Talk:Monty Hall problem/Arguments/Archive 2

Simple game
Suppose you're on a game show, and you're given the choice of three doors: behind one door the host has placed a car; behind the others, nothing. You pick a door, say No. 1, and the host then says to you what he always says, "Do you want to swap and have the contents of both doors 2 and 3?" Is it to your advantage to switch your choice?

Also, what is your chances of winning if you swap and if you do not? Martin Hogbin (talk) 20:48, 24 February 2009 (UTC)


 * I really cannot tell, because nothing is said about the distribution of the car. A Bayesian approach may be to call p the probability of the car behind door 1 and assume some distribution for p, and then integrate over it. Hence P(car behind 1) = E[p]. Any outcome is possible. But someone might argue that ignorance about p means a uniform distribution. Then P(car behind 1) = 1/2, reflecting the idea: "to be or not to be". Others may argue hat the expectation of p will be more likely 1/3, etc. Nijdam (talk) 23:27, 24 February 2009 (UTC)


 * That was exactly my point Nijdam, see reply to Rick below. Martin Hogbin (talk) 08:55, 25 February 2009 (UTC)


 * Surely Martin intends this be the same scenario as Monty Hall problem and simply forgot to say the door the car is placed behind is randomly picked, which means it is advantageous to switch and the probabilities are 2/3 if you switch and 1/3 if you don't. Note that this is equivalent to the "unconditional Monty Hall problem", not the typically stated version where the player decides after a door has been opened (another way to make it unconditional is to force the player to decide before the host opens a door).  In the article the paragraph starting with "The game assumptions play a role here" is alluding to this.  Also note that this version doesn't elicit the same "it should be 1/2" reaction - which I think is present in the standard version precisely because it's a conditional problem (I'll type up my thoughts on this at some point).  Most of the "unconditional" solutions make the 1/3 vs. 2/3 split completely obvious.  What they don't do is put the player squarely in front of two doors with a goat and car behind them, but with unequal probabilities (see, for example, the thread immediately above). -- Rick Block (talk) 00:57, 25 February 2009 (UTC)


 * No, I did mean the problem as I stated it and as Nijdam understood it. I think I am beginning to get the hang of this now.  I deliberately did not say how the host chose where to put the car, thus there is no defined probability distribution for the initial. position of the car.  The same as in the MHP where we have no defined distribution of the host door choice (unless we state that it is random).  If you agree with Nijdam's answer, I would like to move on to an even simpler game. Martin Hogbin (talk) 08:55, 25 February 2009 (UTC)


 * I know this. Somehow, it's gotta be conditional. Glkanter (talk) 01:26, 25 February 2009 (UTC)
 * ;-) Martin Hogbin (talk) 08:55, 25 February 2009 (UTC)


 * For the record: in the MHP the distribution of the car is given, of the choice of the door not, but this latter distribution is irrelevant, as it is stated that the player picks door 1.Nijdam (talk) 14:43, 25 February 2009 (UTC)


 * Yes, I know that. The problem with Morgan's solution of the MHP is there are two intermixed issues, firstly there is the basic, unconditional, problem which is very unintuitive and most people get it wrong; then there is the issue of the host choice of door, which is not stated to be random, leading to another layer of complication.  What I am trying to do is to compose a problem where the basic solution is obvious so that I can discuss the issue of a non-random choice of the host more clearly.

I have changed my mind and will stick to this game. Can I just confirm the proper answer to this question. Is it that we cannot give a definite answer without a defined probability distribution for the car's initial position? Is it that we have to take an arbitrary distribution and integrate over it to get a valid solution? Martin Hogbin (talk) 09:15, 25 February 2009 (UTC)

In this case I wouldn't choose a Bayesian solution, so as for me I cannot give an answer. NB. Mark what I said about the distribution of the choice of door. Nijdam (talk) 14:43, 25 February 2009 (UTC)


 * To get to the point you're talking about, you have to first assign probabilities of the car being behind doors 1, 2, and 3 to variables p1, p2, and p3 (0 <= pi <= 1) such that:
 * p1 + p2 + p3 = 1
 * Then we can observe if the player has picked door 1 her probability of winning by not switching is p1 and her probability of winning by switching is p2+p3 = 1-p1. The only probability that matters to the question asked is therefore p1 which is the player's probability of winning by staying (1-p1 if switching) and the player should switch if p1 < .5.  Most analyses would stop here.  If you insist on continuing to get a definite numeric answer you must assume how p1 is distributed between 0 and 1, and then in a Bayesian way you can compute a number based on this assumption (which, as Nijdam mentions above, turns out to be 1/2 if you assume p is uniformly distributed) but proceeding along this line without mentioning this assumption would be as unjustified as assuming p1 is 1/3 (both are assumptions).


 * If you haven't managed to find a copy of Morgan et al, I'll just note that they continue their analysis of the Parade version of the MHP (explicitly granting the "obvious" assumptions that the car is initially randomly placed and the host must open a door to show a goat and the host must make the offer to switch) in this vein. They mention that by assuming the host's preference p is uniformly distributed between 0 and 1, we can say the probability of winning is the integral of 1 / (1+p) from 0 to 1, which is ln(2) (about .693).  I haven't suggested we include this in the article because it requires more than simple probable theory to explain (and we wouldn't want Glkanter's head to explode, would we?).  -- Rick Block (talk) 15:11, 25 February 2009 (UTC)


 * OK, so we agree that strictly speaking you cannot just say that the player has a 2/3 advantage of winning if she swaps. Let us now say that the prize is not a car but $1200 cash. Now assuming the player takes the swap and you have the option of buying the two boxes from her for $800, would you do it?  Martin Hogbin (talk) 18:34, 25 February 2009 (UTC)


 * One more question, does it make any difference if the player is stated to choose her initial door randomly. Martin Hogbin (talk) 18:34, 25 February 2009 (UTC)


 * Are you going to circle back to the MHP eventually? Please just cut to the chase. -- Rick Block (talk) 19:20, 25 February 2009 (UTC)


 * As I explained above I am just trying to separate out two different issues. Is there any problem in answering the questions that I have asked? Just a simple yes/no for each would do. Martin Hogbin (talk) 21:22, 25 February 2009 (UTC)


 * The problem is you're bringing in more and more assumptions without specifying what you're talking about. What does "have the option to buy the two boxes" mean?  Is this a game show that has been broadcast repeatedly and we have some expectation for how often this chance is offered, or do you mean to be posing this as a probability problem (in which case what does "would you do it" mean?), or is this a huckster on the street, or some bar bet?  As a probability problem you've made it so ill-defined that any sound analysis should consider all the details you're (apparently deliberately) omitting. Maybe you can find someone else, but I don't care to play this game. -- Rick Block (talk) 22:47, 25 February 2009 (UTC)


 * What do you mean with: that strictly speaking you cannot just say that the player has a 2/3 advantage of winning if she swaps.? Who is she? Is she the player in the MHP or in your game? And for the other part of your question, you just seem to replace probabilities by expectations. What's the use? Nijdam (talk) 23:49, 25 February 2009 (UTC)

Paper by Morgan et al.
I have just purchased the paper by Morgan et al. but before I criticize it I would like to check one thing. Was there more that one statement of the problem published in Parade?

The statement given in Morgan, which claims to be taken from Parade, is cut-and-pasted directly from the paper below:

In a trio of recent columns, titled "Ask Marilyn," in Parade Magazine (vos Savant 1990a,b, 1991) the fol- lowing question was posed: "Suppose you're on a game show and given a choice of three doors. Behind one is a car; behind the others are goats. You pick door No. 1, and the host, who knows what's behind them, opens No. 3, which has a goat. He then asks if you want to pick No. 2. Should you switch?"

The statement that I find everywhere else is:

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?

Am I missing something or is the first thing that Morgan have done is to change the question? Martin Hogbin (talk) 10:56, 1 March 2009 (UTC)


 * The latter is also found on Marilyn's own homepage. To me it doesn't matter much, but my guess is to you it will be a whole worlds difference. I wait and see. Nijdam (talk) 15:07, 1 March 2009 (UTC)


 * I have a copy of the Sept 9, 1990 Parade column and in this column the wording is exactly (down the punctuation) as quoted in the article here (the latter of the above versions). I don't have copies of the other two columns (they are somewhat difficult to find).  The wording differences are pretty clearly editorial in nature and at least in my opinion don't change the meaning of the problem.  Whether it's a minor paraphrase (in which case they shouldn't have used quotation marks) or an exact quote from one of the other columns I'm curious whether you think these statements describe different problems.  -- Rick Block (talk) 15:11, 1 March 2009 (UTC)


 * Firstly, your point that they should not have use quotations marks for a paraphrased version seem fairly serious to me. If it is indeed true that Morgan have paraphrased what is claimed to be a direct quotation then, in my opinion, that fact in itself throws some doubt on their integrity and credibility.  The first question that I would want to ask is whether the change was accidental, just slipshod workmanship, in which case you would have to ask whether these are the right people to be dealing with such a tricky problem; or deliberate, in which case the question arises of what their motives were for doing this.


 * You ask if it describes a different problem. I would start by quoting what Seymann says in his commentary, 'Without a clear understanding of the precise intent of the questioner, there can be no single correct solution to any problem. Thus, with respect to the three door problem, the answer is dependent on the assumptions one makes about the intent of the one who initially posed the question'.  In considering this intent is is essential to stick to the exact words of the questioner.  It is quite staggering that in a situation where so much rests on slight changes of meaning Morgan should see fit to change the wording of the question.


 * Finally what changes? Quite a lot in my opinion.  In the Morgan version it is quite clear that the doors are numbered (which they were in the show) and that the player has actually chosen door 1.  The questioner actually say 'Pick a door, say number 1 [my emphasis].  Now when we look at Morgan's one line dismissal of what they call F5, they say that it is incorrect, '...because it does not use the information in the number of the door shown'.  It seems to me that one reasonable interpretation of the original questioners intent would be that no information should be revealed by which door was opened by the host.  That possibility is ruled out by Morgan's restatement of the question.


 * The original question could also be interpreted to mean the question that Morgan give as the answer to F1. The Morgan restatement rules out this interpretation.  In my opinion, Morgan have changed the question to suit their preferred answer. Martin Hogbin (talk) 16:06, 1 March 2009 (UTC)


 * If the problem is misquoted I agree it's editorially sloppy, however there is no mathematical sloppiness (Gillman paraphrases the problem in his paper - which, BTW, reaches all the same conclusions).


 * Here are the differences in the Morgan et al. version:
 * Suppose you're on a game show, and you're given the a choice of three doors: Behind one door is a car; behind the others, are goats. You pick a door , say No. 1, and the host, who knows what's behind the doors them, opens another door, say No. 3, which has a goat. He then says to you, "Do asks if you want to pick door No. 2? " Is it to your advantage to Should you switch your choice ?
 * The only changes that are not obviously logically identical are the changes to the third sentence, from (for example) "a door, say No. 1" to the seemingly more definite "No. 1". In both versions the doors are clearly numbered and use "No. 1" to refer to the door the player picks and "No. 3" to refer to the door the host opens.  I know the point you're making, but to a mathematician the label "No. 1" is merely a notational convenience.  Both of these are saying "let's call whatever door the player picked No. 1 and whatever door the host opens No. 3".  The Morgan et al. analysis has nothing to do with the specific numbers on the doors that are picked, only that there are 3 doors and they can be clearly identified as the one the player picked (which we'll call No. 1), the one the host opens (which we'll call No. 3) and the other one (which we'll call No. 2).  Mathematically, we could just as well call them A, B, and C, or "the door the player initially picked", "the door the host opens" etc.  The information content is not the number, per se, but that there are three distinct doors (we're not doing quantum physics here, just a simple probability problem).


 * Both Morgan et al. and Gillman explicitly describe the difference between how the problem is usually phrased and how it would have to be phrased to be sensibly interpreted as an unconditional problem. You think an unconditional interpretation is valid, and want the article to prominently feature an unconditional solution (right?).  I suggest this thread may more properly belong at talk:Monty Hall problem not here, since it's about the article rather than the math behind the problem.  However, my response (I think I've told you this before) is that it truly doesn't matter whether you think an unconditional interpretation is valid (per WP:OR) - only whether there are sources supporting this view.  Trying to convince anyone an unconditional interpretation is valid is a complete waste of time without sources to back up this view (and lest anyone reading this is misinterpreting this as WP:OWN, what I mean is that Wikipedia policies prohibit WP:OR and require verifiability).  Furthermore, since the sources (plural, Morgan et al. and Gillman) saying the problem should be interpreted as a conditional problem are published in academic math journals, sources supporting an unconditional interpretation should also be in math journals (ideally referring to one or both of Morgan et al. and Gillman).  I don't know if you have convenient access to a university library, but given such access chasing down all academic sources that cite Morgan and Gillman should not be that hard (and would likely take less time than you've spent arguing about this here).


 * I think this should be a simple question to resolve. There are sources supporting your view or there aren't.  If they exist, what are they?  -- Rick Block (talk) 18:35, 1 March 2009 (UTC)


 * I have to accept that, on the face of it, Morgan is a reliable source, that is why I have not been making radical changes to the article. All I am trying to do here is to get some acceptance of my view that Morgan should not be taken as the last word on the subject. As you no doubt know, I find the paper poorly written, unconvincing, patronizing and arrogant.  There is a hint that I am not alone in this view in Seymann's polite comment at the end of the paper.

It is all about how you choose to interpret the question. There is no reason that Morgan should be considered any more expert at this that vos Savant or you or me.Martin Hogbin (talk) 09:47, 2 March 2009 (UTC)
 * Well, there actually is. The Morgan et al. paper was accepted for publication in The American Statistician.  Morgan is now a professor of statistics at Virginia Tech university .  The paper's conclusions are exactly consistent with another paper separately written by Leonard Gillman published in American Mathematical Monthly.  From a Wikipedia perspective (see WP:RS), reliability is an attribute of a published source rather than an individual.  If you haven't published anything on this other than your writings here (which Wikipedia doesn't consider a reliable source) your thoughts on the topic are irrelevant (mine, too).  This is precisely what WP:OR is all about.  At the level of individuals, I don't know whether you're a dog, nor you, me (the intent here is to be mildly amusing, not insulting).  -- Rick Block (talk) 14:39, 2 March 2009 (UTC)


 * So is this Wikipedia article wrong? I read the Morgan et al. paper and I agree with their treatment of the problem as a conditional problem.  The Wikipedia article says that the probability of winning when switching is 2/3, Morgan's paper concludes that we cannot even calculate this probability without knowing the host's strategy, but we can conclude that it is at least 1/2 thus we should switch as we can do no worse.  Either the Wikipedia article is wrong, or is solving the unconditional problem (which is also wrong). --67.193.128.233 (talk) 18:24, 2 March 2009 (UTC)


 * Morgan et al. analyze the Parade version of the problem which does not include the constraint that the host pick equally between two goats (if given the chance). The fully explicit statement here (from Krauss and Wang) includes this constraint - which makes the conditional probability 2/3.  It's the p=1/2 case in the Morgan et al. paper.  -- Rick Block (talk) 19:42, 2 March 2009 (UTC)


 * Rick, my point was not that Morgan is not an expert in statistics; that he clearly is and I am not. I hope you will agree with me that the best approach to the MHP is to follow the following procedure.  Start with some statement of the problem, the Parade version is usual, decide on what this question means in terms of formulating a well defined probability problem, then solve the problem.  Maybe repeat this process for other interpretations.  When it comes to formulating the problem, we should ask ourselves, as Seymann suggests, what was it that the original questioner actually wanted to know.  Who is the expert in doing that?  Not necessarily a professor in statistics.  If anyone, I would say that vos Savant was the expert there.  She ran a column in Parade in which she regularly answered readers' questions.  She knew the readers well and would have had the best idea what they were getting at when they asked rather vague questions.


 * My understanding of her view is that she believed the reader to be intending to ask a simple, well defined, probability question, in which anything undefined should be taken as random, ' the host acts as the agent of chance' as she put it. If you start with that interpretation then I think even Morgan would not see the need for taking into account which door the host (randomly) opened.  What they do in fact do is start with a more complicated interpretation, where the host action does make a difference, and then insist that the simple case must be treated as a special case of the more general one.  Of course you can do that but, myself, 67, and many others do not accept that this is absolutely necessary and I am not sure that even Morgan say that.  They intentionally leave the formulation issue slightly vague so they can justify their more complex solution. Martin Hogbin (talk) 19:59, 2 March 2009 (UTC)


 * Martin - I'm sorry I missed your point. Please accept my apologies.  Responding to your actual point, the organization of this article seems to be nearly as tricky as the problem itself.  I suggested a while ago (see Talk:Monty_Hall_problem) that we might treat this in much the same way as intense POV magnets like intelligent design.  I'm starting to suspect no organization will satisfy everyone, but find myself leaning to starting with a full out conditional solution presented as intelligibly as possible followed with other solutions, both simpler (unconditional) and more complex (Bayes).  The reason I'm drawn to this is that any variant of the problem can be solved with a proper conditional approach - which seems to make it the right tool for the job.  My issue with an unconditional approach is that it would present an analysis highly subject to misuse in situations where it would not apply (like the "Monty forgets" scenario).  Unconditional approaches don't lead to an actual understanding of the problem.  -- Rick Block (talk) 03:33, 3 March 2009 (UTC)


 * Thanks for you apologies but they may not be fully deserved. I have two arguments.  The first is that the unconditional solution that we already have should be better presented in the article.  I do not think that you have too much problem with that in principle but there may be some argument as to where and how it best fits in.


 * My second argument is with the Morgan paper in that I believe that it seriously misrepresents the problem. I have not yet managed to adequately articulate exactly what my problem is and I appreciate her that this may be a battle that I cannot win but I intend to keep trying. Martin Hogbin (talk) 09:31, 3 March 2009 (UTC)


 * To make things clear, the so called 'unconditional solution', can only be presented as a simplified aid of understanding, not as a solution. That's what I pointed out, and what started the recent eruption of arguments.Nijdam (talk) 09:50, 3 March 2009 (UTC)


 * For which formulation of the problem? Martin Hogbin (talk) 17:47, 3 March 2009 (UTC)
 * Of course for the more exact statement of the problem is as it is stated right before the solution section. Nijdam (talk) 22:31, 4 March 2009 (UTC)

What is the sample space?
I have moved a discussion about what the correct sample space is to this new section for ease of editing. I hope that is OK with you all.


 * I have to interject here, there is a lot of incorrect use of probability theory and the term sample space. There is only ever one sample space.  It does not change and is not "downsized".  If you change the sample space in the middle of a problem, you are NOT doing probability theory. Here are the definitions: A sample space is merely a set S.  A probability space is a sample space, S, along with a probability measure, p, which assigns a probability (when the sample space is countable or finite) to every subset of the sample space, which we call events.  The "downsizing" you refer to is conditioning.  When we condition on an event A in S, the sample space remains the same, but we have a different probability measure which assigns zero to events which do not intersect A and is given by the usual formula. ( p(B|A)=p(A intersect B)/p(A) )


 * Now let's look at this problem. The sample space contains only 3 elements, S={D1,D2,D3}, corresponding to which door the goat is behind.  Our probability measure is p(Di)=1/3 for all i=1,2,3.  The game is: we choose a door, then the host shows us one of the other doors, containing the goat, and we have to decide if we should swap or not.  So, we'd like to find the probability of winning when we swap and when we don't.  To find a probability, you first need to find the event associated with that probability and then you can compute the probability of that event using the probability measure.  I will condition on the door that we choose.  Let X be the door we choose (i.e.X=1,2, or 3).  Let Ex be the event that we swap and win given that we choose door x (i.e. X=x).  Then let's consider each case individually.  For X=1 (corresponding to us choosing door 1), the event is E1={D2,D3} since we will win if the car is behind door 2 or 3, thus p(Ex)=p(D2) + p(D3) = 2/3.  It is clearly the same for all the other cases (X=2, E2={D1,D3}, X=3, E3={D1,D2})


 * Now you're probably going to say something like, "you've neglected the fact that the host gives you information by opening a door." Well, actually, the host tells you nothing.  You already know that one of the 2 doors that you did not guess has a goat behind it.  And the host will always show you that door, so it tells you absolutely nothing.  Mathematically, the event that "the host shows you a door with a goat behind it" is the entire sample space (because it is a certain event i.e. it has probability 1).  So when we condition on this event nothing changes.  The game is actually equivalent to this:  You choose a door, and the host says, "You can keep your door, or you can switch and guess BOTH of the other two doors.  If either of the other two are correct, then you win!".  Here you'll obviously choose to switch, but this is the exact same game, just disguised differently.  Hope this helps you understand. --67.193.128.233 (talk) 23:07, 1 March 2009 (UTC)


 * 67 - I disagree. Wikipedia's published definition of Sample space includes only what is "possible" ("is the set of all possible outcomes"). It's simply not possible that the removed door has the car behind it, nor is it possible that the removed door will be chosen. For these reasons, the removed door is not in the sample space at the point of the 2nd choice. Until this point is cleared up, my line of reasoning will not be agreed with by your viewpoint. 216.153.214.89 (talk) 23:43, 1 March 2009 (UTC)


 * Do not place so much emphasis on the word "possible" in the definition of a sample space. A sample space is merely a set. It has the interpretation of being the outcomes of a trial and could even include events which have probability zero (i.e. events which are not possible).  In the case where S is finite this is usually avoided, but when S is uncountable (e.g. the interval [0,1]) then there can be (are) infinitely many events with probability zero (e.g. not possible events). Also, events do not get "removed" from sample spaces.  There can be only one sample space in a given problem.  Choose it and stick with it. --67.193.128.233 (talk) 03:07, 2 March 2009 (UTC)


 * I agree with what you say but the reliable sources quoted in the article do not. They seem to insist that you must consider which door the host opens, see ingoing discussion on Morgan et al. for example. Martin Hogbin (talk) 23:28, 1 March 2009 (UTC)


 * Martin - Right.


