experiment. And I think that a large set of games and hosts with randomly distributed p's is such a row of experiments (in mind) which also delivers the value 2/3 for the situation where the contestant chooses door 1 and the host opens door 3. And this is also the value which people "expect" who have understood the crucial aspects of MHP.--Albtal (talk) 18:32, 11 June 2013 (UTC)


 * Wonderful. Sounds like we're still agreeing.  Although in more concrete terms, isn't the expected value you're talking about the fraction of times a player who picks door 1 and wins by switching (to either door 2 or door 3, whichever one the host does not open)?  Isn't this conventionally denoted as P(win by switching|player picks door 1), not P(win by switching|player picks door 1 and host opens door 3)?  In particular, to observe this result we must not tally only those cases where the player picks door 1 and the host opens door 3, but all cases where the player picks door 1 (regardless of whether the host opens door 2 or door 3).  Just to be clear, I agree if we do this we'll see this fraction approach 2/3 regardless of p.  The "randomly distributed p" is harder to observe, but could be observed perhaps as follows.  Before opening a door the host obtains a random number (p) between 0 and 1, and a second random number between 0 and 1.  In the event there are two losers to choose from, the host opens the higher numbered door if the second random number is larger than the first and otherwise opens the lower numbered door.  For all intents and purposes this is simply a more complicated method of flipping a coin to decide which door to open between two losers (but it more closely matches our model of what's actually going on - i.e. p is random rather than p being 1/2).  If we do this, then I also agree we're ensuring the fraction of times a player who picks door 1 and sees the host open door 3 wins by switching will approach 2/3.The question now becomes, how did we know to run the simulation this way?  It seems to me subjectivists are willing to insist until they're the only ones left talking that p doesn't, cannot, will not, won't in a million years matter - i.e. if you don't know p your chances of winning (regardless of which door you pick and which door the host opens) are 2/3.  However, if we tally only cases where the player picks door 1 and the host opens door 3 (or any of the other 5 possible combinations of player pick and host reveal) the fraction of times switching wins doesn't necessarily approach 2/3 unless we carefully correct for p.  This means to me that p does matter - and that the subjectivist's statement about the probability of winning by switching being 2/3 (given ignorance of p) does not mean what most people would expect it means. -- Rick Block (talk) 19:55, 11 June 2013 (UTC)


 * Sorry, but as it is too time-consuming I have to stop talking for a time. I think, as I wrote, that the ongoing problem arises in equal measure with the simpler game Do I have a coin in my hand?. It seems that a "pure frequentist" is not able to answer the question "What is the chance for "Yes"?", whereas a "subjectivist" takes (1/2)*p + (1/2)*(1-p) = 1/2. (As usual we recommend to the player that he should choose his answer randomly by tossing a coin. But we cannot recommend to the host that he decides in this way whether he takes a coin or not.) All what the "pure frequentist" could say is that the chance is between 0 and 1. Whether this coin game or the n doors game: The "subjectivist" seems to have the concept of probability which fits best. It is a concept which "preserves the centroid".--Albtal (talk) 10:58, 12 June 2013 (UTC)

One In Three or One In Two?
The Monty Hall problem has been much discussed throughout the world, and there is an almost universal acceptance that the correct choice is to "change your choice".

The problem, as described, states that:

"Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?"

with the qualification that:

"the host always opens a different door from the door chosen by the player and always reveals a goat by this action".

Before the game starts, and with the knowledge that the host will show you a goat, then this effectively makes the game a one in two choice game, and not a one in three choice game.

Much of the discussion surrounding the probabilities involves an opening argument that the door chosen has a one in three chance of winning but this is incorrect. With the knowledge that the host will remove one of the wrong choices by showing you a goat, then the game is reduced to a one in two chance game.

Thus, the probability of you having selected the correct door is 1/2. There is only one other choice (remembering that the host has removed one choice by showing you a goat), and the chance of the car being behind this door is one in two.

The so called paradox is created by ignoring the fact that before the game starts, there is the knowledge that one of the choices is removed by the host.

Craigkb (talk) 15:19, 25 June 2013 (UTC)


 * Are you seriously trying to tell us that if you pick one of three doors at random at the start, knowing that the host will then open a door to reveal a goat, the probability that your originally chosen door hides the car is 1/2? Martin Hogbin (talk) 18:26, 25 June 2013 (UTC)
 * User:Craigkb, you've illustrated the compelling cognitive illusion that people fall into when thinking about this problem, which is why so much is written about it. Just because there are two options does not mean the probability is distributed evenly between them. As Martin above points out, your position implies that faced with options A,B and C the probability of each of them being correct is 1/2, which shows something wrong. To see the error more clearly, imagine that there are one million doors rather than three, and that once you make your choice the host will open 999,998 doors that have goats behind them. Does it still seem to you to be equal probabilities of 1/2? MartinPoulter (talk) 22:32, 25 June 2013 (UTC)

Think of it binary numbers.

There are 3 possibilities:

100 (car is behind door 1), 010 (car is behind door 2), 001 (car is behind door 3),

The first part of the game - you picking the door - is pure showmanship. The real game begins when Monty removes a goat (or a zero in our simulation), which he is obliged to do by the rules.

Thus:

100 becomes either 10 or 10, 010 beomes either 01 or 10, 001 becomes either 01 or 01

Note that there are three 10's and three 01's

Thus a random sequence of the three binary digits representing the start of the game could be say:

010 100 100 010 001 001 001 100 010

and becomes a sequence of two binary digits after Monty removes one of the zeros, and as we have shown, with equal probability of a 10 or a 01 occuring.

It should be intuitive at this point that the odds of the real game i.e. the part after he removes a goat or a zero, is the equivalent of a toss of the coin. It matters not whether you switch or keep your original selection.


 * Indeed this fact seems intuitively obvious to almost every newcomer to Monty Hall Problem but it's wrong! Hence the paradox. Richard Gill (talk) 14:01, 4 July 2013 (UTC)

The vast majority of the discussion I have read on this problem is flawed because it fails to recognize that there is a change to the game once Monty removes a goat. All the probabilities are calculated when there are three choices (thus you see references to "one in three" or "two in three"), but the real game - the part after one of the three doors is removed - has only two choices. Three is not a valid denominator in the probability calculations.

In simpler language, you start off choosing one of three, Monty removes one of the two remaining choices and returns your selection to "the deck" and effectively gives you a choice of one in two.

Craigkb (talk) —Preceding undated comment added 06:53, 26 June 2013 (UTC)

Statistical Independence Of Two Events

The whole game consists of two events.

1. You select one of three doors

The outcome of this event or "game" is that Monty shows you one of the other doors that has a goat behind it. There is no prize and it effectively eliminates one of the doors from the game - let us assume here that the contestant will not pick the door that he or she has been just shown to contain a goat!

2. Monty asks you if you wish to keep your choice, or choose the remaining door. By this time he has reduced the original three down to two by showing the negative result of one of the doors.

This second event is statistically independent of the first event, and thus the probabilities are calculated independently of what has happened in the first event.

Failure to recognize statistical independence is a common error.


 * Suppose you pick a door completely at random. Suppose the car was hidden completely at random, independently of your choice. Suppose that given the location of the car and the door chosen by you, the host opens a door at random when he has a choice. These are the statistical independencies which we may assume. Any others need to be deduced. The door opened by the host is *not* statistically independent of the door chosen by the player, nor of the location of the car. Suppose the player initially chose door 1. Then the host is half as likely to open door 3 if the car is behind door 1 as when it is behind door 2, whereas a priori the host is equally likely to open any door.


 * Failure to recognize statistical dependence is a common error.  Richard Gill (talk) 09:47, 4 July 2013 (UTC)

The best example comes when you visit a casino. If you first go to the roulette table, the croupier will give you a card and a complimentary pencil with which to record previous spins. This is because the outcome of one spin is statistically independent of anything that has previously occured. Number 26 black might have come out the previous seven spins, but as soon as the croupier lifts the ball and sends it around for a new game, 26 black has exactly the same chance of coming out as any other number - assuming a fair table.

If you tire of roulette and wander over to the Blackjack table with your card and pencil, and start writing down the cards that are coming out, in all probability you will be asked to leave the premises. This is because one Blackjack game is statistically dependent upon the outcome of the previous game i.e. the cards are not replaced in the deck, and if you have knowledge of the previous cards that have emerged then you have a mathematical edge over the casino.

The Monty Hall problem consists of two events, both statistically independent upon each other. Failure to recognize:

i) the existence of two events and ii) the statistical independence of the two events

has led to many, many hours of calculations made under flawed assumptions. — Preceding unsigned comment added by Craigkb (talk • contribs) 07:25, 26 June 2013 (UTC)

Craigkb (talk)
 * Craig, you have not actually answered my question. Here it is again.  A car is placed behind one of three doors, and goats behind the other two. You pick one at random.  Given that, after you have picked a door the host will open a different door to reveal a goat, what is the probability that your originally-chosen door hides the car?  Martin Hogbin (talk) 11:59, 26 June 2013 (UTC)

Martin. Door 1 has a 33.3% chance of containing the car. Door 2 has a 33.3% chance of containing the car. Door 3 has a 33.3% chance of containing the car.

The action of Monty Hall showing one of the three doors to be a goat (and this action is a very important consideration) means that there are now two doors, both with equal chance of containing the car - from the perspective of the contestant. The contestant is now given the choice of keeping his original choice (50% chance) or discarding it and taking the other (50%).

A better understanding of the optimum strategy might be known if you consider this sequence of events - and you are the contestant.

You are faced with three doors and you select one of them - let us say Door 2. Note that there is no prize at this point. So the probability of you winning a car at your first selection is zero. Monty never says "you have selected the car - congratulations and that is the end of the game".

At this point, and sticking with your perspective, Monty opens Door 1 and reveals a goat (remember you selected Door 2). Monty then gets the floor manager to take the goat away and dismantle Door 1.

You are standing there, and what has changed is that you are now looking at Door 2 and Door 3. Remember you have no idea which door the car is behind.

I put it to you, that from your perspective, there is a 50% chance that the car lies behind Door 2 and a 50% chance that the car lies behind Door 3. If you have no idea of the correct selection, then it follows deductively that changing your selection makes no difference.

The paradox arises because people fail to realize that the first part of the show, selecting one of three doors is totally irrelevant to the final game, which is the selection of one door of two.

Monty could have just as easily not asked you to make an initial selection, and just showed a door with a goat. He would then ask you to make a selection of the remaining two doors. It is exactly the same thing.

There is a bit of logical trickery there, but the first selection is simply an exciting means of reducing a one-in-three chance game into a one-in-two chance game, and there is no mathematical advantage to switching your selection.

Craigkb (talk) 15:30, 26 June 2013 (UTC)


 * Craigkb - you are absolutely correct that most explanations are flawed in precisely the way you say (they aren't addressing the situation after the host has opened a door). However, you are incorrect in your analysis of the probabilities that arise after the host has opened a door.  Please read this section of the article, which examines the situation after the host has opened a door using conditional probability.  The probabilities for the two remaining doors are not each 50% (as they might seem), but rather 1/3 for the originally chosen door and 2/3 for the other one.  If you knew nothing about how this situation involving two doors arose, the chances the car is behind either door would be 50%.  But you know that after you picked a door the host cannot open your door and cannot open a door showing the car - which means the host's actions are not statistically independent of your original choice.  Please read the section of the article I've linked to, and look at either the decision tree or the large figure.  Post again here afterward.  -- Rick Block (talk) 15:43, 26 June 2013 (UTC)

The key to the paradox: It's not random

 * And Craig, please read what Leonard Mlodinow says in section "The paradox" (he worked on a book together with Stephen Hawking).
 * Mlodinow calls it conditional probability. He says most people are wrong and think "two still closed doors, so it doesn't matter if you switch or don't switch". They think it's random, they think it's 50:50, they think it doesn't matter. But that's wrong. He says we should be careful:
 * Monty asked you to select one of three doors, and you selected a door (with only a "1 in 3" chance to hide the car), and then Monty would open one of the two you didn't choose, in order "to show a goat". ( ! )  –  But he won't show you the car!
 * So it's not longer random. Most people think it's still random, but it isn't random any more! Because:
 * Only in case your guess was right and you have picked the car right off, and accordingly you are in the lucky guess scenario (only a "1 in 3 chance"), only then the host truly opens one of the two other doors at random, as both hide goats.
 * Mlodinow says "then he does not care which booby-price he shows you, or we are going to assume that he doesn't."
 * And I am adding: in that lucky guess scenario, where both non-selected doors hide goats, and the host opens "one of them", it's completely futile to ponder whether the host "could be biased to prefer a special goat to show, or to prefer a special door to open" – or not, resp. "to have been biased", or not. – Because you would succumb to a hidden cryptic fallacy: any such futile "assumptions" are completly inapplicable as long as you're only just assuming but not exactly "knowing" (see Ruma Falk). Because, by just "assuming", you necessarily would be misguided and necessarily would come to actually wrong conclusions. So it's completley futile "to condition on wrong assumptions".


