Talk:Monty Hall problem/Arguments/Archive 13

Half Balls problem
To highlight the difference between conditioning on the identity of the goats and the properties of the goats.
 * There are three hemispheres flat side down on a table under a cloth such that their color is concealed but their location is not.
 * The curved side of each hemisphere is painted one of 2 colors (Red, Black).
 * Hemisphere1 has a 1 painted on the flat side,
 * Hemisphere2 has a 2 painted on the flat side,
 * Hemisphere3 has a 3 painted on the flat side.
 * Before the game The host randomises which of 1, 2 or 3 will be the winning hemisphere.
 * The host also randomises the colors of the hemispheres.
 * At the start of the game the contestant picks one of the three hemispheres under the cloth.
 * Then the host removes one of the other two that is not the winning one, in doing so revealing its color.
 * The contestant may then stick or switch between the remaining two hemispheres and if they end up on the hemisphere corresponding to the random winning number, they win a car.

The odds of winning are 2/3 by switching. However lets look at the possible conditional probabilities. In all cases we're looking at the probability of winning by switching and in all cases there is insufficient information for them to affect the final value of 2/3


 * "Given the contestant selects the leftmost hemisphere". The contestant could have picked any 1 of 3 and can identify which one he picked.
 * "Given the host reveals the rightmost hemisphere". The host can theoretically pick between the remaining two and the contestant can always tell which one he picked.
 * "Given the revealed hemisphere is Red". The color could have been red or black and the contestant can easily tell which it is.

Now for some that are based on identities the contestant can't know.


 * "Given the contestant picked the winning hemisphere" While it trivially makes the probability 0 and is objectively true or false in every possible game, the contestant can never identify it prior to making the decision so it cannot be informative.
 * "Given the host reveals Hemisphere3" Again objectively true in every game. However the contestant never finds out the ball numbers, they're only ever available to the host. To the players POV the game is identical if the numbers on the bottoms of the hemispheres are switched or even removed.

The point I'm making about the goat's identities is the same I've just made about the numbers on the hemispheres. The identities of the goats, like the numbers of the hemispheres do not exist in the contestant's sample space.

From the hosts POV there are 144 different set ups for the above game. 6 permutations of number in position, 8 permutations of color and 3 permutations of winning location/hemisphere. In each of these the contestant can pick 3 so that's 432 permutations. In 1/3 games the host has a choice of ball to reveal giving a total of 576 possible games.

From the contestants POV there are fewer. 3 winning ball locations * 3 chosen hemisphere locations is 9 in 1/3 of those 9 the host has a choice of ball to reveal, so 12. The revealed ball can be two colors so a total of 24 permutations. This number is so much lower because of the missing information the player has.


 * You can condition on the color of the hemisphere
 * You cannot condition on the number of the hemisphere
 * In MHP you can condition on the properties of the goat
 * In MHP you cannot condition on the identity of the goat

Is that any clearer? SPACKlick (talk) 14:54, 28 August 2014 (UTC)


 * I am not sure how objects have identities that are not based in some way on their properties. Bosons have identical properties and are therefore regarded as fundamentally indistinguishable.  For objects to be distinguishable they must have at least one property that varies between different objects.


 * Yes. You seem to be assuming that the property of having a number written on it is somehow what gives an object its identity.  The identity of an object is determined bt the sum of all its properties.  For ordinary objects these remain fixed (goats being living are a little different but I assume that this is not what you are your basing your argument on).


 * In your example you have arranged for the player not to be able to distinguish between two hemispheres on the basis of the property perceptible to him (colour). If you were to say that the player is unable to distinguish between two of the hemispheres during the game but the host is able to do so I would agree with you.


 * In the case of the goats there will also be some properties of the goats that the host can distinguish but the player cannot (their names for example). For the player to (non-trivially) revise the chances of winning by switching after he has seen a goat he must be able to percieve the properties of the goats that the host uses to decide which to reveal.  I agree that he might not always be able to do this, but he might be able to glean something, and that is my point.  By having a fixed goat policy the host breaks the rule that he must never do anything that might give the player a clue, however small, as to where the car is.  It is Gerhard's principle again.  Martin Hogbin (talk) 09:43, 29 August 2014 (UTC)


 * I am not sure how objects have identities that are not based in some way on their properties. It is related to their properties but it is unique. An object may change colour, but remain the same object. An object may change size, shape, location, mass, charge or any number of properties and remain the same object and no other object may become that object.
 * You seem to be asserting some philosphical concept that all objects have some kind of immutable 'essence' or 'soul'. There is no evidence for this, neither is it a widely held belief as far as I know. One of your hemispheres could be continuously transmuted into a different one.  We could start by painting it a different colour, then we could change the number on its base, finally we could swap each atom, one by one, with one from a different hemisphere.  At what point, if any, would it become a different hemisphere?


 * Luckily, in probability, we do not need to address philosophical problems of this nature. What matters is our knowledge, and the exact question that we are asking.


 * Suppose in your example, we know there were two black and one red hemispheres. The two black hemispheres are numbered 1 and 2 and the red one 3.  What is the probability that you will pick the red hemisphere?   We would agree 1/3.  Now we paint number 2 red, muddle them up and pick again.  What is the probability now?


 * The answer depends entirely on the exact question that we ask. Do we ask for the probability of picking the hemisphere that was originally read, or of picking number 3, or do we ask for the probability of picking a red hemisphere or a hemisphere that is red.


 * If we change the hemispheres in more ways and more subtly the answer to our question still depends only on exactly what question we ask. We have no need to concern ourselves with supposed immutable identities of objects. Martin Hogbin (talk) 09:18, 30 August 2014 (UTC)


 * I agree that the answers depend on what question we ask. This persistent identity of objects exists only is as long as they satisy the definition of the object and yes it is an in depth philosophical concept which we don't need to fully address. What I'm trying to express, albeit ham-fistedly is the difference between the persistent identity of the goats, such as may be labelled "The goat Albie" or "Goat1 And the temporary nature of the label "The revealed goat". The first of which by its nature exists throughout the problem the second of which exists only once the action of revealing the goat exists and could potentially apply to either of the two goats on any given run&#32;SPACKlick (talk) 22:37, 30 August 2014 (UTC)
 * Edited to add, as far as transmuting from one hemisphere to another, or one goat to another. Surely we can agree that the object, known the the host, as the goat "Albie" or the one known to the host as hemisphere2 persists throughout the puzzle?&#32;SPACKlick (talk) 22:39, 30 August 2014 (UTC)


 * You see to be assuming that the property of having a number written on it is somehow what gives an object its identity. Actually I was using that property as a marker of identity but that's by the by, identity is merely a special property. Only one object can have any given identity property making it intrinsically unique. Hemisphere3 is still Hemisphere3 if you scratch the number off, paint it yellow and blue polka dot, move it to another studio. It will always be Hemisphere3 and nothing else will ever be Hemisphere3. The goats don't have such an obvious label for their identity. The best we can do is name them or give them a label and explicitly tie that label to their identity. So "Albie" or "Goat1"
 * It is not just the best that we can do, it is the only thing that we can do. We can label objects or we can refer to their properties (internal and external) but I am cannotsee what other kind of identity they might have.  Luckily, in probability, this does not matter. Martin Hogbin (talk) 09:55, 30 August 2014 (UTC)
 * It entirely does matter. There are properties which uniquely identify the object. "The door on the left of the three doors" or (assuming differing names for both goats) "The goat named albie". There are properties which do not uniquely identify objects "A brown door", "A white goat". It's the labels with unique referents that I'm using as identifying labels. You seem to also want to use the second set as uniquely identifying labels.&#32;SPACKlick (talk) 22:37, 30 August 2014 (UTC)


 * In your example you have arranged for the player not to be able to distinguish between two hemispheres on the basis of the property perceptible to him (colour). True, although the argument would remain the same if they were all differing colours and the player didn't know which. It doesn't matter whether the player could distinguish between them if he saw all of them because the player doesn't see all of them. Even If the player saw all of them he still couldn't condition on Hemisphere1 being chosen.
 * You can label objects by their role on a specific instance of a game. And, mathematically, you can condition a sample space on that basis.  Whether this is of any value depends on your state of knowledge. Martin Hogbin (talk) 09:58, 30 August 2014 (UTC)
 * You can but thinking through the process. You have decided to label whichever goat is revealed, "The revealed goat" so you would then be conditioning on "the revealed goat" being revealed. But you knew this would be the case before either goat is revealed so it doesn't limit the sample space at all. &#32;SPACKlick (talk) 22:37, 30 August 2014 (UTC)
 * There is no problem in labelling an object using its role in a specific instance of the game.  In the frequentist model of probability we have to divide things that happen into two types: those that are fixed for every repetition of the game and those which change each time we play it.  (This in my opinion does represent a potential weakness of this model but that is a discussion for another time).  In the MHP, the general assumption (I think this is generally agreed) is that there are always two goats, one car, and three doors and that the rules of the game are fixed.  I am working on the basis that that the doors and goats always remain the same, even though in a real game it is posssible for the props to change from time to time as the show evolves.  For puzzles of this nature these would certainly be the normal assumptions.


 * For initial placement of the car and goats, the player's initial choice, and the host's choice we must either be given some probaility distribution or we must take the distribution to be uniform at random. If we take these as fixed the frequentist model of probability becomes pointless.  As we are given no information in the problem statement on these distributions we should take them all as random.


 * When we observe a single game, we see a specific door opened and a specific goat revealed. We can note the properties of these items and could, if we were allowed to, attach physical labels to them.  The probability of winning by switching for a single game is modelled by repeating the same game.  Clearly we do not repeat exactly the same game, with the same car and goat replacement, the same initial door choice and the same host choice.  These are changed randomly every time.   When the game is repeated, it is possible that different doors will be involved and a different goat revealed.  In these conceptual games, I accept that although the player would see the other goat, in the instance of the game in question the player only sees one goat and does not know what properties the other goat has, however we do know that the other goat could be revealed and we know some of the properties of one of the goats.  Martin Hogbin (talk) 08:57, 3 September 2014 (UTC)


 * Agreed to all of the above. However, the player doesn't know that he will be able to tell the events in the sample space where the same goat is revealed from the events in the sample space where a different goat is revealed so the sample space doesn't divide by the specific goats it divdes only by the properties of the revealed goat. Using the tree formulation the two branches separating at the relevant node would be "a short white goat is revealed, it is one of the two goats" and "a not short white goat is revealed it is not the goat that was revealed this time" and the contestant can then meaningfully condition on being on the first branch. You can structure the sample space to divide into "the goat that was revealed this time is revealed" and "the other goat is revealed" however the contestant cannot meaningfully condition here because in any other run of the game while he can know he's on the second branch in cases where the other goat is distinct enough to tell apart he can't know he's on the first branch. &#32;SPACKlick (talk) 09:18, 3 September 2014 (UTC)
 * I agree that the player may not be able to distinguish between events involving the goat that he saw from events involving the other goat and that in the case of the doors he probably will be able to distinguish between events involving that door that he saw opened from events involving the remaining door but conceptually he can see a difference.
 * Below we talked about the question that was being asked. These are two questions that I want to answer.


 * 1) Taking the problem as stated and only considering choices mentioned in the question and making the standard Bayesian inferences about the unknown distributions, what is the probability of winning by switching. My answer, your answer, and Morgan's final answer: 2/3.


 * Now for my second question I agree that I am getting little perverse but that is in response to what I see as the perversity of Morgan's original solution.  This is what I think they were trying to answer:


 * 2) Taking the problem as stated and only considering choices mentioned in the question, making the standard Bayesian inferences about the original car and goat placement but assuming that we have some (unspecified) extra information about the host's preferences what is the probability of winning by switching.


 * Their answer considered only the host's door preference but I believe it should include the host's goat preference. Note that the unspecified information mentioned above could be a general preference, for example 'The host prefers to open the highest numbered door', or 'The host prefers to open the lighter coloured goat', or it could be morse specific information like, 'The host always opens door 3 when he can', or even, 'The host always reveals that goat that he did do this time when posssible'.  I do not distinguish between these different types of information because Morgand did not do so in their paper. Martin Hogbin (talk) 10:27, 3 September 2014 (UTC)


 * For the player to (non-trivially) revise the chances of winning ... he must be able to percieve the properties of the goats that the host uses to decide which to reveal No, he can revise his assesment of probability by any and all properties that he can percieve about the goat and his understanding of how likely they are to affect the hosts decision. What the host actually uses is irrelevant.
 * Yes, it is the players knowledge of the host's preference that counts. It is usual in these problems to assume a rational player who either know something about the host's preferences (and knows that he knows it) or does not know anything (and knows this) and therefore makes a Bayesian assumption that probability is equally distributed betweeen the posssible options. Martin Hogbin (talk) 10:26, 30 August 2014 (UTC)
 * I would say the second, does not know anything (and knows this) is the only sensible option for the MHP&#32;SPACKlick (talk) 22:37, 30 August 2014 (UTC)
 * I course, so would I but that applies equally to the doors. Martin Hogbin (talk) 10:29, 3 September 2014 (UTC)
 * I agree that he might not always be able to do this, but he might be able to glean something, and that is my point. No, he never can glean anything. He can know the goat is one of the set of goats with property "white". He can know the goat is one of the set of goats with property "alive" and so on until we have listed every property of the goat. This subset of potential goats is still greater than 1 (in fact i believe it's infinite but that's neither here nor there) and therefore the goat hasn't been uniquely identified. So the contestant can only condition on the properties of the goat not on the identity of the specific goat.SPACKlick (talk) 09:59, 29 August 2014 (UTC)
 * If you want to say that it is fine with me. I do not see any meaning to the word 'identity' except as the collection of properties of an object.  In any case, identity as such is, irrelevant.  I do not make any claims about identity.  My argument has always been that, on the event that a goat is revealed (meaning that some of it properties are discernable to the player) the player may be able to revise his estimate of the probability of winning by switching if he has some information (not of any specified kind) about the host's goat preference.  In the case of the door opened by the host the player may be able to revise his estimate of the probability of winning by switching if he has some information (not of any specified kind) about the host's door preference.  Note that 'preference' refers to the properties of the object that the player knows the host uses to make his choice. In the actual problem the player knows nothing in both cases.  In both cases the player can mathematically condition on the host's choice (using functional labels) and in both cases this does not change anything. Martin Hogbin (talk) 10:26, 30 August 2014 (UTC)


 * Regarding identity, the goat revealed in a specific instance of a game can be labelled by the player. By your own argument it retains its immutable identity (maybe goats do have souls) and mathematically it is possible to condition using this label.  How valuable that is depends on what else you know and what is being asked. Martin Hogbin (talk) 10:26, 30 August 2014 (UTC)
 * No we disagree here. There are three sets of labelling under consideration.
 * At the start of the game we know a goat will be revealed. We choose to label that goat "The revealed goat". This label uniquely identifies one goat in any instance of the game but could potentially apply to many goats. We cannot condition on "The revealed goat being revealed", because in every instance of the game that would be the label applied to the goat that was revealed.
 * Yes, I agree that we know a goat will be revealed but we do not know which one. We watch just one game and we see a goat, we can label this,' the goat that I saw in the game that I watched '.  When we calculate a probability, we must not just take account of what did happen but also of what might have happened.  This is the basis of probability and, in frequentist probability, it is modelled by repeating the same experiment, where a different goat might be revealed.