 * You are correct about this. We are actually both correct in a way.  I read the paper by Morgan et al. It's very nice.  This isn't the right section for discussion on the paper, but my understanding is the reason you need to consider the door the host opens is that you don't know the host's strategy and the probability that you will win if you switch depends on the host's strategy and which door he opens.  However, if you know that the host picks a random door (when there are more than one to choose from), then I don't think you need to consider which door he opens.  So maybe I was right in a limited case, but for the general case where the host's strategy is not known, you must consider which door he opens.  I'll elaborate in the Morgan et al. section. --67.193.128.233 (talk) 18:12, 2 March 2009 (UTC)


 * 67 - The full sample space is a 3x3x3 matrix of car location, initial player pick, and door the host opens (many of which have 0 probability). We're talking about various events in this sample space, which (since they are simply subsets) can in turn be considered to be conditional sample spaces.  Rather than use precisely formal terms (since that seems to confuse the non-mathematicians in the crowd), I'm echoing terms folks are using in the way they appear to be used.  -- Rick Block (talk) 02:07, 2 March 2009 (UTC)


 * We can certainly consider the sample space 3x3x3. But there is no unique sample space for any given problem in probability.  Since the location of the car and the initial player pick do not influence the probability we wish to calculate, there is no use in modeling them so a sample space with 3 elements is suitable. --67.193.128.233 (talk) 03:07, 2 March 2009 (UTC)


 * I think you need at least 4, although 6 is probably more typical. Calling the door the player initially picks door 1, the six are CGG2 (Car-goat-goat, host opens door 2), CGG3, GCG2, GCG3, GGC2, GGC3.  Under the normal rules p(GCG2) and p(GGC3) have probability 0 and p(CGG2)+p(CGG3) = p(GCG3) = p(GGC2) = 1/3.  The Parade version doesn't specify the relationship between p(CGG2) and p(CGG3), leading to an answer of the form 1 / (1 +p).  -- Rick Block (talk) 06:56, 2 March 2009 (UTC)

What is wrong with the Morgan paper
The problem with the original Parade question is that it leaves many things undefined, the major ones being:

1 The way the car and goads are distributed behind the doors initially.

2 The way the player chooses a door initially.

3 Whether the host always offers the swap.

4 Whether the host always opens a door to reveal a goat.

5 How the host decides which door to open when he has a choice.

In order to produce a well posed and unambiguous probability problem we must answer all these questions (and maybe more). So how do various people do it?

At one extreme, vos Savant takes it that the original questioner intended to ask a simple, well define, probability problem in which the host always offers the swap, always reveals a goat and the other probabilities not defined in the question are taken as random.

At the other extreme we could consider it as a 'real world' problem, in other words what should you do if you find yourself on a show as described by the question with no other information being given, not even assuming that the question refers to 'Let's Make a Deal'. The mathematical answer in that case is quite simple. The probability of winning by switching is between 0 and 1, we can say no more than that as there are so many unknowns. Maybe the car is always behind door 2 for example. If we want to go beyond that, we need to make some assumptions about the unanswered questions.

So, what information is present in the original question regarding point 1? None. How can we deal with this? I suggest three ways in principle: a) say the question is so vague that it cannot be answered, b) assign probabilities that the car is behind each door (say C1, C2, C3) or c) take the distribution to be random. There is no right answer.

What about point 2? The question says that you initially pick a door, say 1. What does this mean? Now we have the three options above plus one more, which is to assume that, because the question mentions the number 1 the player always chooses door 1. Again, there is no right answer.

Does the host always offer the swap. Many people here seem to take it that he does but we need no do so. We could assign an overall probability that he will do so, or even probabilities depending on the players original choice or the position of the car. There is no correct answer.

Similarly for the host always revealing a goat. Most people assume that he does. (I believe that there is no record of Monty ever revealing the car.) On the other hand, we could assign probabilities that he reveals a goat dependent on the players original choice etc.

Finally we come to the matter of the host's choice of door. This is exactly the same as 1 and 2. No information is given in the question as to how the host might choose but a door number is mentioned. We therefore have the same options as in 2 above. All the options are valid, it is a matter of choice as to how we formulate the precise question.

So, what do Morgan et al. do?

They seem to ignore point 1 and and tacitly take the initial probability distribution of the car between the doors to be random, that is to say evenly distributed between the doors, my option 'c' above. They start by assuming information not given in the question, as they put it.

If we assume the initial distribution to be random, then the player's initial choice of door is not important although Morgan neither state nor demonstrate this fact. They consider only the case in which the player initially picks door 1, letting us assume that the results would always be the same if the player chose another door.

Regarding point 3 Morgan only consider the case in which the host always offers the swap but suggest that the interested reader might like to consider the case that he may not.

On point 4 Morgan start with a general case and assign probabilities to each door that the host may open. They then exclude the possibility of revealing the car to get what they (incorrectly in my opinion) call the vos Savant scenario.

For point 5 Morgan take my option 'b' above by using the probabilities assigned in 2 above to the various possibilities of the door that the host may open.

What is wrong with Morgan's formulation then? Nothing at all. Morgan have produced a precise formulation of the problem from the original vague question.

What they do wrong is assume that their formulation is the only correct one. Once this is done it then follows that their solution is the only correct one. If we assume the vos Savant formulation there is no need for a complex calculation involving the probabilities of various host actions and the simple solution becomes correct.

I would ask anyone who may think that this is just my individual view to look at the comment by Seymann at the end of the Morgan, where he rather more subtly says exactly the same as I have said. He says [my italics] 'Without a clear understanding of the precise intent of the questioner, there can be no single correct solution to any problem'. Martin Hogbin (talk) 23:24, 5 March 2009 (UTC)


 * I don't understand what your point is. Are you claiming the Monty Hall problem is commonly understood as something different from the problem Morgan et al. analyze (if so, do you have any references for what the commonly understood version would be)?  The refinements they make to the problem statement (notably not including anything about your point 5) are the ones necessary to justify vos Savant's unconditional answer (she clarifies how she interpreted the problem in the subsequent columns, see ).  And, although you don't seem to think so (WP:OR comes to mind), saying nothing about how the host picks between two doors if he has the chance does not make the problem unconditional in nature.  You keep suggesting the problem should (or at least can) be interpreted unconditionally and that Morgan et al. (and, BTW, Gillman who you have so far ignored) unnecessarily insist the problem as stated in Parade is a conditional problem.  If this is anything other than simply your own opinion, (again) please provide a reference.


 * I presented my whole argument above (maybe not very well) so that you could see where I was going rather that try to ambush you. I will now go through it in stages to see where, if anywhere, we disagree.


 * Can I take it that you agree that the Parade statement is vague and that to get any solution at all we must formulate precise statement of the problem?


 * Do you agree that we must decide what the original questioner meant in respect of my points 1 to 5 above? Martin Hogbin (talk) 18:14, 6 March 2009 (UTC)


 * Yes, the Parade statement is vague although in subsequent columns vos Savant clarified your points 3 and 4 above. Since the car is hidden at the point of the player's initial pick, points 1 and 2 don't actually matter (a blind choice among 3 alternatives has a 1/3 chance of being right, regardless of the underlying probability of the 3 alternatives - we can go through the math if you'd like but the bottom line is a random choice is equivalent to locating the car randomly in the first place).  Rather than decide what the original questioner of the Parade version meant, I think what we actually need to do is present the problem as it is usually meant to be interpreted (by the preponderance of sources), and that the unambiguous Krauss and Wang version (in the article) captures what most sources actually mean. -- Rick Block (talk) 03:59, 7 March 2009 (UTC)

Morgan et al.'s assumptions

 * Firstly, can we suspend the reliable sources bit just for the moment. I accept that these would be needed for changes to the article itself but they are not needed for me to make a point.


 * My problem is in the way Morgan treats my points 1 and 5. Strictly speaking, as you and others have pointed out earlier, to solve a problem like this we must use only the information actually given in the question.


 * What information is actually given in the question as to my point 1, the way the car is initially placed, and my point 5, the way the host chooses which door to open? In both cases the answer is quite clear, none.


 * How can we deal with this lack of information?


 * a) We can say that it is impossible to give an answer because the problem is not well enough defined


 * OK, but if we do that in both cases that is the end of the problem.


 * b) We can assign unknown probabilities to the various possibilities and proceed with the calculation from there, maybe considering as special cases the case where the car has an equal probability of being behind any door or the host chooses with equal probability which door to open. For some unexplained reason Morgan do that for the host choice but not the initial position of the car. You might say that because the player's choice is random (which is not actually given to be the case) the initial car placement does not matter, but exactly the same argument applies to the host choice of door.


 * c) We can take what we have no knowledge of to be random. In this case the initial disposition of the car and the host choice of door should both be taken as random.  This could be taken as obligatory (as I have suggested before) for the reasons that you give above.  Both ourselves and the player are blind to the original car placement and the host's door opening strategy.  In other words if the car were to be always placed behind door 2 and the host were always to open door 3 where possible, the player would not know this so we could take it to be effectively random. There is no difference between the two cases.


 * e) We can use some real world knowledge to suggest likely scenarios. Almost anything is a possibility here, most likely, I would say was that the initial car placement would be roughly random but I doubt truly so, and the host's strategy would be approximately random also since if it were not someone in the audience would have noticed.  But all this is just idle speculation.  We can easily invent plausible scenarios for any possibility.  This is equally true for the host door choice and the initial car placement


 * f) We can, and should according to Seymann, ask ourselves what the original questioner meant. I personally doubt that he expected an answer based on the concept that the car would always be behind door 2 or that he meant us to assume that the host might have some dastardly plot to prevent the player from winning. If there is an expert here it is vos Savant not Morgan.


 * Do you see my point now? Morgan treat my points 1 and 5 quite differently although the information given in the question, our knowledge, and the player's knowledge are exactly the same in both cases.    Martin Hogbin (talk) 17:36, 7 March 2009 (UTC)

(outindent) I see your point, but I disagree. Morgan et al., and Gillman both attempt to interpret the problem the same way vos Savant did. Nothing more, nothing less. vos Savant makes it very clear in her third column (see ) she considers the initial placement (or the player's initial pick which has the same effect) to be random. We're not lacking information in this case. It's not explicitly in the problem statement, but neither are the "must open a door" and "always shows a goat" constraints. And, as I've said, assuming the car and goats are behind doors logically forces the initial choice to be random (which is equivalent to randomly placing the car).

They assign an unknown probability to the host's preference for one goat over another because vos Savant never attempted to clarify this. Assigning a variable to this probability in this case is not inexplicable, but rather exactly what needs to be done to solve the problem. Your points 1 and 5 are completely different. vos Savant clarified #1, but never said anything about #5. To remain true to vos Savant's interpretation they make the same assumptions she did. BTW - the original questioner (of the version in Parade) was Craig Whitaker, not vos Savant. -- Rick Block (talk) 00:12, 8 March 2009 (UTC)

Changing gears

 * Changing gears somewhat, do you find the explanation I provided to 216 above (in this thread) convincingly clear (this could be directly referenced to the Morgan et al. analysis)? The follow on that gets to the conditional probabilities is not there (basically, the next step is to show the conditional probability of the car being behind door 2 if the host opens door 3 is p(2) / (p(1b) + p(2)), which is 2/3 - for the unambiguous Krauss and Wang problem statement).  This style of analysis works for any variant (host forgets, host picks leftmost door if he can, etc.), so although it's a little less simple than the unconditional 1-1/3 type of solutions, seems far better to me.  You've never really responded to my criticism that the unconditional solutions don't address common variants (the "Monty forgets" scenario has been brought up by vos Savant in Parade).  I think one of the main reasons people don't get the right answer to this problem (I mean 1/2 rather than 1/3) is because they don't understand the difference between conditional and unconditional probability - do you not think making that difference clear in this article would be a Good Thing™? -- Rick Block (talk) 14:34, 6 March 2009 (UTC)


 * I think you are underestimating just how clear the solution needs to be to convince most people. It need to be abundantly clear, virtually obvious, to get the message across - not easy.


 * Regarding the ability to address various variants as you call them, I would say that there are no variants there is just one question with a variety of formulations, which are attempts to clarify what the original questioner meant. This is really what my point about Morgan is all about.  The various possible formulations are part of the answer not the question.   If I were asked to answer the question, I would start by stating what I took the original Parade statement to mean.  In my case that would be the same as vos Savant.  Having (hopefully) stated exactly the assumptions that I was making, I would the proceed to a simple answer, without the need for conditional probability.  To attempt to answer all possible meanings of the Parade statement in one hit is impossible. What Morgan do is to take their own particular formulation and then insist that this is the only on possible. Martin Hogbin (talk) 19:50, 6 March 2009 (UTC)


 * Heretic! Glkanter (talk) 23:48, 6 March 2009 (UTC)


 * There are clearly variants - many sources discuss them (host forgets is one). There are also different formulations, but I think there is definitely a "standard" formulation - and it is identical to the problem Morgan et al. and Gillman analyze (although at this point, I think most sources also include the "host must pick randomly if given the choice" constraint to force the 2/3 win by switching answer). -- Rick Block (talk) 03:59, 7 March 2009 (UTC)


 * Rick, this statement of yours "I think one of the main reasons people don't get the right answer to this problem (I mean 1/2 rather than 1/3) is because they don't understand the difference between conditional and unconditional probability" betrays your mathematicians brain bias and a completely erroneous understanding of the non-math mind thinking process. Furthermore, your comments exhibit a POV bent which, if unchecked would likely result in WP:POV violations should you edit the article. Here's why: I spent the better part of several hours writing posts explaining how the non-math mind looks at this, and you've gone right ahead and reverted to technical jargon. There's only one reason why the non-math mind gets MHP wrong: They don't know to compose an accurate formula which takes into account the correct data and correct rules of Probability. Simply put, most people think choice two is a guess, so they don't know that choice one provides valuable information and most people don't know that the removed door is still relevant to the calculation. When you say "I think one of the main reasons...", it's sheer speculation. You haven't vetted your thinking by trying to understand the non-math mind. All you've done is extrapolate your limited knowledge of most people and leap to conclusions. 216.153.214.89 (talk) 15:19, 6 March 2009 (UTC)


 * This is the talk page, not the article. Technical jargon is OK here.  In the sources of confusion section, there's a reference to psychology papers that explore why people don't get the right answer.  This is a famous problem among psychologists because people get it wrong so often.  People have a very strong tendency to think probability is evenly distributed among as many unknowns as are present  ("partition-edit-count" is what Fox and Levav call it, Falk calls it the "equal probability" assumption).  Mathematically, this is a confusion between unconditional and conditional probabilities.  People do not recognize that using your DR (unconditional probability) is not always appropriate.  Most conditional problems that create a difference between the DR (unconditional probability) and a conditional probability lead to massive confusion (Boy or Girl paradox is another).  These are mathematically trivial, but extremely hard for people to solve correctly.  People simply suck at conditional probability evaluations. -- Rick Block (talk) 03:59, 7 March 2009 (UTC)

Martin, so you don't get dominated math-mind bias here, I've decided to chime in. The MHP as explained in this article bears on your 5 points thusly:

1 The way the car and goats are distributed behind the doors initially.
 * Irrelevant; Player is playing only once, so distribution method bias has no chance to creep in over time.
 * Also (per my comment above), if the choice is random the distribution does not matter (and might as well be random). -- Rick Block (talk) 03:59, 7 March 2009 (UTC)

2 The way the player chooses a door initially.
 * Irrelevant; The Player's Probability on his 1st choice can't be anything other than 1/3 - unless he has information about what's behind any or all doors. There's nothing in MHP which indicates he has any such information and presuming he does violates the fact set presented to us.
 * Exactly. Since the doors are closed the player's choice is inherently random. -- Rick Block (talk) 03:59, 7 March 2009 (UTC)

3 Whether the host always offers the swap.
 * Irrelevant; For the episode of the show we are watching and with this contestant he does. What happens other times doesn't affect our player.
 * This one is relevant, and vos Savant was criticized for not clarifying this. It is a notable criticism, but in the "standard" version the host must make the offer. -- Rick Block (talk) 03:59, 7 March 2009 (UTC)

4 Whether the host always opens a door to reveal a goat.
 * Irrelevant; The problem as written tells us that he does: "[The host] opens another door, say No. 3, which has a goat." The "say", as a matter of logic, can only refer to any door which has a goat - the sentence clearly states this to be so via "which has a goat" and the "which" is referring to the term "another door" not the agreed presumption of "No. 3" which is applied by virtue of us not being present to object to the "say". If you take the "say No 3," out of the sentence, the rule for the choice becomes clear. The "say No. 3" only refers to how the rule for the choice is applied in this instance. Re-write it like this "[The host] opens another door, which has a goat (say No. 3)" and there's no question that the "say" is a presumption of execution, not a variable of host choice. The sentence was clearly written as it was to distract people who are trying to logic diagram it. It must be accepted at face value. It's grammatically correct and logically sound as written and it doesn't leave open any alternatives other than the host picks a goat door.
 * This one is also relevant, and vos Savant was criticized for this as well - but it is also clear in the "standard" version. -- Rick Block (talk) 03:59, 7 March 2009 (UTC)

5 How the host decides which door to open when he has a choice.
 * Irrelevant; The host will always face a choice of either car/door vs. goat/door or goat/door vs. goat/door. Unless he reveals a car, which is impossible, he'll always choose a goat door.
 * Also relevant, per the Morgan et al. and Gillman papers. The probability of winning by switching is typically meant to be 2/3 which requires the addition of the "equal goat" constraint.  This constraint is also in what most sources consider the "standard" version. -- Rick Block (talk) 03:59, 7 March 2009 (UTC)

I think more dialog on this page ought to be spent actually understanding how non-math minds think this through, so as to better clarify the phenomenon associated with this problem - that being the arguing which goes on over the solution. 216.153.214.89 (talk) 15:46, 6 March 2009 (UTC)


 * There's a start on this in the section "Sources of confusion" in the article (with references). If you can provide more and or better references about this, or clarify the discussion in this section of the article, please do so.  -- Rick Block (talk) 03:59, 7 March 2009 (UTC)

216 begs off

 * Rick - I don't think confusion and ignorance are the same thing. Confusion is what you experience when you are in a crowded room during an event and 5 people are shouting to you at the same time. Ignorance is putting JiffyPop in the Microwave. MHP is not confusing in the least. Rather, it's that the rules for composing the right solution are not well known and that the question as posed doesn't prompt people to think deeply which causes people to miss. It's deceptively simple in it's presentation, so much so that even otherwise capable people overlook the solution. Having chatted on this page, I see how MHP as an object lesson provides a good forum for opening a dialog on Probability, but I am nonetheless irked at how most presentations of it gloss over or omit the augmentory information which could make apprehending it accesible. Perhaps I needed to think though my frustrations against Massimo and his "guess" variant, but now that I have talked here, my only interest in the article is to see what I can find about how people can use MHP as a learning tool to understand a workable everyday appreciation for the value of deeper analysis before giving shallow answers. My personal feeling is that "gotcha" questions cause resentment and make people not want to learn, so if this were my article, I'd focus more on explaining how it is that most people get it wrong, and less on the esoteric skill of doing probabilistic calculations. I prefer apprehendable knowledge which is available to many on a wide dispersion and I am interested in building bridges of understanding between ignorance and the ability to self-teach. I'm going to reconsider, in that light, my interest level towards making contribtuons here. Thanks for your feedback. 216.153.214.89 (talk) 05:42, 7 March 2009 (UTC)


 * I agree confusion and ignorance are different, however MHP is one of several problems that nearly all people get wrong - even people who are not ignorant of conditional probability. For example, Paul Erdős (one of the most brilliant mathematicians of the 20th century) got it wrong.  Massimo's quotes about this in the article "... no other statistical puzzle comes so close to fooling all the people all the time" and "that even Nobel physicists systematically give the wrong answer, and that they insist on it, and they are ready to berate in print those who propose the right answer." are dead on.  If he teaches a "guess" variant, he either doesn't understand it himself or is changing it enough to make sure his students are actually paying attention to the problem statement and not offering up a solution they don't actually understand.  If you're objecting to the section title in the article, please change it to something better.  The point of the article here is definitely not to pull a fast one on anyone.  -- Rick Block (talk) 01:58, 8 March 2009 (UTC)

What's funny to me is that even if the odds of switching were only 1/2, that's still better than 1/3, so switching would still be better. MHP as it 1st came out in Parade via MvS's column clearly asks: "Is it to your advantage to switch your choice?". The answer to that question regarding that "original" version of MHP, is always "yes". Even if one calculates (wrongly) that it's 1/2 - that's still an upgrade from 1/3 and does improve your odds. My thinking is that this problem tells us much more about interpersonal authority assumptions that anything else. People are always on the alert to take authority clues from those in charge. When a math-able person gets this wrong, I am confident that it's because they don't realize the information from the excluded door is still allowable in their calculations. It's as if they rule it "off limits" when the host authority modifies it from its choosable state to a non-chooseable state. People mistake that state change as a removal from the location set, but it's not. It's state is changed, but it's still in the location set and subject to all Probability calculations. 216.153.214.89 (talk) 02:21, 8 March 2009 (UTC)


 * The typical wrong conclusion is not that the player's original door stays 1/3 and the other door becomes 1/2, but that both doors become 1/2 after the host opens a door. In terms of the pie chart explanation below, I think what people think happens when the host opens a door is only that the door 3 piece is removed leaving two equal sized pieces.  They overlook the fact that the host's action creates two alternatives (two pieces of the original pie), and that all of the original (unconditional) 1/3 chances must be divvied up between these two alternatives.  The way the problem is set up, all of the probability associated with each unpicked door ends up in one alternative (the entire door 2 piece is in the alternative where the host opens door 3, and vice versa) but the probability associated with the player's door has to be be split between the two alternatives.  If we want the probability to end up the same in these two cases the player's piece ends up getting split exactly in half.  This kind of thinking is precisely what conditional probability models.  People run into this sort of problem all the time, but the division into subcases is generally "fair" which keeps the proportions in the subcases the same as the original.  In the pie perspective, what people are used to is the pieces being cut in half and the two alternatives consisting of 3 pieces just like the original pie, so overlooking that they went from 1/3 - 1/3 - 1/3 to 1/6 - 1/6 - 1/6 doesn't affect their relative probabilities (but this doesn't mean the original probabilities aren't divided into two subcases).  The unusual aspect of this problem is that the division into the two alternatives is not "fair".