 * But in scenario 2 or 3 which both are "wrong guess scenarios" where you selected goat A (1 in 3) or selected goat B (also 1 in 3), what happens then is completely DIFFERENT: Then the host is extremely biased, as he is slave to the act and specifically avoids to show the car, and because of this he never can open a door at random. Mlodinow says: because it's no longer random, "you have been given more information, and that's what changes things". – Because in both wrong guess scenarios, when you selected one of the two goats, the host will not show the car but he always will show the other goat, and in both cases you win by switching.
 * Please consider that in 2 out of 3 the host is extremely biased and never will open a door at random. Never at random. So you have been given more information. That's "the key" to let the paradox arise, and the key to understand the paradox presented by Marilyn vos Savant.
 * And have a short look (3 minutes only) to min.25 – min.28 of 51:35 min, or to the page of university of California, San Diego. They also show, outside the "paradox", another scenario, contradictory to the premise of the standard paradox: "The host does not know where the car is". Gerhardvalentin (talk) 16:48, 26 June 2013 (UTC)

@Craigkb [edited in parallel with Gerhardvalentin's comment], reinforcing the comments of Martin Hogbin, MartinPoulter, and Rick Block: Your argumentation implies that your chance for one of the three doors, say door 1, is 1/2 even without the host opening a door. So pick door 1. You know that the host will open door 2 or door 3 with a goat. Now you might say: "I know that after you have opened door 2 or door 3 with a goat I shall have a one in two chance with door 1. So let the door closed. I have a one in two chance with door 1 anyway." - Or simply close your eyes: Does it matter for your chance with door 1 whether the host really opens a door or not? - I think you didn't consider that the host is restricted to doors 2 and 3 when showing a goat: If he opens door 2, you pick door 3, and if he opens door 3, you pick door 2. So you will win by switching in two of three cases. - The contestant correctly could say: "I have a 1/3 chance with door 1; and I know that you will open door 2 or 3 with a goat. If my chance for door 1 would change then, maybe to 1/2, I would know this already now, which is a contradiction. So my chance for door 1 stays 1/3, and my chance for the other door will be 2/3."--Albtal (talk) 17:09, 26 June 2013 (UTC)

Rick. That link you posted (Conditional probability by direct calculation) has a graphic. This graphic contains the legend "Tree showing the probability of every possible outcome if the player initially picks Door 1". This is incorrect. The probabilities shown (1/6, 1/6, 1/3 and 1/3) are the probabilities that the host will reveal a particular goat. Using my language, they are probabilities of the outcomes of Event 1. They determine the format of Event 2, but are irrelevant to the outcomes of Event 2. A new graphic needs to be produced to show the probabilities of Event 2.

Using the same format, this new graphic would have a legend of (say) "Tree showing the probability of every possible outcome if Door 1 was eliminated". There would be two lines with each 50% chance leading to the first column which has a title of 'Car Location'. This column would contain the values 2 and 3 (1 has been eliminated). The next column would be titled 'Contestant Chooses', but would be limited to one value with a probability of 100% because there are no other choices. The final column, the 'Probability' would be both 50%, representing the true probability of the two outcomes in Event 2.

Craigkb (talk) —Preceding undated comment added 23:14, 26 June 2013 (UTC)

Albtal. You said "Your argumentation implies that your chance for one of the three doors, say door 1, is 1/2 even without the host opening a door."

My argumentation does not imply this at all. I am saying that the selection of one of three doors is irrelevant. This is pure show, and a preparatory measure to the second event.

There are two clearly defined events. Is this fact in dispute? If you dispute this, please argue and show why. If you agree with this, please show me how they are statistically dependent upon one another. — Preceding unsigned comment added by Craigkb (talk • contribs) 23:43, 26 June 2013 (UTC)


 * Craigkb - lets take this one step at a time. And can we use an idealized sample of 300 shows where the player has always picked door 1?  I think you agree that in this idealized sample we expect the car to be behind each door 100 times (which is what it means for the probability of the car to be behind each door to be 1/3).  Out of these 300 shows, please answer what your expectations are for the following (I assume we already agree about the 100 times behind each door):
 * a) Number of times the car is behind door 2 AND the host opens door 3? ____
 * b) Number of times the car is behind door 3 AND the host opens door 2? ____
 * c) Number of times the car is behind door 1 AND the host opens door 3? ____
 * d) Number of times the car is behind door 1 AND the host opens door 2? ____
 * e)Total number of times the host opens door 2? ____
 * f)Total number of times the host opens door 3? ____
 * Please let us know what you think these numbers will be. -- Rick Block (talk) 00:35, 27 June 2013 (UTC)

@Craigkb: It seems that you don't see a difference between "joker 1" and "joker 2" in the following game which I posted earlier:

''Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. The host knows the door with the car. Instead of simply choosing a door you have the choice between two "jokers":

1. You may ask the host to open one door with a goat before you choose your door.

2. You may determine two doors of which the host has to open one with a goat before you choose your door.

Which joker should you choose?--Albtal (talk) 08:50, 27 June 2013 (UTC)--


 * Craigkb has correctly observed that given the door opened by the host, the car is equally likely behind either of the two closed doors (cf. Vos Savant's little green woman from Mars). What he hasn't realized is that given the door opened by the host and the door initially chosen by the player, the car is not equally likely behind either of the two closed doors. He has reduced the game to just two events, but there are actually three which we need to keep track of. He discarded information which intuitively seems irrelevant (which door did you initially choose?) but actually is very useful. Richard Gill (talk) 14:20, 4 July 2013 (UTC)
 * Conclusion: the host's opening of a door never is fully at random. Yes, still randomness regarding the location of the car, after the guest's first pick. But randomness is ending at this point. In the article, it endlessly hasn't been obvious enough for the reader that the host, after the guest made his first selection, in 2 out of 3 is slave to the guest's first pick, so no more randomness after the host's opening of a door "in order to show a goat". Gerhardvalentin (talk) 06:26, 5 July 2013 (UTC)

Please explain why it's not one-in-two?
124.169.61.181 (talk) 13:12, 19 July 2013 (UTC)Please edumicate me. I thought I was quite good at probability some years ago. After the contents of door 3 are disclosed as a goat, why is it not then 50/50 as to which the others contain? I don't get the 2/3 thing unless I need to watch the show to determine the full rule space. Best regards, Pete


 * Can you answer the questions I asked in the thread above about an idealized sample of 300 shows where the player has always picked door 1? I suspect you'd agree that in this idealized sample we expect the car to be behind each door 100 times (which is what it means for the probability of the car to be behind each door to be 1/3).  Out of these 300 shows, please answer what your expectations are for the following (I assume we already agree about the 100 times behind each door).
 * a) Number of times the car is behind door 2 AND the host opens door 3? ____ 100 --124.169.61.181 (talk) 01:23, 20 July 2013 (UTC)
 * b) Number of times the car is behind door 3 AND the host opens door 2? ____ 100 --124.169.61.181 (talk) 01:23, 20 July 2013 (UTC)
 * c) Number of times the car is behind door 1 AND the host opens door 3? ____ 50 --124.169.61.181 (talk) 01:23, 20 July 2013 (UTC)
 * d) Number of times the car is behind door 1 AND the host opens door 2? ____ 50 --124.169.61.181 (talk) 01:23, 20 July 2013 (UTC)
 * e)Total number of times the host opens door 2? ____ 150 --124.169.61.181 (talk) 01:23, 20 July 2013 (UTC)
 * f)Total number of times the host opens door 3? ____ 150 --124.169.61.181 (talk) 01:23, 20 July 2013 (UTC)
 * If you pick door 1 and then see the host open door 3, your chances of winning by switching is your answer for (a) divided by your answer for (f). And, I think it should be fairly obvious that (e)+(f)=300, (a)+(c)=(f), and (c)+(d)=100.  -- Rick Block (talk) 15:36, 19 July 2013 (UTC)

--124.169.61.181 (talk) 01:23, 20 July 2013 (UTC) Thank you Rick Block. I'm grateful for you repeating the important part of your explanation without getting cross with me. Seems I am not watching nearly enough game shows on the telly to realize the complexity of them! Best regards, and thanks, Pete


 * The numbers help make the answer believable, but they don't really convey the "why" - which is that if the host opened door 3 it was either because the car was behind door 2 (which it is 1/3 of the time) and the host had no choice but to open door 3, or because the car was behind door 1 (which it also is 1/3 of the time) and the host randomly chose to open door 3 (halving the original probability it was behind door 1). It is this asymmetry in the reason the host opens door 3 (forced in one case, random choice in the other) that makes the probabilities end up in the ratio 2:1 in favor of door 2, i.e. 2/3 chance the car is behind door 2 but only 1/3 chance the car is behind door 1.  Most explanations basically ignore the door the host opens - saying things like "the chance the car is behind door 1 is 1/3 and if you stay with your original choice this is your chance of winning the car, so your chance of winning by switching must be 2/3".  This is true, but it's not talking about the case where you pick door 1 and the host has then opened door 3.  In particular, it says if you switch to whichever door the host opens (i.e. you decide to switch before seeing which door the host opens) your chance of winning is 2/3 - you win the car by not switching only if it's behind door 1 (1/3 chance), but you win if you switch if the car is behind either door 2 or door 3 (2/3 chance).  Translating this observation (which is pretty simple) to the situation where you've picked door 1 and have seen the host open door 3 is a step most explanations don't take.  If your chances of winning by switching are 2/3 if you decide to switch before seeing which door the host opens, and if your chances of winning by switching are the same whether the host opens door 2 or door 3 (which they are if the host randomly chooses which door to open if the car is behind door 1), then your chances must be 2/3 in both of these cases.  Although an entirely logical (and valid) argument, this still doesn't quite ring true with many people.  I'm in the minority here, but I think most people remain unconvinced unless they understand the specific case the problem statement directs you to - i.e. the case where the player picks door 1 and the host opens door 3.I assume you've read the article in its current form and remained unconvinced.  There's an example with the numbers I suggested you think about in the article.  Did you miss this?  Or had you stopped reading before you ran into it?  Any suggestions for improving the article?  -- Rick Block (talk) 04:08, 20 July 2013 (UTC)

124.169.61.181 (talk) 04:55, 20 July 2013 (UTC)Rick mate, You've clarified it to the point where I'll make it a party problem next Friday between engineers and hospitality! Best regards, Pete

Example Proof of the Monty Hall Problem, using Bayes Theorem
I moved the following text from the article (see talk page). I put it here, for reference. Richard Gill (talk) 12:10, 23 August 2013 (UTC)

Example Proof of the Monty Hall Problem, using Bayes Theorem

As an example, assume you choose Door 1, and then the host opens door 3. We can define the“events” as follows:

Event A1 = Prize is behind Door 1; therefore p(A1) = 1/3 and  p(B|A1)= 1/2, because if you chose Door 1 and that is where the prize is,the host could choose to open either of the other two doors.

Event A2 = Prize is behind Door 2; therefore p(A2) = 1/3 and  p(B|A2) = 1, because if you already picked Door 1, but the prize is behind Door 2, then the host must open Door 3.

Event A3 = Prize is behind Door 3; therefore p(A3) = 1/3 and  p(B|A3) = 0, because he won’t pick Door 3 if that is where the prize is.

Event B = host opens Door 3

p(B) is the probability that he opens door 3. This could happen one of three ways, and you must take all possibilities into account in probability.

Either, (the prize is behind Door 1 AND he opens Door 3) OR (the prize is behind Door 2 and he opens Door 3) OR (the prize is behind Door 3 AND he opens Door 3).