 * At the point of revelation we learn some properties of the goat that is revealed. For example it is "a White Goat". This label does not uniquely identify a goat. We can condition on "a White Goat being revealed". This condition would refer to all branches of a sample space tree where at least one goat is white and it is revealed, including those branches where both goats are white.
 * I do not argue with this. I agree that all we see are some features of one goat and that his does not guarantee that we can uniquely identify it.  The question the player must ask is this, 'What is the probability that, given the choice, the host opens a door with the same features that I saw opened (Arbitrarily labelled 1 by vS'.  This does not necessarilly mean that it has a figure '1' visisble to the player on it) maybe it is the leftmost door, or the tallest door.  The player should also ask what is the probability that, given the choice, the host reveals a goat with the features that I saw, maybe a large white goat with horns.  If the player know nothing about the host's pereferences he can only take the probability to be 1/2 in each case.  If he does obtain some information about the host's probability he may be able to revise his probabilities, depending on exactly what information he has. Martin Hogbin (talk) 15:15, 3 September 2014 (UTC)
 * At the start of the game there are two uniquely identifiable goats, say "Goat1" and "Goat2". These labels uniquely identify one goat each. We cannot condition on "Goat1 being revealed" because the player never has the information which of the two goats it was.
 * There is no way of labelling a unique goat that we can condition on. I suspect where I said Identity I should have been referring to persistent uniquely identifying labels.&#32;SPACKlick (talk) 22:37, 30 August 2014 (UTC)
 * Yes there is, 'the goat that was revealed in the game that I saw'. Martin Hogbin (talk) 15:19, 3 September 2014 (UTC)
 * But, Martin, nothing in the problem statement guarantees you will recognise this goat the next time. Nijdam (talk) 14:42, 11 September 2014 (UTC)
 * I do not need a guarantee. When I see a goat the next time, if the probability (from my state of knowledge) that the goat is the same as the last one is not exactly 1/2, I can use that information to revise my probability of winning by switching.  Even if the goats look identical and I cannot tell which is which, so the probability that I have got the same goat is 1/2, I still need to include this in my probability calculation.  I know that there are two goats and that I have seen one of them, I therefore need to do a conditional probability calculation using this fact.  Of course, in the question as stated, we know nothing about the host's preferences so there is no new information of any value, just as with the doors. Martin Hogbin (talk) 20:34, 11 September 2014 (UTC)

Applying Bayes' rule
Imagine completely unconditional scenario so far. The player has chosen door 1 and is told that the host is going to open a door and reveal a goat. He calculates the probability that the car is behind his originally chosen door (PC=1) to be 1/3.

Now if he sees the host opening door 3, he can apply Bayes' rule to revise his original estimate. He must now calculates P(C=1|H=3)

P(C=1|H=3) = P(C=1) * ( P(H=3|C=1) / P(H=3)) = P(C=1) * (P(H=3|C=1) / ( P(H=3|C=1) * P(C=1) + P(H=3|C=2) * P(C=2) +  P(H=3|C=3) * P(C=3))

Assuming P(C=1)=P(C=2)=P(C=3)=1/3

P( C=1|H=3) = 1/3 * P(H=3|C=1) / ( 1/3 * ( P(H=3|C=1) + 1/3 * 1 + 1/3 * 0) = P(H=3|C=1) / ( P(H=3|C=1) +1 )   Morgan's solution

Suppose the host always opens door 3 when the player has chosen door 1 and car is behind door 1. What is P(H=3|C=1) from the players perspective in the following two cases, using only what we are told in the question:

1 The player does not know that the host always opens door 3 when the player has chosen door 1 and car is behind door 1? My answer: 1/2

2 The player knows that the host always opens door 3 when the player has chosen door 1 and car is behind door 1? My answer: Not 1 ! Martin Hogbin (talk) 21:46, 11 September 2014 (UTC)

Simple solution
The simple solution, which is not a solution, reads: the probability to hit the car at the first choice is 1/3, the door opend by the host has probability 0 on the car, hence the remaining closed door must have probabilty 2/3 to hide the car.Nijdam (talk) 19:05, 4 September 2014 (UTC)
 * I add Morgan's solution F2, which is what Martin calls 'simple solution'. It is essentially the same as I mentioned here. Sample space {AGG, GAG, GGA}, where for instance AGG means car behind door 1, goats behind doors 2 and 3. If the player initially picks door 1, she will get the car when switching in 2 of 3 cases. Nijdam (talk) 11:53, 9 September 2014 (UTC)
 * I don't care if anyone agrees, this is the simple solution as it is presented in the literature, as it is offered by vos Savant and a lot of others, and subject of many discussions. Nijdam (talk) 20:04, 6 September 2014 (UTC)
 * No it isn't. Let's look at Vos Savant's solution. As I can (and will) do this regardless of what you’ve chosen, we’ve learned nothing to allow us to revise the odds So Marylin is actually stating we learn nothing from which goat is revealed. The original simple solution is ambiguous as to whether it says "Always switching wins 2/3 times" or "Whatever the host does you have a 2/3 chance of switching" and in Marylin's column she isn't rigorous about separating them. However she makes clear at several points that the reason that the odds don't change is that the revealed goat doesn't give us new information about the location of the car. It's not because there's 0 chance of it being the car. Cecil at the Straight dope says Because (1) the chances were 51 in 52 that the ace was in the dealer's stack, and (2) the dealer then systematically eliminated all (or most) of the wrong choices. Here he's clearly saying more than "the cards turned over by the host have probability 0 on the Ace". Devlin puts it beautifully showing the information the player is using to calculate the odds. (1) The probability that the prize is behind door B or C (i.e., not behind door A) is 2/3. (2) The prize is not behind door C. The simple solutions relies on the fact that the revealed goat gives you no information. — Preceding unsigned comment added by SPACKlick (talk • contribs) 10:29, 10 September 2014‎
 * Well, MvS as well as Devlin are beautifully … mistaken. Nijdam (talk) 14:39, 11 September 2014 (UTC)
 * Sorry are you saying they're mistaken about what the simple solution is or mistaken about the simple solution being an accurate solution because my response to both is to disagree but in entirely different ways. SPACKlick (talk) 14:44, 11 September 2014 (UTC)
 * They both gave the simple "solution" (or an equivalent formulation) and consider it to be a (correct) solution. Nijdam (talk) 20:49, 11 September 2014 (UTC)
 * Do you agree that their version of the correct solution contains the additional claim, about the lack of information from a revealed goat that your version didn't give? And if so, could you explain why you feel the solution is incorrect? SPACKlick (talk) 20:56, 11 September 2014 (UTC)
 * I've no idea what you mean by the additional claim. As far as the argumentation of MvS and also Devlin concerns, their way of reasoning just is not correct. Nothing to do with my feelings. The best thing is, if you will understand, to formulate any argumentation in correct mathematical formulas, because words are sometimes misleading. Nijdam (talk) 08:45, 13 September 2014 (UTC)
 * When you're intrested, answer my questions. What are the probabilities p1, p2 and p3 for the doors to hide the car? Nijdam (talk) 08:49, 13 September 2014 (UTC)

Ok, we'll give this a go. Given the lack of information the player has about the distribution, p1=1/3, p2=1/3, p3=1/3. SPACKlick (talk) 10:11, 13 September 2014 (UTC)
 * The player first point to door 1 and the host opens door 3 showing a goat. What are the probabilities for the doors to hide the car? Nijdam (talk) 10:17, 14 September 2014 (UTC)
 * Given the lack of any additional information p1 = 1/3 p2=2/3 p3=0. SPACKlick (talk) 07:13, 15 September 2014 (UTC)
 * That's quite inconsistent with what you just wrote above, don't you think? Nijdam (talk) 11:50, 15 September 2014 (UTC)
 * Not in the slightest (although I did use the unstated assumption that the standard rules apply host is obliged to open a door that the contestant hasn't picked and conceals a goat). Could you explain what you feel is inconsistent, or provide an argument, or provide some maths to back up anything you're claiming? SPACKlick (talk) 11:55, 15 September 2014 (UTC)
 * Well, tell me the value of p3. Nijdam (talk) 17:08, 15 September 2014 (UTC)
 * Where p3 is the probability of the car being behind the door the host opened p3=0. What is your point? SPACKlick (talk) 17:49, 15 September 2014 (UTC)
 * Quite peculiar, than just some lines above you wrote: p3=1/3. So, what is it? Nijdam (talk) 09:18, 16 September 2014 (UTC)
 * You'll note p3 refers to two different things in the two cases. p3 in the first case is the probability of the car being behind some specified door at the start of the game. p3 in the second case is the probability of the car being behind the door the host has opened at the point of decision. Just because you have chosen to label them the same doesn't mean their different answers conflict. SPACKlick (talk) 10:04, 16 September 2014 (UTC)
 * Wouldn't it be less comnfusing, and mathematically required, to use different symbols for different quantities?Nijdam (talk) 16:32, 16 September 2014 (UTC)
 * That would be a great idea. For example using P(3t0) Probability the car is behind door 3 initially and P(Ht2) for probability the car is behind the door the host opened at the point of decision or more relevantly P(Ct2) for probability the care is behind the chosen door at the point of decision and P(Rt2) for Probability the car is behind the remaining door at the point of decision. Using functional labels for the doors because the specific numbers collapse under symmetry. SPACKlick (talk) 05:49, 17 September 2014 (UTC)

For the moment I make it easy, and use a 'q' in the situation after the host has opened door 3: q3=0. But now what about q1 and q2? Nijdam (talk) 08:59, 17 September 2014 (UTC)
 * That depends entirely on what you mean by q1 and q2, if they are equivalent to P(Ct2) and P(Rt2) above then they equal 1/3 and 2/3 respectively. SPACKlick (talk) 09:03, 17 September 2014 (UTC)
 * And right they are, and what's more, the numerical values you mention are (with reasonable assumptions) correct. The only thing is, I've no idea how you came to these values.Nijdam (talk) 11:28, 17 September 2014 (UTC)
 * Then allow me to demonstrate.
 * The first event in the puzzle is the selection of a door, given that the contestant posesses identical information about all the doors prior to this event each door is equivalent/symmetric and so the particular door can be ignored. The relevant sample space division is into those cases (A)where the contestant picked a Car and {B} where the contestant picked a Goat. Because in case (A) switching at the end will lose and in case (B) switching at the end will win. Because there are two goats and one car and it is a free choice across an unknown distribution the odds are 1:2 (B):(A)
 * The second event in the puzzle is the host opening some specific door to reveal some specific goat. The 2 remaining doors are symmetric and the goats are symmetric for the same reason as above (the contestant possesses identical information about them immediately prior the event) therefore the specific door selected and specific goat selected provide no new information and thus the distribution of (B) and (A) is unchanged.
 * From this we see p1 = 1/3, p2=2/3 and p3=0 SPACKlick (talk) 11:50, 17 September 2014 (UTC)
 * Well, if you had stuck to my line of questioning, you would have noticed the contestant picked door 1, and the host opened door 3. Both events are not represented in your sample space. Furthermore I guess you mean to say that q1=1/3 and q2=2/3. However, how this follows from your sample space, puzzles me. Nijdam (talk) 12:08, 17 September 2014 (UTC)
 * Sorry I apologise, both terminologies were used above and I used the wrong one here. It should indeed have been q1 and q2.
 * I have stuck to your line of questioning, you will notice that above I have shown that specific door numbers make no difference within the sample space, as specific door numbers interact only with self contained information, and hence do not need to be included in the reasoning. For all 6 combinations of the three doors the above sample space is identical.
 * How the probabilities follows is that
 * The probability the car is behind the chosen door after the host has opened a door with a goat behind it is the same as the probability of being in Scenario (A), defined as having chosen a door with a car behind it, after the host has opened a door with a goat behind it thus q1 = 1/3.
 * The probability of the car being behind the remaining door after the host has opened a door is the same as the probability of being in scenario (B), which is defined as having chosen a door with a goat behind it, after the host has opened a door thus q2 = 2/3
 * there are various lines of reasoning for q3=0 the simplest of which is the car must be behind one and only one door thus q1+q2+q3=1 --> q3 = 1-q1-q2 = 1-1/3-2/3 = 0
 * See now? SPACKlick (talk) 12:22, 17 September 2014 (UTC)
 * The point is you derive the probabilities asked for by me, through a sample space that's not appropriate for the situation. I have difficulty in understanding what you mean by your scenario's A and B. So let's concentrate on q1 and q2. You do not have to show q3=0, that's a given fact. Hence q1+q2=1, so calculating one of them is sufficient. Please show me how you reason in finding, let's say, q1=1/3. Nijdam (talk) 13:30, 17 September 2014 (UTC)
 * q1, as you said above is equivalent to P(Ct2), which was defined as "probability the car is behind the chosen door, at the point of decision". The sample space above is appropriate because it is complete for the given scenario. There are only two outcomes (because the majority of data is informationally isolated). I'm sorry if you can't understand the above, feel free to query any points you don't think apply but the conclusion is sound. The sample space is two state and no information between the start, t0, and the decision, t2, gives any indication of which case you are ("will be" at t0) in. 1/3 of cases are (A) where the chosen door concealed the car. 2/3 cases are (B) where the chosen door concealed a goat. SPACKlick (talk) 13:42, 17 September 2014 (UTC)
 * I know what I'm talking about. Your sample space {A,B} does not allow for the event "first choice door 1" etc., nor for the events "host opens door 1" etc. Maybe you may use a reduction of the appropriate sample space for ease of calculation, but such reduction is not suited to model the problem. Nijdam (talk) 14:46, 17 September 2014 (UTC)
 * Sorry, I don't need the event "Contestant chose door 1" to find the "probability the car is behind the chosen door, at the point of decision". It is simply not a factor. That said, the sample space above shows that for ALL specific door combinations q1=1/3 and q2=2/3, so it STILL covers it. [ aside; A pot contains 50 coins, If you know that for all coins in a pot the probability of heads is 1/2 and tails is 1/2 then do you need a sample space showing every coin to find the probability that when one coin is drawn and tossed the result will be heads? ]
 * If you want specifically to display all 6 arrangements of car and two goats, and all 3 chosen doors and all possible opened doors then the 2 branch tree devolves into 24 branches where 12 of them are case (A) branches and 12 of them are case (B) branches differing only by the labels applied to the goats and the doors. And All (B) branches are twice as likely as all (A) branches at their terminus. And for any set of doors chosen, doors opened and goats revealed as a given q1=1/3, q2=2/3. SPACKlick (talk) 15:02, 17 September 2014 (UTC)

One by one: Yes you do need all the events possible in the problem. If you mean by 'the sample space above' your sample space {A,B}, it says nothing about any combination whatsoever. In your second section you come more to where you have to be. And indeed then you may calculate q1, which is the conditional probability of the car being behind door 2, given the contestant has initially chosen door 1, and the host has opened door 3 wit a goat. A mouthfull, but that's what q2 is. Now for something completely different: ….. do you know the way of arguing of MvS and Devlin?Nijdam (talk) 09:34, 18 September 2014 (UTC)