 * I'll venture a guess based on what you've said in this thread that the way you really think about this problem (now) is that you think the player's original 1/3 piece remains the same and the host's action doesn't affect it. A better model is that the unpicked piece remains the same and the player's piece is cut in half.  Since we're now talking about half the original pie, the player's "new" 1/6 piece ends up being 1/3 of what's left but it has changed from a 1/3 sized slice to a 1/6 sized slice.  -- Rick Block (talk) 16:48, 8 March 2009 (UTC)

Why people say the answer is 1/2
The basic reason many people when presented with this problem think switching does not matter is because they unknowingly confuse conditional and unconditional probabilities.

The unconditional probability of initially picking the car or of the car being behind either of the unpicked doors is obviously 1/3.


 * No, the probability depends on how the car and goats are placed. This is not given in the statement of the question. We can chose to take this as random if we wish. Martin Hogbin (talk) 18:31, 6 March 2009 (UTC)


 * I'm assuming we're talking the "standard" version of the problem, which I claim is the fully unambiguous Krauss and Wang version in the article. In this version the car and goats are distributed randomly. -- Rick Block (talk) 05:54, 7 March 2009 (UTC)

After the host opens a door showing a goat, many people mentally evaluate the new situation with reasoning like "the door the host opened is 0, my door is 1/3, the other door is 1/3, 1/3 = 1/3 so switching doesn't matter".

The problem with this reasoning is that the original 1/3 probabilities are unconditional. After the host has opened a door, the conditional probabilities in this case must sum to 1. The question the problem asks is how do the original unconditional probabilities transfer to the two conditional cases where the host opens one door or the other.

The answer to the question of where the probability goes is that some of it goes to the case where the host opens door 2 and some of it goes to the case where the host opens door 3. At the beginning, unconditionally, the three probabilities are equal and sum to 1:


 * 1/3Door 1 + 1/3Door 2 + 1/3Door 3 = 1

After the host opens a door there are two basic terms:
 * (probabilities in the case door 2 is opened) + (probabilities in the case door 3 is opened) = 1

Where to put the door 2 and door 3 probabilities is obvious, but what to do with the door 1 probability is not so obvious. Leaving the door 1 probability as an unknown results in


 * ( ?Door 1 + 0 + 1/3Door 3) + (?Door 1 + 1/3Door 2 + 0) = 1

This equation is still talking about unconditional probabilities, meaning probabilities that cover all cases and must sum to 1. Whatever the two ? probabilities are they must sum to 1/3. What these correspond to is the joint probability the car is behind door 1 and the host opens door 2, and the joint probability the car is behind door 1 and the host opens door 3. If one of these is 1/3, the other must be 0. If the host opens door 2 and door 3 equally often when the player has initially picked the car, then these are both 1/6. In many versions of the problem (for example, the well known Parade version) which door the host picks if the player initially selects the car is not specified. Assuming, or specifying, the host makes a random choice in this case makes these two joint probabilities 1/6:


 * (1/6Door 1 + 0 + 1/3Door 3) + (1/6Door 1 + 1/3Door 2 + 0) = 1

These are still unconditional probabilities. The term on the right is the unconditional probability that the host opens door 3, which is the door the problem statement typically says the host opens. 1/6 + 1/3 is 1/2, which means the host opens this door half the time. If the host has opened this door, switching wins in the 1/3 case where the car is behind door 2. The conditional probability of winning by switching is therefore (1/3) / (1/2) which is 2/3.

The question remains what to do if it is unknown how the host decides what door to open, like in the Parade version of the problem. What Morgan et al. and Gillman do in this case is assign the host a preference p (0 <= p <= 1) for one door over the other. The original 1/3Door 1 can then be split according to this preference, without making any assumptions about it or knowing exactly what it is. The opened door gets p/3 and the other door gets (1-p)/3. Assuming p refers to the preference for door 3, the unconditional probabilities are:


 * ( (1-p)/3Door 1 + 0 + 1/3Door 3) + (p/3Door 1 + 1/3Door 2 + 0) = 1

Computing the conditional probability from the door 3 term results in a probability of winning by switching of (1/3) / (1/3 + p/3). Factoring out the 1/3 from this equation leaves 1 / (1 + p). Remembering p is a number between 0 and 1, this means the conditional probability of winning by switching is between 1/2 and 1. In the case where the host is constrained to pick randomly if the player initially picks the car p is 1/2, so in this case the chance of winning by switching is 1 / (1 + 1/2) which is 2/3.


 * I do not contend that any of this is wrong but it does not help me to understand. It may help some people, but I still think that the concept of conditional probability is not necessary to solve the (basic) problem. Martin Hogbin (talk) 18:31, 6 March 2009 (UTC)

Comments welcome. -- Rick Block (talk) 16:04, 6 March 2009 (UTC)


 * Rick, you are talking about "many people". If by "many" you mean the large number of people who actually know Probability math, yet do the math wrong - then yes that's "many people". However, the fact remains that "most people" don't know Probability math at all and "most" people therefore do exactly what I've beeen telling you they do. They say, "Hmmm, there's now one car and two doors, that's 1/2 or 50/50". You are presuming too much if you think "most people" do anything more than that. 216.153.214.89 (talk) 16:10, 6 March 2009 (UTC)


 * I agree, that is how most people see the problem, and get the wrong answer. Martin Hogbin (talk) 18:31, 6 March 2009 (UTC)


 * The wording is mathematical in nature, but it means the same thing that you're saying. People evaluate the original situation as all equal, and can even say the probability of initially picking the car is 1/3 (which is correct).  After the host opens a door people see two doors (and would be able to correctly say the probability of the car being behind the open door is 0) and since there's only one car say the chances are now 50/50.  They know the original probabilities for the two remaining doors started equal, at 1/3 and 1/3, and think after the host opens a door they should remain equal now making them 1/2 and 1/2.


 * What people don't understand is that the 1/3 - 1/3 - 1/3 probabilities apply only before the host opens a door (i.e. unconditionally), have never heard of conditional probability (or simply forgot everything they might have once known about it), and simply lack the tools to think about how the host opening a door affects the situation. Many people are taught only the very basics of probability and learn that n objects randomly distributed among m locations have a n/m probability of being at each location.  It's like posing a quantum physics problem to people only exposed to classical Newtonian physics.  Perhaps unlike quantum physics, conditional probability situations are frequently encountered in real life and my guess is people often mis-evaluate probabilities they encounter but never realize it (and think they've been doing it correctly).


 * Hence, my claim is the stubborn refusal to budge from the 1/2 each conclusion is rooted in being utterly oblivious to the conditional nature of the problem. The player is being asked to decide whether to switch at the point where she can clearly see two closed doors and one open door and knows there's only one car, which makes it a conditional problem. We're not interested in an "average" probability across all players, but just this player who can see the open door showing a goat.  The exquisitely disturbing nature of the problem depends on it being presented in this conditional form.  Those who see the trick behind the "unconditional solution" (a player who ignores what the host does and decides to stay with her original choice must have a 1/3 chance of winning the car, so switching must have a 1-1/3 = 2/3 chance of winning) become possibly more stubbornly attached to this solution, my guess is because they've been converted to this view from an earlier belief that the probability was 1/2 and because the 2/3 answer is actually correct for an "unconditional" version of the problem.  However, deciding to switch beforehand makes it a different problem, essentially equivalent to "what is the chance of picking a car behind 3 doors?".


 * So, we're left with a problem that has three levels of understanding. The naive, "must be 1/2" view.  These folks stubbornly cling to their view because they can see two doors and know there's one car.  The second level is the "unconditional" view, which avoids the complexity of a conditional analysis but doesn't address the problem as asked (and relies on what seems like a trick to people with the first view).  These folks stubbornly cling to their view because they know they're "correct" (but they're correct about the wrong problem).  The third level is the conditional view, which takes more probability theory than most people are exposed to.  People from any two of these sets can end up arguing with each other, but when people from the first two sets argue they're both arguing from positions of (more or less) ignorance.  -- Rick Block (talk) 05:54, 7 March 2009 (UTC)


 * And YOU called ME 'arrogant'?    Glkanter (talk) 07:54, 7 March 2009 (UTC)


 * Rick, the only question being asked is: Is it better to stay or switch? Anyone who says "switch" gets the right answer. My version of the formula could be analogized as Spanglish math or possibly better as Pidgin math. And though it's got mangled syntax, I understand it, it works for me and allows me to figure out this problem. 1-1/3(removed door)=2/3><1/3(stay);better to switch. 216.153.214.89 (talk) 07:47, 7 March 2009 (UTC)


 * [[image:PieChartFractionThirds.svg|120px|float|right]]Imagine the pie chart to the right represents the probabilities of where the car is and is labeled door 1 at the top, door 2 on the left, and door 3 on the right. The player picks door 1.  The host now opens door 2 or door 3.  By opening a door the host has to cut this pie into 2 pieces.  If the host adheres to the "equal goat door" rule, the pie is cut exactly in half (with a vertical slice).  The half on the left is the probabilities if the host has opened door 3 and the half on the right is the probabilities if the host has opened door 2.  At the point the player is choosing to switch we're only looking at one of these halves of the pie (not the whole pie).  In the "host opened door 3 case" (the left half) the player's piece (which has been cut in half) is exactly half the size of the door 2 piece, so the player's piece is 1/3 of this half and the door 2 piece is 2/3 of this half.  Similarly on the other side.


 * If the host doesn't have to adhere to the "equal goat door" rule, he can divide the pie into two pieces by starting the same vertical cut from the bottom, but then cut the player's piece however he'd like. He has to end up with two pieces, but he could at the extreme make one piece be 2/3 of the whole pie (including all of both the door 2 and door 1 probabilities), but if he does this the other piece will be only 1/3 of the pie and will include only the door 3 probabilities. If he cuts the pie this way, in the case where he's opened door 3 the player's piece is the same size as the door 2 piece (so the player only has a 50/50 chance of winning by switching), but if he does this then in the case where he opens door 2 the player wins 100% by switching.  The point is if you only look at the whole pie (which is what any unconditional solution does) you can't see this effect.  When the host opens a door (cuts the pie into 2 pieces) the "must not reveal the car" rule says only how the door 2 and door 3 pieces have to be divided.  The Parade version doesn't say how the player's piece is cut, so (in this version) the host can cut the player's piece anywhere.


 * The description above in terms of the pie chart is exactly the same as the the "math" description I started this thread with, and exactly the same as what Morgan et al. and Gillman are saying. Does the pie chart make this more clear?  The large figure in the article shows these same proportions in a table (not pie chart) format.  -- Rick Block (talk) 11:23, 7 March 2009 (UTC)


 * This is becoming increasingly far-fetched. The article should be the MHP as understood via the Marilyn vos Savant Parade articles. Which is how it starts. Then her solutiuons gets dis-credited, and Morgan takes over. But since you seem to have the passive-aggressive support of some like-minded subject matter experts, I don't see this changing. Glkanter (talk) 12:10, 7 March 2009 (UTC)


 * The Parade version does not specify the how the host picks between two goats if given the chance, so he can cut the player's piece of the pie however he'd like. Monty could be host #1 who chooses randomly between two goats (by, say, secretly flipping a coin), or he could be host #2 who always opens the leftmost door if he gets the chance.  Both of these hosts always open a door and always show a goat, so both meet the criteria given in the Parade version.  Please look at the, above.  If we've picked door 1 and the host has opened door 3 the winning percentages by switching for these two hosts (across the same 600 simulated games) are 65.46% and 100%, respectively.  If the host has opened door 2 the percentages are 65.88% and 48.63%, respectively.  Across all players, ignoring which door the host opens, the percentages are identical at 65.67%.  Again, both of these hosts follow the Parade rules.  Across 600 trials, the player's chance of winning is between 48.63% (this is the one that is theoretically 1/2) and 100%.  Just like Morgan et al., and Gillman say. -- Rick Block (talk) 17:15, 7 March 2009 (UTC)

More criticism of Morgan's solution

 * The Parade version does not specify how the car is initially placed or how the player chooses a door. The car might always be placed behind door 2 and the player might always choose door 1. Morgan ignore that possibility. Martin Hogbin (talk) 17:45, 7 March 2009 (UTC)
 * Morgan et al. ignore that possibility and so does vos Savant, and so does every other analysis of this problem that has ever been published (although there are published variants where the initial distribution is non-random and the player knows it - the question then becomes what strategy of initial choice and switching or not optimizes the player's chance of finding the car). I think ignoring the case you raise is actually legitimate since if the car and goats are behind doors, and the doors are opaque, the player's initial choice has to be random.  If the initial choice is random, then regardless of the actual initial distribution the problem is equivalent to one where the initial placement is random (the answer is exactly the same, but the journey to get there is a little more complicated).  Conceptually, the randomness of the player's choice cancels out any non-randomness of the initial placement of the car (just like guessing whether to switch at the end results in a 50/50 chance of winning).   Do you need to see the math here?  Or is your point something different, like perhaps that you consider not assuming the host picks between two goats randomly to be as bizarre as assuming the car is always placed behind door 2 and the player is confined to choosing door 1?  -- Rick Block (talk) 19:00, 7 March 2009 (UTC)


 * I gave an extreme example, but it is not unreasonable to suppose that the car might not be placed completely randomly behind the doors and that the player might not chose completely randomly. If a real test were done in which one person placed a car behind one of three doors and another picked a door, I doubt the picker would get the car 1/3 of the time on average.


 * But you missed my point a bit, I am not saying that Morgan's assumptions were unreasonable just that they are not the only ones possible or even the most general ones given the question. Morgan give a perfectly good solution which I see nothing wrong with.  What I object to is their calling vos Savant's (and other) solutions false.  They are not false, they are just based on a different assumptions resulting in a different formulation of the problem.   Martin Hogbin (talk) 23:21, 7 March 2009 (UTC)


 * They call vos Savant's (and other) solutions false because these solutions either present an unconditional solution to what they view is clearly a conditional problem (F1, F2, F3, F5), or are manifestly incorrect (F4), or (F6) include an additional assumption (the equal goat constraint) not in the Parade statement of the problem (as amended by vos Savant's later clarifications). In the context of their paper, they're talking about the (amended) Parade version and various solutions extent at the time - all allegedly addressing this same problem. -- Rick Block (talk) 00:40, 8 March 2009 (UTC)


 * As you say, Morgan view the problem as necessarily conditional, but only because of the way they choose to formulate it. Suppose we take it, for the moment, that the original questioner, meant the problem in the way that vos Savant understood it (as later revealed by her).  In other words, the original placement and choice are random, the host always offers the swap and always reveals a random goat (plus anything I have forgotten).


 * In the above case the problem need not be treated as conditional. Not even Morgan claim that it must, in fact they strongly imply that it need not. I agree that, in a pedantic sense, it is conditional in that we must answer the question, given that the host has opened a particular door, but in what I will call the vS formulation which door manifestly makes no difference to the probability of winning if you swap.


 * On the other hand, I can call Morgan's solution false. If I choose to formulate the problem such that the initial placement is not defined to be random (which is in fact the case), then there is a condition that the player picks a door.  We should really say, given an unknown car distribution, and given that the player picks a door,...


 * I am sure that you have read it before, but please have a look at Seymann's comment again, and ask yourself why the journal considered it appropriate to add a comment to a published paper. I suggest that it was to soften the overly strongly worded and patronizing claims of Morgan. Martin Hogbin (talk) 10:31, 8 March 2009 (UTC)

(outindent) I have read Seymann's comment and, more directly pertinent to what we're talking about, vos Savant's response to the Morgan et al. paper and Morgan's rejoinder (in the letters to the editor, American Statistician vol 45, no. 4, Nov 1991, pp 347-348) in which Morgan says "... even if one accepts the restrictions that [vos Savant] places on the reader's question, it is still a conditional probability problem. One may argue that the information necessary to use the conditional solution is not available to the player, or that given natural symmetry conditions, the unconditional approach necessarily leads to the same result, but this does not change the aforementioned fact." I don't think this is a matter of formulation, but is an inherent attribute of the problem. We're not asked about chances pertaining to all players, but chances pertaining to a player who can see which door the host opened and is then deciding whether to switch. The decision point is clearly after the host has opened a door, not before. We've been through this before, but I think it is this very aspect of the problem that makes the solution so counterintuitive. Symmetry may force the conditional answer to be the same as the unconditional, but then why is the problem symmetric? And, as I've said countless times by now, if we don't say or assume the host picks between two goats randomly the problem is NOT necessarily symmetric. Have you read my latest reply to 216, in above? Morgan's point (and Gillman's and User:C S's and user:Nijdam's and mine) is that the problem as phrased is not asking about the whole pie (which an unconditional solution addresses) but the piece that remains after the host has opened a door. This is the essence of a conditional problem. -- Rick Block (talk) 17:44, 8 March 2009 (UTC)


 * BTW - If you only have the electronic version of the Morgan paper, you may not realize Seymann's comment continues on page 288 and is followed by Morgan's rejoinder on page 289. -- Rick Block (talk) 18:19, 8 March 2009 (UTC) Thanks that is indeed the case. Martin Hogbin (talk)


 * To a respond to your last point first, as I have replied before, of course, if you change the formulation of the question you need a different solution, but I am talking about the vS formulation. The main point that I am trying to make at this stage is that for the vS formulation it is not mandatory to use the the method of conditional probability to obtain a valid solution.


 * As I say above, if you change the formulation yet again then Morgan's solution is a false one because it does not take account of the possible initial non-random placement of the car.


 * I am not sure that there is such a thing as a 'conditional probability problem' or an 'inherent attribute' of a problem that forces a particular method of solution. With any problem in mathematics we are free to choose any method of solution open to us.    In the case of the vS formulation of the MHP we can observe that the revealing of a goat does not change the probability that we have chosen a car and thus say that the unconditional solution will be the same as the conditional one. Having made that observation we can then calculate the unconditional probability.  This type of approach is very common in many branches of mathematics.


 * You say, 'We're not asked about chances pertaining to all players, but chances pertaining to a player who can see which door the host opened and is then deciding whether to switch'. That is correct but every player will see a door opened in in the vS formulation and no information is gained in this process, thus it makes no difference when the decision to switch is made.  Martin Hogbin (talk) 18:50, 8 March 2009 (UTC)


 * Also the term 'false solution' is one I have never heard before. As far as I know any solution to a problem which must give the correct answer is a valid one. Martin Hogbin (talk) 18:53, 8 March 2009 (UTC)


 * Martin, I hope this line of reasoning (the 5 paragraphs immediately above) is successful. You capture most of my arguments regarding the current article. Glkanter (talk) 19:04, 8 March 2009 (UTC)


 * You say "In the case of the vS formulation of the MHP we can observe that the revealing of a goat does not change the probability that we have chosen a car and thus say that the unconditional solution will be the same as the conditional one." Let's talk about this. First, what do you mean by the vS formulation - the problem as stated in Parade, without the "equal goat constraint"?  And, what exactly are we observing?  My claim is that what we're observing is that the host opening a door does not affect the original (unconditional) probability of having chosen a car, but this doesn't mean opening a door can't affect the conditonal probability - and this is the point I think you (and probably Glkanter and 216) are not understanding.


 * I thought that I made it clear above that I am using the term vS formulation to mean that the host chooses randomly which door to open when he has a choice.


 * Consider door 3. The host opening a door (even door 3!), does not affect its original (unconditional) probability of hiding the car which remains 1/3 - but opening door 3 clearly affects its conditional probability (in the case door 3 is opened it's clearly 0 and in the case door 2 is opened it's not necessarily 1/3 anymore).  If opening a door can affect door 3's conditional probability, why can't the same action affect the conditional probability of door 1 (and, while we're at it, door 2) as well?  In fact, in the normal interpretation of the problem it does affect the (conditional) probability of door 2 (right?) - so what exactly is so special about door 1?  The answer is . . . . nothing (!).  The probability that opening a door doesn't affect is the unconditional probability, and it doesn't affect the unconditional probability of ANY door (not just the player's door).


 * Opening a door CAN actually affect the probability of all three doors! Consider a host whose preference among the two doors the player doesn't pick is split 1/3 vs. 2/3.  In this case (which is not the problem as usually interpreted), after the host opens a door the probabilities are either 1/4 : 3/4 : 0 or 2/5 : 0 : 3/5.  This is a host who always opens a door, and always shows a goat, just like in the normal interpretation, and never affects the player's initial (unconditional) chance of having picked a car.  1/3 of players who don't switch will win with this host, just like in the normal interpretation.  The point is that the assertion that opening a door does not change the player's initial chance of having picked the door, although absolutely correct, is talking about the unconditional probability and says nothing about the conditional probability of the car being behind that door after the host opens a door.


 * Sorry Rick. We may be arguing to cross purposes here.  I meant the vS formulation to mean that the host chooses randomly, which is what vos Savant assumed.  If you do not like this terminology then please suggest another, but at the moment I am considering this formulation only.