Or as an equation, using the Multiplication Rule of Probability and the Additional Rule of Probability,

P(B) = p(A1)p(B|A1) + p(A2)p(B|A2)+ p(A3)p(B|A3)

P(B) =(1/3)(1/2) + (1/3)(1) + (1/3)(0)

P(B) = ½

So you want to know which is greater,

1. The probability that the prize is behind Door 1 given that he opened Door 3, p(A1|B), or

2. The probability that the prize is behind Door 2 given that he opened Door 3, p(A2|B)

This is where Bayes Theorem comes in...

The probability that the prize is behind Door 1, given that the host opened Door 3 = p(A1|B) = p(A1)p(B|A1)/p(B) = (1/3)(1/2)/(1/2) = 1/3

The probability that the prize is behind Door 2, given that the host opened Door 3 = p(A2|B) = p(A2)p(B|A2)/p(B) = (1/3)(1)/(1/2) = 2/3

Therefore, in this case you are better off switching to Door 2. And this will work out the same no matter which door you pick and which door you assume he’ll open, because then the three conditional probabilities p(B|A1), p(B|A2), p(B|A3), will switch around.


 * In the past there has been a mathematical solution using Bayes. Who removed it, and why?? Nijdam (talk) 12:16, 24 August 2013 (UTC)


 * It was removed a long time ago, I think by me, but only after discussion on the talk pages. It was removed because it is superfluous. It consists of introducing some mathematical notation, writing out a long formula, substituting model ingredient probabilities, and then calculating the answer. The calculation has already been done in words which everyone can understand earlier in the article. We don't learn anything new about MHP by redoing the same solution in a language which much fewer people understand. (In fact, the article on Bayes' theorem had a solution of this kind as an illustration of Bayes theorem. That's a sensible place to do such an exercise.) Richard Gill (talk) 13:38, 3 September 2013 (UTC)

Falsifiable
(This paragraph has been moved from the article talk page to the Arguments page)

The "explanation" for this theory bases itself on the claim that the middle option out of three is statistically likely to be a much, much higher chance of being the car. If the options are completely random, this is completely false.

The only way for the options not to revert to 50/50 is if, somehow, a completely random door generation leaves the middle door as having the car much higher than is statistically possible. 124.168.241.91 (talk) 11:32, 6 September 2013 (UTC) Sutter Cane


 * Please try to understand the intended scenario of this world famous unintuitive paradox that is based on the premise that the host knows what's behind the doors, and in opening of a door will always show a goat but never the car. I repeat: the host will never open a door that hides the prize. So it is NOT on the 2/3-subset only, where the host by randomly opening of an unselected door happened to show a goat simply by chance. The remaining 1/3-subset of ALL events, representing 1/2 of all "winning events", may not be discarded.
 * We have to consider that only in the "lucky guess scenario", where the contestant by chance initially picked the door with the car (be it door #1, #2 or #3) the host is really free to open any of this two remaining doors at random, as both non-selected doors hide goats. Only in this 1-out-of-3 case where the host may open an unselected door at random to show a goat, sticking wins the car and swapping to the door offered will hurt, as it hides the second goat.
 * But in the remaining two "wrong guess scenarios", where the contestant initially picked a wrong door hiding either goat A or a wrong door hiding goat B, the host never can nor will open one of the two unselected doors at random, as one of them definitely will hide the car. In that 2/3 of all events where swapping will win the car for sure, the host is forced to show nothing but the other goat and must offer a swap to his door that hides the car. In these 2-out-of-3 cases swapping wins the car for sure.
 * The correct answer to this famous paradox is that players who stick have a chance of 1/3 only to win the car, whereas players who swap to the door offered as an alternative have double chance of 2/3 to win the car. Please read section "The paradox" and have a look to University of California San Diego, Monty Knows Version and Monty Does Not Know Version, An Explanation of the Game.
 * Regards, Gerhardvalentin (talk) 22:21, 6 September 2013 (UTC)

Sigh. It is irrelevant whether or not the "host" knows what's behind which door. The problem claims that the host always opens the third door after the contestant picks the first door. The host then asks if the contestant wants to switch doors. The claim is that, somehow (magically apparently), the first door will have a 66% chance of being a goat, so the player should switch.

The host isn't switching the "doors" around. So given that the 2 goats and a car are RANDOMLY placed in their perspective areas, if the third door is ALWAYS a goat then the first and second doors have a 50/50 chance of being a goat or a car.

In fact I mention this as I have recently finished doing this problem after reading about the ridiculous, illogical reasons people come up with as to how a randomly selected three door pick, in which the first door is always picked and the third door is always a goat, would magically significantly statistically come up with the first pick as a goat every time. I've just finished doing this 500 times over the last few days. I couldn't be bothered making it to a thousand, even after the ridiculous amount of people claimed (after hilariously changing it from 100) it "totally proves it".

I'm sorry but it's effectively 50/50. In fact, as i'll point out below, the claim you should switch is statistically incorrect.

Taking 3 cards, a black (car) and two reds (goats), I shuffled them repeatedly (5 times quickly per deck to create a truly random system). I would then lay them out into a row of three.

- Card(1) Card(2) Card(3)

I then lifted the third card, which if black I put back and repeatedly shuffle again until I lift a red, and when red I flip over facing up.

- (Card(1) Card(2) Red(3)

Now, following the scenario, I choose the first card. I ask myself if I want to choose the second card, obviously picking no, and I flip both cards 1 and 2.

Example: Black(1) Red(2) Red(3)

Now a slight discrepancy is just a statistical likelihood, it explains what I found, but this ridiculous claim of 66% likelihood of hitting goat if you don't switch is moronic.

Out of 500, FIVE HUNDRED, attempts I ended up with 54 percent of attempts of not switching coming up with black (the car).

54% FIFTY FOUR PERCENT. In which NOT switching comes up with the car.

The claim that the choice between the first and second card after the third is shown to be a goat is 66% (1/3) in favour of the first choice being a goat if you don't switch is flat out false. It's 50/50. 124.168.241.91 (talk) 01:16, 7 September 2013 (UTC) Sutter Cane


 * If Monty opens a remaining door at RANDOM to show a goat, then things are different. This time you will be left with two doors, each with 50% chance of the car. Here is why.  In 2/3 of all cases Monty will show you a goat (by symmetry) and in 1/3 of all cases you have picked the car correctly initially (also by symmetry).  P(door 1 has car given that a goat was shown) =P(door 1 has a car and a goat is shown) / P(a goat is shown) = (1/3) / (2/3) = 1/2.  Note that P(door 1 has a car and a goat is shown) = 1/3 because the chances of door 1 having a car is 1, and in that case the chances of showing a goat has to be 1 (since there are then only goats to show).
 * So if he shows you a goat, you are on a 50/50, if he shows you a car, you are on a 1 (100% chance of winning) - you only get the car revealed 1/3 of times, so it works out at a 2/3 chance of winning. This 2/3 chance of winning makes sense because two doors are opened, but in this 'monty has no knowledge' version, the final door is effectively chosen at random (with a 1/3 chance of the game terminating with the car behind it).  Gomez2002 (talk) 15:27, 17 October 2013 (UTC)


 * Sorry, you are not talking about the intended scenario of the famous *paradox*, where the host strictly avoids to show the prize in the 2-out-of-3 cases where the contestant picked one of the two goats. Although it is a clear premise of the famous paradox that the host can only show a goat but never the prize, because he knows what is behind the doors and he intentionally avoids to show the prize.
 * But in contrast, by your three cards you only show the remaining subset of 2/3 of cases where the host just only "by luck" opened a door with a goat behind, after you have discarded all events (1/3 !) where the host by accident happened to open a door with the car behind, instead of showing the second goat as per the the clear premise of the famous paradox, where switching necessarily will win. You have just silently discarded "those 1/3 winning events" of ALL events and are referring only to the remaining subset of 2/3 of ALL events, meaning that in effect you discarded exactly "one half of all winning events". See the link that I gave above to UCSD. They show that it is quite another problem but no more the famous paradox, if the host does NOT know where the car is located.
 * If Monty (the host) in a plain winning event should happen to open the door with the car behind by accident, instead of a goat, and you interpret this to mean that – if the car is revealed – then the game is over (winning event discarded) and the next contestant plays the game, then this is no more the scenario of the famous paradox. Then it is quite "another problem", where the chances "staying:switching" are only "1:1". Regards, Gerhardvalentin (talk) 08:18, 7 September 2013 (UTC)

Sutter, the Monty Hall problem is famous because so many people, like yourself, believe that the odds of winning by switching are 50:50. The correct answer is that the odds are 2:1 in your favour if you switch. This fact has been proved by many people using various forms of mathematical analysis and it has been conclusively verified by numerous simulations, some using computers some not.

The question to be asked is why you are not convinced by any of the arguments given in the article. If you have a distrust of mathematicians and statisticians you can indeed to a simple simulation of the problem with a pack of cards or three cups and a pea but, as Gerhard points out, you must do it correctly, that is to say, in a way that corresponds to what happens in the problem.

Here is how to do the simulation correctly with cards.


 * 1) Take three cards, two red, representing goats, and one black representing the car.
 * 2) Have a friend shuffle the cards and place then face down on the table so that you both have absolutely no idea which card is which.
 * 3) Chose one card but do not turn it over.
 * 4) Have your friend look at the two remaining cards, without letting you see either of them, and always turn a red card face up.
 * 5) After your friend has done this, decide whether to stick with the card you originally chose or to switch to the other face-down card on the table,
 * 6) Repeat the experiment many times and keep a record of how many times you win by sticking and how many by switching.
 * 7) Let us know your results. Martin Hogbin (talk) 11:33, 7 September 2013 (UTC)


 * Ha! Martin Gardner's three shells problem, which Marylin vos Savant clearly knew about! Marilyn seems to know instinctively that the three shells problem and the three doors problem are the same. Actually they are different, but the latter can be reduced to the former by arguing that by symmetry, the actual door numbers in any particular case don't matter.


 * I would suggest that one explains the easier problem (the three shells problem) first. After all, the first problem on wikipedia is to get people to realize that it's not 50-50. If you can explain people through a simulation experiment, start with the most simple possible simulation experiment. Don't keep track of any door numbers. After they have got that, you can start to look at the problem in different ways.


 * Three cards get placed face down next to three numbers. The host takes a peek. The player may choose any door, the host secretly tosses a coin to determine which different card to turn over in case he has a choice. (He should make a big show of tossing his coin every time, also on those occasions when he doesn't need it). You can study the success rate of switching for each of the six possible situations the player can be in, at the last moment before they would have to decide between switch or stay.


 * It's a more complicated experiment but it might be helpful if you want to explain to people the difference between unconditional and conditional probabilities. A difference which in the standard MHP turns out not to be very important. Something which one could have realized in advance. Richard Gill (talk) 17:20, 10 September 2013 (UTC)
 * As you know we disagree about the MHP and its 'correct' interpretation and solution but I do very much agree with your approach. Start by explaining the solution to the shell (unconditional or whatever else you want to call this) in order to convince the listner that the answer is 2/3 not 1/2 then, for those interested, the complications of conditional probability can be discussed ad infinitum. Martin Hogbin (talk) 20:01, 10 September 2013 (UTC)
 * Martin - as you well know, this doesn't quite match the description of the problem. The cards need to be put next to numbers 1,2,3 on the table, and you should record only those cases where the initial pick is #1 and the card the host turns up is #3.  This will go faster if in step 3 the initial pick is always #1.  Gerhard's point is that in step 4 the host MUST turn over a red card (which means if the initial pick is #1 the host CANNOT always turn over #3).  And now, if you do the experiment you still might not see the 2:1 odds in favor of switching - particularly if the host looks at #3 first and turns it over if it's a red card (in the same spirit of speeding things up as the player always picking #1).  The issue here is you need another constraint in addition to the host always turning over a red card - specifically, that the host MUST pick which card of the remaining two to turn over randomly if they are both red cards (i.e. the host can't just look at #3 and turn it over if it's a red card but must look at both and if both are red cards then must flip a coin or something to decide which one to turn over).So, here's the revised experiment:
 * Take three cards, two red, representing goats, and one black representing the car.
 * Shuffle the cards and place them face down on the table next to numbers 1,2,3 so that you both have absolutely no idea which card is which.
 * Choose one card but do not turn it over (always pick #1).
 * Have your friend look at both of the two remaining cards, without letting you see either of them, flip a coin, and always turn a red card face up. Either there is only one red card, in which case that is the one your friend must turn up, or there are two red cards.  If there are two, which one is turned up is determined by the coin flip (for example, turn up the lower numbered one if the result of the coin flip is heads).
 * After your friend has done this, decide whether to stick with the card you originally chose or to switch to the other face-down card on the table,
 * Repeat the experiment many times and keep a record of how many times you win by sticking and how many by switching in the cases where you picked #1 and your friend turned over #3.
 * Let us know your results.