—>  . . . e.g. Number XXXVIII, October 2012 or newer ones? – or an older one including host doesn't know? Gerhardvalentin (talk) 23:39, 17 September 2014 (UTC)
 * Sorry Nijdam, you're simply wrong here. It's like with the coin bowl aside. In order to calculate the probability of the coin toss being heads once the host has drawn a coin and tossed it you don't need to include every coin being drawn in the event space because all those events are identical to the players information. Similarly with the MHP you don't need to include every combination of doors in the event space because all the doors (and the goats they contain) are identical to the players information. The probability that the car is behind the chosen door, given the contestant chose door C and the host revealed Goat G from behind door H is 1/3 for all legitimate combinations of C,H&G (I specify legitimate to avoid C=H). Those specific values do not enter into any necessary calculation for the player.
 * To repeat the specifics of the argument above (again with the apology about any misused terminology)
 * 1) Before the contestant choice there is no informational difference between the three doors. Therefore the problem is symmetric/equivalent about the value of C (the door chosen by the contestant)
 * 2) As the location of the car is determined by an unknown distribution unrelated to the contestants choice, in 1/3 of cases (A) the contestant chose the door with the car behind it. in 2/3 cases (B) the contestant chose a door with a goat behind it.
 * 3) In both those cases, there is no informational distinction between the two remaining doors or the goats they conceal therefore the problem is symmetric/equivalent about the values of H (the door the host opened) and G(the goat thereby revealed).
 * 4)A) In the case where the contestant picked the car, the host opens one of the remaining doors via an unknown distribution, the car is behind the contestants door therefore probability C conceals car [P(C)] = 1 and the probability the remaining door [R] conceals the car [P(R)] = 0
 * 4)B) In the case where the contestant picked a goat, the host can only open one of the remaining doors and does so. P(C)=0, P(R)=1
 * No information above can indicate which case you are in (A) or (B) so at the point of decision it is always 1:2 you are in A:B case giving the values P(C)=1/3, P(R)=2/3 for all combinations of C,H,R and G.
 * Where doyou feel this is incomplete, specifically? SPACKlick (talk) 04:14, 18 September 2014 (UTC)
 * We seem to be back from where we started. Let me ask you: do you somehow consider the car to be randomly distributed behind the 3 doors? If your answer is 'yes', then, how do you describe this; if 'no', how then is the porition of the car distributed. Nijdam (talk) 09:34, 18 September 2014 (UTC)
 * I make no assumption about the distribution because it is unnecessary to do so. The car is distributed behind the 3 doors with an unknown distribution that is not dependent on the values of C,H,R and G. It really doesn't matter if the car is evenly distributed or always behind door 1 because the player has no knowledge of the distribution. Helpfully because of this and other similar lacks of information the doors are symmetric and specific door number can (effectively) be ignored. SPACKlick (talk) 09:45, 18 September 2014 (UTC)
 * Then what did you mean with your former answer: Given the lack of information the player has about the distribution, p1=1/3, p2=1/3, p3=1/3? Nijdam (talk) 10:20, 18 September 2014 (UTC)
 * Ok, I'm confused by this question, your quote and my previous response say almost exactly the same thing. Given that the contestant doesn't know the distribution (or any other probablistic information about the doors for that matter) each choice of door is indistinguishably likely to conceal the car therefore the probability is 1/3 that the contestants door choice will be conceal the car irrespective of the door they choose. I also note you're not quoting from my answer summary above, which I'd prefer you use as the presentation has been slightly altered by this discussion. SPACKlick (talk) 10:36, 18 September 2014 (UTC)
 * You seem to suggest I didn't quote you exactly as you've written. I just used 'copy and paste', so I wonder what you imply to suggest. In your comment above you need a lot of words trying to avoid specific door numbers to be mentioned, isn't it? Didn't you just mean to say: P(C=1)=P(C=2)=P(C=3)=1/3, where the variable C denotes the number of the door hiding the car? If not, what in the world did you mean then?Nijdam (talk) 11:17, 18 September 2014 (UTC)
 * I wasn't suggesting your quote was inaccurate merely that you quoted my first comment in the discussion and not the summary who's phrasing has been tailored to the discussion so far. I am perfectly happy with your phrasing above that "P(C=1)=P(C=2)=P(C=3)=1/3, where the variable C denotes the number of the door hiding the car" except to add that this only refers to those probabilities at t0 (At the start of the game) and t1 (between the contestant choosing the door and the host opening a door) and not t2 (after the host has opened a door). Now I ask you again to tell me where you feel there is a mistake in the summary answer detailed above at 1-4b.
 * The lot of words is to try to explain why mentioning door numbers is an irrelevance. I'm perfectly happy to talk about door numbers but the sticking point here is that you don't seem to agree to the logical step.
 * "The probability is shown to be 2:1 R:C for every individual combination of RCH" to "The probability is shown to be 2:1 R:C for one individual combination of RCH values" Unless that's not the sticking point which I've been trying to clarify for several posts now. SPACKlick (talk) 11:23, 18 September 2014 (UTC)

Okay, I'm glad you weren't suggesting I misquoted you, and I rather would not call what you wrote 'your first comment', but instead 'your answer to a very specific question of mine'! Of course these probabilities refer to the start of the game. As soon as something (an event) has happened we have to use conditional probabilities. Now I prefer to continu slowly, so explain to me how for instance the event {C=1} is found in your sample space.Nijdam (talk) 11:46, 18 September 2014 (UTC)
 * First "As soon as something (an event) has happened we have to use conditional probabilities." I fundamentally disagree, many things that happen are part of the initial information for that time point. For example "Given that you were offered a switch" is an event that occurs but would be a meaningless condition.
 * Second, Where (C=1) means all those cases where the car is behind door 1. I have not separated those in the sample space because as the sample space shows, the probability of any end position of the car is unrelated to the values of C, H and R so the sample space for C=1 identical to the sample space for all C. I've explained this many times now. I do not need to model for C=1, C=2, C=3 because it is demonstrated that the specific value of C affects only the specific value of C and not the probability that the car is behind the Chosen Door, the Opened door or the Remaining door. (please note that at several points above I have used C to mean the Chosen Door I will from now on refer to that as P for Picked).
 * Third, the modelling you are attempting to do is irrelevant to the question asked by the puzzle, which is why I have repeatedly asked you to respond directly to the 5 point answer I have given above. Please do so.SPACKlick (talk) 12:06, 18 September 2014 (UTC)
 * Apparently your'e not very familiar with probabilistic jargon. I especially added '(an event)' to indicate that 'something happening' means an event (happening). And an 'event' is not just what is commonly understood, but a (measurable) subset of the sample space, i.e. the occurence of a meaningfull 'something' in the light of the experiment. Now about (the event?) {C=1}. When you accept that in the game the car might be behind door 1, this is an event in the model, just as you said the total of all 'cases' where the car is behind door 1. The rest of your remarks about this item I do not understand. You speak about separating them, and about the end position of the car, I've no idea what you mean. Better to accept that anything that's mentioned in the description of the MHP has to be reflected in the model. I.e. the door with the car, the door initially chosen and the door opened by the host. All you mention about symmetry, may be used to simplify calculations, but cannot lead to the omission of terminology to formulate the solution. Nijdam (talk) 13:53, 18 September 2014 (UTC)
 * I'm familiar with some terminology and not others. I wasn't certain event meant the same in mathematical modelling as the philosophical logical modelling I'm more familliar with. And I'm beginning to think they're not quite the same, so again apologies for any lack of clarity from misuse of terms. C=1 would not be an event [sometimes and hereafter relevant event] in the terminology I am familiar with because the value of the property doesn't have a material effect on anything else in the problem, elaborated below. Similarly symmetry, in the form of modelling I am used, to means that where a model branches due to the variance of a property if all branches are identical, bar the value of that property, then the property is symmetric and not a relevant event.
 * Where I think we firmly disagree is that I would defend that something only needs to be included in a model if it is a relevant event, that is if its variant forms change the outcome of the model, or specifically change a material outcome. For instance, if we include the hosts Tie colors, then the model would look identical except all branches would be duplicated for each possible color of tie or set of colors of tie you chose to include, but adding that distinction would not improve the accuracy or utility of the model. The same can be said of door numbers and goat identity. Even though it is mentioned in the puzzle itself, the specific door number values of P,H&R have no effect on any part of the model except their own and eachother's values. As there is no information, anywhere within the universe of the MHP, which is dependent on door number, other than door number itself, it can be safely ignored when modelling the sample space.
 * If you don't accept the above I would ask you to justify including door numbers and not tie colors or any of the myriad other variables which have no effect beyond their own value. SPACKlick (talk) 14:26, 18 September 2014 (UTC)

why 2/3?
Moved from talk page by SPACKlick (talk) 17:13, 11 September 2014 (UTC) I guess this isn't the place, as this is a page to discuss the article, not the topic, but why does everyone say the chance is 2/3 after opening a door? There are now 2 doors, so isn't the chance out of 2? Namely 1/2 and 1/2?--Richardson mcphillips (talk) 16:13, 11 September 2014 (UTC)


 * Hi there . There are various versions of the answer to this question in the article itself and different ones tend to be eureka moments for different people. For me, the explanation that gave me my eureka moment was this


 * When you pick a door you separate the doors into two sets Your Door and The Other Two Doors. The host then takes The Other Two Doors and reveals a goat. Because he knows where the goats and cars and can always reveal a goat, what he is in essence doing at this point is giving you a choice between staying with All the cars behind Your Door and switching to All the cars that were behind The Other Two Doors. Which highlights the fact that the switch group is twice as likely to have the car.


 * Another explanation that helps sometimes is to consider that When you first pick a door you either picked a goat (2/3) or a car (1/3). If you picked a goat, the host has to reveal the other goat and the remaining door has a car. If you picked the car then the host can reveal either goat but either way the remaining door has a goat. So 1/3 times you picked the car first and the switch door is a goat. and 2/3 times you picked a goat first and the remaining door has a car. SPACKlick (talk) 17:13, 11 September 2014 (UTC)

Thanks for moving my post to a better place, and thanks for your reply. It still seems to me we have 2 different scenarios, one with 3 possibilities, and a subsequent one with 2. I don't see why they can be mixed up. But most people, many of whom are much smarter than I am, see it, so I defer. --Richardson mcphillips (talk) 18:02, 11 September 2014 (UTC)


 * You're perfectly correct that there are 2 different scenarios - one with 3 possibilities and a subsequent one with 2. Many explanations address the scenario with 3 possibilities (if I pick door 1 and switch to whichever door the host doesn't open which might be either door 2 or door 3, then I lose only if the car was behind door 1 to start with - which happens with a 1/3 chance).  To get from here to the 2 possibility case you need understand that which door the host opens is not completely random.  If you pick door 1 and the car is behind door 2, the host MUST open door 3.  However, if you pick door 1 and the car is behind door 1, the host can open EITHER door 2 or door 3.  At this point it might help to think about what might happen in 300 shows (where you've picked door 1).


 * How many times (out of 300) will the car be behind door 1? _______
 * a) Of these, how many times will the host open door 2? ______
 * b) Also of these, how many times will the host open door 3? ______


 * Similarly, how many times (out of 300) will the car be behind door 2? ______
 * c) Of these, how many times will the host open door 2? ______
 * d) Also of these, how many times will the host open door 3? ______


 * And, just to be complete, how many times (out of 300) will the car be behind door 3? ______
 * e) Of these, how many times will the host open door 2? ______
 * f) Also of these, how many times will the host open door 3? ______


 * These are the only possibilities, so now we can answer some other questions.


 * g) How many times (out of 300) does the host open door 3? ______ (the answer is b+d+f)
 * Of these, how many times is the car behind door 1? ______ (the answer is b)
 * Of these, how many times is the car behind door 2? ______ (the answer is d)
 * (and just to be complete) Of these, how many times is the car behind door 3? ______ (the answer is f)
 * If you've seen the host open door 3, what is the probability the car is behind door 1? ______ (the answer is b/g)
 * If you've seen the host open door 3, what is the probability the car is behind door 2? ______ (the answer is d/g)
 * (and just to be complete) If you've seen the host open door 3, what is the probability the car is behind door 3? ______ (the answer is f/g)


 * Does this help? -- Rick Block (talk) 19:58, 11 September 2014 (UTC)

Yes, she had a chance of 1/3 to have first selected the door hiding the prize (having arrived in the lucky guess scenario). In this 1/3 the host may randomly open any one of his two unselected doors, both hiding goats, and switching doors hurts in that 1/3.
 * In short: The contestant already has made her first choice.
 * But in 2/3 she will have selected one of the two goats (being in the wrong guess scenario), so the two unselected doors hiding either car+goat or goat+car.
 * As the paradox is based on the extremely biased role of the host to never show the prize but in any case to show a goat only in order to offer a switch, in that 2/3 he is bound not to open his door that hides the car, but to show the second goat and to offer his door that hides the prize, so switching doors mandatorily wins the car in that 2/3. – Was this helpful and illustrative?


 * But consider a different problem that contradicts the assumptions of the standard paradox:
 * The host's role, instead of intentionally revealing a goat, was to haphazardly open one of the remaining two doors (goat+car or car+goat). If the host opened a door at random and revealed a goat simply by chance, then the odds are reduced from the standard paradox's 2:1 in favour of switching to 1:1 only.  The latter odds track the common intuitive wrong answer because half of the potential winning cases are wasted when the host accidentally reveals the car and by that discards such plain winning event.
 * See also University of California San Diego: University of California San Diego, "Monty Knows Version and Monty Does Not Know Version, An Explanation of the Game". Regards, Gerhardvalentin (talk) 11:36, 15 September 2014 (UTC)
 * That is an excellent page which shows the essential paradox plus why in matters that the host knows where the car is. Martin Hogbin (talk) 21:31, 18 September 2014 (UTC)


 * "The essential paradox" always includes the "knowing host"; moreover: The host is committed to open a not chosen door with a goat and then to offer a switch. The 2/3 solution without this (or an equivalent) assumption is wrong. On the "excellent page" you mentioned we can read: Obviously the car is not behind door 3. But before I open door 1, the door you selected, I'm going to let you switch to door 2 if you like. Why does the host there say this which is "nonsense" (Steinbach) if the contestant already knows what the host has to do?--Albtal (talk) 17:46, 19 September 2014 (UTC)
 * That's, of course, an interesting point. Let us assume the contestant did not know this was gonna happen. And, just like with the Three Prisoners, they ask the host how he chooses the door to be opened. Nijdam (talk) 18:14, 19 September 2014 (UTC)
 * Could we do better than hear what the author of the "Three Prisoners Problem", Martin Gardner, himself said about Marilyn vos Savant's "Monty Hall Problem"? - Gardner: The problem is not well-formed, unless it makes clear that the host must always open an empty door and offer the switch. Otherwise, if the host is malevolent, he may open another door only when it's to his advantage to let the player switch, and the probability of being right by switching could be as low as zero. Moreover he said the ambiguity could be eliminated if the host promised ahead of time to open another door and then offer a switch (see Tierney, New York Times, 1991).--Albtal (talk) 19:12, 19 September 2014 (UTC)

@Richardson mcphillips: The crucial rule leading to a 2/3 solution is that the host is committed to open a not chosen door with a goat and then to offer a switch. So, for example, if you first "choose" door 1, you know in advance that the host will open door 2 or 3 with a goat. So by switching you will win whether the car is behind door 2 or door 3.

Also you may argue as follows: The chance for door 2 and 3 together is 2/3. If the host now is committed to show you a goat behind door 2 or 3, the remaining of these two doors keeps the 2/3 chance.

But if you choose door 1 and the host opens door 2 or 3 with a goat offering you a switch, it is a joke to claim a 2/3 solution, if the host had not been committed to do so.

So in the game which really has a 2/3 solution, it is the contestant who governs the process, not the host.--Albtal (talk) 15:45, 15 September 2014 (UTC)
 * Comment:
 * And what when the host opens door 3?
 * Why should the remaining door 'keep' the 2/3 chance?
 * Right about the joke: the required probability depends on the host's strategy.
 * The last section is abracadabra to me.