 * What forces the unconditional and conditional probabilities to be the same is NOT that the host's actions cannot change the (unconditional) probability. Nothing can change the player's initial chance of having selected the car (or  the unconditional probability the car is behind either of the other doors as well) - it is the definition of "unconditional probability".  The entire question is what is the relationship of the unconditional probability (which is clearly 1/3) to the conditional probability.  They indeed (assuming the equal goat constraint) turn out to be equal.  But why? -- Rick Block (talk) 20:17, 8 March 2009 (UTC)


 * Mu.
 * 'In his 1974 novel Zen and the Art of Motorcycle Maintenance, Robert M. Pirsig translated mu as "no thing", saying that it meant "unask the question".'
 * http://en.wikipedia.org/wiki/Mu_(negative)  Glkanter (talk) 20:44, 8 March 2009 (UTC) Glkanter (talk) 05:24, 9 March 2009 (UTC)


 * Firstly, let me make clear that I am talking only about the random goat door formulation. The conditional probability that the player has chosen a car, after the host has revealed a goat is the same as the unconditional probability because the host always reveals a goat and the revealing of the goat does not give any more information as to the probability that the player has chosen a car. If you do not accept this fact then I can demonstrate its truth in a number of ways.  So the condition that you talk about is a null condition, a condition that cannot change the conditional probability from the unconditional probability. Mu as Glkanter calls it below. Martin Hogbin (talk) 21:16, 8 March 2009 (UTC)

Mu diversion

 * Not mu. This same question comes up in the Boy or Girl paradox and in Bertrand's box paradox, and in pretty much any probability problem involving conditional probability.  The insight is always the same, which is that the unconditional probabilities and the conditional probabilities are different and confusing them leads to incorrect answers.  I know you don't like this analogy, but using an unconditional solution to this problem is exactly like saying 22 is 4 because 2+2=4.  Right answer.  True statement. But wrong explanation.  -- Rick Block (talk) 21:06, 8 March 2009 (UTC)


 * This is a better analogy, an is not in general equal to a * a but in the special case where n=2 we can say that for all a, an =a*a.  Martin Hogbin (talk) 22:47, 8 March 2009 (UTC)
 * Absolutely not! And I suspect you to know that. How would you judge a student who solves the exercise: solve x = 22, by stating: x = 22 = 2 + 2 = 4?? Nijdam (talk) 23:39, 8 March 2009 (UTC)
 * Right. The equivalent rule would be "unconditional probabilities are not in general equal to conditional probabilities, but in the special case where ____ we can say that they are".  How to fill in the blank is the same as the question below - certainly the general rule is not, "but in the special case of the MHP we can say that they are"!.  The student saying 22 = 2 + 2 might know what she's talking about (since 22 = 2 * 2, and 2 * 2 really does equal 2+2), but without the intermediate step (i.e. without some justification that the unconditional and conditional probabilities are the same) it looks like the student is simply confused about the definition of exponentiation.  -- Rick Block (talk) 02:34, 9 March 2009 (UTC)
 * I am not going to respond to this point unless you really want me to. The reason being that we will then be arguing about the aptness of the analogy rather than the original question.  All I can say is that I do not see your analogy as being a good one and you do not see mine as a good one.  Best call it off here.  Martin Hogbin (talk) 09:09, 9 March 2009 (UTC)

More on vos Savant's interpretation

 * Martin - above you say vos Savant assumed the host chooses randomly between two goats. As far as I know, she's never said this.  In the experiment she suggested in what I think was her third column about this (see ) she says the steps are to:
 * 1. Randomize the initial placement.
 * 2. Randomize the player's initial choice.
 * 3. Host "purposely" (with no randomization between two losing choices) shows a losing choice.
 * 4. Repeat 200 times with the "stay" strategy and then 200 times with the "switch" strategy.
 * We don't know for sure, but it appears she thinks of this as an unconditional problem (as apparently do you and Glkanter).


 * vos Savant said that she considered the host to be purely the agent of chance. I take this to mean the he chooses randomly. Martin Hogbin (talk) 22:47, 8 March 2009 (UTC)


 * Even with the random goat constraint (which I suspect is commonly included because of Morgan et al.'s analysis), the problem is a conditional problem. You say with this constraint the unconditional and conditional probalities are the same "because the host always reveals a goat and the revealing of the goat does not give any more information as to the probability that the player has chosen a car.  If you do not accept this fact then I can demonstrate its truth in a number of ways."  I do accept this fact, but I'm curious what your explanation would be.  Assume for the moment I don't accept this as a fact.  How would you respond? -- Rick Block (talk) 21:50, 8 March 2009 (UTC)


 * Oh dear, I thought I had a few but they are failing me at the moment (I meant to say random revealing of the goat by the way) . All I can say for now is that the host can always reveal a goat and the only information given is in the choice of door.  As this is random it provides no information.  I will think more and come back later.Martin Hogbin (talk) 22:47, 8 March 2009 (UTC)


 * Having thought a little about this point, it reminds me of the well known mathematics story. A lecturer is going through a particular derivation when a student asks how a particular expression was obtained. 'It's obvious', says the lecturer.  The lecturer then stops... scratches his head.. and the walks out of the lecture.


 * Twenty minutes later he returns and briskly says, 'Yes, it is obvious', and continues the lecture from where he left off. Martin Hogbin (talk) 09:22, 9 March 2009 (UTC)


 * Yes, it is obvious! We all know at the start of the game that the host will open a door to reveal a goat and that he will choose randomly if he has a choice.  Therefore we get no new information when he does so, he opens a door, we knew he would, he reveals a goat, we already knew that, he picks a door, we know that no information is contained in that choice because we know it was a random choice.  Martin Hogbin (talk) 20:40, 9 March 2009 (UTC)


 * This may go on forever. Let me ask Martin: Are you aware that opening the door puts a restriction on the sample space? And if your answer is "yes", the unavoidable implication is: from then the probabilities are conditional. That's all.Nijdam (talk) 23:46, 8 March 2009 (UTC)
 * Martin: "Restriction on the sample space" is the probability jargon way of saying the same thing I've been saying about the pie chart. We're not talking about the whole pie, but the piece that's left after the host opens a door.  It doesn't matter which door - the question is about the piece that's left after a door is opened.


 * Nijdam: Martin is wanting to view the MHP as equivalent to an urn problem along these lines - host puts three identical balls in an urn, two marked "loser" and one marked "winner". The player picks one without looking at it.  The host looks in the urn and withdraws one marked "loser" and shows it to the player.  Should the player switch the one in her hand for the one left in the urn - i.e what are the chances the one in the player's hand and the one left in the urn are marked "winner"?  The problem is that in the MHP the doors are inherently distinguishable since they have an apparent physical arrangement on the stage.


 * The assumption the host picks randomly if given the chance is so natural that many (perhaps many, many, many) people have great trouble even seeing this might not be the case. Bayesian analyses (!) published before Morgan (and probably some published after) include this constraint even without it being stated (making these "false" solutions in Morgan's view).  The "standard" version now includes this constraint, which makes the solution 2/3 rather than 1/(1+p) - but this still doesn't make it an unconditional problem!  -- Rick Block (talk) 02:34, 9 March 2009 (UTC)


 * Thanks for your responses. I am new the the concept of a sample space but believe that I understand it.  I do have an issue with Morgan's limited choice of sample space, but that is another matter which I will leave for now.  Morgan's sample space is: CGG2, CGG3, GCG, GGC (using their notation).  Now if we are are given that the host always opens a random goat door (which is the only formulation that we are discussing at the moment) as far as I can see this does not reduce the sample space, all options remain - with their original probabilities.  Thus we need not have treated the problem as conditional. Martin Hogbin (talk) 09:22, 9 March 2009 (UTC)


 * Well, in the formulation we are discussing, door 3 has been opened! Do you notice the restriction? Nijdam (talk) 17:55, 9 March 2009 (UTC)


 * No, in the formulation that I am discussing, the host opens any door that he knows hides a goat, if he has a choice, he chooses randomly. One way to deal with this is to consider each door in turn but as Rick points about above, as we know the choice is random we can treat the doors as indistinguishable. Martin Hogbin (talk)


 * Then I don't know which formulation you mean. Normally the problem is stated in such a way, that after the host has opened a door, the player is offered the opportunity to switch. Nijdam (talk) 10:53, 10 March 2009 (UTC)

A more complete solution
To solve the problem more generally than Morgan et al. we should use a sample space that at least includes all the possibilities of original car position, player initial choice and the door that the host opens. As I think Rick has said above, this gives us a 3*3*3 matrix. A typical event could be 1,2,3 to mean that the car is behind door 1, the player initially chooses door 2 and the host subsequently opens door 3. To calculate the probabilities of these events we need to start with the probabilities that the car is behind each door say C1,C2,C3. Nothing in the statement of the question tells us that these are all equal. We also need to consider the probabilities that the player chooses each door say (P1,P2,P3). Finally we need to assign probabilities to the door that the host opens given the car position, the player choice, and what we decide that the rules of the game are, as Morgan do.

Of course we could informally reduce this sample space with some observations. As Rick has said, the player does not know the disposition of the cars and a random choice from a non-random distribution is effectively random, thus we might informally argue that we need not include both C1..3 and P1..3, just one would do. But what if the host's choice of door was door dependent as Morgan suggest? Then we cannot make this simplification. Door 1 could be inherently different from door 2 due to fact that the car is more likely to be behind door 1 than door 2. To do the job properly we must take the complete sample space without assuming anything not given in the question as Morgan say. Anything less is a false solution. Martin Hogbin (talk) 21:39, 9 March 2009 (UTC)


 * Addition. I have just noticed that near the end of their paper Morgan admit to failing to include my first two possibilities.  Their excuse is that doing so is  '...unlikely to correspond to a real playing of this particular game'.  So much for answering the stated question. Martin Hogbin (talk) 22:11, 11 March 2009 (UTC)


 * They don't "admit to failing to include" your first two possibilities. They say "Other generalizations appear to be of less interest.  One possibility is to incorporate prior information on the part of the player as to the location of the car, or, related to this, to allow nonuniform probabilities of assignment of the car to the three doors ...".  This is not an excuse, but an explicit acknowledgment that such generalizations are possible.  The specific problem they analyze is clearly meant to be the problem they think vos Savant was addressing. -- Rick Block (talk) 01:14, 13 March 2009 (UTC)
 * Are you wanting to develop such a solution here? Or include one in the article?  Or are you simply continuing to express your dislike for the Morgan et al. article?  Is the Gillman article more to your liking?  Here are some representative quotes:
 * I am mainly continuing to express my dislike for Morgan. I believe that statements which call other solutions 'false' are uncalled for and unscholarly (if there is such a word) and I believe that we need not be so dogmatic in following Morgans view within the article.  My real point is simply that the Morgan article is not perfect and that some of the issues that they have ignored in formulating their solution are precisely the same ones that they so dogmatically say we must not ignore.
 * [speaking of vos Savant's solution] This is an elegant proof, but it does not address the problem posed, in which the host has shown you a goat at #3. Instead it is still considering the possibility that the car is at #3—whence the host cannot have already opened that door (much less to reveal a goat!).  In this game—Game II [the one corresponding to vos Savant's solution]—you have to announce before a door has been opened whether you plan to switch.


 * Game I is more complicated: What is the probability P that you win if you switch, given that the host has opened door #3? This is a conditional probability, which takes account of this extra condition. When the car is actually at #2, the host will open #3. But when it is at #1, he may open either #2 or #3. The answer to the question just asked depends on his selection strategy when he has this choice—on the probability q that he will then open door #3. (Marilyn did not address this question.)


 * The italics in the above are in the original. I've said this before, but your opinion about this doesn't matter and neither does mine.  What matters is what reliable sources have to say.  There are two (actually there are more - for example Falk) who say unequivocally that this problem is a conditional problem.    -- Rick Block (talk) 03:33, 10 March 2009 (UTC)


 * I understand that any changes to the article should be based on reliable sources but I still believe that there is room for maneuver. I might include, for example Seymann's published remarks at the and of the Morgan paper, 'Without a clear understanding of the precise intent of the questioner, there can be no single correct solution to any problem'.


 * The Parade statement is written all in the present tense so there is no actual indicator of where in the proceedings we are. In other words, is the question asked after the host has opened a door? Most might say it is, but there is some room for doubt. However, the real question that we need to ask ourselves is what did the questioner actually want to know.  Was it along the lines of, 'Suppose I were to find myself in the position of being on a game show where the host has already opened a door...' or might he have meant, 'Suppose I were going to enter a game in which the following is the normal sequence of events, what would be my best strategy?.  The reliable sources (except vos Savant) assume the former meaning but it is by no means certain. Martin Hogbin (talk) 09:55, 10 March 2009 (UTC)


 * The reliable sources assume that the 'answerer' comes along after Monty has opened a door? No. Monty can't open a door until the contestant has already chosen one. The 'answerer' is the contestant who has been there all along. Glkanter (talk) 12:37, 10 March 2009 (UTC)


 * Glkanter: Martin is not suggesting the answerer "comes along" after Monty has opened a door, but that the contestant is meant to choose at this point as opposed to deciding whether to switch at the beginning of the game.


 * Martin: This is also a question we (as Wikipedia editors) should seek the answer to in reliable sources. How you or I or any other editor interprets the problem doesn't matter.  Wikipedia does not include its own formulations of or solutions for problems, but instead says what can be verified through reliable sources.  What this means is that in the article the predominant interpretation of the MHP in published sources is the one that should be the primary topic.  Other (minority) interpretations can certainly be discussed in the article, so long as they are also published in reliable sources.  Many reliable sources publish unconditional solutions without mentioning anything about whether it should be considered conditional or unconditional.  However, since the academic sources (which Wikipedia considers to be the most reliable) say the problem is inherently conditional and that unconditional solutions are not addressing the problem as stated then the article should say this, too.  The question is not "do you agree with these sources", but "do these sources say what the article says they say".  Is there something about this you're not understanding?


 * If you want to argue that the sources are wrong, this page (the "arguments" page) is a fine place and I'll happily discuss this with you. But (in the extreme) even if we were to agree the sources are wrong we couldn't then go edit the article to say what we've agreed to.  It might motivate us to try to find sources consistent with our agreement that are strong enough to override what  the existing sources say, but fundamentally what we think about what the sources say doesn't matter.


 * I'll put this more bluntly. If you can't offer up a source that acknowledges the two fundamental interpretations of the problem that we're talking about (conditional and unconditional) and says that either interpretation is valid, or perhaps that the conditional approach is not necessary, then there's really nothing more to say.  If you find such a source, then we might want to talk about how reliable it is.  Since Morgan, and Gillman (and Falk) are academic sources, I think it would have to be an academic source to even consider including what it has to say. -- Rick Block (talk) 14:15, 10 March 2009 (UTC)


 * I have already come up with a reliable source that softens some of the dogmatic assertions made by Morgan, and that is the commentary by Seymann about the Morgan paper. I am not suggesting that we rewrite the whole article but I believe that it should not be dominated by just one academic source which calls all solutions other than their own false (or a handful of others). Over 30 reliable sources are currently cited in the article and many of these treat the problem unconditionally.


 * My suggestion is only that we start the article with a simpler and more convincing solution to the problem treated unconditionally. This might be after we include my quote from Seymann above which, in my opinion, is one of the the wisest commentaries on the subject by anyone.  No one knows exactly what the questioner had in mind, possibly not even the questioner themselves, although in terms of interpreting exactly what a Parade reader had in mind vos Savant clearly is the real expert.


 * It might then to be reasonable to introduce the unconditional solution along the lines that many sources have taken the problem to be unconditional. This section could then be followed by the academic opinion on the subject and their proposed solutions much as now.


 * All I am really trying to do is address the issue that makes this problem notable and worthy of inclusion in WP, which is that it is a really simple problem (in its unconditional form) that most people get wrong. A complicated problem that most people cannot understand is far less notable and less interesting.  Martin Hogbin (talk) 19:37, 10 March 2009 (UTC)


 * How would that be different from (say) this version, which, as I recollect, none of Glkanter, Nijdam, or you liked (but for wildly different reasons)? The problem is notable because dozens (maybe hundreds by now) of articles have been written about it.  I wouldn't say many sources take the problem to be unconditional, but many do indeed present an unconditional "solution" without mentioning anything about whether it's a conditional or unconditional problem - which leaves one wondering whether they understand the difference.


 * I've never found a reference that acknowledges the difference that doesn't say something similar to what Morgan says. For example, the Grinstead and Snell book that is referenced (available online in PDF format), presents an unconditional solution but then goes on to say "This very simple analysis, though correct, does not quite solve the problem that Craig posed. Craig asked for the conditional probability that you win if you switch, given that you have chosen door 1 and that Monty has chosen door 3. To solve this problem ..." followed by a conditional analysis.  Then they say "At this point,  the reader may think that the two problems above are the same, since they have the same answers. Recall that we assumed in the original problem if the contestant chooses the door with the car, so that Monty has a choice of two doors, he chooses each of them with probability 1/2. Now suppose instead that in the case that he has a choice, he chooses the door with the larger number with probability 3/4. In the “switch” vs. “stay” problem, the probability of winning with the “switch” strategy is still 2/3. However, in the original problem, if the contestant switches, he wins with probability 4/7."  This structure is nearly identical to the structure of the article as it was in the version linked above, although the structure here I would argue was better since the unconditional solution that was presented flowed naturally into the conditional solution.


 * If we present an unconditional solution without immediately discounting it we run into both Glkanter's problem (the rest of the article is superfluous) and Nijdam's problem (it is "obviously" [to a probability theorist] not the solution to the problem that's asked). But if we do immediately discount it we run into your problem (you want to start with a simple, convincing solution - but immediately discounting it makes it somewhat less than convincing).  I've said this before, too, but my current thought is that the best approach may be to present a conditional solution first (I mean, it's not really that hard is it?) followed up with other alternate solutions (qualified however we'd like).  I think the back and forth between Morgan and vos Savant might be worth mentioning.  I think the 1/(1+p) solution is definitely worth mentioning.  -- Rick Block (talk) 00:45, 11 March 2009 (UTC)
 * You keep referring to 'the problem that was asked'. In reply to this I can only repeat two things.  The first is Seymann's statement with my emphasis, 'Without a clear understanding of the precise intent of the questioner, there can be no single correct solution to any problem'.  This statement is absolutely unassailable, it was written by an academic expert and was published in a peer reviewed journal, it also is in agreement with everybody's common sense and understanding, I seriously doubt that anybody disagrees with it.  The other thing, that nobody really disagrees with, is that the Parade statement of the problem is not at all clear about exactly what is meant (I am sure that there will be many reliable sources to support this view as well).   Taking the two together (I hope you will not call this WP:synth) we can say that there is no 'the problem that was asked'.  My main disagreement with Morgan is that they do not make clear that they are defining the 'problem that was asked' themselves. Martin Hogbin (talk) 22:06, 11 March 2009 (UTC)
 * I'm not sure to understand what you're aiming at, but my guess is that any formulation of a variant of the MHP only make sense if the player is asked to switch after a door has been opened. And hence it implies a condition. Nijdam (talk) 00:07, 12 March 2009 (UTC)
 * In the most common academic interpretation of the problem there is indeed something that occurs between the player's initial choice of door and the and the player's final decision on which door to open. No one has yet given me a reason why this particular occurance must be taken as a condition in order to get a valid solution to the problem in the 'random goat door' case.


 * Saying there is no "the problem that was asked" doesn't seem very helpful. Certainly the Parade version and many others are so imprecise that many interpretations are possible, but I believe nearly all sources intend the answer to the "standard" MHP (however they phrase it) to be that switching wins with probability 2/3.  The fully explicit version from Krauss and Wang which is quoted in the article forces 2/3 to be the answer both conditionally and unconditionally, so if the intent is for the answer to be 2/3 I think there's a pretty strong argument that this is the question most sources mean to be asking.  Furthermore, I think this version is now the "standard" version.  With the exception of the additional constraint forcing a random selection between two goats, the fully explicit K&R version is the same as the problem analyzed by Morgan (and Gillman and ...).  Based on her solution and the simulation she specifies, I think there's also a pretty strong argument that this version (without the equal goat constraint) is the same as vos Savant's interpretation and she simply overlooked the possibility of the host choosing unequally between two goats.  Before Morgan, most of the controversy about the problem definition was about the "host must show a goat" and "host must make the offer to switch" constraints and, although I can't read his mind, I suspect this is what Seymann was referring to as well.  All versions start with 3 doors, have the player pick one, have the host open one, and then ask whether the player should switch.  The point Morgan, and Gillman, etc make is that this basic framework is the framework of a conditional problem, whether or not the unconditional solution produces the correct answer.


 * I disagree. Saying there is no "the problem that was asked" right at the start is extremely helpful. Most of the arguments about the solution can be traced back to uncertainties in the problem formulation.  The starting point in a problem should not be an intended answer but discussion of what the questions should be taken to mean.  Vos Savant, in my opinion, quickly grasped the essence of what her reader actually wanted to know but she unfortunately did not make this explicit in her reply.


 * Regarding your statement, '...this basic framework is the framework of a conditional problem', I do not accept that solutions are somehow embedded the problem in mathematics. Any solution that can only give the correct answer is valid.  In the specific case of the fully defined problem with random goat door, I believe that other mathematical techniques than conditional probability can be used to  arrive at a correct answer.  Not that I am not saying that conditional probability cannot or should not be used or even that it might not be the best approach in this case, just that it is not the only way to answer the question if certain assumptions are made.


 * I have a question for you. If the "standard version" is meant to be taken unconditionally, why not the "host forgets" variant (that vos Savant discussed in Nov 2006, see ) as well?  Of course, maybe vos Savant considers the "standard version" an unconditional problem but the "host forgets" variant, which is the same problem except the host always opens a random door, a conditional problem.  Or, maybe, they're both conditional problems and the (now standard) assumptions of the K&R version force the unconditional solution to end up with the right answer. -- Rick Block (talk) 04:17, 12 March 2009 (UTC)


 * You say that I say that the "standard version" is meant to be taken unconditionally. That is not quite correct.  What I say first and foremost is that nobody knows what the 'standard version' is.  That may be unfortunate and unhelpful, but it is true. (If you simply mean by 'standard version' the fully-defined random-goat-door formulation then it is just a probability problem, which we are free to answer by any valid method, there is no way that it is meant to be taken.)