 * Rick Block (talk) 17:04, 7 September 2013 (UTC)
 * Rick, you're describing not the Monty Hall problem but something crucially different. Martin above has a more accurate description of the Monty Hall problem. Where in any description of the MH problem does it say that the host/ the friend is constrained to turn over #3? MartinPoulter (talk) 17:36, 7 September 2013 (UTC)
 * The host is not constrained to turn over #3 (as the original poster is apparently thinking - which I agree is not the Monty Hall problem), but we're told the case to think about is where the player has picked #1 and the host has revealed #3. The point is you are deciding whether to stay or switch in one of the 6 possible cases of two closed doors (one you originally picked and the other one) and one open door - and we're using the case where you pick #1 and the host opens #3 as the example (and presumably the intent is that the odds will be the same in all 6 of these cases).  In Martin's problem, you're effectively deciding to switch before seeing which door the host opens (i.e. if you initially pick #1, you win by switching if the car is behind either #2 or #3 - in which case, of course switching has a 2:1 advantage!).  It's not until after the host opens a door that the car must be behind only 2 doors you're deciding between (your original pick, say #1, and the other unopened door, say #2).  The difference between these two problems is fairly subtle, and with the constraints that the host MUST always open a door and MUST pick randomly between two goats (if this comes up) they have the same answer - but they're still slightly different problems (distinguished by numerous sources, see the "Criticism of the simple solutions" section of the article).  -- Rick Block (talk) 18:37, 7 September 2013 (UTC)
 * Once again the whole point of the problem (that people think the answer is 1/2 when really it is 2/3) is lost because of a conjuring trick by a bunch of probability professors. This has been discussed her endlessly before and the consensus on every occasion has been to ignore the Morgan scenario until after the basic problem has been solved.  Martin Hogbin (talk) 21:11, 7 September 2013 (UTC)
 * The original poster here is clearly thinking about the conditional probability the car is behind door 2 given the player originally picked door 1 and the host opened door 3, as opposed to the overall probability of winning by a strategy of staying vs. a strategy of switching (look at the experiment he claims to have done - which indeed shows a 50% chance of winning by switching). Why slyly switch problems on him (without even mentioning you're doing this)?  Why not, instead, show him how to correctly simulate the problem he's actually thinking about - which necessitates bringing up the additional constraint on the host?  -- Rick Block (talk) 22:29, 7 September 2013 (UTC)
 * Perhaps the 124.168.241.91| would be kind enough to let us know how they interpreted the question.


 * You will, no doubt recollect that the original question does not specify that the host opened door 3. It says, 'opens another door, say No. 3'. This might be ambiguous but for the fact that we know for sure that the original composer of the question (vos Savant) did not intend to specify door numbers. Martin Hogbin (talk) 08:59, 8 September 2013 (UTC)
 * We know for sure vos Savant's published wording includes door numbers and that this is the most well known statement of the problem. Are you perhaps referring to Whitaker's original question which is all but unknown? -- Rick Block (talk) 17:19, 8 September 2013 (UTC)
 * No, vos Savant said that she added the door numbers and that this was a big mistake since she did not want to specify door numbers. Martin Hogbin (talk) 18:19, 8 September 2013 (UTC)
 * OK - how about if we wait for the original poster here to say how he interpreted the question. -- Rick Block (talk) 19:04, 8 September 2013 (UTC)

To be more specific, everyone here is saying the general rules of the show are: At this point there are at least two possible interpretations for what question is being asked. Is it
 * 1) 2 goats and a car are hidden behind three doors, with the location of the car selected randomly (e.g. the host rolls a die and if it comes up 1 or 4 hides the car behind door 1, if it comes up 2 or 5 hides the car behind door 2, and if it comes 3 or 6 hides the car behind door 3)
 * 2) the player initially selects a door, which is noted but not opened
 * 3) the host must now (after the player selects a door) open another door showing a goat, and must make the offer to switch


 * a) What is best, a strategy of switching or a strategy of staying, i.e. if you are intending to go on this show should your strategy be to pick a door and stay with your original choice, or pick a door and then switch?

Or is it


 * b) Consider the case where you pick door #1 and then see the host open door #3. Should you stay with your original choice of door #1 or switch to door #2?

The question to the original poster of this thread is which of these questions do you think is being asked? -- Rick Block (talk) 15:58, 10 September 2013 (UTC)
 * I am perfectly happy for you to ask this question and I would be very interested to hear the answer. The problem is though that most people on seeing the question will not see any difference between the two cases you give (and if course we know the numerical answer is in fact the same for both cases).  I doubt that we will hear from the OP again since his original point has now been lost in what many see as an irrelevant detail.  As I have said before, I have nothing against studying this detail but this can only be done after a person has come to terms with the fact that the answer is 2/3 not 1/2.  Martin Hogbin (talk) 16:22, 10 September 2013 (UTC)
 * Given the experiment he describes doing (way above at this point), it seems fairly obvious to me that he's attempting to answer (b). I agree he may be unlikely to respond, but let's give it a while.  -- Rick Block (talk) 00:08, 11 September 2013 (UTC)
 * As I said to Richard above, I have nothing against discussing the details of conditional probability with anyone, indeed I would be happy to help explain the Morgan argument to newcomers. I think though that to pile straight into this complication before individuals have come to terms with the fact that the answer is 2/3 not 1/2 in unhelpful.  We know that the belief that the probabilities for the original and unchosen doors are 50:50 is extraordinarily pervasive and until that belief can be dispelled no progress can be made in issues of conditional probability.  I think the additional confusion that it creates probably drives many away.  I therefore suggest that you refrain from discussing conditional probability with newcomers until they have accepted the 2/3 answer.  After they have understood why this answer is correct I am perfectly happy to leave you to explain the Morgan solution to them without interruption.  Martin Hogbin (talk) 09:01, 11 September 2013 (UTC)
 * Martin, Martin, after all these years, and all the discussion, you still don't get it. Nijdam (talk) 09:51, 11 September 2013 (UTC)
 * On the contrary...Martin Hogbin (talk) 14:16, 11 September 2013 (UTC)
 * Per this section above, it seems to me discussing the conditional probability with newcomers leads directly to accepting the 2/3 answer. While (per this section) attempting to get someone to accept the 2/3 answer by ignoring the conditional situation (which was your approach) fails miserably.  I therefore could just as legitimately (more so, since we have two data points supporting my approach neither of which supports yours) suggest that you refrain from discussing "simple" solutions with newcomers (on this, the Arguments, page) until they have understood the conditional answer is 2/3.  And, are you continuing to respond here in a deliberate attempt to prevent the original poster from responding?   I've suggested we wait for a response from the OP twice now.  Are you willing to wait yet, or are you going to continue to argue that your approach (reflecting your POV about how the problem should be understood) is the one and only approach that newcomers will understand?  -- Rick Block (talk) 16:10, 11 September 2013 (UTC)
 * Rick, I was waiting for the OP to respond. Martin Hogbin (talk) 21:53, 11 September 2013 (UTC)

What problem are we talking about?
"Chose one card but do not turn it over" what? Just...what? That would indeed be a 1-in-3 chance since none of the variables have changed. It's still 3 cards unflipped. That's not the question I posed.

The question i'm specifying is literally what I stated earlier.

--

Taking 3 cards, a black (car) and two reds (goats), I shuffled them repeatedly (5 times quickly per deck to create a truly random system). I would then lay them out into a row of three.

- Card(1) Card(2) Card(3)

I then lifted the third card, which if black I put back and repeatedly shuffle again until I lift a red, and when red I flip over facing up.

- (Card(1) Card(2) Red(3)

Now, following the scenario, I choose the first card. I ask myself if I want to choose the second card, obviously picking no, and I flip both cards 1 and 2.

Example: Black(1) Red(2) Red(3)

--

In that question the outcome between whether or not you should change cards between the two card options left IS 50/50. In that question, you may change the cards around and STILL get the 50/50 outcome. Be the 2 card picked first, followed by the 1 and 3 or the 1 card picked first followed by the 3 and 2 etc. etc.

This is the question posed both on here and other sources (including the "pop culture" references section that uses this) and the exact problem specified on a variety of sources stating the Monty Hall problem. Which does not, demonstratively, result in a "66% chance" but a 50/50 chance on whether the "door" you originally chose will be a goat or a car regardless of it flipping (which i'm just going to blatantly assume here that people are not trying to claim that the choices will magically change places based on your decision to stay or change).

Which question of the previous stated 1/2/a/b options does the above problem i've stated pertain to, Rick?

If one and it is, somehow, not 50/50 in the proposed explanation then the proposed explanation is demonstratively false. If it applies to none of them, then we have a few issues (which I sought to address with my original post) that I shall specify:

1) The problem is poorly worded. It views exactly how I have stated above (in fact a variety of people defending the 66% chance equation stating that the method I used would demonstrate how it works, which as i've shown is false) and the issue here is addressing it's wording to REMOVE the suggested equation (which I used above) and replace it with something more specific.

2) The problem is unsound and is relying on poor research to come to it's erroneous conclusions (which I doubt is what occurred here).

3) The problem is poorly worded but doesn't rely on the equation shown but mitigating, vague factors that can't be processed specifically in text.

4) The problem is poorly worded but is ignoring real word applications in a situation in which psychological factors and other such non-objective equations are at play, which defies the claim of 66% (it could be a 1/4 chance, 1/37 chance, given the mitigating factors).

My monies on number one, in that the equation is poorly worded and is relying on different scenarios while using a SPECIFIC example to test with that ISN'T (somehow) meant to be repeated. 124.168.241.91 (talk) 09:28, 19 September 2013 (UTC) Sutter Cane


 * Sutter Cane, you are right in saying that the article still is a mess, despite years of efforts to show the clean *paradox*:


 * One open door showing a goat, and two doors still closed, one of them with certainty hiding the second goat, and one of them with certainty hiding the car. But the odds in favour of switching to the door offered are not 1:1, but they are 2:1 in favour of switching doors. That's the *paradox*, but the misty article still gets bogged down in details that are not addressing this famous paradox. Thank you for your comments, it will help to make the article more intelligible. Despite some authors here who are of the opinion that this Wikipedia article is NOT to make the paradox better intelligible. Gerhardvalentin (talk) 10:08, 19 September 2013 (UTC)