Nijdam (talk) 15:04, 19 September 2014 (UTC)


 * No comment.--Albtal (talk) 16:59, 19 September 2014 (UTC)


 * Nijdam is correct, the paradox is based on the host's role, e.g. extremely biased to never show the price, and to open ONE door only, etc. It is not a "joke", but the paradox only arises due to its implicit inherent restrictions that are easily and intuitively detectable for everyone. Otherwise the intended paradox will not occur. My question to Nijdam: As to this intended paradox, and as to the knowledge of a well informed contestant, is it possible / conceivable that the chance on the prize by switching doors can ever differ from 2/3? I do not ask for all possible methods of reasoning, just for yes or no. But if your answer should be "yes", I would welcome any justification. Thank you and kind regards --Gerhardvalentin (talk) 17:40, 19 September 2014 (UTC)
 * What do you mean by a well informed contestant? If the behaviour of the host may differ from random, then, as you yourself indicated, the answer is yes. Nijdam (talk) 18:02, 19 September 2014 (UTC)
 * Thank you for your speedy reply. But I was talking about the intended famous paradox with a well informed contestant. She knows that she had a chance of exactly 1/3 to having picked the car, but of exactly 2/3 to having picked one of the two goats. And she knows that the host's role was to open an unselected door in order to show a goat but not the car, and finally to offer a switch to the second unselected door.
 * As said, I am talking about the intended paradox, so I do not understand your above statement:
 * "If the behaviour of the host may differ from random, ..."
 * because in their rejoinder of Nov., 1991 (The American Statistician, Vol. 45, No. 4, p. 289) Morgan et al. say about this intended paradox:
 * "... that 2/3 is the answer to the relevant conditional problem only if p = q = 1/2. Certainly the condition p = q = 1/2 should have been put on via a randomization device at this point. It could also have been mentioned that this means that which of the unchosen doors is shown is irrelevant. ..."
 * As I understand Morgan et al., the behaviour of the host of the intended paradox may NOT differ from random. Am I wrong? Gerhardvalentin (talk) 19:02, 19 September 2014 (UTC)
 * Morgan et al. analyse the MHP, not only in the case p=q=1/2, but also for other values, i.e. non random behavior of the host. You will have read it all, so, what do you want to know from me? Nijdam (talk) 22:11, 19 September 2014 (UTC)
 * Thank you so much, am away but will be back soon, and I strongly feel we are getting on. Regards, Gerhardvalentin (talk) 09:16, 20 September 2014 (UTC)


 * Yes, I surely hope we are getting on. I was addressing the intended paradox: only the host knows the actual location of the car. After having picked one door out of three, say #1 e.g., and before any further action, the contestant's chance to having picked the car is 1/3, and the chance to having picked one of the two goats is 2/3. – In this given state, the contestant's KNOWLEDGE is limited to this chance only. Is this correct? And my supposition regarding the intended paradox (in compliance with the hitherto existing academic literature) evidently, obviously and implicitly includes that by showing a goat afterwards behind one of the two unselected doors (disclosing that the car is NOT behind that unselected door) the host will not simultaneously disclose that the actual location of the car currently definitely is NOT behind door #1 also, as simultaneously disclosing the contents of TWO doors would be considered rigging the intended game, as Martin calls it, and I would call it a clear inadmissibility. I'm now just talking about the contestant's chance, i.e. about the chance of the door she had picked. Before the host's showing of a goat, her chance of having picked the car was 1/3. My question: As to the intended paradox, after the host did show a goat, what additional information can have helped to "update" her said estimation of originally 1/3 to some closer value now (zero, e.g.)? What exact additional info (illicit info?)is NECESSARY in the intended paradox to update her estimation now to a closer rate than 1/3? – As I understand Ruma Falk, it is necessary that the host IS biased to a certain extent, and that the contestant KNOWS about this bias, otherwise NOT. Now, what additional information is NECESSARY to update her knowledge? Can you answer this question? Thank you and regards, Gerhardvalentin (talk) 15:06, 24 September 2014 (UTC)

Continuation
Well, I had a hunch you didn't understand "my" concept of 'event', but on the other hand you had no objection using terms like 'the probability the car is behind door 1'. This looks inconsistent to me. Anyway, the model to be adequate to give the probability of any relevant occurence, the events should cover all aspects mentioned in the problem statement. Allthough you may use the term 'event' in a different way, you still use concepts like 'the probability the car is behind door 1', etc. In my opinion, the sound approach is to do all kind of calculations in the model, and not state the model after all kind of primarily calculations. But then, calculations have to be done, be it that you just do them in a loosely way without properly formulate them. Nijdam (talk) 16:11, 18 September 2014 (UTC)

What about the color of the host's tie? Also a favourite item of Martin. It is quite simple, the problem statement does not have any reference to it. Hence there is no need to include it in the model. Note that it is not the fact that the color does not influence any probabilities, why it is not incuded! How should we know, as calculations are performed after the model is formulated. Nijdam (talk) 16:18, 18 September 2014 (UTC)

Returning to my question: do you agree something like "the car is behind door 1" may occur? Nijdam (talk) 16:21, 18 September 2014 (UTC)


 * To How should we know, as calculations are performed after the model is formulated. I fully disagree, calculations are performed at all stages in the formation, whilst creating the maximal model (in order to avoid accidental inclusion of impossible cases) and then in turning the maximal model into a model of maximum utility. That said, the door numbers can be discarded before we do any calculations by simply observing that they do not fall into any value dependent relations other than with eachother. Whereas what prize is behind the door and what function the door fulfils do fall into relations dependent upon their value.
 * Yes I agree that the Car can be behind a door numbered 1 but I also agree the host can wear a coloured tie and as such I find your rules for building the model inconsistent. If the same question was asked but included the phrase "the host, wearing a tie (red or blue) asks you to choose a door" at the beginning you would have a different sample space with a different model with a greater variety of cases and yet it is undoubtedly a question about the same events (either definition I think) in the same universe. The irrelevant properties that can be added to the description are limitless and many of them can be left with infinite values which would make it impossible to model a sample space at all. Whereas wherever a detail is included in the question which is an irrelevance to all elements of the question excluding itself a useful model of the universe will discard the irrelevance as soon as it can be determined it is irrelevant. SPACKlick (talk) 17:08, 18 September 2014 (UTC)

Well, it's not up to you to rewrite probability theory. Like others, you easily say that the door numbers can be discarded. But what do you mean by discarding them? Indeed is it possible the car is behind door 1, and I'm 100% sure that it is one of the possibilities. On the other hand one may say it's in general possible for the host to wear a colored tie, but it's not 100% sure it is one of the possible situations in the problem, actually nothing is said about it. So this is not a good argument to defend leaving out of the door numbers. Nijdam (talk) 18:05, 18 September 2014 (UTC)


 * I haven't just said the door numbers can be discarded I have explained why they can be discarded. There is no relation in the information available to the contestant dependant on the value of the door number. Simple as that. This can easily be demonstrated by drawing the maximal tree with the three branches (or 6 if you consider individual goats) for the arrangements of the prizes behind doors. Then branch at the door number selection of the player (which also selects the chosen door prize) and then for host revelation. Then you can observe that all branches which branch solely on the door number are identical barring that door number. Having observed this surely you agree that it demonstrates the door number is not having an effect on the probability of win/lose by switch/stick. The next step, not invented by me and certainly not re-writing any new mathematics by seeking to model only relevant variables, would be to explain why that is the case and it's evident that it's because of the lack of relations dependant on the value of the door numbers. This, as it turns out, is a general rule for modelling universes, elements without relations dependent on the values of the properties within are not relevant events within a universe, whether mentioned in the question or not. Throghout you have asserted where my process differs from yours but you have not yet indicated what in my process you feel makes it inadequate to answer the question within the MHP that is "should the contestant switch or stick" or in full form "across all cases what rules of switching and sticking will maximise car wins". Could you elucidate that point? SPACKlick (talk) 18:18, 18 September 2014 (UTC)

Is it possible for you to answer any of the following two questions: (1) What is the probability the car is behind door 1? (2) What is the probability the host wears a colored tie? Nijdam (talk) 07:56, 19 September 2014 (UTC)
 * I will answer but I'm still asking why you feel the model I have presented is inadequate to answer the question asked. I can't answer the second question. The reason the second can't be answered is the higher number of possible values requiring greater processing, if you will allow me to switch the color of tie for, say, the hand used to open the door (or any similar property of the universe with a limited number of possible values). Then answering the question is identical to the position of the car. With the probability distribution of the cars behind the doors the contestant has no information so all options are equiprobable. With the hand the host uses to open the door there is more information (10% of the population are left handed, Left handed people perform 90% of simple tasks with their dominant hand and right handers 95%) so this information weighs in to the problem giving us 86.5% chance the door is opened with the right hand and 1/3 chance the car is behind door number 1. — Preceding unsigned comment added by SPACKlick (talk • contribs) 08:16, 19 September 2014 ‎

Okay, if there is 1/3 chance the car is behind door number 1, I may immediately answer your question about the inadequasy of your sample space (model?). The required probability to solve the problem is the conditonal probability that the car is behind door 2, given the contestant has initially chosen door 1, and the host has opened door 3 showing a goat. As you may notice: (1) by telling me the value of the chance the car is behind door 1 ({C=1}), you implicitly admit the existence of the event {C=1}, which is not a subset of your sample space, i.e. your sample space is not suitable to answer the question; (2) your sample space does not allow for the required probability.Nijdam (talk) 08:48, 19 September 2014 (UTC)
 * But that's not the question the MHP asks. The MHP asks for the strategy for all cases. There's your problem, you're answering the wrong question. The question is in a game with these rules "Is it to your advantage to switch your choice?" not In one case of the game is it to your advantage to switch your choice.
 * As I said above, for a complete answer I would expand the question to "''What rules define the strategy where you maximise the odds of winning the car" which would consist of a rule for choosing the door and a rule for when to switch or not switch. As it turns out the maximal strategies are those with any rule for picking the first door followed by any rule that always results in switch. SPACKlick (talk) 09:21, 19 September 2014 (UTC)

You better reread the problem formulation. Nijdam (talk) 12:25, 19 September 2014 (UTC)
 * I have done, I'm very familiar with it. The problem formulation 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?
 * This is clearly a case of laying out the problem rules and then asking for the strategy. To say otherwise one would have to presume that the "say No. 1" meant consider only cases where number 1 happens, which is far from common use of a "say x" clause which is usually used as a "for example x" clause. If you are trying to answer the question. That being said, if you want to treat it as saying, Here are the rules of the game, in the case where P=1, H=3 and R=2 which has higher probability C=1 or C=2. You can still calculate that from my model above because it shows that for all values of P, H and R P(C=P) = 1/3 and P(C=R) = 2/3. Plugging in your values to be considered C=2 is more likely, twice as likely infact. SPACKlick (talk) 13:15, 19 September 2014 (UTC)

I wouldn't say a strategy was asked for, but a decision. And the decision has to be based on some probabilty, which turns out to be the most intriguing part. Of course the door numbers 1 and 3 are mentioned as examples, but undoubtly the player, at the point of decision, knows which door they have chosen, and which door is opened. I'm perfectly happy with you chosing other doors as examples. And no, your sample space. {A,B} as I recall, is not suited for the purpose you mention. I'm anxious to hear, how you, with the probability measure given by P(A)=1/3 and P(B)=2/3, manege to calculate P'(C=2|X=1,H=3), where P' is another probabilty measure, and X is used instead of the confusing P, for the number of the door initially chosen. Be my guest.Nijdam (talk) 14:41, 19 September 2014 (UTC)
 * First of I'd like to say I agree that in each specific run of the game at the point of decision (t2) the contestant will know the values of X, H and R. When I said a strategy by it I mean I think it seems to be asking for a decision in multiple cases, for the entire space of the game. We could theoretically imagine, by adding more restrictions the answer being a set of conditions where you switch and where you stick, say Stick where {H=1}else switch or the more complicated Stick where H = X mod 3+1. Either way it definitely is asking about decision in all cases.
 * As for how cases {A,B} can be used to calculate the probability above. A is simply the chosen door contains the car and B is it doesn't and the remaining door contains the car so {A,B} give us P(A)=P(C=X)=1/3, P(B)=P(C=R)=2/3, for all 6 combinations of {C,X,R}={1,2,3} so for the case you mentioned above; from X=1 and H=3 we determine 2=R and replace. P(C=2|X=1,H=3)=P(C=R)=P(B)=2/3. Simple, question answered. And the general statement of that answer is, "In all cases switching is advantageous. Specifically it doubles the odds of winning." And that is the correct solution to the MHP.SPACKlick (talk) 23:52, 19 September 2014 (UTC)

Well as soon as you say for all combinations … you get into trouble, as none of these combinations is represented in your sample space. How hard you try, it is impossible to derive P'(C=2|X=1,H=3), from P(A) and P(B).Nijdam (talk) 12:19, 20 September 2014 (UTC)
 * There's no trouble in saying for all combinations I have already demonstrated P(C=R) is as independent of the value of X,H&R as is which hand the door is opened with. If you are disputing that independence is demonstrated then just say so and we'll go back to that. I just demonstrated the maths for how to find p(C=2|X-1,H=3). Please point to the error in the maths. All probabilities below are at the point of decision

[1]P(A) = P(C=X) for each X [Definition] [2]P(B) = P(~C=X) for each X [Definition] [3]P(C=X) + P(C=R) + P(C=H) = 1 for each {X,R,H} [Sufficiency and exclusivity in problem] [4]P(C=H) = 0 for each H [From definition of H] [5]{X,H,R} = {1,2,3} [As defined the labels must be unique and map to door numbers] [6]P(C=R) = 1-P(C=X) for each {X,R} [3]&[4] [7]P(B) = P(C=R) for each R [2],[5]&(P(x)= 1-P(~x)) [A] Consider P(C=2|X=1,H=3) [8] P(C=2|X=1,H=3) = P(C=2|R=2) + P(C=R|R=2) [A]&[5] [9] P(C=R|R=2) = P(B) specific case of [7] [B] Therefore P(C=2|X=1,H=3) = P(B)


 * I'll stop at that conclusion to see where you think the maths is wrong. SPACKlick (talk) 13:47, 20 September 2014 (UTC)
 * It goes wrong from the start, when you say for all X. Either there is only one X, a stochastic variable, but unfortunately not on your probability space, or you mean X just stands for one of the values 1, 2 and 3, and then {C=X} is no subset of your sample space, and hence no probabilty assigned to it. It should be not to difficult to understand that when you only have P(A) at your disposal (P(B) being 1-P(A)), it is impossible to derive all those different probabilities from it. Simple logic.Nijdam (talk) 15:07, 20 September 2014 (UTC)
 * IT is perfectly possible to derive all the probabilities. See collapsed below. Before the consider, [A], the statements hold true for all cases. The reason I highlight for all X. is that P(C=X)=1/3 means ( P(C=X|X=1) + P(C=X|X=2) + P(C=X|X=3) )/3 = 1/3. Whereas P(C=X) for any X =1/3 means P(C=1)=1/3, P(C=2)=1/3 and P(C=3)=1/3. I could reword by adding a variable. P(C=X|X=x)=1/3 when x $$\in$$ {1,2,3}. If you know the correct symbology for saying that P(C=X)=1/3 for every value of X rather than as an average across all cases feel free to let me know, however as a concept it is sound P(C=X) is independent of the value of X is the concept I'm trying to convey. And if you'd answer to the concepts I'm communicating rather than criticising the terminology that's be grand. SPACKlick (talk) 15:27, 20 September 2014 (UTC)
 * Clearly you have great difficulty with mathematics. You seemingly have not understood a bit of what I wrote above. I'll give it one go. Be sure you understand every syllabe. I asked you what your sample space was. You insisted it was the space {A,B}. This means the only events you can talk about are A and its complement B. Think carefully about this. They are the only events you have at your disposal. And the only probabilities you're allowed to talk about is P(A), P(B), 1 and 0. No more. Yet you talk about {C=1}. But in your model this has no meaning, is not defined. Got it sofar? Nijdam (talk) 16:44, 20 September 2014 (UTC)
 * I have no difficulty with mathematics just limited formal training, as I said before my experience in determining event spaces comes through philosophy. First, you seem to have missed my explanation of the difference between events and relevant events. A (C=X) and B (C<>X) are the only events with relevant properties and they form a complete space wrt to the question at hand. In each individual run C,H,X and R still have a numeric value, just as the host's tie has a colour and he opens the door with a specific hand, these are not included as irrelevancies. There are myriad instantiates which take all kinds of properties over the universe but still, the only event which divides the sample space wrt the question is whether C=X or C<>X. You seem to believe I should be including an irrelevant event in the mathematics, fine. compare it to the following;
 * What is the probability that of winning by switching when the host wears a blue tie(T=B) ? Well. The probability of winning is independent of tie colour so it's 2/3 for all colours of tie so P(C=R|T=B)=2/3. You inserted an irrelevant event to the sample space, I answered given that it doesn't matter. The same logic is true for door numbers. So while I can't put it in mathematical language the logic is sound. The probability of winning by switching is independent of the number value of any of the doors because the number of the doors has no relationship to any other property in the universe. We seem to be at an impasse due to my inability to use the terminology your using. That's fine but you should be aware that while you struggle with my terminology, the logical point that the individual door number bears no relation to the probability of winning by switching is valid. It's evident from the lack of relation to any other factor in the problem, it is evident from the fact that even in the conditional models P(C=X|X=a, H=b, R=c)=1/3 whatever values you plug in for A,B and C and it's evident from the fact that if you remove the door numbers from the problem and the contestant's knowledge of specific doors beyond their sustained function the problem is identical in its scope and the answer remains the same. I don't know how else to explain it to you. SPACKlick (talk) 20:50, 20 September 2014 (UTC)