 * Whether the original questioner intended the problem to be taken as conditional or not we cannot know, my guess is that he had not even considered the distinction. If a person went to a statistician and actually asked problem stated in Parade, the reply would have to start with a series of questions, to establish what was actually meant.  What do you really want to know? Have you been invited to play on the show?  Do you want to know if players have in general been making the right decisions on past shows?  Is it just a simple well-defined (but very unintuitive) probability problem that you want answered, perhaps to win bets with your friends?  In both the cases you ask about, the answers to the above questions would determine how the statistician would proceed with the reply.  Martin Hogbin (talk) 09:52, 12 March 2009 (UTC)


 * Martin, you seem to have arrived where many other editors, including myself have been. That is, that the unconditional solution is valid for this problem. How do you propose to change Rick's (and what I think is at least one Wikipedia Math Guru's) insistance that it must be solved as a conditional problem? Glkanter (talk) 14:34, 12 March 2009 (UTC)


 * Having purchased a copy of the Morgan paper I am now of the opinion that that it is aabitrary mixture of slopiness and dogmatic pedantry. If anyone wants to argue that case I am happy to do so. Unfortunately, Morgan et al are reputable academics who have published in a peer-reviewed journal and we are dogs so there will be some battles that we cannot win.  My suggestion still is that we should try to give a good unconditional treatment at the start of the article, supported by reliable sources (which I think can be done) followed by the more academic conditional solution as now, even though my opinion has moved against it. Martin Hogbin (talk) 18:43, 12 March 2009 (UTC)


 * I'm not certain that 'The American Statistician' is a peer-review publication. That might actually be the 'Journal of the American Statistical Association' which comes from the same organization.


 * http://www.amstat.org/publications/jasa.cfm
 * You may well be right, WP describes it as an 'American magazine'. Another blow for its credibility. Martin Hogbin (talk) 22:10, 12 March 2009 (UTC)


 * It is (at least currently) definitely peer reviewed, see http://pubs.amstat.org/page/tas/information-for-authors . -- Rick Block (talk) 00:56, 13 March 2009 (UTC)


 * Just for fun, I'm pasting this snippet from long ago...
 * Mr. Hogbin, I don't see how this solution is not 'rigorously complete'. Simple, yes. Incomplete, no. Glkanter (talk) 18:02, 26 October 2008 (UTC)
 * I am, for the moment, simply accepting the opinion of others regarding rigorous completeness. Martin Hogbin (talk) 19:43, 26 October 2008 (UTC


 * Glkanter (talk) 21:43, 12 March 2009 (UTC)


 * The framework I'm talking about does not refer to a solution that's embedded in the problem, but the meaning of the question - which in this case clearly (according to the reliable sources) refers to the conditional probability of winning by switching after the host opens a door. An unconditional solution for this sort of problem is valid if (and only if) the conditional and unconditional answers are the same.  This is often the case in "pure" probability problems (e.g. urn problems), but in general an unconditional solution needs to be accompanied by an argument for why this is true in the problem being addressed.  In the case of the MHP, Morgan, and Gillman and others show that this is true only if the host is constrained to pick between two goats equally (with the distinct implication that sources that present an unconditional solution that ignore this constraint essentially don't know what they're talking about).  Again, this gets back to sources.  If you want to "feature" an unconditional solution in the article you simply need to reference this to a source that knows what it's talking about (preferably academic, but certainly one that acknowledges that the question is about conditional probability). -- Rick Block (talk) 15:11, 12 March 2009 (UTC)


 * It seems to me that you are agreeing that an unconditional solution is valid for the fully defined problem with random goat door (can we agree on a terminology for this to save writing it out all the time, I would like to call it the vS formulation but I do not care so long as we agree on something and stick to it).


 * I fully support the WP concept of reliable sources, but they still have limitations. No source is entitled to claim to know exactly what the Parade questioner actually meant and the problem can only be answered once we have decided that (as the reliable source Seymann makes clear).  I believe that we are entitled to at least consider the possibility that the original questioner may have intended this to be taken essentially as an urn problem. If even that degree of independent thought is considered unacceptable maybe we could describe it as the formulation that vos Savant understood the original question to be, or if you really want to be picky later claimed that she had understood the question to be. Martin Hogbin (talk) 18:43, 12 March 2009 (UTC)


 * I'd prefer to call it the K&R version rather than the vS version. And, I agree a hypothetical unconditional solution that justifies why it is the same as the conditional solution would be valid (and, if you can find one from a reliable source would be good to include in the article).  I also agree no source is entitled to claim to know what Craig Whitaker meant, except Craig Whitaker.  On the other hand, that's mostly irrelevant now since the problem has taken on a life of its own and even has a "standard" formulation, well known in probability theory which is equivalent to the K&R version (this claim is in the article, with a reference to Barbeau's Mathematical Fallacies, Flaws and Flimflam book) - the article is about this well known probability problem, not the problem posed by Craig Whitaker as answered by vos Savant or even the original problem as posed by Selvin.


 * If you are saying that there is a well-known formulation of the problem in which goat door is randomly chosen then I can see no objection to providing a clear and convincing unconditional solution at the start of the the article.


 * Just as we can't claim to know what Whitaker meant, we can't really claim to know how vos Savant understood the problem either but we do know what she wrote about it (and whether we want to or not, we need to be picky about it). What she said is in her original columns (which I think are presented verbatim on her website here) and in a letter to the editor published in American Statistician (vol 45, no 4, p 347).  The particular points made by Morgan (and Gillman and others), that the problem is about conditional probabilities and that the host might have a preference for one door over the other affecting the result, are never explicitly mentioned.  She does claim (in her letter to the editor) that "circumstances in default are reasonably considered random". However, she never said this in her column, never made any apparent attempt to justify the validity of the unconditional solution, never acknowledged the specific issue that the host might have a preference, and never acknowledged the main criticism that the question is talking about a conditional probability.  For our purposes here, I think all this means her columns are not necessarily the most reliable sources. -- Rick Block (talk) 00:45, 13 March 2009 (UTC)


 * I think you are being very unfair on vos Savant. Knowing the likely lack of statistical sophistication of her readers, she took the problem in its simplest form, as a urn problem.  She did fail to make this clear when she originally answered but considering her readership and the space available this is not an unreasonable decision.  She later made clear the assumptions that she made in her answer.  Vos Savant's assumptions can also be informally justified on the basis that the player in unlikely to have had any knowledge of the initial setup and the hosts policy, thus they might as well be taken as random.


 * Morgan on the other hand make a terrible mess of the whole thing. They start my misquoting the question, they then (by their own admission) fail to take account of the initial probability distribution of the car and of the player's initial choice of door but then criticize others for doing exactly the same thing with regard to the host's choice of door, calling solutions which take this to be random 'false'.  I believe that some of the balance needs to be restored to this article and suggest that this discussion is now moved on to the article talk page. Martin Hogbin (talk) 09:54, 13 March 2009 (UTC)


 * What vos Savant made clear was that the initial placement is random, that the initial choice is random, and that the host is forced to open a door (revealing a goat) and make the offer to switch. An informal justification that the player has no knowledge of the host's preference between two goat doors so this might as well be taken as random is completely unsound, and reflects a misunderstanding of what's going on in this case.  We're not talking about the probability as perceived by the player, but the probability as can be determined by the problem statement.


 * Morgan et al. don't "fail to take account of the initial probability distribution".  Your characterization of this is bizarrely inaccurate - they explicitly acknowledge this is a generalization that could be considered.    They analyze the problem as clarified by vos Savant.  Their not so subtle point is that by treating the problem unconditionally you miss the significance of the host's potential preference for one goat door (or even goat!) over the other.  Yes, you can make a claim that you were assuming everything that might possibly be random is random but the real issue is you don't know what is significant unless you solve the conditional probabilities.  It's possible to approach it unconditionally, but you have to know what you're doing - and if you don't, your "solution" is no solution at all because it applies to variants of the problem  where it produces the wrong answer.


 * I agree this discussion is more appropriate for talk:Monty Hall problem. -- Rick Block (talk) 15:42, 13 March 2009 (UTC)


 * You seem to be applying completely different criteria to Morgan's decision not to consider the effects of the initial car placement and player door choice, to vos Savant's decision not to consider the effect the host's non-random choice of goat door.


 * If you want to solve the problem based only on the information given in the Parade statement then you must consider the distribution of probability for the car's placement. You must also consider the distribution of probability for the players initial choice of door; this information is not given in the problem statement.  Strictly speaking it is not good enough to say that we do not know how these things were decided so we can take them as random.  The possible effect of these factors on the probability of winning by switching is easily demonstrated by considering extreme cases.  If the car is always placed behind door 2 and the player always picks door 1 then the probability of winning by switching is 1.  On the other hand if the car is always behind door 1 and the player always picks door 1 then the probability of winning by switching is 0.  Quite a dramatic effect caused by a possibility that Morgan have ignored.  As you have said above, 'you don't know what is significant unless you solve the conditional probabilities', in this case for example that the player chooses door 1 given that the car is always behind door 2.


 * In exactly the same way, we are not told that the host chooses randomly which door to open and, even though the player does not know this, we cannot assume that it must be random. The possible effect of this factor on the probability of winning by switching is demonstrated by considering an extreme case, such as Morgan do in the case that the host always opens door 3 when possible.


 * I am now considering the possibility that Morgan are even more wrong that this, but more about that later Martin Hogbin (talk) 18:06, 13 March 2009 (UTC)


 * Morgan et al. never claimed to be solving the problem based only on the information given in the Parade statement. They make the same assumptions vos Savant did (initial random distribution, host must open a door showing a goat, and host must make the offer to switch).  They discuss the possibilities of variants on these assumptions as well (including the one you seem to think they "overlooked").  They solve this problem without making any assumption about whether the host has a preference for one goat over another (a possibility vos Savant had never mentioned - and without reading her mind we don't know whether she was aware this might make a difference).  The insight that this might matter comes from solving the conditional probabilities involved in the problem, which is obviously (at least to probability theorists) what the problem is all about.


 * I again encourage you to try to find sources to back up your claims. Without sources your opinions (on the problem, on the validity of an unconditional solution, on Morgan's paper, on vos Savant's solution) are simply WP:OR. -- Rick Block (talk) 04:31, 14 March 2009 (UTC)

What exactly did Morgan do?
It is not at all clear what Morgan are claiming, they start with a misquotation of the Parade statement and then state that vos Savant defended her original claim with a 'false proof'. They do not make clear at this stage exactly what formulation they believe vos Savant assumed or what formulation they are assuming. It is not possible to claim that a solution is false without first agreeing what the question is. Morgan have not done that.

Morgan then go on to talk about 'one form' of the problem having a solution that assumes no additional information. Additional to what? They have not said what the original information is. Next they say that this '...little problem' can be solved in a number of erroneous ways but still provide no definition of what the problem is.

In section 1, Morgan start by limiting their formulation to refer only to the specific case where the player initially chooses door 1 and the host reveals a goat behind door 3, but they still make no comment about the initial car placement. I am not sure what problem they think this refers to. I am pretty sure that Whitaker did not intend to ask a question about only the case where the player opens door 1, and I do not believe that vos Savant took this to be the case either. Curiously it corresponds quite well to the misquotation from Parade given at the start. Morgan then discuss six solutions provided by other authors for the problem that they have just formulated. These do not do well, but that is not so surprising considering that they may have been attempting to solve a different question.

Later Morgan show us how to answer their (still not fully defined) question properly, starting with a limited sample space, which only includes the specific case referred to above. I do have to agree, this is now a conditional problem, simply because Morgan have made it so. Morgan then calculate the various probabilities based on the unstated assumption that the car is randomly placed.

Maybe Morgan have not actually overlooked the issue of initial car placement but they have not stated their assumption either. This is not good enough in a problem in which the answer depends heavily on the assumptions made as to what the question actually is. Martin Hogbin (talk) 18:43, 14 March 2009 (UTC)


 * Martin - What are you hoping to accomplish with your criticisms of the Morgan et al. paper? I don't know if you noticed, but I responded to the thread above about The American Statistician being a peer reviewed journal (it is now, and presumably was in 1991 as well).  If it's good enough for a peer reviewed journal, it meets Wikipedia's standards for reliable sources.  It is a primary reference about the MHP, cited (as I've said before) by 52 subsequent scholarly works (according to this search on Google Scholar).  The basic conclusions, i.e. that the MHP is about the conditional probability of winning by switching and that an unconditional approach does not address the actual problem, are supported by numerous other references - for example the Gillman article and the Grinstead and Snell book which are both also referenced in the article.  Rather than continue this pointless attack on the Morgan et al. paper, it would be vastly more productive if you go find a reference supporting whatever view you want to put forward.  -- Rick Block (talk) 01:12, 15 March 2009 (UTC)


 * Thank you for your reply which I missed, it does seem that The American Statistician is indeed a peer-reviewed general interest journal.


 * I am trying to achieve two things the first it to improve the article so that it better addresses the popular and most notable understanding of the Monty hall problem (the unconditional problem).  I accept that Morgan is nominally a high quality reliable source but think that its impact could be reduced for part of the article in order to provide a simpler and more convincing solution.  Other reliable sources could be used to provide references for my proposed changes.  I think you agree with this to some degree and all I need to do is get on with it.


 * Secondly, the more I read of the Morgan paper the worse I think it is as a good source for the article. I appreciate that it is a published source, but this page is not the article or even the article talk page but the arguments page, where I think I should be able to criticize the Morgan paper along with other interested parties.  With regards to the article itself, Morgan is a reliable source and I doubt that I can do anything thing to change that.  However, I believe that it is seriously misleading in that it does not answer any reasonable, or commonly understood, formulation of the problem and that is what I would like to demonstrate here.  I thank anyone who is willing to argue their case with me but point out that the published source argument not so relevant here where the facts should be allowed to speak for themselves. Martin Hogbin (talk) 11:59, 15 March 2009 (UTC)

Why Is It Conditional?
Maybe we can focus on 'why a conditional solution is the only true solution'.

First of all, is the above statement correct? You may recall a guru came on here a month ago saying it was so. Then he waffled.

Secondly, the current article is written that way. In the Solutions section it calls the unconditional solution inadequate.

So, here's the simplest unconditional formulation so far:


 * 1/3 of all contestant door selections will be cars
 * 1 - 1/3 = 2/3
 * =>The contestant increase his likelihood of winning the car by switching.

Since there is no mention of Monty or goats, I think any criticism should focus on the solution provided, not different problems which introduce new constraints. How do any of the above 3 statements cease to be true from the time the contestant sees the doors for the first time, through the point at which he decides to switch or not?

I'm reaching here just a little bit, as I do not know the puzzle, but how does this differ from the 'Urn Puzzle', which I think I understand reaches the same conclusion using an unconditional proof? Glkanter (talk) 15:19, 15 March 2009 (UTC)


 * I think we're actually talking about 4 things here (216 might enjoy this conversation), and it might help us communicate if we're clear when we're talking about each:
 * What you or I individually believe to be true.
 * What (we think) most people believe.
 * What is factually true.
 * What reliable sources say.
 * When Martin talks about the "popular and most notable understanding", I think he's in the realm of what most people believe (a #2 sort of issue). We can talk about this, but any claims added to the article about this would then need reliable sources - just as the claim that is already in the article about the "standard analysis" is sourced (because what we individually believe doesn't matter, the only thing that matters is what reliable sources say).  I understand Martin to be saying (with no sources yet) that he thinks the popular and most notable understanding of the MHP (a #2 sort of issue) is as an unconditional problem.  I'm saying the reliable sources (Morgan et al., Gillman, and Grinstead and Snell for starters) say the problem is about conditional probabilities (a #4 sort of issue) which I think also means if most people believe otherwise they're misunderstanding the problem (a #2 sort of issue) which in my opinion the article here should help clarify.


 * I personally think (in a #1 sort of way) that most people understand the problem in exactly the same way Morgan et al. analyze it. By this I mean people understand the problem, whether explicitly described this way as in the K&R version, or not as in the Parade version, to be (a #2 claim):
 * car is randomly placed
 * initial pick is random
 * host must make the offer to switch
 * host must show a goat (and, hence, knows where the car is)
 * player decides after the host opens a door
 * This understanding of the problem, specifically without the constraint that the host pick randomly between two goats if given the chance, is precisely the version Morgan et al. and Gillman analyze - and in this version, Morgan et al. and Gillman both (in a #4 sort of way) say the player should switch but the probability of winning by switching is not the 2/3 unconditional answer but rather 1/(1+p) which ranges between 1/2 and 1 depending on p (the host's preference between the two unpicked doors if the player initially picks the car). I also think (in a #1 sort of way) that while most people who put forth an unconditional solution understand the problem to be that the player decides after the host opens a door, they do not understand that the unconditional solution directly addresses a problem where the player must decide before the host opens a door -  which may or may not have the same solution (i.e. they do not understand that this makes a difference).


 * Would it help to try to agree on what we think most people think the problem means (focus on the #2 issue)? Does anyone disagree with any of the constraints in the bullet list above?  We'll no doubt argue about the host's preference between goats, but can we agree on this much first? -- Rick Block (talk) 17:17, 15 March 2009 (UTC)


 * The case that Morgan analyze is the specific case where,'The player has chosen door 1, the host has then revealed a goat behind door 3 ...' . Their analysis applies only to this case.  Had the question been, 'The player has chosen door 1, the host has then revealed a goat behind door 2 ...' then the probabilities they calculate for any given host preference would be reversed.


 * Had the question been taken to mean, 'The player has chosen door 1, the host has then revealed a goat behind one of the other doors ...', the hosts preference becomes unimportant and the problem can be treated unconditionally.


 * So to answer your question above, yes, I think we should try agree to on what we thing the original questioner meant. I think that he clearly meant the last of the options that I give above. Martin Hogbin (talk) 18:31, 15 March 2009 (UTC)


 * I think (a #1 claim) you are misunderstanding, and that Morgan et al. use "door 1" and "door 3" in the same manner as vos Savant, which is (effectively) as arbitrary names for "the door the player picked" (door 1) and "the door the host opened" (door 3). Before we hash this out, can you say whether you agree most people (a #2 consideration) understand the problem with the constraints listed in the bullet list above, or do you think it's critical to add a #5 which would be what Craig Whitaker meant?  As I've said, at this point, I don't think we specifically care about this (and, unless we ask him, I don't think we can resolve this).  We can ask what most people think he meant (which is the same #2 issue I'm suggesting we talk about), but if we disagree even this is outside the realm of what we'll be able to factually resolve by discussing it here.  If it turns out we disagree about this, we can try to find references supporting our respective viewpoints.  Sound fair enough?


 * And, again, I'd like to start at the bullet list above. Do you agree this is how most people understand the problem (ignoring the "host preference" issue for the moment)?  -- Rick Block (talk) 19:07, 15 March 2009 (UTC)


 * I would say that most people do not think about the formulation that much but if pressed they would probably agree with your bullet list, with the addition of host chooses randomly. Martin Hogbin (talk) 20:12, 15 March 2009 (UTC)


 * We need to talk about the host chooses randomly bit, but should we wait for Glkanter to chime in before proceeding? -- Rick Block (talk) 20:20, 15 March 2009 (UTC)


 * Thanks for waiting. The bullet points are OK with me, but I am not in agreement with the subsequent paragraph. Does anyone care to argue over the parenthesis in bullet #4, maybe Monty is told via a cue which door to open? Glkanter (talk) 03:15, 16 March 2009 (UTC)


 * Can we also agree that most people do not interpret the problem to mean literally only the case where the player initially picks door #1 and the host opens door #3, but that the question is meant to be asking about the chances given any initial player pick (whether it's door 1, 2 or 3) and any door the host opens? By this I mean that the "full" answer needs to consider any door the player picks and any door the host opens and we're using "door 1" to symbolically refer to "door the player initially picks" and "door 3" to symbolically refer to "door the host opens"?  I'll assume we all agree on this (and please let me know if I'm wrong).


 * Now, the only question that really remains is this pesky "host preference" issue. Glkanter raises the possibility that the host is told (by the producer or somebody) which door to pick.  I think we can slightly generalize "host" to include this case (would that be ok?).  Assuming this is also OK, I'll state what I think (in a #1 sort of way).  I think most people who say the answer is 2/3 probability of winning by switching think that since the player starts with a 1/3 chance of having picked the car, and since the host must open a door to reveal a goat (which he can always do) that the host cannot change this 1/3 chance, so any player who switches has a 2/3 chance of winning - whether the host might have a preference for one door or the other does not matter in the slightest.  Moreover, I think (still my opinion) that most of these people think this is the factual truth (in a #3 sort of way).  I'd like Glkanter to reply first if that's OK with you Martin.  -- Rick Block (talk) 04:35, 16 March 2009 (UTC)


 * Yep. 1/3 of the time the contestant picks the car. 1 - 1/3 = 2/3 he does not pick a car. I don't see how Monty's deciding which losing door he shows makes any difference. I'm relying on the 5 bullet points here, as there is no mention of the contestant being aware of a 'host's bias'.Glkanter (talk) 05:49, 16 March 2009 (UTC) Glkanter (talk) 05:52, 16 March 2009 (UTC)


 * My entire argument rests on the contestant's original probability of 1/3 not changing due to Monty's revealing a goat behind a door. The only way he could affect that 1/3 is to provide new information. At the moment he reveals a goat, he provides no new information. Revealing a car is the only way he can provide information that affects the original 1/3. This new info would drive that probability to 0. But premise #4 says he will never do that. BTW, are premises #3 and #4 in the correct chronological order? I don't think so. And I don't agree that Monty has to know where the car is. That is not one of the 5 bullets I agree to. Glkanter (talk) 06:49, 16 March 2009 (UTC)


 * I agree that most people do not take the problem to be door specific. Martin Hogbin (talk) 09:55, 16 March 2009 (UTC)


 * Yes, bullet #3 comes after bullet #4. "Not door specific" meaning it doesn't (and can't possibly) matter, which is not the same as "host randomly chooses between two goats if given the chance".  Correct?  -- Rick Block (talk) 13:15, 16 March 2009 (UTC)


 * Just to make quite clear what I was saying, I was answering your question, 'Can we also agree that most people do not interpret the problem to mean literally only the case where the player initially picks door #1 and the host opens door #3'. I agree with your statement.