 * @Sutter Cane: Where did you get the notion that the host reveals a goat before the player's initial pick? Here's the wording as vos Savant published it in Parade:
 * Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
 * Isn't the sequence of events pretty clearly, in order:
 * 1. Car is hidden behind one of 3 closed doors.
 * 2. Player makes an initial pick (example given is player picks door #1)
 * 3. Host opens a door showing a goat (example given is host opens door #3)
 * 4. Player is asked whether he/she wants to switch to the other door (example given is switching from #1 to #2).
 * In your problem, you've flipped steps 2 and 3. I agree the problem is not worded very well (for example it implies, but does not explicitly state, that in step 3 the host must open a door showing a goat, which means if the player has picked door #1 the host cannot always open door #3), but I think it at least makes the sequence of events clear and this sequence is different from your interpretation.  Is this interpretation from some source, or is it your own analysis (here on the "Arguments" page we can talk about original analysis - I'm just curious where this came from)? -- Rick Block (talk) 16:08, 19 September 2013 (UTC)
 * @Rick Block: Please read his comment again, he did not say "before", he just said that in his "test" the contestant always first selects door #1, and that – in his test – he makes sure that the host is "avoiding to show the car" in just discarding 1/2 of all winning events where the unselected door #2 contains the second goat and the unselected door #3 contains the prize. By his method the the host discards all those winning events where door #3 hides the car. And I say that's because the still messy article is not clear enough in distinguishing the two scenarios "host knows" and "host doesn't know". In the latter case the chances indeed are 1:1. But MvS said that there’s no way the host can always open a losing door by chance! Gerhardvalentin (talk) 08:01, 20 September 2013 (UTC)
 * Actually, he did say "before". His "host" flips card 3 first (and, if it's the winner this causes a re-deal), and then his "player" selects card 1. I'm expecting him to say it doesn't matter whether the player picks first or not, because his host is never flipping card 1 (if card 3 is the winner then this is a case where the host could have flipped card 2, so wouldn't fit the given example where the player picks #1 and the host reveals #3).  To me it's obvious he's attempting to simulate the conditional case posed as an example by vos Savant where the player initially picks door #1 and the host opens door #3.  However, he's doing it in a way where the host always opens door 3 if possible, i.e. does not pick randomly between #2 and #3 if #1 is the winner (and the way he's doing it indeed makes the odds 50/50).  I think you and I agree he's not following the "normal" rules where the player picks first and then the host must reveal a loser from the other unselected two, and must pick randomly between the remaining two if the player initially selects the winner.  I also think he and I agree that the problem is poorly worded - for example the requirement that the host pick randomly between two losers (if this comes up) is not mentioned.<P>There are actually two "fixes" to his experiment.  One is to do as I suggest above.  After the player picks #1, have the host look at both #2 and #3, reveal #3 if #2 is the winner (and vice versa), and flip a coin to decide which of #2 or #3 to reveal if #1 is the winner.  If #3 is revealed, count how many times staying with #1 wins vs. switching to #2.  The other "fix" (suggested by Martin) is to ignore the example case and count every "deal" regardless of which card the host ends up flipping (and, with this approach, it doesn't matter how the host decides which of #2 of #3 to reveal if #1 is the winner).  I think the OP has already acknowledged that in this case (player picks #1, wins by staying if this is the winner and wins by switching if either #2 or #3 is the winner) staying wins 1/3 of the time and switching wins 2/3 - so it's clear to me he's thinking about the example case described in the problem (player picks #1 and host reveals #3).  The 2:1 odds in favor of switching necessarily occur in this example case only if the host picks randomly which of #2 or #3 to reveal if #1 is the winner (as you know, the odds of winning by switching in this case can vary between 1/2 [as the OP has shown] to 1 depending on how the host picks which loser to reveal if both #2 and #3 are losers).  Picking which loser to reveal randomly (if there's a choice) is as crucial a condition as that the host knows where the winner is (and avoids revealing the winner). -- Rick Block (talk) 15:27, 20 September 2013 (UTC)
 * No Rick, you got it wrong. On Sept. 7 above he clearly said "The problem claims that the host always opens the third door after the contestant picks the first door." Meaning the host opens door #3 after the contestant in any case picked door #1 before. And he clearly said "The host then asks if the contestant wants to switch doors." Meaning to swap from door #1 to door #2.
 * In his "test" the contestant ALWAYS (!) picks door #1 (card #1), but the host does NOT ALWAYS open #3, but only if #3 is a looser. So the host "afterwards" is checking the contents of #3, but will show the contents of #3 and offer a swap to #2 ONLY if door #3 (card #3) is NOT THE PRIZE. If #3 hides the prize: game over in such clear winning event. In fact that's the pure "host does not know" scenario. Please read it again. The article still isn't clear enough in this respect. Gerhardvalentin (talk) 16:23, 20 September 2013 (UTC)
 * Per the description at the beginning of this section ("I lay out the cards ... I then lifted the 3rd card ... Now, following the scenario, I choose the first card"). If the host always reveals #3 (if possible) it doesn't matter if the player picks before or after the host reveals #3, the chances end up 50/50.  Yes, it eliminates a winning "switch" if #3 is the winner, but if #3 is the winner then if the player picks #1 the host cannot reveal #3 but must reveal #2.  If you're interested in the odds in the case where the player picks #1 AND the host reveals #3 (as he clearly is), then this case (where #3 is the winner) is irrelevant (which is why he simply re-deals in this case).  His problem is that he's NOT randomizing which loser to show in the case #1 is the winner.  He needs to re-deal if #3 is the winner (in which case #2 must be revealed) and also if the random choice when #1 is the winner selects #2 (i.e. he's counting all cases where #1 is the winner rather than only half of them). -- Rick Block (talk) 18:32, 20 September 2013 (UTC)

@Sutter Cane: In fact we should know what we are talking about. Above there is my example with "joker 1" and "joker 2". "Joker 2" is the well formed problem with a 2/3 solution, "joker 1" has a 1/2 solution. I think you'll agree.--Albtal (talk) 17:36, 21 September 2013 (UTC)
 * It seems obvious to me he's interested in the specific case where the player has picked #1 and the host has revealed #3 (at which point there are only two possibilities for the location of the car, not three). @Sutter Cane: is this correct? -- Rick Block (talk) 19:28, 21 September 2013 (UTC)


 * We all are interested in the (not specific) situation where the host has opened a door with a goat and where two doors remain. And the question in the joker example is: Will "joker 1" yield to another chance in the two doors situation than "joker 2"?--Albtal (talk) 10:40, 22 September 2013 (UTC)


 * You may be interested in the non-specific situation, but the question is what is Sutter Cane interested in? Once again, it seems obvious to me he's interested in the specific case where the player has picked #1 and the host has opened #3.  The odds in this specific case with your joker 1 and joker 2 can be equivalent (if joker 2 always opens #3 if possible - indeed, this is what Sutter Cane's procedure above is showing).  With joker 2 if you pick #1 you win by staying with probability 1/3 (if the car is behind #1) and you win by switching with probability 2/3 (if the car is behind either #2 or #3).  But these are probabilities measured at the beginning of the game, before the host has opened a door.  It's something rather different than saying if you do this 100 times or so, you'll see #1 win about 1/3 of the time and #2 win about 2/3 of the time after the host has revealed #3.  Saying this requires the host to pick which loser to show randomly if the player initially picks the winner. -- Rick Block (talk) 17:57, 22 September 2013 (UTC)

--

The difference between Sutter Cane's experiment and the usual interpretation is in which outcomes are discarded. Sutter's version discards those times when the car is behind door #3 and then has the host always open door #3 in the remaining cases. The standard interpretation is that, once the contestant picks a door, the host picks either door #2 or door #3 to open, choosing uniformly at random if both have goats; otherwise choosing the one with the goat when the other has the car. We then discard the times when the host picked door #2 - which are all the times when the car was behind door #3 and (by assumption) half the times the car was behind door #1.

There are two reasons why Sutter's interpretation isn't the standard one: the answer of 1:1 is less interesting than 1:2; and, probably more importantly, by guaranteeing up front that the car could never have been behind door #3, it contradicts the usual assumption that the car is equally likely to be behind any of the three doors.

The standard interpretation assumes that the door numbers are not special during the game, and it's only when we come along and choose which games to count for our statistics that it's narrowed down to games where the contestant happened to choose door #1, and the host happened to open door #3 - in particular, we want the answers to be the same (to within experimental error) if we, from the same simulation run, choose the games where the contestant chose door #3 and the host opened door #2 - something that never happens in Sutter's experiments.

Rmsgrey (talk) 00:03, 2 November 2013 (UTC)


 * I completely agree. However, there are folks arguing here that the "right" way to fix the experiment is to force the host to always open a "goat door" (selected between #2 and #3, without specifying how the host chooses which of #2 or #3 to open if the car is behind #1) and then count all outcomes, not just the times the host opens #3.  This "fix" changes the probabilities the experiment is simulating from
 * A: P(car behind door #1|player picks #1 and host opens #3) vs. P(car behind door #2|player picks #1 and host opens #3)
 * to
 * B: P(car behind door #1|player picks #1 and host opens either #2 or #3) vs. P(car behind #2 or #3|player picks #1 and host opens either #2 or #3)
 * Since the host must always open either #2 or #3 if the player picks #1, B can be mathematically simplified (regardless of how the host decides to pick between #2 and #3 if the car is behind #1) to
 * C: P(car behind door #1|player picks #1) vs. P(car behind door #2 or #3|player picks #1)
 * Note that C (and since B can mathematically simplified to C, B as well) is simply the probability the car is behind #1 vs. the probability it is behind #2 or #3, before the host opens a door - which is obviously 1/3 vs. 2/3. My contention is that it seems fairly clear Sutter Cane (and, IMO, nearly anyone reading the standard problem description) is interested in A (the probability after the host opens a specific, known, door that the car is behind door #1 vs the probability it is behind one other door, not both other doors), not B (or C). -- Rick Block (talk) 19:18, 2 November 2013 (UTC)


 * I think a lot of people assume that it's natural to say that A and A' (like A but where the host opens door #2) should have equivalent answers so it doesn't matter whether you look at A, A', or combine them to give B. And, of course, B is easier to solve since it doesn't require you to make any assumptions about how the host chooses between two goats. 81.156.216.158 (talk) 23:02, 2 November 2013 (UTC)
 * Did someone just say 'chooses between two goats'? Martin Hogbin (talk) 00:13, 3 November 2013 (UTC)
 * Clearly meaning choosing which of #2 or #3 to open if the car is behind #1. -- Rick Block (talk) 00:28, 3 November 2013 (UTC)
 * It is entirely natural to say A and A' should have equivalent answers. The insight that this means they both are numerically the same as B is (I think) beyond most people - indeed, I believe most people (including some who regularly comment here) do not have a clear idea of what B is actually saying.  On the other hand, I think nearly everyone understands there is a difference between C and A - just not the difference they usually assume. -- Rick Block (talk) 00:28, 3 November 2013 (UTC)


 * I really like comparing the approaches by comparing


 * A: P(car behind door #1|player picks #1 and host opens #3)


 * A': P(car behind door #1|player picks #1 and host opens #2)


 * C: P(car behind door #1|player picks #1)


 * If one simulates MHP then the probabilities A, A' and C will be estimated by different relative frequencies, because they talk about conditional probabilities given different events. It is only if we assume that in the simulations, the host chooses a goat door completely at random when he does have a choice, that the three relative frequencies will be more or less equal. Simulation will confirm that C is 1/3 however the host makes his choice, but A and A' will only be equal to one another and equal to 1/3 when the host's choice if completely at random.


 * Regarding the question, which probability are we supposed to calculate, it is clear that opinions in the literature are divided: the maths professors tend to go for A and A', the amateurs tend to go for C. Since the numerical answers are the same (assuming an unbiased host) I'm afraid that the lay people simply aren't going to care much. I don't see why this should be a problem for a wikipedia article. In a wikipedia article one reports neutrally what is "out there" and one of the things that is out there is a lot of disagreement about what constitutes a solution to MHP. Richard Gill (talk) 14:16, 6 November 2013 (UTC)


 * We're not talking about the article here - we're talking about understanding. And the question is what probability is Sutter Cane (presumably an amateur, also presumably unconvinced by the usual lay explanations showing C is 1/3) thinking about?  -- Rick Block (talk) 17:57, 6 November 2013 (UTC)

Sample space
Lets talk here (rather than on the main talk page) about the sample space. There's one car and two goats hidden behind three doors. The player initially picks a door. The host subsequently opens a door revealing a goat. So, the set of outcomes consist of triples indicating which door the car is hidden behind, which door the player picks and which door the host opens. As has been mentioned before, there are 12 possible outcomes (with the rule that the host cannot open the same door the player picks and must open a door showing a goat). These are as follows where (x,y,z) indicates the car is behind door x, the player initially picks door y, and the host opens door z. (1,1,2)  (1,2,3)   (1,3,2) (1,1,3) (2,1,3)  (2,2,1)   (2,3,1)          (2,2,3)   (3,1,2)   (3,2,1)   (3,3,1)                    (3,3,2) I've arranged these in three rows corresponding to where the car is located and three columns corresponding to the player's initial pick of door - so (for example) the first row shows all possible outcomes for when the car is behind door 1 (the player picks door 1 and the host opens door 2, or the player picks door 2 and the host opens door 3, and so on), while the first column shows all possible outcomes if the player initially picks door 1 (the car is behind door 1 and the host opens door 2, the car is behind door 1 and the host opens door 3, and so on). Note that there are 4 possible outcomes in each row and in each column.