What you referring at is the solution using symmetry. Indeed is: P(C=1|X=1, H=2)=P(C=1|X=1, H=3)=P(C=2|X=2, H=1)=P(C=2|X=2, H=3)=P(C=3|X=3, H=1)=P(C=3|X=3, H=2). Notice in the first place we need events {C=c}, {X=x} and {H=h) to describe this. Further is is not directly clear that the commn value should be 1/3. You probably come to this value due to 'common sense'. The tricky point is that many people say 'common sense' leads them to believe this value is 1/2. Nijdam (talk) 11:58, 24 September 2014 (UTC)

My two pennyworth again
I think you must both end up with the same conclusion that suggested a while ago. If you decide to formulate the problem (as Nijdam has done) in terms of door numbers, then you have a conditional problem, with one condition being the door number opened by the host and the other being the, often ignored, door number initially chosen by the player. This is really a mathematical fact and there is nothing that can change that. You can, using symmetry or other arguments, quickly come to the conclusion that these conditions do not change the probability of interest but they do exist in that formulation.

On the other hand, it is possible, having recognised the fact that the door numbers (or other identities) give us no useful information, to formulate the problem without reference to door numbers, for example using the sample space {C, G1, G2} (which presents a problem in which the goat revealed is a condition) or, recognising the insignificance of goat identity (discernible properties), just {C,G}. Of course the doors still play a physical role in the game in that they are moved opened to reveal the prizes, but their numbers can be ignored completely.

I disagree with Nijdam in that I assert that there is no algorithmic way of turning a natural language problem into a precisely defined mathematical one. There are always judgements to be made about what is important and what is not. Martin Hogbin (talk) 13:17, 24 September 2014 (UTC)

Deriving all the values
P(C=1|X=1) = P(C=X|X=1) = P(A) = 1/3 P(C=1|R=1) = P(C=R|R=1) = P(B) = 2/3 P(C=2|X=2) = P(C=X|X=2) = P(A) = 1/3 P(C=2|R=2) = P(C=R|R=2) = P(B) = 2/3 P(C=3|X=3) = P(C=X|X=3) = P(A) = 1/3 P(C=3|R=3) = P(C=R|R=3) = P(B) = 2/3 P(C=1|H=1) = P(C=H|H=1) = 0 P(C=2|H=2) = P(C=H|H=2) = 0 P(C=3|H=3) = P(C=H|H=3) = 0

Statement of February 2006 ante battle
An unambiguous statement of the problem, with explicit constraints on the host as described by Mueser and Granberg:
 * Behind each of three doors is either a goat or a car (two goats, one car), with the car behind each door with equal probability.
 * The player picks one of the three doors. The contents are not revealed.
 * The game host knows what is behind each door.
 * The game host must open one of the remaining doors and must make the offer to switch.
 * The game host will always open a door with a goat.
 * That is, if the player picks a door with a goat, the game host picks the other door with a goat.
 * And if the player picks the door with a car, the game host randomly picks either of the two doors with a goat.
 * The host offers the player the chance to either claim what is behind the originally-chosen door, or to switch, claiming what is behind the one remaining door.

Do the player's odds of getting the car increase by switching? --- this was the problem formulation in 2006. Gerhardvalentin (talk) 13:37, 19 September 2014 (UTC)
 * I find that the 7th point ""the game host randomly picks either of the two doors with a goat"" is an addition by a later author, there is nothing to suggest this in the original problem (and it's also not relevant to the maths at hand. SPACKlick (talk) 13:39, 19 September 2014 (UTC)


 * This famous paradox is spread all over the world, and it arises due to its implicit inherent and intuitively detectable restrictions (see Mueser and Granberg, Keith Devlin, Leonard Mlodinow - the host's role -, Norbert Henze - the host's secrecy - and others). MvS and the misinterpretation of the host's role by MCDD are incidental. Basis of the paradox is "of course" (Henze) that the host never gives any hint as to the current scenario the contestant actually is in: lucky or wrong guess scenario. No "update" by the host, the ratio remains "1/3:2/3". Gerhardvalentin (talk) 14:13, 19 September 2014 (UTC)


 * I fully agree that the paradox involves the host giving no hint. However, whatever the host's strategy, as long as it remains unknown the player the probabilities stay the same. The original problem doesn't specify a host strategy so adding one in is changing the problem. You can't complain about adaptations of the problem being focussed on and then use an adaptation. In the original problem the player is ignorant of the host's strategy and so are we. SPACKlick (talk) 14:20, 19 September 2014 (UTC)


 * Thank you, but my view is different. In the first line, the article is on a world famous paradox. Before telling the convoluted history of omissions, misunderstandings, recriminations and special interpretations, the article should present the clear context that it is about. And as said, it is about a world famous paradox that should be presented understandable and intelligible. Not every reader is in a position to judge within the first seconds of reading the implicit inherent restrictions that are easily and intuitively detectable. As said, not in the first second while reading the article. So the intended paradox has to be presented first, clearly laid out and plain, to avoid misinterpretations. The theater of war may be explained later on. Gerhardvalentin (talk) 19:27, 19 September 2014 (UTC)


 * I agree that a reasonable discussion should discuss the paradox before discussing misunderstandings and variants however adding the "Host much choose at random" rule is a special interpretation just as much as a "the host avoids opening Door 2 where possible" rule is. The point being to explain the paradox one needn't ever mention how the host chooses in the case where the contestant chose the car, it isn't relevant whether it's equiprobable or highly biased and introducing the rule adds unnecessary complexity. SPACKlick (talk) 23:52, 19 September 2014 (UTC)
 * Thank you. As to the article, newest literature's fact-finding – when presenting the intended paradox in the beginning – can help the reader to avoid confused meandering. Gerhardvalentin (talk) 09:32, 20 September 2014 (UTC)
 * @SPACKlick, in short: The chance that the contestant – by picking one of three doors – did arrive in the lucky guess scenario obviously is 1/3, and obviously the intended paradox impossibly can nor does nor may include / allow / comprise / tolerate that the host can nor may indicate that "actualy" this rate effectively is zero. Absurd. An inadmissibility. After MCDD suggested such aberrant variant, the intended paradox has to take a firm stand that such approach is impermissible. To avoid such meander, it is necessary to say that this is impermissible, hence we "HAVE" to take the host's choice of two goats "AS IF" he had chosen evenly. Or as Prof. Norbert Henze says: "Of course, the host is obliged to absolute secrecy as to the actual location of the car". Just to avoid unnecessary meanders . . . Gerhardvalentin (talk) 20:28, 24 September 2014 (UTC)

The host's tie is not my favourite item
My favourite complaints are two:

The door initially chosen by the player is not included in any of the solutions. Using a mathematical formulation based on door numbers, we should calculate P(C=2|P=1.H=3) not P(C=2|H=3). Nijdam and many others agree this, yet nowhere in the article do we see the complete calculation. In the article, the 'proper' solutions are all based on Morgan's incomplete solution. All we have are hand-waving arguments as to why we can ignore the door number originally chosen by the player.

The fact that there are two different goats but we only see one of them is mentioned in the problem statement yet no solution in the article deals with this important fact.

Of course, as the problem was intended to be, and, as it is understood by most people, my two complaints are, along with Morgan's solution, absurd and unnecessary pedantry. The problem was clearly intended to consist of two identical goats and a car. Martin Hogbin (talk) 19:01, 18 September 2014 (UTC)


 * Do you mean the door initially chosen is not included in any of the solutions in the article? Both the pictural and the Bayes' solution refer to the conditional probability given the door initially chosen is door 1.Nijdam (talk) 08:01, 19 September 2014 (UTC)


 * Both those solutions treat the problem as if the player always chooses door 1, and then proceed to calculate P(C=2|H=3). The player can, under the rules of the game, pick any door to start with, just as the host has a choice of door.  In the instance in question the player has chosen door 1 and the host opened door 3, thus we must calculate P(C=2|X=1.H=3). Martin Hogbin (talk) 09:39, 20 September 2014 (UTC)
 * Better to use X instead of P as P already indicates 'probability'(Thanks, I changed that). But notice the Bayes solution uses P1 as short for P(.|X=1), and the pictorial also shows the conditional P(.|X=1). Nijdam (talk) 12:14, 20 September 2014 (UTC)
 * Is P1 standard notation for P(.|X=1)?
 * I see no sign in the pictures that the player might have chosen door 2 or door 3 initially. The caption says just, 'Player initially picks Door 1'. Martin Hogbin (talk) 12:41, 20 September 2014 (UTC)
 * Nijdam, I see that you have changed the notation in the article, and this is a great improvement in clarity in my opinion. Unfortunately this makes the incompleteness of the solution even clearer. Why do we not consider any outcomes in which the player initially chose door 2 or door 3?   These are clearly permitted under the rules of the game.  The current calculation implies that the player must always pick door 1.


 * If we are to provide a complete solution we must show the three Bayesian inferences: that the player is equally likely to choose any door (although we are asked to put ourselves in the position of a player we are given no information on what our initial door preference might be), that the car is equally likely to be initially placed behind any door, and that the host is equally likely to open any doors available to him.


 * Now, how about the pictures. Martin Hogbin (talk) 10:11, 21 September 2014 (UTC)


 * To be more specific, despite the fact that the host only opens door 3, the pictorial solution shows the host opening door 2. On the other hand, although the player might have picked door 2 or door 3 initially we do not see that represented in the pictures. We should be consistent. Martin Hogbin (talk) 13:34, 23 September 2014 (UTC)

The article neglects the famous paradox
Unfortunately, the article does not pay attention to the emergence of the famous paradox. The emergence of the paradox is due to its implicit inherent and intuitively detectable restrictions (Keith Devlin, Leonard Mlodinow, Norbert Henze and others). MCDD, who completely misinterpreted the host's role, thereupon have accused the author MvS of alleged "errors". For years, the article consists of the slavish recitation of such alleged Savant's errors, as named by MCDD. The classical significance of the famous paradox however remains completely neglected. Gerhardvalentin (talk) 20:34, 18 September 2014 (UTC)
 * And that is the real shame, I simple problem that most people get wrong becomes a complicated (but incomplete) one that no one cares about. Martin Hogbin (talk) 21:09, 18 September 2014 (UTC)

Comment by Rick Block

 * What Nijdam is getting at here is that if you're going to say p3=0 you HAVE to be talking about a conditional case where you have clearly been given additional information (if you're revising the odds of p3 from 1/3 to 0, you must have done this with addition information). This additional information affects not just p3, but p1 and p2 as well.  Either you show this effect directly (perhaps by assuming the host has no preference in the case the car is behind door 1), or you argue that p1 remains unchanged in the conditional case because ...  If you do neither of these (which many "simple" solutions don't) you're using the posterior probability p3=0 with the prior probability p1=1/3 (which you have not shown to be the same as the posterior probability) to compute the posterior probability p2 - which is what he's saying is nonsense.  Note that the argument cannot be "you know the host will always open a door, so opening a door does not change the probability of your selected door" (because by this argument none of p1, p2, or p3 are changed by the host opening "a door").Even vos Savant later  commented about this confusion (which arguably SHE is responsible for creating!), saying (about the "host forgets scenario") "Back in 1990, everyone was convinced that it didn’t help to switch, whether the host opened a losing door on purpose or not. ... Now everyone is convinced that it always helps to switch, regardless of what the host knows. But this is just as incorrect!" This is absolutely true. And, IMO, it's precisely because the popular sources do NOT address the "classic" MHP using conditional probability. By avoiding addressing the problem in this way, the popular sources have simply replaced one incorrect notion (two unopened doors always means each has equal probability) with another (whatever the host does he can never change the 1/3 probability of the player's initially selected door). -- Rick Block (talk) 15:27, 15 September 2014 (UTC)
 * Sorry Rick but you can call it conditional if you want. It's how I know the probability of the door the host opened containing the car refers specifically to door 3 but I needn't ever refer to it as door 3. From the very start of the problem you know that in 1 in 3 cases the car will be behind the door you picked and in 2 in 3 cases it won't. You know nothing about the doors or goats which can change this at any point in the problem. Therefore the remaining 2 in 3 cases it will be in the only other location. You're conflating arguments for the simple solution being incomplete with arguments for the simple solution being incorrect. SPACKlick (talk) 07:53, 16 September 2014 (UTC)

"Conditional probability" is a math term, and as to the paradox, the relevant "condition" is RESTRICTED to the host's never revealing of the prize. You are not correct in saying "This additional information affects not just p3, but p1 and p2 as well", because we KNOW from the outset that the host's revealing of a goat doesn't have ANY influence on the probability of door 1 to hide the car, as for the paradox you are bound to consider the host's secrecy, that emerges by our accepting it: In case he has a choice of two goats, we KNOW that he shows one of them "as if" he had tossed a coin. Gerhardvalentin (talk) 20:29, 15 September 2014 (UTC)
 * No Rick, we finally should stop to pellmell.
 * Exactly! The idea that the host could do anything, not required by the rules of the game, that would give the player a clue as to where the car was is absurd. No real game show would allow this.  Do you think the host would ever be allowed to say things like, 'The car is probably behind this door.', or 'Better not pick that door.', or 'I only open this door when I absolutely have to.'?  Of course not, that would be considered rigging the game. In modern game shows the better hosts often chat with and tease the players, but always in a way that does not change the player's chances of winning; that principle is inherent in the term 'game show'. Martin Hogbin (talk) 21:56, 15 September 2014 (UTC)


 * Yes, see Henze and others. As to p1, Nijdam's and Rick's point is the same, they both differ from the standard paradox as chalk from cheese, both ignoring the host's confidentiality provision. IMO this is a gross negligence, because the host never may nor will tell the contestant whether it is more or less likely than 1/3 that she unalterably arrived in the lucky guess scenario, nor whether it is more or less likely than 2/3 that she unalterably arrived in the wrong guess scenario. For the intended paradox, as well as for any game show, this would be considered rigging the game, as you call it. Yes, a clear inadmissibility. Let us be fully aware of the impact of such "view". Gerhardvalentin (talk) 11:23, 16 September 2014 (UTC)

Comment on the above discussion by Martin
Nijdam, of course you are correct, but only because you insist on using mathematical formulation based on numbered doors to describe the problem. Once the problem is formulated this way, there are two different numbered doors that the host might open and we therefore have two distinct cases to consider, one of which applies to the instance that we are considering, so the problem is strictly speaking conditional.