 * This is a really important point because Morgan do take the problem literally to be that player opens door 1 and the host opens door 3. Why do I say this?  Firstly it what they say, to quote, 'To avoid any confusion, here is the situation: the player has chosen door 1, the host has then revealed a goat behind door 3'. There is no 'for example' or anything else to lead us to believe they are referring to the more general case.  Secondly when you look at their analysis you will see that the parameter p is equal to P12 and P32 (the probability that the host will open door 3 given that the car is behind door 1 and 3 respectively) whilst the parameter q (which equals 1-p) is equal to p13 and p33 (which are the probabilities that the host will open door 3).


 * Both Glkanter and I have always felt intuitively that there is something wrong with the Morgan analysis because the probability of winning seems to depend on some information that the player does not have access to (namely the host's preference in opening doors), yet the paper appears to show that the chances of winning by swapping depend on the hosts door preference. It seems as though some conjuring trick is involved.


 * The trick is to only consider one literal case as above. This has the effect of making it a given in the question that the host has opened door 3.  This the host's door preference comes into the calculation. I am in the process of doing some calculations to show how this works more clearly. Martin Hogbin (talk) 22:13, 16 March 2009 (UTC)
 * This is what I have done so far Talk:Monty_Hall_problem/Analysis Martin Hogbin (talk) 00:25, 17 March 2009 (UTC)


 * Martin - Once again, Morgan et al. analyze the case where the player picks door 1 and the host opens door 3 as representative cases (in exactly the same sense that the problem uses the wording "You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3").


 * No they do not. You are reading something into their paper that simply is not there.  If you think that I am wrong, please point me to any wording or calculation in the Morgan paper that says that they take their case to a representative one. Martin Hogbin (talk) 08:41, 17 March 2009 (UTC)


 * Of course they do. It's inherent in the wording of the problem.  Door 1 is clearly the player's random pick, which could just as well have been Door 2 or Door 3.  For convenience only, it is called Door 1 rather than "the door the player picked" in the problem statement.  Similarly, Door 3 is simply a convenience for referring to "the door the host opens".  Morgan et al.'s analysis uses these names for the doors only because it is what the problem says to do.  What is problematic are solutions that don't consider a specific case where the player picks some door (which we might as well call door 1, since it is what the problem statement calls it) and where the host opens some other specific door (which we similarly might as well call door 3).  The sense of the problem (which I'm trying to discuss here - would you please answer the question below rather than continuing to quibble about the Morgan paper?) is that the player is put into a situation where she's made an initial choice and can now see only two closed doors and is given the opportunity to switch, i.e. in one of the following 6 conditions:
 * picked 1, host opened 3 (the exemplary one from the problem statement)
 * picked 1, host opened 2
 * picked 2, host opened 1
 * picked 2, host opened 3
 * picked 3, host opened 1
 * picked 3, host opened 2
 * Any given player is deciding to switch in one and only one of these states. Because it is assumed the car is initially randomly distributed, analyzing the chances for one of these cases is clearly sufficient - which is why the problem statement can simply pick one of these cases. -- Rick Block (talk) 14:45, 17 March 2009 (UTC)


 * I am happy to answer the question below but this is not a quibble, it is the cause of the whole confusion. Your statement that Morgan mean one of six conditions is purely an assertion on your part that is not supported by any wording in the paper.  If you disagree, please tell me where in the paper they give any indication that their case is to be taken as an example of a more general one. Also look at their calculation, it applies only to the one case, that the payer opens door 1 and the host door 3.  If the host were to choose door 2 their calculation would be reversed. If Morgan are, in fact, intending to refer to the general case, their calculation is wrong.


 * So as not to disrupt your flow I will continue my discussion of this point in a separate section below. Martin Hogbin (talk) 17:17, 17 March 2009 (UTC)


 * I would like to comment here, that in my opinion, Morgan indeed solves the particular case "picked 1 and opened 3", but also considers this case to represent all the other 5 combinations. As the car is placed randomly, they are equivalent. The initial picking doesn't have to be random for the solution. Nijdam (talk) 23:30, 17 March 2009 (UTC)
 * See my point below. You cannot always use a specific case as an example. Martin Hogbin (talk) 23:57, 17 March 2009 (UTC)


 * If we all agree on the bullet points above, specifically that the player decides to switch after the host opens a door, and that the problem does not mean to refer literally to only the case where the player chooses door 1 and the host opens door 3, then there's really only one more thing we need to clarify (which relates to the above point Martin is making and his analysis). Does the problem mean the chances of winning by switching in general, or the chances of winning by switching given some specific door the player picks (like, say, door 1) and some specific door the host opens (like, say, door 3)?  I think (in a #1 sort of way) that it's the latter and (#2) that's how most people interpret the problem, but I'm curious what you guys think.  Please ignore (for the moment) whether you think these have the same numeric value - but focus on what the question is actually asking.  If you think there might be a difference, please let me know both what you think (#1) and what you think most people think (#2). -- Rick Block (talk) 02:44, 17 March 2009 (UTC)


 * I believe that the problem should be interpreted to refer to the general case and think that most people would believe the same.

Martin Hogbin (talk) 17:17, 17 March 2009 (UTC)

What is the effect of referring to one specific case?
To explain what I mean, let us consider a very simple game show where the producer places a car behind one of two doors (and nothing behind the other) and the player chooses one door and immediately get the prize behind it. At first sight you might think that it makes no difference whether we describe the question in general or take a specific case as an example.

Let us describe the question in two ways and then calculate the chances that the player wins a car.

1 A car is placed behind one of two doors and the player then randomly picks a door.

In this case the answer is 0.5 It does not matter what the initial distribution of the car behind the doors was because the player is known to  have chosen randomly.

2 A car is placed behind one of two doors and the player chooses door 1.

In this case the answer clearly depends on the probability with which the car was placed behind door 1. If we call that p then the answer to the question is simply p.

This second case is similar to that which Morgan have answered in the MHP, if that is not clear from their wording then it should be clear from the calculation that they do. Although the situation is not exactly the same, it is rather more complicated, they introduce a parameter p (in their case the host's choice of door) in just the same way. Martin Hogbin (talk) 17:33, 17 March 2009 (UTC)


 * I don't see what your problem is here. The MHP (as typically worded) is actually far more similar to #2 than #1.  Except, instead of "the player chooses door 1" we say "what is the player's chance of winning if she picks, say door 1".  With this wording, just as in the MHP the question is not what is the "overall" chance of winning (1/2 for this problem, 2/3 for MHP) but what is the chance of winning per door - i.e. it shifts the question from being about the player's choice to being about the how the car is distributed behind the doors.  If we call the probability that the car was placed behind door 1 "p", and the probability that it is behind door 2 "q", then q = 1 - p and both p and q are in the range 0-1, so it really doesn't matter which one we say is which.  The player wins with probability p, where 0<=p<=1, is a completely general solution.  If, from the problem statement, we knew the probability of the car being behind door 1 was always 1/3, the solution would be 1/3 if the player picks door 1 and we would note this means a player picking door 2 would have a 2/3 chance of winning.  We wouldn't say the answer is 1/2 - even though if this is a guessing game, over all players 1/2 would win.


 * Regarding my simple game, how would you answer this question then: A car is placed behind one of two doors and the player then picks a door, say door 1.  What is the probability that he will win a car? Martin Hogbin (talk) 13:30, 18 March 2009 (UTC)
 * It equals the probability of the car being behind door 1. No more can be said. Nijdam (talk) 14:00, 18 March 2009 (UTC)
 * OK it looks as though you have chosen to interpret that problem statement as my statement 2 above, 'A car is placed behind one of two doors and the player chooses door 1'.
 * What would the answer be if the problem were stated as in 1 above, 'A car is placed behind one of two doors and the player then randomly picks a door'? Martin Hogbin (talk) 21:08, 18 March 2009 (UTC)
 * The outcome of a random pick between two choices is always 50/50. This is an attribute of the choice, not an attribute of the distribution of objects being chosen.  For example, if the host rolls a die and places the car behind door 1 if the die is 1 or 2, and behind door 2 otherwise, the two questions you pose have different answers.  The answer to your first question is still 50%, but the second question asks about the conditional probability given that the player has picked door 1 - this is 1/3 (and 2/3 if the player picks door 2).  This is similar (not an exact analogy) to the distinction between the unconditional and conditional questions about the MHP.  One setup, but different questions have different answers.  -- Rick Block (talk) 01:25, 19 March 2009 (UTC)
 * Exactly! So the answer to the representative question '.. the player picks door 1' is not necessarily 1/2 and the answer to the representative question, '.. the player picks door 2' is also not necessarily 1/2 but the answer to the question, '...the player randomly picks 1 or 2' is 1/2.  By taking the two possibilities individually you do not get the answer for the general case.  It is exactly the same for the MHP except that it is not even necessary for the hosts choice to be random to achieve the same effect. Martin Hogbin (talk) 19:57, 19 March 2009 (UTC)


 * In the Morgan et al. paper, the expression at the top or page 286 is the completely general solution, i.e. the probability of winning by switching is p23 / (p13+p23) - i.e. the probability the host opens door 3 given the car is behind door 2 divided by the sum of this plus the probability the host opens door 3 when the car is behind door 1. By changing the door numbers, this formula applies to any of the 6 possible conditions any player might be in. Since all the terms are variables the expression would have exactly the same form, i.e. a/(a+b), regardless of which door the player picks and which door the host opens - and this form covers all plausible game rules as well.  In this formula "a" is the probability the host opens "door 3" (i.e. the door he actually opened) when the car is behind "door 2" (i.e. the other door he might have opened,) and "b" is the probability the host opens "door 3" (again, the door he opened) when the car is behind "door 1" (i.e. the door the player picked) .  By the specific rules of the game, a is 1 (if the player didn't pick the car the host's choice is forced) and b is unspecified in the Parade version (if the player picked the car, the host can open whatever door he'd like).  Completely general.  Exactly correct.  Leads directly to the not so obvious observation that the probability of which door the host opens when the player initially picks the car is totally overlooked in the Parade statement of the problem (and most of the quibbling about how underspecified this version was) and has a significant role in the answer.


 * The Morgan formula involving the parameter p applies only to the case where the host opens door 3. If the host opens door 2 then the parameter will have a different value (in fact 1-p). As we are assuming that the host may open either door 2 or door 3, we do not know whether to use p or (1-p).  Please have a look at my analysis of the MHP for a more general case that Morgan.  My notation may not be quite right but I think the analysis hold good.  You can see from this that, after applying a set of game rules and assumptions (basically what Morgan call the vos Savant case) The probability of winning by switching is independent of the host action unless we restrict ourselves only to the case where the host opens door 3. Martin Hogbin (talk) 13:30, 18 March 2009 (UTC)


 * What you're saying is that the unconditional probability is independent of the host action (which, of course, it is - this is the definition of unconditional probability). The conditional probability is not.  -- Rick Block (talk) 01:25, 19 March 2009 (UTC)


 * No what I am saying is that if our condition is that the host opens one of the unchosen doors to always reveal a goat rather then  the host opens door 3 to reveal a goat, the conditional probability is independent of the host's choice of door. Martin Hogbin (talk) 09:10, 20 March 2009 (UTC)


 * This a/(a+b) equation expresses pretty much everything anyone needs to know about the Monty Hall problem. It cleanly explains all the variants you might ever want to consider:


 * Standard analysis? No problem. a=1, b=1/2.  Answer = 2/3.


 * Forgetful Monty? No problem.  a=1/2, b=1/2.  Answer = 1/2.


 * Leftmost Monty? No problem.  a=1, b=0 (or b=1, depending).  Answer = 1 or 1/2.


 * See how this works? -- Rick Block (talk) 04:27, 18 March 2009 (UTC)


 * I fully understand how it works but it is wrong. Martin Hogbin (talk) 13:30, 18 March 2009 (UTC)


 * Why do you think their analysis is wrong? I found that the case where p=1 is the most enlightening.  Suppose the host's strategy is to always show you door 2 unless he cannot because there is a car behind it.  Then if you select door 1 and he shows you door 3, then the car must be behind door 2, so the probability of winning when switching is 1, not the 2/3 which is given by the unconditional solution.    --67.193.128.233 (talk) 00:03, 19 March 2009 (UTC)


 * It is wrong because it only applies in the case where the host has opened door 3.  If the host has has exactly the same strategy but has opened door 2 your chance of winning by swapping is now only 1/2.  Martin Hogbin (talk) 19:36, 19 March 2009 (UTC)


 * So you agree that the probability can depend on which door the host opens? If you agree with this then how can you advocate an unconditional solution which by definition does not consider the door the player selected OR the door the host opened? --67.193.128.233 (talk) 13:29, 20 March 2009 (UTC)


 * Martin - the conditional probability P(switching wins|player picks door i, host opens door j) is dependent on both the door the player picks and the door the host opens. The condition the host opens one of the unchosen doors to always reveal a goat is independent of the host's choice of door only if the host is constrained to choose randomly between doors if the player initially picks the car.  The general solution is 1/(1+p) where p is the probability the host opens door j if the player picks door i and the car is behind door i.  If we don't constrain this probability by forcing the host to pick randomly if given the chance (making p=1/2), then the solution is not 2/3 but some number between 0.5 and 1.  If unconstrained, p may vary based on the door the player picks or the door the host opens.  If you don't restrict the analysis to a player pick of door i and the host opening door j, you're solving for the unconditional probability (in which case, p is irrelevant). -- Rick Block (talk) 12:23, 20 March 2009 (UTC)

Math
It is time to introduce some notation and stick to it. Let C be the random variable indicating the number of the door with the car and X the number of the door picked initially by the player. C and X are independant. A is the number of the door opened by the host. Probabilities: P(C=c), P(X=x), P(A=a|X=x,C=c). Solution in the case {X=1,A=3}: calculate: P(C=1|X=1,A=3). Etc. Normal case: P(C=c)=1/3; P(A=3|X=1,C=1)=1/2, P(A=3|X=1,C=2)=1 and P(A=3|X=1,C=3)=0. Hence:
 * P(C=1|X=1,A=3) = P(A=3|X=1,C=1)P(C=1|X=1)/P(A=3|X=1)=(1/2.1/3)/[1/2.1/3 + 1.1/3 + 0.1/3] = 1/3.

I suggest we formulate what we want to argue in these terms. Nijdam (talk) 14:24, 18 March 2009 (UTC)


 * I am happy to use whatever notation you suggest as you obviously know more about the subject than I do. Can you please comment on this page Talk:Monty_Hall_problem/Analysis  I will be happy to convert it to your notation if that helps. Martin Hogbin (talk) 21:14, 18 March 2009 (UTC)


 * Nijdam, I have started to try and convert my page but I am not sure how to proceed, can you help please. Martin Hogbin (talk) 21:34, 18 March 2009 (UTC)
 * Martin - I've converted roughly half of your analysis to Nijdam's notation. In the part that remains I'm not sure exactly what you mean, so have left it as it was. -- Rick Block (talk) 02:01, 19 March 2009 (UTC)
 * Martin, I converted a lot more, but some unclear stuff remains. Nijdam (talk) 13:45, 19 March 2009 (UTC)


 * Nijdam and Rick, thanks for your help. I think I have now completed my argument using your notation. Martin Hogbin (talk) 19:25, 19 March 2009 (UTC)

Resuming discussion about why it is conditional
This is a continuation of the discussion started above, at. I'd still like to make sure we're clear when we're each talking about I think we agreed most people understand the problem to be that:
 * 1) What we individually believe to be true.
 * 2) What (we think) most people believe.
 * 3) What is factually true.
 * 4) What reliable sources say.
 * car is randomly placed
 * initial pick is random
 * host must show a goat (and, hence, knows where the car is or is told which door to open by someone who knows this)
 * host must make the offer to switch
 * player decides after the host opens a door

I think (please let me know if I'm wrong) Glkanter has agreed these constraints are the only ones necessary, specifically that it doesn't matter if the host has a preference for one goat door over another. I don't think Martin has directly answered this question yet, but has agreed most people don't take the question to be door-specific (meaning we're not interested in only the case where the player picks door 1 and the host opens door 3). We got sidetracked at the point I was asking whether we think most people think the problem refers to the general chances of winning, or to the chances of winning by switching given some specific door the player picks (like, say, door 1) and some specific door the host opens (like, say, door 3). This is an extremely important point, and I'd like to get back to it.

I think it is factually true that there are 6 possible conditions a player might find herself in and we can ask about the probability of winning by switching in each of these cases individually. I.e. we can ask
 * 1) What is the probability of winning by switching if the player picks door 1, host opens door 3?
 * 2) What is the probability of winning by switching if the player picks door 1, host opens door 2?
 * 3) What is the probability of winning by switching if the player picks door 2, host opens door 3?
 * 4) What is the probability of winning by switching if the player picks door 2, host opens door 1?
 * 5) What is the probability of winning by switching if the player picks door 3, host opens door 1?
 * 6) What is the probability of winning by switching if the player picks door 3, host opens door 2?

I think this is where Martin's Talk:Monty_Hall_problem/Analysis page is meant to be heading.

I think it is also factually true that in addition to the above six questions we can ask the general question
 * 0. What is the aggregate probability of winning by switching - meaning across all players regardless of which specific door they pick and which door the host opens?

So, here's the $64,000 question. Do you (Glkanter and Martin) think the question as usually phrased is #1 from above (on the assumption that this answer serves as a representative answer for any of questions 1-6), or is the actual question being asked #0? If you think there's a difference between what you think and what most people think that would be nice to know, too.

I suspect you probably know what I think, but I'll say it anyway just to make my stance clear. I think:
 * it is factually true (in a #3 sort of way) that the question is question #1 from above
 * the problem as generally posed intends the answer to be taken as representative for the answer to any of questions 1-6
 * most people probably haven't thought about it, but given a more detailed explanation would say they understand the question this way
 * most people assume the answer to questions 1-6 must be the same (or, turning this around, most people asking the question intend the answer to questions 1-6 to be the same), and must be the same as the answer to question 0

Please let me know what you think about this. -- Rick Block (talk) 14:23, 19 March 2009 (UTC)
 * I fully agree. Nijdam (talk) 23:03, 25 March 2009 (UTC)

I think: Martin Hogbin (talk) 19:51, 19 March 2009 (UTC) Corrected Martin Hogbin (talk) 10:09, 21 March 2009 (UTC)
 * it is not factually true that the question is question #1 from above. The question says 'say door 1' and 'say door 3' indicating that we are to consider other possibilities.
 * the problem as generally posed intends the answer to be taken as representative for the answer to the general case of 1-6.
 * most people probably haven't thought about it, but given a more detailed explanation would say they understand the question this way
 * most people assume the answer to questions 0-6 must be the same
 * the answer to any individual question 1-6 is not the same as the answer to the question of what is the probability given that any one of 1-6 occurs. Strange but true.  It is exactly the same as in my simple example.  Martin Hogbin (talk) 19:51, 19 March 2009 (UTC)


 * Rick, thanks for picking up the discussion...
 * I think there must be a codicil somewhere in the Probability Handbook that says, 'at all times we are talking about aggregate probability'.
 * Regardless of how eloquently I state it, or whether I use proper notation, we all agree that the probability of 'car' + the probability of 'not car' = 1.
 * I have never heard anyone disagree that the contestant's probability of selecting the car is 1/3.
 * No one has demonstrated how Monty's actions change the original 1/3 probability. In fact, the 'Combining Doors' solution shows why Monty's actions do NOT change the 1/3. (Those pesky parenthesis).
 * Therefore, there is no beneficial purpose or requirement to emphasising a more complicated solution to this problem. Further, it is incorrect to state that the unconditional solution is in any way incomplete, invalid, lacks rigor, etc. Glkanter (talk) 15:03, 19 March 2009 (UTC)


 * I would just like to point out that there are at least 2 separate discussions taking place regarding this article. And I understand there is a 3rd page as well. If our goal is to reach consensus on editing the Article, I think we would benefit from all editors being in 1 discussion. Yes? Glkanter (talk) 15:15, 19 March 2009 (UTC)


 * Here we go again: The contestant's initial chance to pick the car is indeed 1/3, due to the random placing of the car. But ... there are several variations of the problem where this probability changes! And no one needs to proof that Monty's actions changes this probability, on the contrary we have to proof that Monty's actions do NOT change it. This is exactly the whole issue. And the "combining doors solution" doesn't proof anything. Nijdam (talk) 18:30, 19 March 2009 (UTC)


 * I'm starting to sense a 'double standard' here. The Combining Doors Solution is in the article, and it has 4 separate sources. How is it relavent (in a positive manner) to the discussion that your opinion is 'the "combining doors solution" doesn't proof anything'? It's the published sources that matter, right? Glkanter (talk) 15:22, 30 March 2009 (UTC)

Somebody wrote the other day about how much the MHP article bothered him in an intellectual way. Me too. I just looked at Martin's Analysis page. Wow! Maybe all that stuff adds value to a different problem. But not the MHP via MvS.

I'm not sure who I'm trying to win over to my way of seeing things anymore, but here goes...

So, for the MHP as described by MsV (see Rick's 5 bullet points), and not some variant with new constraints/premises:

We agree that the probability of 'car' + 'not car' = 1. When there are 3 doors, the range of possible values for the contestant's door are 0, 1/3, 2/3 and 1. When there are two doors, we add a new possibility, which is 1/2 (plus 0 and 1). When there is only 1 door, the only possible values are 0 and 1. So, outside of 0, 1/3, 1/2, 2/3 and 1, are there any other possible values? No.