Unlike many probability problems, in this one the outcomes are not all equally likely. This is because the host sometimes but not always has a choice for which door to open. In particular, if the player's initial pick happens to be the same as the door hiding the car the host has a choice (usually assumed to be 50/50) but if the player's initial pick is not the door hiding the car the host has no choice. When the host has a choice, there are two outcomes that together have the same probability as one of the outcomes where the host has no choice.

At the start of the game, before the player initially picks a door, the probabilities of the six "no choice" outcomes are all 1/9 while the probabilities of the six "choice" outcomes are all 1/18. The total probability must be 1, and indeed 6/9 + 6/18 = 1.

After the player initially picks a door, there are always 4 possible outcomes (one of the columns from above). For example, if the player picks door 1 these are (1,1,2), (1,1,3), (2,1,3), and (3,1,2). These outcomes still are not equally probable with the "choice" outcomes (the first two) half as likely (probability 1/6) as the latter two (probability 1/3). Again, these sum to 1 (1/6 + 1/6 + 1/3 + 1/3 = 1).

After the host finally opens a door, there are only 2 possible outcomes. For example, if the player picks door 1 and the host opens door 3 the only possibilities are (1,1,3) and (2,1,3). As before, these outcomes are not equally probable with the "choice" outcome (the outcome where the car is behind door 1) half as likely (probability 1/3) as the "no choice" outcome - the outcome where the car is behind door 2 (probability 2/3). Again, these sum to 1 (1/3 + 2/3) = 1. -- Rick Block (talk) 01:54, 5 January 2014 (UTC)
 * Thank you for advancing the dialog. However, the diagram which I want to start with, as one of two potential diagrams, is this: What does the sample space look like BEFORE any choice is made? for this first diagram, I do not want to include any information about the possible permutations available as a consequence of that choice. Rather, I only want to illustrate a visual representation, in the form of blocks, as to what the sample space looks like before the first choice. If you agree that's what we're discussing there, then ok. If not, then why are we here?Tweedledee2011 (talk) 04:00, 5 January 2014 (UTC)


 * This is the block diagram I want to discuss. If it's not accurate, state the reasons why. If it is accurate, let's discuss how to use it in the article. Tweedledee2011 (talk) 04:02, 5 January 2014 (UTC)
 * The "1st choice" diagram does not include any information about what door the host opens, which is in turn influenced by what door the player initially picks - so it is not the sample space for the MHP (it's a sample space for "I've hidden a car and two goats behind three doors, pick one"). The "2nd choice" diagram might be a continuation of the first ("now I've opened a door showing a goat, pick one of the two that are left") - also having little, if anything, to do with the MHP.


 * The probabilities involved in the MHP arise because of the rules pertaining to the host which are related to the door the player initially picks - specifically
 * the host can't open the door the player picks
 * the host must show a goat (i.e. the host knows what is behind the doors and deliberately reveals a goat)
 * the host must pick between 2 goat doors evenly if the player's initial pick is the door hiding the car
 * If any of these conditions are not met, the probabilities of winning by staying and switching may not be 1/3 and 2/3.  If the sample space ignores the host's actions as influenced by the player's initial pick, it isn't relevant to the MHP.  In particular, the fact that the door the player picks initially has a 1/3 chance of hiding the car does not mean that the probability this door hides the car remains 1/3 regardless of what happens (of course assuming the car is not moved to a different door).  As Ruma Falk  puts it (emphasis in the original): "Truly, Monty can always open one of the two other doors to show a goat, and the probability of door No. 1 remains unchanged subsequent to observing that goat, still, it not because of the former that the latter is true."  With this comment she's criticizing the slightly more sophisticated argument (put forward by vos Savant and many others) that the probability the car is behind door 1 does not change when the host opens door 3 because with two doors to choose from the host can always open one of them to reveal a goat.  Even this is not accurate.  The probability of where the car is depends on how the host decides what door to open, which in the usual interpretation includes all three of the conditions mentioned above.  Omit any one, and you have a different problem with different probabilities. -- Rick Block (talk) 05:34, 5 January 2014 (UTC)
 * Rick -For a minute please think about and try to answer only this question: Yes or no, is the 9 blocks an accurate visual representation of the game's sample space at the moment just prior to the first choice? I need a clear answer to that point before I can effectively dialog with you. Tweedledee2011 (talk) 06:31, 5 January 2014 (UTC)
 * No. As I've explained above the sample space for the MHP just prior to the player's initial choice consists of 12 potential outcomes, not 3 as your figure shows. -- Rick Block (talk) 07:30, 5 January 2014 (UTC)
 * My block diagram shows nine possible results to the first choice, not three. Now presuming you are correct, can you make a block diagram the same as I have, showing the 12? Tweedledee2011 (talk) 07:58, 5 January 2014 (UTC)
 * Note: any diagram Rick presents is a Block diagram. Nijdam (talk) 14:03, 5 January 2014 (UTC)
 * No, your diagram shows only 3 outcomes, i.e. the rows labeled "Layout 1", "Layout 2", "Layout 3". Each one of these rows is a possible outcome of the initial step of hiding the car and two goats behind 3 doors - which is the situation the player finds herself in at the beginning of the game (not after having made her initial selection or after the host opens a door).  I think I may have tumbled on to what one of the issues here might be.  Can you let me know what problem you're attempting to address (both involving 3 doors, 1 car and 2 goats, a player who makes an initial selection, and a host who must show a goat and make the offer to switch)?


 * A. You are going on this show and want to know whether you are more likely to win the car by picking a door and staying with it, or by picking a door and switching to whichever door the host does not open. For example, should you pick door 1 and stay with it, or pick door 1 and switch to whichever of door 2 or door 3 the host does not open.


 * B. You are on this show, made your initial selection, have seen which door the host has opened, and are now deciding whether you are more likely to win the car by staying with your initial selection or by switching to the other unopened door. For example, you picked door 1 and the host opened door 3 so you are deciding whether to stay with door 1 or switch to door 2 while looking at a goat behind door 3.


 * These sound quite similar, and they are indeed related but they are not the same problem. Are you attempting to address (A) or (B)? -- Rick Block (talk) 17:36, 5 January 2014 (UTC)
 * Rick - You keep over-thinking and getting too far ahead. What I am trying to do is resolve (1) simple question. Then, after doing that, I want to think some more and come back for more discussion. As I see it, there are only three places for the car to be located. Based on that fact, there are only three possible ways to configure the layouts, if the goats are fungible. Of they are not, then we must distinguish between them. And if so, that adds some additional possible initial layouts. And mind you, I am talking about the initial layout of the doors. Again, I think the vernacular is what's tripping up this dialog. Allow me to be as precise as a I can, using links to emphasize terms which have particular meaning in probability. That said, here goes: I see nothing in the description of the game which says anything about how soon, after the player makes the first choice, that the host will open a door. This being the case, it's possible to freeze time and examine where are we in the logic of this process at an exact point in time. From the player's perspective, because the instructions do not say that the player knows ahead of time that the host will open a door, or that the player knows (ahead of time) he will be offered a second choice, then as far as the player knows, he is facing (1) car (2) goats and (3) doors. The instructions do not say if the goats are distinguishable, therefore, because this is an unknown, we can examine it either way. This particular walk-through assumes they are fungible. And if we get through this, we can always go back and look as if the goats are not fungible. But for now, with two indistinguishable goats, what the player knows at the start is on this: There are three Doors. There is one Car. There are two Goats. Based on that knowledge alone - which is all the player has access to at the start, there are only three possible ways to configure the doors (as per my diagram). Now the player is told to choose a door, but at the point in time of that choosing, that player would reasonably expect that the chosen door is his choice - because as of yet, he's not been told he'll get a second choice. Therefore, at the point just before the first choice, the first choice can only be seen by the player as an experiment, which consists of a Singleton event. And that should, from the players perspective, lead to the outcome of the door he chooses becoming an opened door. However, from our perspective (outside the game) we can see that the player doesn't get to the point of immediately opening the door. Rather, that information which would result from that event outcome is stopped by the host saying something like "before you open that door, would you like to keep it or switch...". So then, the outcome of the first choice is not immediately known. In other words, the player does not at that point find out the result of his door selection. The experiment was to find out what's behind the door. The door is chosen. But the player does not at that point find out. That information is not yet revealed. So then, up to this point, before the host opens a door and offers a new choice, the only possible outcomes of the first choice, were nine - just like my block diagram shows. This is what I want you to discuss with me. DO NOT YET introduce the host's opening of a door or offer to switch - and any effects those have on the composition of the sample space. Instead, look at it from the perspective of the player. Up until the door is opened and the offer to switch is made, I see the player's sample space as containing nine possible outcomes. It's not three, because we don't know which door the car is behind, so we have to think it could be any. And if it's any, then there are nine possible outcomes to player's the first choice. Three possible outcomes at any of three doors. Now please, think about this and tell me if what I am saying is true. At the point in time I am specifying, is the total number of the player's first choice possibilities nine? Tweedledee2011 (talk) 22:41, 5 January 2014 (UTC)


 * So, you're at this point wanting a sample space describing only the possible outcomes after the car is hidden and after the player's initial choice? As you say, there are 3 possible locations for the car and 3 possible doors the player may pick which results in a set of 9 possible outcomes.  Generally, outcomes are specified by showing the values of the variables involved.  In this case there are two relevant variables, the location of the car and the door the player picks (this assumes we don't care at all about the goats - they're simply zonks after all).  If you want to show this as a 2-dimensional array it looks like this (the cross product of the possible locations for the car and the possible player picks):


 * Note that this differs from your table in that each cell shows the combination of car location and player pick rather than the car/goat "outcome". It's important to keep track of these if we're next going to talk about the door the host opens (which I assume we are).  If you want, you could annotate these with the car/goat result, perhaps like this:


 * Is this more or less what you're looking for? -- Rick Block (talk) 06:25, 6 January 2014 (UTC)
 * Is your 9 block diagram correct for the point in time we are talking about? Tweedledee2011 (talk) 06:29, 6 January 2014 (UTC)
 * What about this look, is this correct? Tweedledee2011 (talk) 06:38, 6 January 2014 (UTC)


 * Correct? Well, not really.  The text "This is door #n" in each cell is distinctly odd.  The cells are not doors.  They are combinations of the two variables "where is the car hidden" and "what is the player's initial pick of door".  It's like you're rolling two dice with the cells corresponding to a particular combination of the value of one die and the value on the other die.  And you're omitting information in the cells that will be important later.  For example, the cell in the first column and second row says it's door #1 and it has a goat and the player picked it - but not that the car is (in this case) behind door #2 - in fact it looks identical to the cell below in which case the car is behind door #3.  These are not the same outcome even though they look identical in your table.  The location of the car is critically important and must not be lost.  For example, for all the cells in the second row the car is behind door 2 - each cell in this row should say this.  Why are you resistant to the table I suggested? -- Rick Block (talk) 07:17, 6 January 2014 (UTC)
 * Rick - I am not "resistant" to your diagram. Rather, I am trying to develop a diagram which explicitly explains the significance of each possible choice to a total novice in this subject. As for what the cells "are", they can indeed been seen as doors. The first row represents the first possible configuration, the second row the next and the third, the final. There are only three possible ways to set the initial configuration of the doors, and this diagram shows them all. Look above at my first version on this page - that version called the rows "layouts". This current version (modified from your diagram) is both a logical map and fully fleshed-out physical diagram of the three possible layouts. And by my count, there are only nine (9) possibilities at the start of the game. Do you see that? Do you agree that at the start of the game, there are only nine (9) possibilities available to the player? Tweedledee2011 (talk) 07:44, 6 January 2014 (UTC)
 * I'm honestly trying to help you, but if you're not going to listen to me I'm not sure why I should bother. If you're talking about the "start" of the game (before the player's initial pick) there are 3 possible layouts of 1 car and 2 goats - not 9.  One layout is (C,G,G).  If you'd like you can call this a sample set consisting of 3 (not 9) outcomes {(C,G,G), (G,C,G), (G,G,C)}.  The expansion of the sample set to 9 outcomes happens because of the player's pick.  Each of these 9 outcomes is the combination of one of the 3 possible layouts and one of the 3 possible player picks.  For example the outcome "player picks door 1 in the layout where the car is behind door 1" might be represented as ((C,G,G),1) (the layout is the (C,G,G) layout and the player pick is door 1).  You can arrange these outcomes visually in a 3x3 array with the rows being all the possibilities with the same layout (same car location) and with the columns being all the possibilities with the same player pick - but the individual cells are not "doors".  Do you understand this? -- Rick Block (talk) 16:30, 6 January 2014 (UTC)
 * Rick - you are not even trying to hear me. My diagram represents ALL THREE of the possible initial layouts of the doors. Yes or no, do you concede that at the start of the game, the car can be behind any of the doors? If yes, then the possibilities I am talking about are the possibilities of WHERE THE CAR IS AT THE START. I am illustrating this in my nine block diagram. Do you understand what I am saying, yes or no? Tweedledee2011 (talk) 18:55, 6 January 2014 (UTC)
 * Rick - you are leaping ahead to "outcomes" of the first choice again. STOP before you get there. My diagram is a diagram of all the possibilities of the playing field BEFORE the first choice is made. Tweedledee2011 (talk) 18:58, 6 January 2014 (UTC)