Using numbered doors is certainly one obvious way to analyse the problem but it is not the only way. By noting, right at the start, that the door numbers cannot affect the probability of winning by switching, because we have no information on them, we can formulate the problem without door numbers. The doors continue to play a physical role in the game, but only that of hiding the chosen objects from the player. Some simple solutions would appear to use this approach and, in my opinion, must therefore be considered perfectly correct. Example of such solutions are Morgan's F1, F3,F5. Martin Hogbin (talk) 11:27, 15 September 2014 (UTC)

Comment by Gerhard

 * Correct, it relies on the implicit understanding that the host may not say "I show this goat behind this unselected door, because the second unselected door, my preferred one, actually hides the car". This implicit and intuitive restriction (the host's secrecy) secures not only that the revealed goat gives us no additional information on the current location of the car, moreover it is the fundamental basis of the famous intended paradox. Gerhardvalentin (talk) 15:05, 10 September 2014 (UTC)


 * Literature: Henze and others show the intuitive restrictions of the intended standard paradox: the host's secrecy.  Gerhardvalentin (talk) 09:30, 7 September 2014 (UTC)
 * Why is that not a solution? Martin Hogbin (talk) 14:12, 5 September 2014 (UTC)
 * You know why! Nijdam (talk) 20:04, 6 September 2014 (UTC)
 * OK, I agree that if you include (distinguishable) doors in your mathematical model, the problem becoms one of conditional probability. We can label the doors by their role in a particular instance of the game. As you mentioned above 'the door opend by the host' we can conceptually label that door, 'H' for example, even though we do not see that door or know anything at all about it.  Martin Hogbin (talk) 08:02, 7 September 2014 (UTC)
 * Of course, if your model includes distinguishable doors, you must also condition on the door number originally chosen by the player.
 * On the other hand, there is no need to mention doors at all in your solution (but you still need to say that you treat the goats as indistinguishable). Martin Hogbin (talk) 21:28, 10 September 2014 (UTC)

An unequivocally unconditional scenario
Would anyone like to propose such a scenario that still uses doors. Martin Hogbin (talk) 08:02, 7 September 2014 (UTC)

My unconditional version
The game is played on the radio. The player is told that there are three doors hiding two goats and a car and that the host must always open an unchosen door to reveal a gaot and always offer the swap. The player has a door initially chosen for him at random by computer. He is informed when a door has been chosen but given no other information at all, he knows that if he were to play the game again he would not even know if he had had the same door chosen for him. He is then informed that the host has revealed a goat, as required by the rules, and has to choose whether to swap or not. Martin Hogbin (talk) 09:32, 7 September 2014 (UTC)


 * Why do you believe in the pompous nonsense which Morgan et al. and their followers produced? Why do you let you lead beyond the old principle of "no loss of generality"? The only important addition in your (don't call it unconditional) version to the version of MvS is that the host must always open an unchosen door to reveal a goat and always offer the swap, the rule which is crucial for the 2/3 solution. Curious enough that Morgan et al. and their followers - contrary to MvS - did not detect this essential lack.


 * Question: At the beginning of the game, a gambler (who knows the rules) bets money according exactly to a 2/3 chance, and "picks" door 1. Then the host opens door 3 with a goat. Should the gambler now change his stake?''--Albtal (talk) 20:57, 19 September 2014 (UTC)

The standard paradox and its intuitive restrictions
Three doors with equal probability of 1/3 to hide the car, so by her first selection of a door the contestant unchangeably is in the lucky guess scenario with probability of 1/3 (switching door will hurt in that 1/3) and she will be in the wrong guess scenario with probability 2/3 (switching door will win in that 2/3). The actual scenario is unalterable, as well as her immutable knowledge of the prevailing scenario she actually is in. The paradox is based on the intuitive assumption that she knows about this unchangeable ratio of 1/3:2/3  because the actual location of the car is and remains secret. If you like, you can conclude that p = q = 1/2 and you can conclude that this may be achieved by a randomization device. In any case, the standard paradox means that which one of the unchosen doors is shown is irrelevant, as it is the basis for the conditional and for the unconditional problem as well. In both cases, the actual location of the car is and remains secret. For the standard paradox, there is no room to ever consider p ≠ q, as this is not addressing the intended standard paradox. Gerhardvalentin (talk) 08:11, 7 September 2014 (UTC)
 * I agree that it should be considered part of the standard problem that the host does nothing, that is not required by the rules of the game, that might give the player any clue as to the location of the car. This is implicit in the term 'game show'.  We would hardly expect the host to say, 'Look on the notice on the wall behind you and it will show you which door probably hides the car', or say to the player after he has initially chosen a door, 'I would not pick that one if I were you', or to say on opening a door, ' I only reveal this goat when I absolutely have to'.   Without such natural assumptions the problem is insoluble (the car might always be behind door 2 for example) and with the natural assumptions there is nothing to condition upon. Martin Hogbin (talk) 09:31, 7 September 2014 (UTC)
 * Yes, the constitutive basis of the paradox is framed by its intuitive restrictions: No clairvoyance, no betrayal of secrets, but a host who never may show the car, his role being extremely biased to open a door in order to show a goat only and to offer a switch. The contestant knows that she is in the lucky guess scenario with probability 1/3 and in the wrong guess scenario with probability 2/3. She knows this from the beginning to the end, no irregular update by any irregular evidence as to the actual location of the car. You are free to deduct from the given constitutive basis of the host's obligation of secrecy that he tossed a coin, by implication. Yes, you can. You may, if you wish. But in any case you have to respect the host's confidentiality. No betrayal by the host and no clairvoyance by the contestant. The ratio lucky:wrong of 1:2 remains unchanged, from the beginning to the end, because the actual location of the car remains absolutely secret (Henze and others). Anything else is another problem. Gerhardvalentin (talk) 12:54, 7 September 2014 (UTC)

After reading all of the explanations and discussions... I'm still not agreeing with the 2/3 answer. My reasoning and a question.
Thanks to anyone that can help clear up. I love the conversation here, by the way....

Best description I can come up with of why I disagree
After revealing a goat, Monty offers me the option of calling a friend for help. My friend doesn't have a clue as to which door I initially picked, or the fact that there was only one more door. All that my friend knows is that there are two doors to pick from (left and right), and that there's a car behind one and a goat behind the other. The probability of my friend choosing the right door must be 50%... left or right... whether 1 &2, 1 & 3, or 2 &3. I can't see how the probability can be something entirely different to him or her than it is to me. Also, if I decide a priori to go with what my friend decides, I've just changed my probability to 50%? By introducing a person with less (more?) knowledge than me?

Friend: "I choose the door on the right".

Me: "Nope! That door is the same one I had picked. Therefore, it only has a 33% chance of having a car!"

Friend: "There's a third door?"

Me: "There was one!"

Friend: "So what?"

(I don't intend for the "call a friend" option to be factored into the probability... one can think of this as the probability to an audience member just tuning in.)

My opinion
In my opinion, there is only one "choice" being made and the odds of winning are 50/50. That the contestant did something to arrive at a choice of two doors is unrelated and quite selective. Why not factor in the probabilities of being picked from the audience to participate? ... or of obtaining a ticket to the game show? ... or of being a citizen of the country in which the game show takes place or of Obama granting you citizenship by executive order since Congress will never pass... ok, you get the point... ;) But, seriously... neither I, nor anyone other than the Monty simulator, knows which door has the prize I want and which has the one I don't want. It's true that I can determine one of the doors that will be included in the "choice", but I can also determine which coin I pull out of my pocket to flip, and that hardly affects the odds of the coin I selected coming up heads or tails... 50%

The hundred doors explanation
... does nothing for me. The odds of the car being behind the 38th door because Monty didn't open that one already assume that it wasn't the 38th door that I picked in the setup to the two door problem.

I already have a car. What I really want is a goat!
50% probability. Seriously... who cares how many cars and how many goats were in the problem setup? If there were 37 cars and 63 goats in the initial 100 doors, don't we still get to 1 car, 1 goat?

My mathematical "proof"
Every time I choose from two doors, the car is behind one or the other... 50%. My mathematical proof both works and is intuitive.

Many thanks in advance for anyone willing to indulge me in this!

NicolasOliva (talk) 12:20, 10 November 2014 (UTC)

Replies
Thanks for your comments. It is always useful to get feedback from readers so that we can improve the article.

The Monty Hall problem is probably the hardest simple probability puzzle in the world and many people find it hard to accept the 2/3 answer but it really is correct.

Can I start by just checking that you fully understand the intended setup. There are three doors, two goats, and a car. You want to win the car. The goats and car are all randomly placed behind the doors so that you have no idea where the car is. The host does know where the car is.

You initially pick a door. The host the must then open a different door from the one that you have picked and he must always reveal a goat. He must always then offer you the swap between your originally chosen door and the remaining door.

To avoid any unnecessary complications, if you have originally chosen the car so that the host has a choice of two goats which he might reveal, we will take it that he chooses one at random so that the particular door opened or goat revealed cannot give you any clue as to where the car might be.

Is this how you understand the problem? Martin Hogbin (talk) 12:53, 10 November 2014 (UTC)

Confirming understanding
Many thanks for your reply, Martin.

Yes, that is precisely how I understand the problem. It just seems to me that all of the pageantry that it takes to get to one choice of two doors behind one of which is a goat and the other a car is just a distraction. I can see where the "probability of switching" arrives at a mathematical proof, but I think it's incorrect to frame the question that way, and that's what I tried to articulate in my comments.

NicolasOliva (talk) 14:07, 10 November 2014 (UTC)


 * Yes it may seem that way but it is not so. The reasoning that convinces people of that fact varies from person to person.  Obviously you did not find vos Savant'w 100 doors argument persuasive.  I have to say that I do not find it particularly good.


 * In order to help improve the article, could I ask you to read through the two paragraphs in the 'Simple solutions' section starting, 'Another way to understand the solution is to consider the two original unchosen doors together.' and say why this argument does not convince you. Martin Hogbin (talk) 18:29, 10 November 2014 (UTC)

Probability of a random choice winning the car vs. probability of best choice
You are perfectly correct that there are two and only choices for where the car is at the point the player is deciding whether to switch. And, you're perfectly correct that a random choice between them has a 50% chance of winning the car.

For convenience, let's call the doors #1 & #2, and say your initial choice was #1. The "simple" explanations (strongly favored by Martin, BTW) ask you to reframe the problem and instead of thinking about the probabilities at this point back up and think about the overall probability of winning or losing from the start of the game.

So, back up with me, to the beginning of the game - when there are 3 doors. You pick a door, let's say #1. The probability of the car being behind this door is obviously 1/3 (right?). Let's imagine that instead of waiting to decide to switch until after the host opens a door, you must commit to staying or switching right now (just after picking your initial door). You turn in your answer now, saying "stay" or "switch", before the host has opened a door, leave your phone number, and go home - and the producers call you later and tell you what you won (car or goat).

Looking at the problem this way, what can happen? If you say "stay", you win the car only if it is behind #1. If you say "switch", you win the car only if it is not behind #1, i.e. behind #2 OR #3. At this point (before the host has opened a door) we don't know whether the host will open #2 or #3, but we know for sure he's going to open one of them. If you stay with #1, you win the car the 1/3 of the time it is behind #1 and if you switch, you win the car the 2/3 of the time it is behind #2 or #3.

But, but, but (you say) .... that's not the problem!

Once again, you're perfectly correct. It's not quite the problem.

The problem asks you to think about the situation after the host opens #2 or #3 - let's say #3 has been opened. What, exactly, has changed? Obviously the probability the car is behind #3 is 0 (even though it used to be 1/3). The trick is, the probabilities the car is behind the other two doors also changes, but not evenly. If the car is behind #2 the host MUST open #3, so the probability the car is behind #2 stays 1/3. On the other hand, if the car is behind #1 the host can open either #2 or #3 so the probability the car is behind #1 is cut in half to 1/6.

After picking door 1, the probabilities are 1/3 + 1/3 + 1/3.

If the host then opens #2 the probabilities are 1/6 + 0 + 1/3.

If the host then opens #3 the probabilities are 1/6 + 1/3 + 0.

If you add these together, you get back to 1/3 + 1/3 + 1/3.

Whichever door the host opens, the probability the car is behind #1 is exactly half the probability it's behind the other door. -- Rick Block (talk) 15:34, 10 November 2014 (UTC)

I follow... but I still don't agree
Rick, I appreciate your explanation. I stepped through it in detail, and I like your method of suggesting you have to decide to stay or switch in advance. However, I still think you, and the others continue to view this in a way I believe to be incorrect. I don't disagree with your calculations. I disagree that they apply to the problem.

For example, when you say, "the probabilities the car is behind the other two doors also changes".... If I wanted to take the sequence of steps into consideration rather than just the decision between two doors (and I'm not saying I agree with that), I would have say something along the lines of the following:


 * 1) The probability of you selecting the car (your objective) from the outset is 50%
 * 2) When you make your choice of the two remaining doors, you will pick from the door you picked in step 1 (50% chance at that point of being a goat or car ), and the door that Monty leaves closed - which also has a 50% chance of being a goat or car at that point.
 * 3) There never was a 1/3 probability of anything because your first "choice" never revealed what was behind the door. All you did in the first choice was ensure that that door would be one of the two remaining. This never impacts your "probability"... it just makes you happy to have a door you can choose to bring to the party.
 * 4) Monty has no choice of any significance at all. Monty is forced to select a car if you selected a goat. If you selected the car on your first, then he makes the completely insignificant (to you) choice of goats to reveal. You already knew that at least one of those doors had a goat, and now you know that at least one of those doors had a goat. I don't see that you learned anything about the door you chose, or the other door. You knew before he selected that your door and the door he doesn't select, each have a 50% probability of having a car.

And if you'll allow me to "back up".... The Monty Hall Problem could be said to be the n=3 version of a "choose from n doors" where on door is a car and n-1 doors are goats. If there were n doors and Monty revealed one after each one I pick until he leaves me with 2 to pick from, when we got to 3 doors left, we would have something completely indistinguishable from the classic "Monty Hall Problem". The other doors that don't have goats were noise... they just don't matter and never did. By my formulation, this works all the way down from any positive value of n until you get to n=2. By your formulation, if we started with 4 doors, then 1/4 of something comes into play in a way that means it's not 1/3 by the time you are choosing from 3 doors. Therefore, the n=3 problem is unique, or the probability really does depend on the initial value of n and the choices you made... neither of which are acceptable answers.

With my formulation, you can pick doors randomly (or by any method you prefer) and Monty can pick doors randomly (if a choice among goats) or by any other method that ensures he doesn't reveal a car... until you get to two doors. Nothing in your previous selections of doors in any way impacts the 50/50 probability of you selecting a car when you get to two doors.

I would love if you or someone else answered the questions I'm posing. There seems to be an assumption that I'm not understanding the problem or getting the math, but I think I've done the work in those and am asking different things. In particular, I want to understand why someone stepping in when there are two doors left gets to make a 50% choice, when I don't. Also, I'd like to understand how the third door is different than the second coin in my pocket when I pick a coin to flip. Persuasive answers to these would really help me come to grips with this!