We all agree the contestant's door starts at 1/3. And, for varying reasons, we all agree it is still 1/3 after Monty has shown a goat. So when was it equal to 0, or 1/2, or 2/3, or 1? And how did it change back? The answer, of course, is that it never changed. Glkanter (talk) 21:35, 19 March 2009 (UTC)


 * Glkanter - from your response above I assume you think the question that's being asked is question #0 (right?). I can't exactly tell, but I sort of also assume you think most people think this as well (also right?).  Your assertion  that there must be a "codicil somewhere in the Probability Handbook that says, 'at all times we are talking about aggregate probability'" is absolutely NOT the case.  Take insurance rates for example.  Car insurance for my 17 year old kid is vastly more expensive than mine because insurance companies don't ask just what is the probability of a driver being in an accident, but what is the probability given how old the driver is.  This is a conditional probability.  There's an aggregate probability across all drivers, but there are conditional probabilities (for all kinds of conditions - males vs females, age, prior accidents, etc. etc).  My point is only that question #0 and questions 1-6 are all different (1-6 are the conditional questions), and we can clearly ask what the answer is for any of them.  I don't see how you can say that the way the problem is phrased:
 * 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?
 * can mean anything other than question #1. It certainly sounds like the answer should also be the answer to question #0, but isn't the exact question obviously question #1?


 * Martin - how can you read the question and NOT think it is asking question #1? Certainly the question seems to imply the answer should apply to questions 2-6 (and therefore 0 as well), which I would take to mean if (in some bizarre universe) the answers for 1-6 were all different that the questioner might be interested in all 7 different answers (if all 6 were somehow different the 7th might not match any of the other 6).  For example, if the game were I'm going to roll an n-sided die and give you $100 times the number that shows up and the question is:
 * I roll, say, a 6-sided die, how much money would you win?
 * The "average" number is $350. Since I'm always giving away some even multiple of $100 clearly no individual ever wins this amount even though it is the average.  The answer to the question (as phrased) is clearly an even $100 between $100 and $600.  Is $350 responsive to this question, or is it the answer to a different question - like "I roll, say, a 6-sided die.  How much money would I give away on average?"


 * Both of you - the phrasing of the question matters. It doesn't say anything about "on average" or "across all players". It's talking about a single player who has picked a door and has seen the host open a different door.  This is obviously a player like a player in one of the question 1-6 cases.  The question is specifically about a player in the question #1 case.  You're making a mental leap from question #1 to question #0 on the assumption that the answers are the same (because the answer to conditional and unconditional questions like these are nearly always the same).  We can talk about the answers later, but I don't understand how you can read the same problem I'm reading and not see that it's question #1. -- Rick Block (talk) 00:33, 20 March 2009 (UTC)


 * Basically, I disagree with everything you just wrote.


 * I've figured it out, finally. I'm actually on some version of 'The Truman Show'. That's the one where Jim Carrey was unwittingly the sole non-actor on a long running TV show. Everyone else was in on the act. All this stuff on Martin's Analysis page, all the conditional stuff, all the Bayesian stuff, the referrals to subject matter experts, Rick's frequent and incredibly lengthy 'proofs' of different puzzles, etc. are there to screw with my mind. As all of them are unnecessary to prove the 1/3 vs 1/2 question, there can be no other explanation. Glkanter (talk) 01:13, 20 March 2009 (UTC)


 * OK. Let's amplify the probabilities a bit.  Get out a deck of cards.  Shuffle.  Deal one to me face down.  You get the rest.  Look at yours.  I'll bet I have a 1 in 52 chance of having the ace of spades and you have a 51 in 52 chance of having the ace of spades (right?).  You can always show me 50 cards that aren't the ace of spades and no matter which 50 of your 51 cards you show me (as long as you don't show me the ace of spades) you still have a 51 in 52 chance of having it and I have a 1 in 52 chance.  Right?  Which means if we're betting who has the ace you don't care whether I place my bet before or after you show me the 50 cards.  Does 10-1 odds sound fair (if I bet $1 that I have the ace I pay you $1 if I lose, but you pay me 10 times that if I win)?  Since it doesn't matter which 50 of your cards you show me, as long as none of them are the ace, you don't care if I tell you which cards to show me.  Right?  So, now, show me 50 cards but keep the ace, but if you don't have the ace keep the two of hearts.  Now you either have the ace or the two of hearts but you still have a 51/52 chance of having the ace and I still have a 1/52 chance (right?).  If you've shown me the two of hearts my bet is a penny.  If you haven't shown me the two of hearts my bet is $100.  I'll happily do this for real money if you're willing to play, say, at least 500 times (I mean, I want to get the ace at least occasionally).  The way I figure this (using the mind screwing magic of conditional probabilities) is if I bet a penny you'll win every time.  But if I bet $100 we're in a 50/50 situation.  Yes, you'll win 51/52 times overall.  But if you show me the two of hearts I'm certain you'll win, and if you don't we're even.  I never exactly know who has the ace - but if you don't show me the two of hearts my chances zoom up from 1/52 to 1/2.  What you do changes my chances.  With this setup, they usually go from 1/52 to 0.  But 2/52 times they go to 1/2.


 * Similarly, how the host "discards" (decides which door he opens) matters! -- Rick Block (talk) 03:02, 20 March 2009 (UTC)


 * Based on your 5 bullet points that we've agreed on, the card deck has 3 cards. An Ace and 2 Jokers. As you know, the Jokers are indistinguishable from each other. The contestant blindly picks one card, then Monty shows a Joker. Glkanter (talk) 12:34, 20 March 2009 (UTC)


 * Nope - the contestant can plainly see which one is door 2 and which one is door 3, so not indistinguishable jokers but perhaps the two of hearts and three of hearts. I made it 52 cards just to amplify the probabilities involved.  We can do the same thing with 3 cards, it will just take a little longer for me to take all your money.  If you deal me one, and you keep two then I have a 1/3 chance and you have a 2/3 of having the ace - right?  So you have a 2-1 advantage.  Would you be willing to play giving me only a 1.5 advantage (if I bet $2 and end up with ace you give me $3)?  Same as before, but I'd like you to discard the three of hearts if you can.  If you discard the two of hearts, my bet is a penny.  If you discard the three of hearts my bet is $100.  If you're willing to play at least 60 times I'll do this for real money.  We can have Martin deal if you'd like.  -- Rick Block (talk) 13:36, 20 March 2009 (UTC)


 * Again, I disagree with just about everything you wrote. Explain to me, using only the 5 bullet points that you wrote, and we all agreed upon, how showing a goat behind any door provides new information that changes the odds. I believe goats and doors provide no new information, just like the identical Jokers. And I believe that is the intent of the problem. And that this is how probability and random events work. In fact, my solution relies only on the 1/3 original selection, plus the rule that 'car' + 'not car' = 1. I would REALLY appreciate it if you would, for once, address my solution, rather than a different problem with new constraints/premises, like the 2 or 3 of hearts stuff above. Glkanter (talk) 14:13, 20 March 2009 (UTC)
 * You know what Glkanter: give me 36 realisations of the game, let's say in the form cxh, meaning doornumbers of the car c, the initial pick x end the opened one by the host h. Then we'll continue. Nijdam (talk) 10:31, 21 March 2009 (UTC)


 * Why not just tell me at what point the original selection ceased to be equal to 1/3? Besides, you guys have already done that on the Analysis page. Certainly there are methods of solving this problem other than listing every possible outcome? You know, like the one I've been a proponent of since October? Why else would it be a famous paradox? The article is intended to provide the solution to the problem, not demonstrate anyone's mastery of probability. Glkanter (talk) 11:54, 21 March 2009 (UTC)
 * And you're still a victim of the paradoxal nature of the problem. Why not give me 36 realisations? Afraid? Nijdam (talk) 12:50, 21 March 2009 (UTC)


 * No, the paradox for which the problem is famous is '1/3 vs 1/2'. Not 'the equal got door' constraint. I've already had a Certified Wikipedia Probability Subject Matter Expert say that my solution is valid for an unconditional problem. So, tell me, why is that not sufficient for the MHP with Rick's 5 bullet points? Just tell me at what point it wasn't 1/3, and then when it switched back. Glkanter (talk) 13:20, 21 March 2009 (UTC)


 * You're right in one sense, but a deeper paradox is the unconditional vs. conditional solution. And guess what: I'm definitely a Qualified and Certified Wikipedia, Wikimedia, Wikibooks,real World Probability, Mathematics, Statistical and a lot more Subject Matter Expert. Impressed? You see, to any solution it's possible to find the problem it solves, but what is the use?Nijdam (talk) 14:09, 21 March 2009 (UTC)


 * No, the MHP is not a Wikipedia article because of 'unconditional vs conditional'. That is false. So, no, I'm not impressed. At this point, quite the opposite. So, at what points does the probability change from 1/3 to something else, and back again? What did it temporarily become? It's a simple question. Why not answer it? Glkanter (talk) 16:14, 21 March 2009 (UTC)

(outindent) Glkanter - I've told you this lots of times, but I'll try to explain it again. The bullet points above do not include anything about how the host decides what to do if the player initially picks the car (right?). The probability can change from the original 1/3 because of this. Consider the host's actions in "super slow motion" (of course, the host only does one thing - i.e. open a door, so the following two steps actually happen at exactly the same time - i.e. the player's original 1/3 probability changes only once, when the host opens the door). Step 1 (primary choice) is if one of the doors has the car he must open the other. At this point, when the player picked a goat and the host has been forced to reveal the other goat, the player's odds of winning by switching are 100% - the original 1/3 has dropped to 0. Step 2 (secondary choice) is when both doors have goats. In this case, the host is free to do whatever he'd like. He could flip a coin to decide which door to open, but with the bullet list of game rules above he doesn't have to. The doors could be spread out on the stage and he always walks across it to open a door, starting from the door 1 side (in which case perhaps he prefers to open door 2 because it's closer). Perhaps which one he prefers depends on which side of the stage he starts from. The point is unless we explicitly say he MUST choose between the two doors randomly he's free to have a preference for one or the other, which Morgan et al. assign the probability p. If he picks randomly (p=1/2) this is when the original chances "bounce back" to 1/3. If he doesn't pick randomly the chances can do anything from stay at 0 (if p=0) to "bounce up" to 1/2 (if p=1).

You might say, but if the player doesn't know what p is it doesn't matter. I say it does matter because the question is about the actual probability of winning by switching, not what the player thinks her probability is. For example in Martin's simple game above (two doors), if the host rolls a die and the die is one puts the car behind door 1 and otherwise puts the car behind door 2, whether a player knows this or not a player who picks door 2 has a 5/6 chance of winning the car (right?).

Back to our card game. If you want it exactly like MHP, we can do that. Use an ace and two jokers. I've repeated the bullet list here with the card version appended to each line.


 * car is randomly placed Shuffle the 3 cards. Deal them face down left to right.  The leftmost card corresponds to "door 1", the middle one "door 2" and the rightmost one "door 3".
 * initial pick is random I'll pick the card on the left (door 1).
 * host must show a goat (and, hence, knows where the car is or is told which door to open by someone who knows this) Here's where it gets tricky. Look at the card in the middle and then the card on the right (the two I didn't pick).  Immediately turn over the other one if either is the ace.  If you haven't done this, now you have a choice to make.  You can comply with the rules by turning either card over.  It doesn't matter if I know how you're deciding, what matters is whether you decide (with preference p) or whether you're forced to take a random action (i.e. your preference p=1/2).  To make the effect of this the most obvious (p=1), please turn over the rightmost card (door 3).  I think you've said repeatedly that it doesn't matter how the host decides in this case and can't affect the initial 1/3 chance, so you should be willing to accept this method of deciding.
 * host must make the offer to switch Exactly the same.
 * player decides after the host opens a door Player decides after you flip a card.

We've scrupulously followed the bullets (right?), so your claim is I have a 1/3 chance of having the ace and you have a 2/3 chance. Again, you have a 2-1 advantage so should be willing to give me 1.5 to 1 odds. I'll bet a penny if you turn over the middle card (door 2) and $100 if you turn over the card on the right (door 3). Repeat at least 60 times and I'll do this for real money. My guess is I'll end up roughly $500 up. -- Rick Block (talk) 15:23, 21 March 2009 (UTC)


 * I have it on good authority that Monty always looks at both doors, then flips a coin, then opens a door. He says nary a word during this process. Or his producer tells him which door to open via an earphone. So, no, you haven't followed the 5 bullet points. Your card game as described above adds a new constraint, making it some other problem. Which is not the focus of the paradox. Which may require a conditional solution. But that's not the MHP.


 * Without attributing any 'host behaviour' (that is, creating a new problem with a new constraint), what does Morgan say the solution is? Glkanter (talk) 15:59, 21 March 2009 (UTC)


 * Which of the 5 bullets says Monty flips a coin?  You're making it a different problem by adding this constraint.  -- Rick Block (talk) 16:14, 21 March 2009 (UTC)


 * Dramatic license. He looks at both doors, then opens one. Or, the producer tells him.


 * Without attributing any 'host behaviour' (that is, creating a new problem with a new constraint), what does Morgan say the solution is? Glkanter (talk) 16:24, 21 March 2009 (UTC)


 * Morgan says if we don't know whether he's choosing randomly between two goats, the best we can do is leave his (or his producer's) preference as a variable p. Leaving this as a variable is precisely what it means to not attribute any "host behaviour".  Assuming p=1/2 is assuming the host is picking randomly in this case.  If we don't know p we don't know an exact numeric answer - only an answer as a function of p.  Whatever p is, the answer is 1/(1+p).   The unconditional solution, which says the answer is 2/3 regardless of p, is effectively saying p must be 1/2.  If this condition is not in the problem statement (it's not in our 5 bullets above, and it's not in the Parade version of the problem) the unconditional solution is wrong for any case where p is not 1/2 (like the card version above). -- Rick Block (talk) 16:44, 21 March 2009 (UTC)


 * Thank you, Rick. So Morgan says, without her stating that Monty's behaviour is random, MvS has provided an invalid solution. It sounds like they're saying I can't calculate the 'not chosen and not revealed door' probability without this premise. So let's calculate the probability of 'chosen door' after Monty has revealed a door. There's only 2 doors left, and added together, the probability of 1 of them hiding a car = 1, right? Well, the last we knew, the probability of 'chosen door' was 1/3. (I don't think 'p' has any affect on this, as Monty is only considering the doors that were not chosen.) So, If MvS has provided an invalid solution, then using the 5 bullet points, specifically without assigning any host behaviour, when did 'chosen door = 1/3" change, and what did it change to (0, 1/2, 2/3, 1)? Glkanter (talk) 17:33, 21 March 2009 (UTC)


 * They're saying you can't calculate the probability of either remaining door unless you know (or assume a value) for p - the probability of both remaining doors depends on p. The door the player hasn't picked changes from 1/3 to 1/(1+p), and (at the same time) the player's chosen door changes from 1/3 to p/(1+p).   If you switch, your chance of winning is 1/(1+p) and if you stay your chances of winning are p/(1+p).  Add these together and you always get 1 (regardless of p).  The only way the player's initial 1/3 doesn't change is if p is 1/2.  -- Rick Block (talk) 18:15, 21 March 2009 (UTC)


 * Then neither MsV or Morgan can assign a numeric value without adding the 'Monty chooses randomly' constraint? Glkanter (talk) 20:27, 21 March 2009 (UTC)


 * Right. If you don't know p all you can say is that "switching wins with some probability between 1/2 and 1 (1/(1+p)), and staying wins with some probability between 0 and 1/2 (p/(1+p))".  Most people probably mean you to take p as 1/2, but most people (including MvS) don't actually say this (which means we can't tell whether it's what they actually meant, or whether they simply overlooked this condition, or whether they mean to allow p to take on any value).  What Morgan points out is that it's always a conditional probability question (no matter what p is) and approaching it unconditionally is basically an ass-backwards way to look at it (although they're much more polite than this) because this approach is inherently based on the assumption that p must be 1/2 (whether it's stated as a constraint or not).  The "fact" that the host opening a door does not change the player's initial 1/3 probability of having selected the car is only true if p is 1/2 (AND if the host must show a goat and must make the offer to switch).  Omit any one of these constraints, and the answer isn't necessarily 2/3 (meaning the host's actions may be changing the player's initial chance of selecting the car).  The "host forgets" variant changes the must show a goat condition (so, not 2/3 anymore).  The "real life" variant where the host makes the offer to switch more often if you've initially picked the car changes the must make the offer to switch condition (so, not 2/3 anymore).  Letting the host decide however he wants between two goats changes the p=1/2 (must picks randomly between two goats) constraint - so, not 2/3 anymore.


 * Another little nugget in the Morgan paper is that the totally general solution is a/(a+p), where p is what we've been talking about, i.e. the probability of opening whatever door the host opens (door 3 in the usual setup) given the player's initial pick is the car and a is the probability the host opens this door given the car is behind the 3rd door (door 2 in the usual setup). In the usual setup, a is 1 (must open a door showing a goat).  In the "host forgets" variant, a is 1/2.  In the "real life" variant, a might be anything - maybe even 0 (that's the case where he only offers the switch if the player initially selects the car).   -- Rick Block (talk) 22:21, 21 March 2009 (UTC)


 * Thank you again, Rick. It seems to me, your answer jumped from 'why conditional' to 'why 'p' must be 1/2 for the conditional solution to result in 1/3 vs 2/3'. You wrote this:


 * What Morgan points out is that it's always a conditional probability question (no matter what p is) and approaching it unconditionally is basically an ass-backwards way to look at it (although they're much more polite than this) because this approach is inherently based on the assumption that p must be 1/2 (whether it's stated as a constraint or not).


 * So I ask, why must it be conditional? Please respond using the agreed upon 5 bullet points, and specifically, no 'host behaviour'. Also, since the 5 bullet points do not mention door numbers, please limit your answer to the scenario you described above:


 * 0. What is the aggregate probability of winning by switching - meaning across all players regardless of which specific door they pick and which door the host opens?


 * Thank you. Glkanter (talk) 23:59, 21 March 2009 (UTC)


 * It's conditional because the problem involves an event where the host opens a door. Before this event we have unconditional probabilities.  After this event we have conditional probabilities.  These are clearly different - the door the host opens goes from probability 1/3 to 0.  We all agree about this.  The question is what happens to the player's door and what happens to the other door the host didn't open.  What happens is what I described above in the "super slow motion" scenario.  If we assume no host behaviour, and want to avoid using a variable (like p) for the host preference, the probability of winning by switching is something between 1/2 and 1 (if we can talk about p it's 1/(1+p)).  Question 0 is a different question than the one ordinarily asked.  The clearest way to phrase this question is "what is the probability of winning by switching if the player decides before the host opens a door"?  Given the constraints that the host must show a goat and must make the offer to switch, the answer is 2/3. But the fifth bullet says we're letting the player pick after the host opens a door, so this is not the question we're talking about.  I assume you're reading the thread below.  I think you and Martin are basically talking about the same point.


 * Do you disagree that the sense of the problem is what is the chance of winning by switching for a player who has picked door i and has seen the host open door j (where 1<=i<=3, 1<=j<=3, and j not equal i)? Even if we don't specifically say what i and j are, this is still a conditional problem.  -- Rick Block (talk) 04:09, 22 March 2009 (UTC)


 * Does Morgan say that it must be solved as a conditional problem even if MsV had added the 'host chooses randomly' as a constraint? Glkanter (talk) 04:48, 22 March 2009 (UTC)


 * And, separately, doesn't the bullet point "* host must show a goat (and, hence, knows where the car is or is told which door to open by someone who knows this)" equate to a probability of 1 that Monty will reveal a goat? Doesn't that eliminate any need for a condition there? Glkanter (talk) 14:49, 22 March 2009 (UTC)


 * Morgan says, in a published rejoinder to vos Savant's reply to their article (note in our "clarity" scheme, this is a #4 sort of item, i.e. not my opinion, not what I think most people think, but what a reliable source, in this case a professor of statistics, says) "... even if one accepts the restrictions that she [MvS] places on the reader's question, it is still a conditional probability problem.  One may argue that the information necessary to use the conditional solution is not available to the player, or that given natural symmetry conditions, the unconditional approach necessarily leads to the same result, but this does not change the aforementioned fact."


 * The host must show a goat constraint means the probability is 1 that the host shows a goat. Here's their "false solution" F5 (in their paper, so again, not my opinion, but what a reliable source says - and, in this case, it is in a paper co-authored by 4 math professors published in a peer reviewed math journal):


 * Solution F5. The probability that a player is shown a goat is 1.  So conditioning on this event cannot change the probability of 1/3 that door 1 is a winner before a goat is shown; that is, the probability of winning by not switching is 1/3, and by switching is 2/3.


 * What they say about this solution is:


 * Solution F5, like F1, is a true statement that answers a different problem. F5 is incorrect because it does not use the information in the number of the door shown.


 * F1 is the "a player who never switches wins 1/3 of the time" unconditional solution, which they have previously discussed, so their comments about F5 are somewhat shortened. About F1, they say:


 * F1 is immediately appealing, and we found its advocates quite reluctant to capitulate. F1's beauty as a false solution is that it is a true statement! It just does not solve the problem at hand.   F1 is a solution to the unconditional problem, which may be stated as follows: "You will be offered the choice of three doors, and after you choose the host will open a different door, revealing a goat.  What is the probability that you win if your strategy is to switch?" The distinction between the conditional and unconditional situations here seems to confound many...