Initial state of game, from player's perspective
This diagram shows all three possible layouts of the doors, and what would happen, if the player picks one of the three doors (each row is one of the three possible starting layouts) and then WE FREEZE TIME at that point to think. At the start of the game, the player is only told to pick a door and that's what this diagram illustrates. It shows what the outcome of that initial pick would be, if that initially picked door was opened. This is a physical diagram map which shows each and every possible result of the player's 1st choice, from the perspective of the player, BUT BEFORE the host opens a door and offers a 2nd pick. This is a moment frozen in time and it's this point in time which I want to discuss. Once the others on this board understand this moment in time, then the discussion can continue. Tweedledee2011 (talk) 19:16, 6 January 2014 (UTC)


 * I hope somebody else can help you. I'm done. -- Rick Block (talk) 19:58, 6 January 2014 (UTC)

@Tweedledee2011: I understand what you mean. Please continue.--Albtal (talk) 20:17, 7 January 2014 (UTC)

@Tweedledee2011: I read in your comment above:

From the player's perspective, because the instructions do not say that the player knows ahead of time that the host will open a door, or that the player knows (ahead of time) he will be offered a second choice, then as far as the player knows, he is facing (1) car (2) goats and (3) doors. ... Now the player is told to choose a door, but at the point in time of that choosing, that player would reasonably expect that the chosen door is his choice - because as of yet, he's not been told he'll get a second choice.

Indeed for the 2/3 solution being correct it is a necessary condition that the player knows ahead of time that the host will open an unchosen door with a goat and will offer a switch. And indeed this condition is not part of Marilyn vos Savant's "Monty Hall Problem" going around the world. But exactly this problem has been published plenty of times together with the asserted 2/3 solution; meaning that the great "Monty Hall Paradox" really is a joke. The problem with this critical rule of the game has a very simple solution: Suppose you first "choose" door 1. Then you will win the car if it is behind door 2 or door 3. For you know that the host now has to open door 2 or door 3 with a goat. If he opens door 2 you finally choose door 3, and if he opens door 3 you finally choose door 2. See my former comments. I have stopped discussing here. But if I understood you correctly this might be a helpful comment for you.--Albtal (talk) 21:35, 7 January 2014 (UTC)


 * @Albtal - I reformatted your post for clarity. No words have been changed. Here's what I am saying: The problem with this problem isn't that the math experts are solving it with probability calculations. Rather, it's that they are solving it by using legitimate math in an illegitimate manner. For instance, the very premise of the problem says "Suppose you're on a game show...". This statement requires a full stop and an adoption of the premise, before we continue. From Merriam-Webster |"Suppose: To think of (something) as happening or being true in order to imagine what might happen". The math experts are explaining this poorly, because they have not yet satisfied the first condition of the problem, which is to suppose you are playing a game show. It does not state that it's Let's Make A Deal, and it's illegitimate to impute that into the premise or that as a player, before the host announces anything, that you know he's going to open a door. Therefore, the decision-tree models which claim the starting sample space to be 12 are illegitimate, because the player, at the start of the game, knows nothing about a door to be opened by the host or a second chance to be given. It's only after the host opens a door and offers a choice, that the player knows anything other than, three doors, one prize. Based on three doors, one prize, there are three possible configurations of doors, and this means that the initial sample space is 9, not 12. Also, as soon as a door is opened, it's not choose-able anymore, so it's no longer part of the sample space. As a result, the sample space is reduced, immediately upon when the door is opened. This leaves a 2x2 (down from 3x3), which is now 4. And when the player sees that, it looks like 1:2 or 50-50. But the 2:3 can still be arrived at legitimately - though not by the means of a 12 position decision-tree calculation. Those calculations are made on a sample space which never existed in the game and are therefore illegitimate by the very definition of sample space: "the set of all possible outcomes or results of that experiment". I can still arrive at the 2:3 result and that result is correct, but it's not correct for the reasons the math experts are saying it is. The math experts are arriving at that conclusion by violating their own premise. Tweedledee2011 (talk) 23:07, 7 January 2014 (UTC)

@ Tweedledee2011: Allow me to offer some thoughts about the confusing issues here, and about why your line of inquiry is meeting so much resistance. I am trying to help. (I actually fault the way introductory courses on Probability & Statistics are typically organized, which leaves many students unclear about fundamental issues even when they have mastered the applied techniques presented.) Note that the opening sentence of the Sample space article you cite specifically refers to "an experiment or random trial". This concept only applies to the Monty Hall problem in an abstract sense (described below) because it is not a problem of statistical sampling. The contestant's initial choice is not a sample or experiment: the door remains closed and no outcome is observed. The revelation of a goat is not a sample, experiment, or random trial: it is selective information. There is no statistical sampling going on. There is an abstract sense in which one may consider a statistical sample from among the set of possible universes, or instances of the game; but calling this a "sample space" rather than simply the set of possible scenarios can be confusing. Firstly, the elementary statistical events in this space are not things like "finding a car behind a particular door", as in ordinary statistical sampling, but are "ways the game might play out" – a probabilistic notion that may not be apparent to the average reader. Secondly, and more importantly, there is no "one right way" to define outcomes for analyzing the problem. Your choice of two possible outcomes for the final question (car behind door 1 and car behind door 2) is not wrong, (any win/lose game may be modeled with exactly two outcomes, "you win or you lose"), but this is not a very informative way to define outcomes. Rick's model for ways the game might play out is much more informative because it is decomposed into more elementary events that can be summed to show the composite win/lose outcomes are not equally probable. In terms of making the solution understandable for our readers, your emphasis on the size of the sample space is not useful when the outcomes you define are not equally probable. Simply counting such outcomes leads people directly to the naïve conclusion that if the car is behind one of two doors then they must be equally likely. As far an analyzing the situation before/during/after the initial choice of door #1, there is no need to belabor the point using sample spaces or any other type of analysis. Everybody understands that this is a blind choice among three alternatives with equal probabilities of success. Finally, and very importantly in terms of Wikipedia policy, if you make the extraordinary claim that all of the math experts are wrong about this then you should cite some very credible expert publications from some field that specifically say they are wrong about the particular point you are criticizing. Otherwise, claiming that you know better than the experts will only give people the definite impression that you don't know what you are talking about. I hope some parts of this have helped you understand why your approach is not being received well, and has given you some food for thought about probabilistic reasoning. ~ Ningauble (talk) 20:35, 12 January 2014 (UTC)


 * @Ningauble - First, let me stipulate there is no argument from me that switching results in a win 2:3 times. That's not the point of my dialog. Rather, my point is that the article is only explaining this math from the orthodox math-proof direction and is not actually thinking it through as a novice would. It's my view that the article could benefit from an examination of the perspective I am trying to offer. And to achieve that, I am trying a three step approach: 1) Explain the perspective, 2) Obtain some consent that the way I want to explain it can fit in the article and 3) Enlist team support to help me find enough reliable sources that the naysayers here will accept the edits I wish to make. Now, since I am only trying to enhance the article, and because I've NOT jumped the gun and rushed my edits into the article, it's obvious that people should try more to dialog and listen, rather than fight me here. So, once again, here goes:
 * 1) Sample space does not mean "sample" the way you have used the term
 * 2) Your misuse of that term in this context proves my point - the vernacular is fraught with traps
 * 3) At the point where the player makes his first choice, from the player's perspective, that choice is indeed an experiment
 * 4) That the payer does not immediately see the result of that experiment, is not relevant to the point I am making.
 * 5) At the point in time when the player makes his first choice, the player has not yet witnessed the host open one of the "lose" doors
 * 6) At the point in time when the player makes his first choice, the player has not yet been told that he will get a 2nd choice
 * 7) Therefore, based on 5 & 6, at the point when the player makes his first choice, the total size of the sample space is 3x3 or 9 - that's it
 * 8) Then, at the very moment that the host opens a door, that door is no longer choose-able by the player, and it is therefore, not part of the sample space of the player's 2nd choice.
 * 10) The player, for his 2nd choice, is allowed to pick only from 2 doors.
 * 11) The reason why switching works is not because of the 12 point decision tree model shown in the MIT .pdf which I linked to on the talk page
 * 12) No, the reason why switching works is because by opening a door, the host reduces the amount of doors behind which the car can be.
 * 13) In the first choice, the player has a best chance of 1:3 - and that never changes because the car's location is fixed
 * 14) When the host opens a door, the odds of switching must be 2:3 because the total must equal 3:3 (1) and the original choice is fixed at 1:3
 * 15) This can indeed be reasoned out via the decision tree model, but only if one cheats and stretches the literal meaning of sample space, which is the set of all possible results of an experiment.
 * When the host opens a door, the sample space shrinks - this is indisputable. The only real argument here is whether or not it starts at 9, 12 or 15. I say it's 9, unless we distinguish between goats, then it's 15. However, the explanations which say it's 12 are simply wrong - because they rely on the logical consequences of what would happen to the array of possible outcomes when the player chooses again. But adopting this logic violates the verbal premise of the problem which requires the problem solver to "Suppose you're on a game show". A player in the game, when facing his very first choice, is not facing a sample space of 12 outcomes. The reasoning which proves 12 relies on the player being given a 2nd choice, and therefore, until the player is actually given that choice, it's logically impossible to build the 12 count sample space decision tree. But, when the player is given the choice, it's only after a door has been opened. And because a door has been opened, that door is not choose-able any more. And because it's not choose-able, it can not hold any results for the player's second choice. And if it can't hold any results, it's not part of the sample space. The sample space starts at 9 and shrinks to 4. But you can still use a decision tree to arrive at 2:3, but you can't call that decision tree a diagram of the sample space of the game from the player's perspective, because it's not. That, is my objection to how this article is written - it omits the fact that the experts are explaining this wrong. They are using correct math, but are sloppily applying the vernacular. But as a result of how we have written this article, anyone who's read an example of the 12 result decision tree model and comes here for more information, will not find anything explaining this point.
 * Tweedledee2011 (talk) 06:32, 14 January 2014 (UTC)