Thanks! NicolasOliva (talk) 18:40, 10 November 2014 (UTC)
 * Why does someone stepping in at the end have a 50% chance and you don't?
 * Because they don't have the same knowledge you do. In particular, if someone has to pick between two alternatives without any knowledge they have a 50% chance of choosing correctly.  This doesn't at all mean that someone with more knowledge also has a 50% chance.  For example, imagine I watch a hustler flip a coin 100 times and it ends up heads 80% of the time.  You walk up now and the hustler flips his (not fair) coin.  You have a 50% chance of guessing whether it's heads or tails (because you don't know the history, you're basically guessing randomly).  I'd guess that it's heads and I'd likely be right about 80% of the time.  Different knowledge, so different probability.
 * Rick, while I appreciate the reply, I don't see how introducing a hustler is a fair explanation, and I don't see how there could be a fair explanation without introducing a hustler. Do you actually believe that the probability of the car being behind door #1 is 33% to me, and the probability of the car being behind door #1 is 50% to someone that just walked in? It just feels to me like everyone is caught up in a logic trap... the only new information you have is about the door that is no longer an option. You don't know anything at all more about which door the car is actually behind. The proof is that if I choose #1 and Monty reveals the goat behind #2, I'm left with #1 v. #3. If I then roll it back and I choose #3, and Monty reveals #2 again... I'm no better off...(edit: I know that Monty could then reveal #1 and give away that the car is behind #3, but I'm assuming that Monty isn't an idiot) and yet, you would have me think that #3 had the car 66% of the time, and #1 now has the car 66% of the time. It doesn't work.


 * Why is the third door different from the second coin in your pocket when you pick a coin to flip?
 * The third door was eliminated for different reasons depending on where the car is. If the car was behind #2 the host HAD to open (100% chance) open #3.  This means the composite probability the car is behind #2 AND the host opens #3 is 1/3 (1/3 * 100%). If the car was behind #1 the host picked randomly between #2 and #3, so the composite probability the car is behind #1 AND the host opens #3 is 1/6 (1/3 * 50%).  The coin in your pocket, or the coin you flipped yesterday, has nothing to do with a coin you're flipping now.
 * In my understanding, the door revealed by Monty has nothing to do with the choice before me now. One could say that doors and coins are inherently different and answer the question... but the substantive question doesn't seem to me to be answered.


 * Another way to think about this is to imagine what happens in 300 shows where you've picked #1 at the beginning. Can you follow through the exercise above at ? -- Rick Block (talk) 19:44, 10 November 2014 (UTC)
 * Yes, I read through that one before. It doesn't clarify it for me at all. I think my ["choose from n doors" where on door is a car and n-1 doors are goats] formulation above is a better representation.

This is a great problem. I have to say that I completely understand why this is so problematic and controversial. It feels to me like the "you just don't get it" argument for dot coms... where there is a stigma attached to disagreeing, but the people that disagree "just don't get it". I'm enjoying the exchange tremendously... and I won't keep droning on about this. Thanks, again! NicolasOliva (talk) 20:18, 10 November 2014 (UTC)


 * Please stay with us for a while longer.


 * Many people remain resistant until they see an actual experiment. If I'm following your explanation I believe you're saying that if we start with (say) 52 cards and I"m trying to pick the Ace of Spades, if I pick one and keep it face down while you look at the other 51 and throw away 50 that aren't the Ace of Spades, then when there are two cards left I'll have a 50/50 chance of the card I originally picked being the Ace of Spades.  Is this correct?  If so, can you try this, say 10 times.  Shuffle a deck of cards.  Deal one to me.  Look at the rest.  Throw away 50, none of which are the Ace of Spaces.  How many times out of 10 do I end up with Ace and how many times do you end up with it?  Played this way, I think I have a 1/52 chance of ending up with the Ace, so I'll be surprised if I end up with the Ace more than once out of ten tries.  Please actually do this, and let us know your results.  -- Rick Block (talk) 02:40, 11 November 2014 (UTC)
 * Rick, your cards explanation got through to me, though I have to admit I still wasn't convinced (but, see below). NicolasOliva (talk) 23:30, 11 November 2014 (UTC)


 * As a direct response to your 4 numbered points above


 * 1) No it isn't. In the initial case you have a 1/3 chance of picking the car. 1 car, 3 doors odds are 1 in 3.
 * 2) Again it isn't but it's a slightly more complicated reason why, and in fact that reason is what other people are elaborating above.
 * 3) Again not quite true. The odds that are 1/3 are that your initially chosen door has the car behind it. Given that your final choice will always be between your initally chosen door and the unchosen door (with Monty's chosen door removed) the key numbers are the odds of your initially chosen door and the unchosen door having the car behind them at the moment you choose.
 * 4) again not quite true (ignoring the typo about Monty picking a car) for the same reasons as 2 above.


 * In the game there are 3 things that happen
 * The prizes are distributed behind the doors. At this point there is a 1/3 chance any door conceals the car
 * You choose one of the doors, this doesn't change the probability of any of the doors concealing the car
 * Monty chooses a door. That door must contain a goat. The odds for that door containing the car go to 0 because of this event. The 1/3 chance that that door contained the car must be redistributed somehow. This redistribution is uneven and that's what makes the puzzle.


 * At step 2 you have chosen a door. in 1/3 cases it contains the car [A] and in 2/3 cases it contains a goat [B]. When it contains a car [A] Monty can open either door reveal a goat and leave a goat. When it contained a goat [B] Monty has to reveal the remaining goat and leave a car. So in [A] you always win by staying and in [B] you always win by switching. Since 2/3 time you are in scenario [B] 2/3 time you win by switching.
 * The relevant information that leads to the asymmetry is you know Monty couldn't have opened the door you picked, the rules of the game don't allow it. However you know that Monty could have opened the door you didn't pick but didn't. This additional information about the remaining door is why (or at least one explanation for why) the remaining door is more likely to be the car than the door you originally picked. SPACKlick (talk) 16:24, 11 November 2014 (UTC)
 * SPACKlick, after Rick got me thinking, your explanation hit the spot. I get it. Thanks, everyone! NicolasOliva (talk) 23:30, 11 November 2014 (UTC)

The way I finally understand this
In case this helps anyone else, the way I finally understand this is as follows:


 * 1) When you choose a door, your probability is 1/3.
 * 2) The probability that the car is behind one of the other two doors is 2/3. Nothing else that happens changes the 2/3 probability of the car being behind one of the doors you did not choose.
 * 3) Monty revealing a goat and offering a switch is exactly equivalent to him saying, "You win the car if it's behind either of the two other doors and you agree to switch." (and "behind the two other doors" has a probability of 2/3)

Thanks again, everyone! NicolasOliva (talk) 17:22, 12 November 2014 (UTC)


 * Nicolas, have you read through the two paragraphs in the 'Simple solutions' section starting, 'Another way to understand the solution is to consider the two original unchosen doors together.'?  This says more or less what you have just said.


 * How do you think it could be explained better in the article? Martin Hogbin (talk) 17:26, 12 November 2014 (UTC)


 * Martin, I had read those before... and had read other things online and seen a couple of youtube videos... so I lost track of that. The table at the top of simple solutions is way confusing; I can't make sense of it. The diagram at right with the two images - in fact, just the second image - is the best "simple solution", in my view. If you show just that at the top of the section, I think that would really capture it best. I'm not sure why I glossed over the diagram before. It may have been because I thought it was related to the table, or because I had trouble with "odds move to 0". It's a nuance, but I like not thinking that the odds "move" at all. The first diagram doesn't distinguish the odds of door 3 from door 2, so saying they moved to 0 makes me wonder... "from what?" I like thinking that the odds for the two doors are still 2/3... and you know now the one of those two doors that doesn't have the car... (anything else I add here to try to close the loop seems to confuse things, so I like leaving it right there). NicolasOliva (talk) 19:12, 12 November 2014 (UTC)
 * Also, I would change the diagram so that instead of door numbers, it just says, "Door you picked" and "Doors you didn't pick". The door number thing is a way to identify doors that is different than identifying them by "you picked" v. "you didn't pick". Having both identification methods leads to "what if I picked #2", etc. The door you picked could be any number... in fact, I would show the doors as points on a triangle rather than side by side to highlight that it doesn't matter which is "#1", etc. NicolasOliva (talk) 19:16, 12 November 2014 (UTC)
 * Thanks for your comments. The problem with door numbers is that they were mentioned by vos Savant in the most well known statement of the problem so they are genrally included. You cannot imagine how much this could complicate things. Martin Hogbin (talk) 19:36, 12 November 2014 (UTC)
 * Door numbers do not complicate the MHP, they form the crux of it. Nijdam (talk) 11:23, 13 November 2014 (UTC)
 * @NicolasOliva: Comment on your points above: 1)The probability $$p_i$$ the car is behind door i is 1/3 for all i. Also the probability $$q_{ij}$$ the car is behind door i if you picked door j is 1/3 for all i and j. 2) It is simple primary school math that for instance $$q_{21}+q_{31}=2/3$$. No big deal. But what you mean by: "Nothing happens etc" is unclear. Of course is $$q_{21}+q_{31}=2/3$$, and this will be so forever. 3) Is just some interpretation from you, but need to be formulated in a much more rigourous way. Nijdam (talk) 11:37, 13 November 2014 (UTC)
 * Door numbers are irrelevant to the MHP and certainly do not form the crux of it, see below SPACKlick (talk) 11:40, 13 November 2014 (UTC)

Door Numbers are the "Crux"?
In response to Nijdam Door numbers do not complicate the MHP, they form the crux of it.. They do not form the crux of it. Un-numbered doors that cannot be distinguished until they have taken their role in the game still form an MHP. Consider the below
 * Three doors which are identical and cannot be distinguished. the contestant sticks a Big yellow star on one to pick it as his
 * The contestant is then blindfolded and the host opens a non-starred door. If that door contains a goat the contestant is unblindfolded. If that door contains a Car then the host opens the other non-starred door and closes the door with the Car then the contestant is unblindfolded.
 * The contestant chooses between the starred door and the closed unstarred door.

So at the start we have 3 identical doors. After the contestant chooses we have a starred door (X) and two identical doors. Then when the host opens a door you have the chosen door (X) and open door (H) and a closed unstarred door (R). At no point do we have door numbers.

1) Do you agree this puzzle is mathematically similar to the MHP 2) Do you therefore agree that door numbers are a distraction in the MHP? — Preceding unsigned comment added by SPACKlick (talk • contribs) 11:40, 13 November 2014


 * No, I do not agree, and this problem is not identical, nor equivalent to the MHP. Only if you understand this this, you understand what the MHP is. Nijdam (talk) 13:32, 13 November 2014 (UTC)


 * Before we reignite an old argument, let me say that there is no official answer to what the MHP is meant to be. We can say some things though:


 * Selvin, Whitaker, and vos Savant did not intend door numbers (or goat identities) to be significant. This can all be clearly verified from reliable sources.


 * Some mathematicians prefer to treat the problem as one in which the door identities (but not the goat identities) play an important part. In this form it can be used to teach some aspects of contitional probability. Martin Hogbin (talk) 14:34, 13 November 2014 (UTC)


 * Nijdam, You have some weird definition of the MHP if you don't think this problem is the same. The paradox of the MHP is in the 2 options with different probabilities because of biased selection and prior knowledge. None of that is reliant on door numbers and nothing in the MHP is reliant on door numbers. This I think is part of why we keep discussing at cross purposes. So please, in the nicest possible way. Define the contents of the MHP that are inherent and relevant to the puzzle. By which I mean, in your definition is it still the MHP if the goats are replaced with Cats? What if there are no prizes behind the losing doors? What if the doors are urns? What is the minimal MHP?
 * — Preceding unsigned comment added by SPACKlick (talk • contribs) 15:02, 13 November 2014


 * Essential is the actual final choice between two alternatives, which seemingly are equally probable. Nijdam (talk) 10:54, 15 November 2014 (UTC)
 * Exactly, and it does not require door numbers to arrive at that. Martin Hogbin (talk) 12:33, 15 November 2014 (UTC)
 * This is a demonstration that words are often multi-interpretable. Anyway: what is the final choice in the MHP?Nijdam (talk) 10:36, 16 November 2014 (UTC)

Yet Another Simple Solution
When I first come across this problem, not only did I get the answer wrong (going for the “common sense” 50/50 one) but I also had trouble fully understanding some of the solutions. However, I did eventually manage to work it out myself as explained below for the benefit of anyone else in the same boat.

Firstly: it’s important to understand all the initial assumptions and have these clear in your mind. These are explained in the main article so I will not bother restating them here.

Following on from the assumptions: when a contestant chooses one of the doors, then there has to be a 2 in 3 chance that there is a goat behind it.

Once a goat has been chosen (as it will be on 2 out of 3 occasions) then the games host has no other choice but to reveal the other goat (he cannot reveal the car or goat picked by the contestant). Therefore, on 2 out of 3 occasions (on average) both goats will be chosen (one by contestant and one by host).

It logically follows that there has to be a 2 out of 3 chance that the remaining unchosen/unopened door must hid the car (and therefore just a 1 in 3 chance of the contestant’s door doing so) and so the contestant should swap to it. — Preceding unsigned comment added by Alanbellis (talk • contribs) 10:28, 18 March 2015 (UTC)


 * Alanbellis, you are fully correct, you managed to descry the paradox and to comprehend this famous paradox: that switching to the door offered by the host to switch on, as an alternative gives the contestant a chance of 2/3 indeed to get the car. Swithing doubles his/her chance.


 * All the contestant knows and will ever know is that with probability of 2/3 he/she actually is in the "wrong guess scenario", in having chosen a goat, and vice versa with probability of only 1/3 he/she actually is in the "lucky guess scenario", in having chosen the car. That's all he/she can ever know and will ever know.


 * Btw: the mishap of the article is that some mathematicians – in ignoring the circumstances, and in deviating from this amazing paradox, declared that – given the host should say:
 * "I just opened this door with a goat behind because I was not able to open my other door that I strictly prefer to open, for regrettably it actually hides the car"
 * - then the contestant suddenly has come to know he actually isn't in the "lucky guess scenario" with probability 1/3, but with probability 0 because actually he/she is in the wrong guess scenario for sure, so switching to the door offered will give him/her the car with probability of actually 1.
 * They use such preposterous deviating scenario of some chatty host to determine the probability to win the car by switching somewhere within the range of 1/2 to 1, and they used such deviating scenario of such chatty host in their maths classooms in teaching the use of formulas in probability theory calculus (see Nijdam's arguments). Confusing by deviating from the original famous clean paradox. Having no bearing on the famous Monty Hall paradox — just only some teaching method for probability theory calculus. Saucily calling such deviant scenario "the crux of the Monty Hall problem". I call this a paradoxical approach, because "the crux" of the paradox is that the contestant only knows that the actual rate of probability "lucky : wrong guess scenario" is "1/3 : 2/3", any update of knowledge strictly impossible, see Prof. Norbert Henze, a famous German mathematician.
 * Regards, Gerhardvalentin (talk) 11:02, 22 April 2015 (UTC)


 * The point the "some mathematicians" (i.e. Morgan et al, plus Gillman, Rosenthal, Lucas et al, Eisenhauer, Falk, Grinstead and Snell - all in the references) make is not that the simple argument produces the wrong answer to the paradox (2/3 chance of winning by switching), but that it's based on sloppy reasoning. Either it


 * a) quite subtly changes the question from "should you switch given you've picked door 1 and the host has opened door 3 showing you a goat" (which is what most people understand the question to be, per Kraus & Wang) to "what is the best strategy, to always pick and stay or to always pick and switch" (i.e. it moves the "switch or not" decision to the point before the host opens a door rather than after, meaning you have to declare your intent to switch before seeing which door you'll be switching to)


 * or


 * b) includes another assumption not generally specified (that the host chooses evenly between two goats if the player initially picks the car, or alternatively that the player has no way to tell the difference between, say, door 2 which remains closed and door 3 which the host opens).