 * I'll add a minor addendum (my opinion) - their unconditional phrasing of the problem implies the player picks before the host opens a door (note the future tense - "you will be offered the choice ..."). -- Rick Block (talk) 16:02, 22 March 2009 (UTC)


 * How does Morgan address the 'Combined Doors' solution? I see that the 'equal goat door' constraint is included in the article as a premise here, but without a reference. There are four references in the section. Devlin, who is referenced twice in the section, does not mention this premise. Williams does. The other two do not. Is it necessary? The published materials favor 'no'. Glkanter (talk) 22:26, 22 March 2009 (UTC) Glkanter (talk) 22:37, 22 March 2009 (UTC)


 * What does this mean? "F5 is incorrect because it does not use the information in the number of the door shown." Glkanter (talk) 21:54, 22 March 2009 (UTC)


 * They don't address the "combined doors" solution, but (my opinion - I could probably find a reference for this if you'd like) this solution is yet another unconditional solution that holds only if p=1/2.


 * Their comment about F5 is that the probabilities of interest are the conditional probabilities given the specific door the player picked and the specific door the host opened. In the case they analyze the player picks door  1 and the host opens door 3 (as in the problem statement), but the criticism holds for any door the player picks and any door the host opens.  The point is the host has not just opened some abstract "door", but a specific one of the two doors the player has not picked and the conditional probabilities may be sensitive to which door the host opens (which they then go on to show is indeed the case unless the p=1/2 assumption is made). -- Rick Block (talk) 22:39, 22 March 2009 (UTC)


 * Let us make clear: even in the case p=1/2, the solution is conditional!! The only simplification (??) is that the player may calculate the conditional probabilities from the unconditional ones, using symmetry arguments. But her decision is based upon the conditional probabilities. And the problem with the simple "solution" is it does not use the symmetry arguments and it does not address the conditional probabilities. Nijdam (talk) 20:31, 28 March 2009 (UTC)

It is all just terminology
Perhaps all this argument is about nothing but terminology. Does anyone disagree with any of the following statements? It is assumed throughout that the car is placed randomly, the players initial pick is random, and that the host always opens an unchosen door to reveal a goat.

1) The probability of winning by switching, given that the player has picked door 1 and the host has opened door 3, is between 1/2 and 1 depending on the hosts door preference. (This is clearly a conditional probability)

2) The probability of winning by switching, given that the player has picked door 1 and the host has opened door 2, is between 1/2 and 1 depending on the hosts door preference. (This is clearly a conditional probability)

3) The probability of winning by switching, given that the player has picked door 1 and the host has opened one of the unchosen doors, is 2/3 for all host preferences. (Morgan call this the unconditional case, maybe this is where the confusion arises).

4) The probability of winning by switching given that the player has picked a door and the host has opened one of the unchosen doors is 2/3 for all host preferences.

5) The probability of winning by switching on average for player who switch is 2/3 for all host preferences.

Martin Hogbin (talk) 10:26, 21 March 2009 (UTC)
 * You're right. And notice that in case 3, opening one of the unchosen doors is not a condition, because it is part of the rules of the game. The only condition, if you would consider the full model, there is the initial picking of door 1. In case 4 there is no condition at all in the full model and hence it is the same as case 5. But ... notice that in the MHP a specific door has been chosen and a door opened.Nijdam (talk) 10:43, 21 March 2009 (UTC)


 * I continue here. The difficulty is what we do mean by probability. If you mean an average over all cases in which any door has been opened, there is, as I mentioned, no condition. If you mean: a door has been opened, but I do not know which one, and I want to average over all cases concerning the same door, the answer depends on the door and is a number between 1/2 and 1. Nijdam (talk) 17:55, 23 March 2009 (UTC)


 * I do not quite agree with you. If we average over all cases where a door has been opened, but you do not know which one, the answer is 2/3 if we include all possible host strategies in our average.  The only way to get an answer other than 2/3 is to average over cases where the host has a fixed strategy and the same door is opened every time. Martin Hogbin (talk) 20:50, 23 March 2009 (UTC)


 * Nijdam, do you see the connection with my simple problem in which a prize is placed behind one of two doors and the player then picks a door? If the player picks randomly the probability of winning is 1/2.  If the player has picked just one specific door then the probability of winning depends on the initial prize placement. Martin Hogbin (talk) 09:07, 24 March 2009 (UTC)


 * Who else agrees? Martin Hogbin (talk) 18:43, 21 March 2009 (UTC)


 * I agree with #1 and #2. I find #3, #4 and #5 somewhat ambiguous without further clarification, for example I don't know precisely what you mean by "for all host preferences".  I would agree with the following revised versions:


 * By 'for all host preferences' I mean regardless of whether the host policiy is to: always open door 2 where possible, always open door 3 where possible, choose randomly where there is a choice, or employ any other method of deciding which door to open within the stated rules. Martin Hogbin (talk) 21:25, 21 March 2009 (UTC)


 * 3) The probability of winning by switching, given that the player has picked door 1, is 2/3.


 * 4) The probability of winning by switching, given that the player has picked a door, is 2/3.


 * 5) The probability of winning by switching on average for players who switch is 2/3.


 * These are clearly "unconditional" statements that say nothing about the player deciding after the host opens a door. You've skipped one, perhaps 2.5, which I think is how most people actually interpret the question, specifically:


 * 2.5) The probability of winning by switching, given that the player has picked some specific door and the host has opened a specific different door, is between 1/2 and 1 depending on the host's door preference.


 * What is a specific door? Do you mean a specified door (ie 1 or 2 or the rightmost door) or something else, in which case what.  The host must always, according to the above rules, open an unchosen door, what is required to make this door a this a 'specific door'?  Martin Hogbin (talk) 21:25, 21 March 2009 (UTC)


 * I mean the probability given the player has picked door i and the host has opened door j, for 1<=i<=3, 1<=j<=3, and i not equal j. -- Rick Block (talk) 22:30, 21 March 2009 (UTC)


 * This is also (clearly!) a conditional probability. -- Rick Block (talk) 20:24, 21 March 2009 (UTC)


 * What is your answer to 3, exactly as I asked it ('for all host preferences' is defined above)?


 * 3) The probability of winning by switching, given that the player has picked door 1 and the host has opened one of the unchosen doors. Martin Hogbin (talk) 21:25, 21 March 2009 (UTC)


 * I still can't tell what you mean. When does the player decide, before or after the host opens a door?   If before, 2/3.  If after, I'd say it depends on the host preference.  I think what you actually mean is "considering all cases where the player picks door 1", which is effectively the "before" case.  -- Rick Block (talk) 22:30, 21 March 2009 (UTC)


 * I mean exactly what I say, 'given that the host has opened one of the unchosen doors. I mean that the host has already opened one of the unchosen doors to reveal a goat but we are given no information as to which door it is. Martin Hogbin (talk) 23:26, 21 March 2009 (UTC)


 * The host has opened a door and you're just not saying which one it is? The the probability of winning by switching is between 1/2 and 1, specifically 1/(1+p) where p is the host's preference for the door he's opened (the probability of the host opening that door when the player initially picks the car).  I think this is equivalent to "what is the probability if the player picks door 1 and the host opens door j, for some j in {2,3} (right?).  I believe the question must refer either to all cases where the player picks door 1 (probability is 2/3, equivalent to deciding to switch before the host opens a door) or to a specific case where the player has picked door 1 and the host has opened some door j.  The question that asks both of these simultaneously is the first one, which is apparently not what you're going for.  What is your answer for 2.5?  -- Rick Block (talk) 03:26, 22 March 2009 (UTC)


 * The door that the host has opened is not identified thus there is no way of quantifying the host's door preference. If the host has opened door 3 then it could be the case that the host always opens door 3 if possible.  If the host has opened the rightmost door it could be that the host preference is always to open the rightmost door.  On the other hand, if all we are told is that the host has opened one of the unchosen doors there is no way to specify a host preference that could have resulted in the actual choice.  The choice is made random by this lack of information.


 * This is a very important point to resolve, do you have any suggestions on ways to do so? Martin Hogbin (talk) 10:17, 22 March 2009 (UTC)


 * Resolve in what sense? In a Wikipedia sense there is no issue, since you've still never offered up a single source supporting anything you've said.  My suggestion is for you to drop this ridiculous foray into WP:OR and pay attention to what the math sources say.


 * If you mean resolve in the "understanding" sense, I think (my opinion) if the player picks a door, and then the host opens a door, the player can clearly see the door she picked and the door the host opened. From the structure of the problem, which door the host opens is plainly obvious.  I think this means we have three choices - consider all cases (the fully unconditional problem), consider all cases where the player picks door 1 (or door i for some i if you must, unconditional with regard to the specific door the host opens), or consider one case where the player has picked door 1 (or door i) and the host has opened door 3 (or door j if you must - fully conditional).  There is no middle ground here.  You can't turn this into an urn problem by coyly refusing to identify which door the host opens since the doors are distinguishable.  The way (the only way) to make it equivalent to an urn problem is to force the host to pick randomly between two goats.  In urn problems the balls are indistinguishable, so any choice among them must be random - in terms of MHP this means the p we keep talking about has to be 1/2. -- Rick Block (talk) 17:03, 22 March 2009 (UTC)


 * You state your view as if it is supported by some evidence. That facts are simple, if the door opened by the host is not identified in the statement of the question then it is not possible to propose a host strategy that makes the chances of winning by switching anything other than 2/3.  There is nothing in Morgan which says otherwise.  Morgan clearly and consistently describes a situation in which one specific and identified door has been opened by the host.  This is what the words words say and what the maths in the paper describes.  There is absolutely nothing in the paper that says the host door preference is relevant to the case where one unknown door has been opened.


 * There clearly is a middle ground here, which is the case that an unchosen and unspecified door has been opened. Morgan call this the unconditional case and express it as 'a mixture of the two conditional cases' and proceed to show that probability of winning by switching in the scenario we are discussing is always 2/3.  I am happy to refer to this particular case as unconditional if you like but it is clear from the maths that Morgan are referring to the case where either door 2 or door 3 has been opened by the host but we do not know which.


 * So my assertion is not OR, it is based on reading a reliable source on the subject. You insist on interpreting it in a way that is not supported by the wording or the maths in the paper itself, if you disagree then please show me where Morgan make clear that the host opening door 3 is intended to represent the more general case where the host has opened either door 2 or door three.  I believe that Nijdam agrees with on this particular point, so I suggest that we do need to resolve it in some way. Martin Hogbin (talk) 18:05, 22 March 2009 (UTC)


 * Martin - your argument here boils down to claiming that a paper co-authored by 4 statistics professors published in a peer reviewed math journal applies to the one and only case where the player picks door 1 and the host opens door 3, and that their solution is not meant to be taken as a solution for the general case of player picks door i and host opens door j (for all i and j, 1<=i<=3, 1<=j<=3, and i not equal j), and that their solution doesn't apply if we don't know the specific number of the door the host opens.


 * In the section you're reading, where they talk about the unconditional problem, they say "The unconditional problem is of interest, too, for it evaluates the proportion of winners out of all games with the player following a switch strategy."  This is almost exactly the same as my wording above ("all cases where the player picks door 1").  It is the probability given only that the player picks door 1 (which might as well be door i - or are you still arguing this one as well?).  In their phrasing of it (quoted in the section above), the unconditional problem has a major difference from the MHP which is that the decision to switch is made before, not after, the host opens a door.


 * Earlier, talking about F1, they say (I've also quoted this above) "The distinction between the conditional and unconditional situations here seems to confound many, ...". You wouldn't by any chance be confounded about this, would you? -- Rick Block (talk) 21:07, 22 March 2009 (UTC)


 * Yes, what I am saying is that in a a paper co-authored by 4 statistics professors published in a peer reviewed math journal you can take it that what is written is what is intended, and what is written is, 'To avoid any confusion, here is the situation: The player has chosen door 1, the host has then revealed a goat behind door 3...'. That seems pretty clear to me. Are you saying that the learned gentlemen actually meant something else?  If so, they make no mention of this fact anywhere in the paper.  Also the maths which follows is based literally on the quoted statement.  Surely, if the authors had intended the calculation to be representative of a more general case (the host has then revealed a goat behind one of the unchosen doors) they would have said so.


 * Regarding the section where they talk about the unconditional problem, I agree that the wording is not entirely clear, but the maths is.  They simply sum the two probabilities for the door 3 and the door 2 cases (weighted according to the chances that the particular door would be chosen).  This is exactly what you would do to get the probability of winning given that that either door was opened.  Nijdam says that the case where one (unspecified) of the doors was opened should not be called conditional because there is no condition other than the game rules.  Perhaps he would like to comment.  I would also say that there is a good case for calling this unconditional as none of the probabilities for the event in the sample space are changed by the host action, but I am no expert.


 * All I was saying when I talked about resolving the issue was that, rather than fight this out, perhaps we could get some other opinions on the matter. Martin Hogbin (talk) 22:23, 22 March 2009 (UTC)


 * I'll wait for Nijdam to reply, but can you clarify what you mean by "the probability of winning given that either door was opened"? Your use of "either" here makes this somewhat ambiguous.  Do you mean the probability of winning given the player picks door i and the rule that the host will open a door revealing a goat (which is the unconditional probability of winning given the rules and that the player picks door i)?  Or, do you mean the general solution for the probability of winning given the player picked door i and the host has opened door j (for some i, j - which I think makes this the conditional probability given the player picked door i and the host opened door j).  Do you see the difference between these, and (if so) which one do you mean?


 * I mean the latter, the the host has already opened an unspecified door consistent with the game rules as described above. I do see the difference but in this particular case it does not affect the probability of winning by switching. Martin Hogbin (talk) 09:54, 23 March 2009 (UTC)


 * BTW - I think it's obvious Morgan et al. intend the solution for the specific case of door 1 and door 3 to be taken as the general solution for player picks door i and host opens door j. One thing that makes this obvious is that their solution ends up expressed as a formula, 1/(1+q), not involving door numbers (and, yes, q is literally defined as probabilities involving the host's preference for door 3).  It is also common when discussing conditional probabilities to use a specific case as representative of all similar cases - consistent with the "say No. x" wording in the problem statement. -- Rick Block (talk) 23:45, 22 March 2009 (UTC)


 * You are entitled to your opinion as to what Morgan mean as I am to mine but they are both opinions, let us see what others think. You point out the flaw in your own argument.  If we do not specify which door has been opened then we cannot define what q means.  Once we specify a door we can give some meaning to q. Note that q (or p)  must be the probability that the host will open a specified door.


 * Let me ask you this question. In the case where the host has opened an unspecified door, what is the probability that the host will win by switching.  You say 1/(1+q), where q=1-p.  Where p is what?  You said earlier that p in that case would be the probability that the host would open the door he did.  After he has opened a door that probability is clearly 1.  Martin Hogbin (talk) 09:54, 23 March 2009 (UTC)


 * p is the host's preference for the unopened door (door 2 in the problem as stated). If the probability of winning by switching is 2/3 for all i,j (where the player picks door i and host opens door j), how can it be something different for i=1 and j=3?  If you want the general probability for any i,j take out Morgan's paper, and replace "door 1" with "door i" and "door 3" with "door j" anywhere you see door 1 and door 3.  q becomes the host's preference for door j.  You've previously said you're a physicist - so surely you understand the meaning of using a variable.  If you object to naming the door the host opens "door j" then you're not talking about a specific case where the host has opened a particular door - you're instead talking about all cases where the player picks door i.  You can't be both specific and unspecific at the same time. -- Rick Block (talk) 13:48, 23 March 2009 (UTC)


 * Yes I do understand what variable is, which is why I can say that the host cannot have a preference for door j, where j is a variable. This is a statement with no meaning.  If the host actually has a preference, then it must be for one identifiable door or another, it cannot be for door j.  Remember we are referring to the host's strategy, not the host's action, so we could ask the host before the game, what their strategy would be should the player choose the door that the car is behind.  The host could answer any of, 'I will choose randomly from the other two doors', 'I will always open door 3 if I can', 'I will always open the center door if I can', or I will always open door 3 if the car is behind door 1', but the host cannot answer, 'I will alway open door i'. Martin Hogbin (talk) 19:01, 23 March 2009 (UTC)

(outindent) I think we need to stop trying to describe this in words. Are you talking about:


 * 1) $$P(\text{win by switching})\,$$

or


 * 2) $$P(\text{win by switching} | \text{player picks door i})\,$$

or


 * 3) $$P(\text{win by switching} | \text{player picks door i and host opens door j})\,$$

or (perhaps)


 * 4) $$P(\text{win by switching} | \text{player picks door i and (host opens door j or host opens door k)})\,$$


 * I am talking about #4. Martin Hogbin (talk) 09:34, 25 March 2009 (UTC)

where in the Parade version of the problem i=1 and j=3 (and k=2). What Nijdam is saying above is that #4 is equivalent to #2. What I'm saying is that if the player has made an initial pick and is now standing in front of two closed doors and one door the host has opened, we can call the player's pick door i and the door the host opened door j, meaning we're talking about #3. Morgan et al. show the probability for #3 (specifically in the case i=1 and j=3) is 1/(1+p) where p is the host's preference for door 3.


 * Agreed. Martin Hogbin (talk) 09:34, 25 March 2009 (UTC)

This obviously generalizes to player pick i and host door j where the probability p refers to the host's preference for door j. -- Rick Block (talk) 03:30, 24 March 2009 (UTC)


 * No it does not, but I am having difficulty at present explaining why. Martin Hogbin (talk) 09:34, 25 March 2009 (UTC)


 * Do we mean something different by generalize? What I mean by #3 (player pick i and host opens j) is simply shorthand for each of the following cases.
 * Player picks door 1 and host opens door 3 (and, in this case, q is the host's preference for door 3) - this is the specific Morgan et al. case
 * Player picks door 1 and host opens door 2 (and, in this case, q is the host's preference for door 2)
 * Player picks door 2 and host opens door 3 (and, in this case, q is the host's preference for door 3)
 * Player picks door 2 and host opens door 1 (and, in this case, q is the host's preference for door 1)
 * Player picks door 3 and host opens door 2 (and, in this case, q is the host's preference for door 2)
 * Player picks door 3 and host opens door 1 (and, in this case, q is the host's preference for door 1)
 * What q is in each case may vary. We could talk about q13 to mean (for example) the q in the first case above and have 6 different q's - but they're all variables between 0 and 1.  In their paper Morgan et al. literally address the first case above, but by substituting "door 2" everywhere they say "door 3" (and vice versa) doesn't this turn their paper into an analysis of the second case above?  My assertion is this means Morgan et al.'s analysis generalizes to show the answer in each case is 1/(1+q), for some q between 0 and 1.  This is what I mean by generalize.  -- Rick Block (talk) 14:07, 25 March 2009 (UTC)

I rest my case until Martin and I have finished our analysis in mathematical terms on the Analysis page. Nijdam (talk) 10:17, 25 March 2009 (UTC)


 * Rick your analysis makes no sense. The parameter q represents the host's preference for a particular door.  This is a fixed preference, representing the host strategy.  To take your first two examples:


 * Player picks door 1 and host opens door 3 (and, in this case, q is the host's preference for door 3)
 * Player picks door 1 and host opens door 2 (and, in this case, q is the host's preference for door 2)


 * The host cannot have a preference of q for door 3 and a preference of q for door 2. This is a pre-existing preference, which exists before the player chooses a door and certainly before the host opens a door.  After the host has opened a door his probability of opening that particular door is clearly 1.


 * To show this, we could consider the example of the malevolent host. With the normal rules (including player has chosen randomly), let us say that the player picks door 1 and the car is behind door 1 (the only case where the host has a choice).  The host wants the player to do as badly as possible, what action does he take?  There is nothing he can do to change the players chance of winning by switching. Martin Hogbin (talk) 09:24, 27 March 2009 (UTC)


 * Martin - did you read the part about specifying these as individually named parameters, e.g. q13? If you'd like we can say the probability of winning by switching is one of 1/(1+q13), 1/(1+q12), etc. for the six different host preferences.  Every one of these is of the general form 1/(1+q) where q is a variable that has a value between 0 and 1.  The answer is always some unknown value between 1/2 and 1.


 * Yes I did read that bit, note that p=1-q in Morgan's example. These are fixed probabilities for the case where the player initially picks door 1 and the host could write them down on a card at the start of the game (the player could have cards for the cases that the player pick door 2 or door 3 but I agree that this is not important).  After the player has chosen a door the host can then look to see where the car is and then using the probability written on his card open a door.


 * In your malevolent host example P(switching wins|player picked the car) is 0, regardless of what the host does. I think what you're really suggesting is that the host try to minimize P(switching wins|player picked door i) for some i (like, say, door 1).  This is always 2/3, regardless of the host's preference so a player who decides to switch before the host opens a door wins 2/3 of the time.  However, if the host has a known strategy this can help a player who gets to decide after the host opens a door, i.e. affects P(switching wins|player picked door i and host opened door j) - which I think is the situation described in the problem.  If you think nothing the host does can change the player's 2/3 chance of winning by switching, the real money offer I made to Glkanter above still stands and I'd be happy to play this game with you as well if you're willing.  -- Rick Block (talk) 14:02, 27 March 2009 (UTC)


 * What you say is true but it changes nothing. One time in three the player will have chosen the car and will lose by switching, whatever the host does and whenever the player makes the choice of what do do.  Two times out of three the host will only have one door to choose and the player will win by switching to the only remaining door.  The host can do nothing to change this.


 * I am not sure what you offered Glkanter but here is what I would offer you. The 'car' is placed randomly behind one of three doors.  I will pick a door (I will always pick door 1, unless you have a problem with that, in which case I will pick randomly).  You must then always tell me the number of a loosing door that I have not picked (2 or 3 if I have picked 1).  You can use any strategy you like to do this. I have the option of switching to the remaining door or sticking to my original door. If I win I will pay you 6 units, if I loose you pay me 5 units. If you want to play this for real money I am game. Martin Hogbin (talk) 20:29, 27 March 2009 (UTC)


 * Again, what the host is not changing is P(win by switching|player picks door i), which is not the question we're trying to answer. Your offer also addresses this probability.  This probability is 2/3.  We all agree about that.  The other probability is the one where the player can decide after the host opens a door - with knowledge of what door the host has opened.  My offer to Glkanter is described in the section immediately above, I'll repeat it here - see section 'Game challenge'.