 * I still do not see how the perspective you are trying to explain will help our readers to understand why it is better to switch, particularly since the inferences you are drawing from that perspective are not correct.
 * 1: Your argument does not support the assertion [your point #11] that the decision tree in the course notes of Meyer and Rubinfeld at MIT (and by implication, the related tree in the "Conditional probability by direct calculation" section of the present Wikipedia article) is invalid.
 * 1.a: If you are suggesting that the tree on page 5 fails to account for the state of affairs at the point of the contestant's initial choice [your points #5 through 7], this is simply not the case: read the diagram from left to right. This is made explicitly clear where Meyer and Rubinfeld construct the tree progressively and show, on page 4, the specific situation of the player's initial choice without reference to subsequent events.
 * 1.b: If you are suggesting that at the point of final decision, i.e. the actual question posed by the Monty Hall problem, the tree improperly includes situations where the contestant chooses the goat that has already been revealed [your points #8 through 10], this is simply not the case: the tree enumerates the possible situations immediately before the final choice. One could add another level to the tree showing the contestant's final "stay or switch" decision, labeled A, B, or C with the constraint that it not have the same label as its parent node (the door that was revealed). Each of these nodes can be tagged with Car or Goat to indicate the resulting prize awarded. In no case is this the goat that was revealed before the question was posed.
 * 1.c: If you simply do not like Meyer and Rubinfeld's use the term "sample space" for the labeling generated by their decision tree, it is really not an issue here because the present Wikipedia article doesn't even use the term. I do not particularly like their use of the term "outcome" in this context to refer to a state prior to observing the result of the contestant's decision, but this is completely immaterial and does not invalidate their logic. I assure you that the decision tree methodology is not a cheat.
 * 2: Your conclusions about the reason switching works are not correct: the fact that switching is better does not follow from your observations about the sample space. To see why this is so, consider this variation of the problem, in which the same observations are true but the resulting probabilities are different:
 * The Mary Hale problem is identical to the Monty Hall problem in all material respects except one: the host does not know or care where the car is. Mary picks one of the unchosen doors to open, and there happens to be a goat behind it. (Monty, on the other hand, picks an unchosen goat, and opens the door it happens to be behind.)
 * 2.a: In both problems "the amount of doors behind which the car can be" [your point #12] is exactly the same. In the Mary Hale problem the probability of winning by switching is only 1/2, not 2/3 as in the Monty Hall problem. Therefore, the  the amount of doors is not, as you state, the reason why switching works.
 * 2.b: In both problems the car does not move [your points #13 through 14]. In the Mary Hale problem the probability that the car is behind the initially chosen door after another door has been opened is 1/2, not 1/3 as it was beforehand. Therefore, the reason it is 1/3 at both points of time in the Monty Hall problem is not, as you state, because the car's location is fixed.
 * Unless you can provide a clear demonstration of how your formulations of the sample space logically determine the correct probabilities, i.e. how the correct result can be deduced from that starting point, I think you should not be focusing on how to include this perspective in the article, but should instead focus on gaining an understanding for yourself of the reason it is better to switch. ~ Ningauble (talk) 20:31, 14 January 2014 (UTC)


 * @Ningauble - There appears to no longer be any point in discussing this with you. The few replies I get, such as yours, are so tangential to what I am saying, that it's an entirely different debate. If you won't concede that the sample space, prior to a door being opened, and prior to the 2nd choice being offered; contains only 3x3 (a total of 9) possibilities, then there is no point in discussing this. Tweedledee2011 (talk) 07:27, 15 January 2014 (UTC)

@Tweedledee2011: The mere fact that you're talking about the sample space, prior to a door being opened, and prior to the 2nd choice being offered, shows you're not very experienced on the matter. Reading a book on probability theory doesn't help. A sample space contains the possible outcomes of a probability experiment. Experiment is the general term for situations where chance plays a role. You have to ask yourself what the possible outcomes are in the MHP. That's not very difficult, although you seem to have some trouble with it. We have the possible positions C of the car, the possible first choices X of the player, and the door H opened by the host. As each of C, X and H may have the values 1, 2 and 3, this leads to 3x3x3=27 possible outcomes. Some of these outcomes will have probability 0. It may simplify the sample space if we leave them out. The outcomes left are (CXH): 112, 113, 123, 132, 213, 221, 223, 231, 312, 321, 331, 332. If you count them well, they are 12. These 12 outcomes are necessary to describe the MHP. The sampole space is not symmetric, i.e. not all outcomes have the same probability. I leave the probability distribution to you. Nijdam (talk) 11:51, 15 January 2014 (UTC)
 * @Nijdam - Have you even read any of my comments, are just going to come in, guns blazing, repeating what we already know? What I am trying to help the others see here, is why so many people get the explanation of this game wrong. Part of that reason is people like you, who do nothing but repeat the obvious - yet ignore the plain language of the problem. The problem says "Suppose you're on a game show, and you're given the choice of three doors". At that exact moment in time, a player on the show would be facing a sample space of 3x3 - this is an irrefutable fact. The 12 space decision tree model is not in existence until after a door is opened and after the player is given another choice. In fact, the very descriptions of the 12 space tree make it very clear that the tree is intended to show the possibilities of switch/not switch - after a door is opened. That you can't see this (or refuse to) makes your comments pointless. What I am trying to help the math experts see, is that the novice player's intuition isn't all that wrong. Rather, it's that they don't know enough about probabilities to know which sets of information they can use. A novice, when told to Suppose you're on a game show does just that - and they go into the mindset of playing a game, a game  which is typically based on guessing (if played by non-math people, which is mostly everyone). The opening of the door does indeed reduce the size of the sample space - and you have not disproved this assertion. The simple fact is that after a door is opened, the player can only make his 2nd choice from a 2x2 set of possibilities - one goat, one car, two doors. And yet, switching still works - but not for the reasons the experts are saying. Removing a door (and it's associated possibilities) reduces the sample space and therefore; the solutions which are based on a 12 space tree misrepresent the true size of the sample space. A door which is not choose-able does not hold any possibilities and therefore; even though the math calculation is correct, calling the 12 space decision tree a model of the sample space for the problem is a false statement. It simply is not one. Rather, it's what the sample space would be, if and only if, the open door (which is not player choose-able) still held possibilities for the player - which it does not. That you can't seem to understand this is truly astounding. Tweedledee2011 (talk) 07:39, 16 January 2014 (UTC)

True answer to Martin's question
Martin - if you first pick a white ball from an Urn #1 mix of BBW, the next pick from Urn #1 can only be B, which is a probability of 0 for W. Do you concede this yes or no? Your mistake is that you think the information developed by the 1st choice is relevant - it's not. If you remove a W from a mix of BBW, there is no longer any possibility that W can come from that Urn. There are two urns and the condition you set was that the first pick "proves to be white". The odds of the first pick being white are irrelevant, because you have predefined the results of the first pick. For that reason, we are only concerned about the possible results of a 2nd pick. The possible results of a 2nd pick from a Urn #1, which starts at BBW, is only B.
 * Urn #1 starts as BBW
 * The first pick "proves to be white"
 * This leaves only BB in Urn #1
 * Pick #2 from Urn #1, can never be anything but B, if pick #1 from Urn #1 is W
 * Tweedledee2011 (talk) 16:32, 16 January 2014 (UTC)

Last try
You say 'Pick #2 from Urn #1, can never be anything but B, if pick #1 from Urn #1 is W'. Yes, of course.

Your problem is with this statement "Half the times, you have W at 0 from URN #1 ... and half the times', you have W at .50 from URN #2...". I have marked the problem in italic. Martin Hogbin (talk) 17:06, 16 January 2014 (UTC)

Because first pick results are predefined ...
Martin - The possible pools of choices results after the first choice are: BB and WB. There are no other combinations of possibilities available for the 2nd choice. Under no circumstances will you find a B in Urn #1 after first drawing out a W. And under no circumstances, will there be anything other than WB in Urn #2, after first drawing out a W. Do you concede this? Tweedledee2011 (talk) 16:52, 16 January 2014 (UTC)
 * Yes, of course. See above. Martin Hogbin (talk) 17:06, 16 January 2014 (UTC)

In a large series of first picks, the first pick will be W .50 of the time
Martin - because there is a total of WWW and BBB mixed between the jars, do you agree that 1:2 times, your first pick will be W, if you do the fist pick a large number of times? Tweedledee2011 (talk) 17:19, 16 January 2014 (UTC)
 * Yes. Martin Hogbin (talk) 17:43, 16 January 2014 (UTC)

In a large series of first picks, the first pick will be from Urn #1 .50 of the time
Martin - do you agree that in a large series of random first picks, your first pick with be from Urn #1 - the BBW Urn .50 of the time? Tweedledee2011 (talk) 17:30, 16 January 2014 (UTC)
 * Yes. That is what I say above. Please read what I have written and try to understand it. Martin Hogbin (talk) 17:44, 16 January 2014 (UTC)
 * Martin - if on average, over a large series of tests, for .50 of the times you run this test you must pick from Urn #1 on the first pick; then it logically follows that since your 2nd pick must be from the same urn (your rule sets this condition) .50 of the times you run this test, your 2nd pick must be from the remaining BB in Urn #1. What you don't understand is that your result of W from Urn #1 is an anomaly from Urn #1 - not the suggested proof you are in Urn #2, which you are interpreting it as. Look at your answer to this section again "Yes, that is what I wrote". Martin, if your first pick is from Urn #1 .50 of the time and if the results of the first pick are W, then for .50 of all first picks, the next result (the 2nd pick) from Urn #1 can only be B - and you have already conceded both of these points. Your predicate conditions create an anomalous subset which you are not correctly thinking about. You are the one who created this - by the way you phrased the series of predicates. Go back and re-read your own question. Tweedledee2011 (talk) 17:52, 16 January 2014 (UTC)

For any a large series of first picks, the 2nd pick will be between BB and WB, with each being .50 of the time
Martin, you can't simply extrapolate out that you are mostly going to be dealing with Urn #2 which is what you think your calculation proves. That's post hoc reasoning. You must first start with the correct mix of urns - as per the above results which are what would occur in a large series of 1st choices. Tweedledee2011 (talk) 17:35, 16 January 2014 (UTC)
 * No, the 2nd pick will not be between BB and WB, with each being .50 of the time. You can call this post hoc reasoning if you like.  Everyone else calls it Conditional probability.


 * Let me try to make this easier for you. Suppose urn A contains 99 white balls and 1 black ball and urn B contains 99 black balls and 1 white ball.  You chose evenly between the urns so you initially have a 1/2 chance of picking urn A and a 1/2 chance of picking urn B.


 * Now you draw a ball randomly from your chosen urn and this ball proves to be white. Do you claim that the probability that your chosen urn is A is still 1/2?


 * If you do claim that I have an idea. Let us play the game for money, real money.  You bet a sum of money.  You pick an urn at random and pick a ball.  If it is a white ball I will give you twice your money if the next ball you pick from the same urn turns out to be black. If the first ball is white and the next ball is white I win the stake.  Would you like to try to set up this game somehow?  Think hard about it!  Do some experiments. Martin Hogbin (talk) 18:12, 16 January 2014 (UTC)
 * Martin, you area again missing my point: Your conditional probability explanation relies upon a false premise - because you fail to accept that you are only dealing with a small subset - one which you cherry-picked in such a way that it could not occur as a consequence of the game you are describing. I've shown you that, but you refuse to accept it. If you concede that .50 of the time your first pick will be W and if you concede that .50 of the time your fist pick will be from Urn #1, then it's inescapable that .50 of the time (counting a full series of the 2 picks as 1 "time"), your second pick with have 0 chance of W and .50 of the time (counting a full series of the 2 picks as 1 "time") your second pick will have .50 chance of W. Tweedledee2011 (talk) 19:47, 16 January 2014 (UTC)


 * But I said that the first pick proves to be white. The means that the cases where the fits pick proves to be black do not count.  That is what Conditional probability means.  Only counting the cases where the condition applies is called conditioning the sample space.  Now re-read what I write at the start.  Martin Hogbin (talk) 20:03, 16 January 2014 (UTC)
 * If of course, you are saying that of the original WWW because WW are in Urn #2, then a larger portion of the original 1:2 chances of W are in Urn #2, then I do see what you are saying. You are saying that because it's more likely to find a W in a pool of WWB than in a pool of BBW, then when you find a W, you are more likely to be in the WWB pool, which means that your next potential results ought to take into account. This is true, but then the odds of getting W in Urn #2 on a 2nd choice are .50 And in any case, then my error on this also shows how people using normal verbal reasoning do not apply the math which you are talking about - which makes my point why how we are explaining MHP is deficient. What should be done is that we should show some ordinary normal vrbal reasoning, then show how that diverges from the conditional probability method have ably illustrated here - and contrast the different results, showing the gaps. With MHP, it's hard to show the gaps - due to the configuration of the problem itself. Also, what is the sample space of pick #2 from Urn #2 - I say it's WB. Tweedledee2011 (talk) 20:07, 16 January 2014 (UTC)

For Martin's answer to be correct...
For Martin's answer to be correct, the question must be phrased this way: If there are two urns, one which holds two white, plus one black ball and the second which holds two black, plus one white ball; when you randomly pick one ball from one of the urns and that ball proves to be white, if you replace that ball back into the urn from which you picked it, what is the probability that an immediately subsequent pick from by you from this same urn will yield a white ball? The answer to this question is 2/3 - but not the way Martin phrased it, it's not. Tweedledee2011 (talk) 16:01, 17 January 2014 (UTC)