 * In particular, the argument that what the host does cannot change the player's initial probability of 1/3 of choosing the car (meaning the chance door 1 hides the car is 1/3 before the host opens door 3 guarantees this chance is always still 1/3 after the host opens a door) is manifestly false (as shown by the "preposterous deviating scenario"). That's simply the way math works.  If you say (a/b)*b = a, and use this in a proof, your proof is at least sloppy (if not wrong) - because if b happens to be 0 then your assumption doesn't hold.  It works most of the time.  Just not ALL the time.  One counter-example, however contrived it might be, is enough to show a proof is incorrect, or in this case that the reasoning behind a solution to the MHP is sloppy. -- Rick Block (talk) 15:55, 23 April 2015 (UTC)

Are the odds ever 50-50 ?
It's been established that the contestant and Monty have produced a pair of doors containing a car and a goat that do not have the same odds of producing a goat. Yet, on Marilyn vos Savant's website where she discusses this teaser, she perhaps yields too much ground to the academic furor that confronted her. She made this concession:

"Suppose we pause at that point [when two doors remain], and a UFO settles down onto the stage. A little green woman emerges, and the host asks her to point to one of the two unopened doors. The chances that she’ll randomly choose the one with the prize are 1/2, all right. But that’s because she lacks the advantage the original contestant had—the help of the host." 

But are the chances 50%? Probabilities don't care much for who does the choosing, and neither the contestant nor the green woman has insider information about the whereabouts of the car behind the doors. The contestant can know that the odds of finding a car behind the switched door are 2/3, but the odds don't change just because someone doesn't know the odds. If I know that four sides of a die are red and two are blue before I paint over the sides, the odds of someone who has never seen the die rolling red are not 1/2 due to her unawareness.

If the odds of randomly selecting a car among the two blind choices are 50-50 for anyone other than the contestant, then they must be 50-50 for everyone. We know that they are not, so the green woman approaches a stacked deck, so to speak. Otherwise, we end up saying that all odds between situations with two outcomes are 50-50 so long as we know nothing about the underlying conditions. That's not correct.

Vos Savant refers to the help of the host as being a missing ingredient, but that help actually settled the odds prior to his final selection, and it must've done so for anyone choosing between the two doors. We are now convinced that this problem is a matter of frequency of appearance of the car behind particular doors after particular actions have settled the odds, so vos Savant needn't have made any concessions to the contrary. Summers999 (talk) 13:38, 26 April 2015 (UTC)


 * The question is what odds are you talking about? If you pick randomly between two choices the odds of your choice being the correct one are 50-50. Note how I worded this - the odds we're talking about are the odds of "your choice". If we roll your red/blue die the odds are 4/6 that it ends up red and 2/6 that it ends up blue. If you roll it 100 times and I randomly guess red or blue, about 50 times I'll guess red and 50 times I'll guess blue. Of the 50 times I guess red I'll be right about 4/6 of the time, so about 33 times. Of the 50 times I guess blue I'll be right about 2/6 of the time, so about 17 times. Altogether I'm right about 50 times out of my 100 guesses, i.e. 50-50. If I don't know 4 sides are red and 2 are blue, my odds of correctly guessing are 50-50 whatever the "actual" odds may be because I have no choice but to randomly guess. If I always guess red (because it's my favorite color), I'll be right about 4/6 of the time, but now we're not talking about the odds of a random guess but the odds that the die ends up red (which is a different question) -- Rick Block


 * I was concerned that the odds of 2/3 and 1/3 were somehow negated in her statement, but your explanation shows me that they aren't, and vos Savant likely meant something similar to that. So, I'll allay my concerns about vos Savant's comment and simply point out for clarity that the odds of the car being behind the 'remaining' door stay at 2/3, even though random chance guessing as you described will land on the car 50% of the time. If the green woman were offered one selection, she'd be facing unequal odds in the two choices, but an army of green women selecting randomly would indeed find the car 50% of the time. Thanks for the discussion! Summers999 (talk) 13:38, 26 April 2015 (UTC)


 * Exactly! "Odds" are a function of perspective, or alternatively a function of the information available to you.  If I pick a random card out of a deck of 52 cards (let's say it's the 8 of spades) you have a 1 in 52 chance of guessing it correctly.  If I tell you it's a spade, you now have a 1 in 13 chance.  If I tell you it's an 8 (without telling you which suit), you have a 1 in 4 chance.  If I pick a card and 3 people get to guess, and one doesn't get any additional information, one gets the suit, and one gets the rank, then those three people have different odds all at the same time.


 * In the MHP, the host knowingly opening a door showing you a goat is giving you some information, but not as much as he has. He knows where the car is so the odds for him are not 1/3:2/3 but either 0:1 or 1:0.  For the player who has picked door 1 and has seen the host open door 3 (with the usual restrictions on the host including that the host picks evenly between two goats if this case comes up), the odds are 1/3:2/3.  For vos Savant's little green alien who doesn't know the setup, the odds are 1/2:1/2.   All of these are true, all at the same time - because the information available is different.  -- Rick Block (talk) 15:53, 26 April 2015 (UTC)


 * The article says it clearly (again): Indeed, if a player believes that sticking and switching are equally successful and therefore equally often decides to switch as to stay, they will win 50% of the time, reinforcing their original belief. Missing the unequal chances of those two doors, and in not considering that (1/3+2/3) / 2 gives a chance of 50%, similar to "the little green woman" example.  –  The little green woman cannot distinguish the "1/3 door" (first selected by the contestant) from the the "2/3 door" (offered by the host as an alternative), so she cannot take advantage of this contestant's knowledge. Gerhardvalentin (talk) 17:43, 26 April 2015 (UTC)

Symmetric host decisions vs. asymmetric host decisions
Another way to look at it:

In the main problem, the host chooses 50:50 which of the goats he reveals if he has a choice.

I found some rest on playing with this probability...

Assume the host is a lazy one, and because he has to walk from left of door one (the one the candidate picks), and he opens the *first* door with a goat behind it.

If he opened door 3, the player knows with absolutely certainty that the car is behind door 2. Why?

The host could in principle open door 3 only in two cases: a) Car, Goat, Goat or b) Goat, Car, Goat. But in case a), lazy host won't open door 3, he'll open door 2!

If on the other hand he opened door 2, the chances are exactly 50:50:

The possibilities of a) Car, Goat, Goat or b) Goat, Goat, Car, are both equally as likely in this specific scenario of "lazy host" (and door 2 open), because for each pattern, the host will pick door 2 in 100% of the cases of the patterns.

Back to the "regular" problem, if a host chooses with 50:50 chance:

If the play sees the host open door 2, he knows that there are only two patterns left (because in the third pattern, goat, car, goat, the host is not allowed to open door 2):

a) car, goat, goat or b) goat, goat, car

Are they equal? No:

In pattern b), the host did NOT have a choice. 100% of all b)-cases can only end up like this, there is no other way.

But in pattern a), the host DOES have a choice. He can just as well choose door 3. So, only 50% of all a)-cases end like this, because there is another option for the host (door 3).

So, if ORIGINALLY both patterns (a) and b)) were equally as likely, then by the host's pick, pattern a) is split half and half between door 2 and door 3.

The other two patterns are not split and retain their original probabilities.

The relative probabily of 50% compared to 100% yields the "official" result: 1/3 to 2/3.

(Or more mathematically: before the host picked, all three patterns were equal. After the host picked, pattern c) jumps to zero. If b) is then twice as likely as a) and those two are all that's left, we have:

a) + b) = 1 and b) = 2 * a) which resolves to a) = 1/3, b) = 2/3.)
 * I think this is all quite well known. There are many ways to look at the problem but no one way seems to work for everyone. Martin Hogbin (talk) 17:23, 15 June 2015 (UTC)

A couple ways to clarify the meaning of the odds
To look at it from a different angle, the contestant has a 2 in 3 chance of choosing a door with a goat. That means the remaining pair of doors have only a 1 in 3 chance of containing two goats. When Monty reveals a goat from the remaining two doors, then the other door will have a goat one-third of the time. Therefore, the contestant halves the odds of getting a goat when switching doors, the same of course of doubling the odds of getting a car.

Note: this also obviates the question as to the importance of Monty Hall revealing a goat.

People get hung up on the idea that the contestant is asked to make a second choice among the remaining two options, leading them to think their odds at that point are 1 in 2. There are two doors. One has a goat, and the other has a car, so a random choice should get the car half the time, they think. It's important to point out that the problem revolves around an initial selection and actions based on that selection. There are two doors, but a car needn't appear behind both of them with the same frequency. The way the problem is structured, you can think of one door having the car more often than the other. It's as simple as that, and the counter-intuitive feel of this problem leads people to think their odds are 50-50.

The choices appear to be equal, but they are not because Monty Hall acts on the contestant's choice. His behavior relies on the contestant's initial choice. This is not the same as being offered a choice between two equivalent options. What happens in a seemingly similar case when there are six contestants in a singing contest and one will be selected to go home? If all things are equal, each contestant puts his/her odds of going home at 1 in 6. But as the MC declares one contestant after another to be safe, the remaining contestants no longer rely on those 1 in 6 odds. Had I placed an initial bet that one of the final two contestants remaining would go home, I would no longer view the odds of my winning that bet as 1 in 6, but I'd think I had a fifty-fifty chance at that point. I wouldn't imagine that the other contestant had a 5 in 6 chance of going home that day just because I made an initial random choice of a contestant. The difference in these scenarios is that the judges and the MC have no knowledge of my bet and take no actions based on it; whereas, Monty preserves the game show contestant's choice when he acts. Summers999 (talk) 13:38, 26 April 2015 (UTC)


 * A much better example to explain the problem would be that a magician lays out 52 cards from a standard deck in front of a member of the audience, asks them to pick one, looks at it, places it in front of them, and then picks up and looks through the remaining cards and places another card in front of himself, then places a gold bar between them. One card on the table, he says, is the ace of spades, and if the person picks the right one they get the gold (since the ace of spades is the most metal of all cards). They can go with the card they initially chose, or pick the one in front ofT the magician instead.


 * In this example it's pretty clear that unless the initial pick was correct, the ace of spades is in the 51 cards the magician picked up, and so he has a 51/52 chance of having the card to pick it and therefore it being in front of him, while you only have a 1 in 52 chance of having it in front of you.


 * I think what confuses people most about the original example is what they're actually being asked. The chance of a car physically being behind a specific door at the point of the second decision is 1 in 2 (it can only be behind one, and was never behind the third) but the chance of their initial guess being correct is 1 in 3. It's also a problem that you're making a decision based on a one-off trial with non-trivial odds that you initial guess was correct, and so even taking the advice you stand a 1/3 chance of losing; increasing the odds and tossing out pointless elements (why goats?) makes it a bit clearer what the point is.


 * After all, if the example confuses the majority of people who hear it, the problem is probably that it's a bad example. Herr Gruber (talk) 07:45, 15 January 2016 (UTC)

Is this the same problem with the same solution?
I want to pose a similar but different problem... Imagine I'm a woman and three different men want to start dating me. I pick one and start dating him. Then I learn from a friend that one of the other two is an absolute nutcase, and I'm really glad I didn't start dating him. Should I now dump my initial choice, and start dating the third man? I'm sure you can see the similarity with the Monty Hall problem. I'm looking for the man who is a nice guy (the car). I want to avoid the guys who are jerks (goats) I want to ignore the moral issues here, and just look at the probabilities. I understand the probabilities of the Monty Hall problem, and I get that the player should switch doors. But I have difficulty seeing how learning that someone I never dated is a jerk means I'll be better off switching partners. Now there are some differences between the problems. I get that. Obviously all three guys could be jerks. ie. There could be three goats, and no car. :( It's simplistic and unfair to classify all men as goats or cars. It's more sensible to say there's one who is most compatible with me, and two who are less compatible. Another problem is knowing a goat when I see one. If all three are jerks, it is possible that the one I heard was a nutcase, is in fact the best of a bad bunch. He could be the car after all. When it comes to the real world of course, we all have squillions of potential partners. The main article talks about what happens if there are more than 3 doors. (N-doors.) As you consider more and more doors, the benefit of switching gets smaller and smaller. So in reality, if I hear that some guy somewhere is a jerk, the benefit of switching is miniscule. It's only if my choices are limited for some reason that there might be a benefit in switching. — Preceding unsigned comment added by 14.2.57.97 (talk) 12:02, 7 May 2015 (UTC)

Reasons why it's not mathematically similar None of the odds are interconnected so the maths of the Monty Hall doesn't even slightly apply. SPACKlick (talk) 12:06, 7 May 2015 (UTC)
 * There was no guarantee of a Nice Guy to begin with
 * There was no guarantee of two Jerks to begin with
 * There was no guarantee your friend would reveal one of the two unchosen jerks to begin with

I think I've worked out the answer to my own question. Yes it's basically the same problem, and yes I should change boyfriends. Imagine this similar game. Instead of doors, Monty has a bag of marbles. The contestant has to put his hand in the bag and pull out a marble without looking. If the marble is red, he wins the car. If it's black, he doesn't. But the contestant isn't told how many or each colour there are in the bag. He just knows there are some of each. So Monty looks in the bag, and pulls out a black marble. The contestant still doesn't know how many black and red marbles there are. All he knows is that there is now one less black one. So the contestant can deduce that his chances of winning have just gotten better than they were before Monty removed the marble. He can't work out what his chances of winning were, what they are now, or how much they have improved, but they _must_ have improved. The only way they could stay the same is if _all_ the marbles were black, in which case my chances were zero to start with, and when Monty removes the ball they stay at zero. If Monty has only three marbles in the bag, the improvement is substantial. If he has lots of marbles, say 300, the improvement is very small. So in the same way, if I draw up a 'short list' of 3 potential partners, and then cross off one from that list, on the basis that he's a jerk, it follows that my new list must have proportionally more nice guys than my old list did, unless I'm really unlucky and they are all jerks. The same logic follows if I have a short list of 3 or 300, but (like with the bag of marbles) the improvement gets less as the list gets longer. — Preceding unsigned comment added by 14.2.57.97 (talk) 09:24, 13 May 2015 (UTC)


 * The thing that surprises many peope about the MHP is that it matters whether the host always shows a goat or picks at random and just happens to show a goat. As SPACKlick has pointed out, many things in your problem are not clear and it is thus impossible to answer. We can probably guess what you meant to ask but it would be easier if you told us. Martin Hogbin (talk) 08:42, 14 May 2015 (UTC)


 * This assumes you know a 1:2 ratio of "nice guys" and "jerks" in the first place, or that the world is cleanly divided into two such groups to begin with (it's not). You could change it to the fact that you could potentially date Andy, Bob or Cody, you date Andy, but then Zelda calls you up and tells you Cody is dead. Your argument would then have you dump Andy on the basis that there is an increased chance that one you initially picked is dead (since you now know Cody is), and there are better odds of Bob being the one who's alive. This probably tips you off to the fact that the way you're using it isn't exactly valid. Herr Gruber (talk) 08:44, 15 January 2016 (UTC)

What is the meaning of "preference"?

 * FourthKind, I recommend you to stop using terms like "choice" or "preference" that are confounding you, and use mathematical terms instead, which have a precise definition in terms of probability theory. A problem in probability can have more than one random variable. Some of these variables represent actions taken by actors that take decisions in the problem, and some other variables are measurements of physical events and properties. You measure probabilities by calculating the chances that a random variable takes a value in a set of values.
 * For example when you say: "In reality this “preference” has a 0 probability of winning as a “preference” in reality", I cannot make sense of that sentence at all, and I think you don't have its meaning clear either. Can you define in mathematical terms what is the event that has 0 probability, so that I can understand it? For this you need to answer these questions:
 * What values can take the variable "preference"?
 * What value or values in the variable "preference" represent a "win"?
 * Do all the values in the variable "preference" have the same chances of happening in reality?
 * If you can answer these three simple questions, you'll have a mathematical model of the first part of the Monty Hall Problem; I think having this model will help you. If you can't give a precise answer to these simple questions, your sentence does not have a precise meaning so it can't be part of a mathematical proof. Diego (talk) 16:59, 3 March 2016 (UTC)