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The Monty Hall problem is a probability puzzle based on the American television game show Let's Make a Deal. The name comes from the show's host, Monty Hall. The problem is also called the Monty Hall paradox, as it is a veridical paradox in that the result appears absurd but is demonstrably true.

Problem[edit]

Steve Selvin wrote a letter to the American Statistician in 1975 describing a problem loosely based on the game show Let's Make a Deal (Selvin 1975a). In a subsequent letter he dubbed it the "Monty Hall problem" (Selvin 1975b).

Selvin's Monty Hall problem was restated in its well-known form in a letter to Marilyn vos Savant's Ask Marilyn column in Parade:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice? (Whitaker 1990)

There are certain ambiguities in this formulation of the problem: it is unclear whether or not the host would always open another door, always offer a choice to switch, or even whether he would ever open the door revealing the car (Mueser and Granberg 1999). The standard analysis of the problem assumes that the host is indeed constrained always to open a door revealing a goat, always to make the offer to switch, and to open one of the remaining two doors randomly if the player initially picked the car (Barbeau 2000:87).

So vos Savant emphasized those elements, all of which she considered implicit, in her solution (Savant 1996, p. 15). When it appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine claiming the published solution was wrong. But virtually none of that controversy was related to these possible ambiguities (Savant 1996, p. 15). (Seymann 1991) says she made her intent quite clear: the host is to be viewed as nothing more than an agent of chance who always opens a losing door, reveals a goat, and offers the contestant the opportunity to switch to the remaining, unselected door. And the standard analysis of the problem assumes that the host is indeed constrained always to open a door revealing a goat, always to make the offer to switch, and to open one of the remaining two doors randomly if the player initially picked the car (Barbeau 2000:87).

The following formulation represents a more formal statement of her intent that is, according to Krauss and Wang (2003:10), what people generally assume the mathematically explicit question to be:

Suppose you’re on a game show and you’re given the choice of three doors. Behind one door is a car; behind the others, goats. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one randomly. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you “Do you want to switch to Door Number 2?” Is it to your advantage to change your choice? (Krauss and Wang 2003:10)

As the player cannot be certain which of the two remaining unopened doors is the winning door, most people assume that each of these doors has an equal probability and conclude that switching does not matter. In fact, the player should switch—doing so doubles the probability of winning the car from 1/3 to 2/3.

The above formulation of the problem is mathematically equivalent to the Three Prisoners Problem described in Martin Gardner's Mathematical Games column in Scientific American in 1959 (Gardner 1959a), and both are related to the much older Bertrand's box paradox. These and other problems involving unequal distributions of probability are notoriously difficult for people to solve correctly, and have led to numerous psychological studies. Even when given a completely unambiguous statement of the Monty Hall problem, explanations, simulations, and formal mathematical proofs, many people still meet the correct answer with disbelief.

Simple explanation[edit]

The player, having chosen a door, has a 1/3 chance of having the car behind the chosen door and a 2/3 chance that it's behind one of the other doors. When the host opens a door to reveal a goat, this action does not give the player any new information about what is behind the door she has chosen, so the probability of there being a car remains 1/3. Hence the probability of a car behind the remaining door must be 2/3 (Wheeler 1991; Schwager 1994). Switching doors thus wins the car with a probability of 2/3, so the player should always switch (Wheeler 1991; Mack 1992; Schwager 1994; vos Savant 1996:8; Martin 2002).

Complete solution[edit]

Mathematical formulation[edit]

The above solution may be formally written in terms of the random variables: C = the door number hiding the car and H = the number of the door opened by the host. As the initial choice of the player is independent of the placing of the car, the solution may be given on the condition of the player having initially chosen door No. 1. Then:

P(C=1) = P(C=2) = P(C=3) = 1/3. (the car is placed randomly)

The strategy of the host is reflected by:

P(H=1|C=1)=0 (the hist does not open the chosen door)

P(H=2|C=1)=P(H=3|C=1)=1/2 (the host acts randomly if needed)

P(H=2|C=3)=1 (the host has no other option)

P(H=3|C=2)=1 (the host has no other option)

The player now may calculate the probability of finding the car behind door No. 2 after the host has opened door No. 3, using Bayes' rule:

P(C=2|H=3)={\frac {P(H=3|C=2)P(C=2)}{P(H=3|C=2)P(C=2)+P(H=3|C=1)P(C=1)}}=

={\frac {1\cdot {\frac {1}{3}}}{1\cdot {\frac {1}{3}}+{\frac {1}{2}}\cdot {\frac {1}{3}}}}={\tfrac {2}{3}}.

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Please, above this line only propositions to alter the above text

Do you want comments here? If not please feel free to move this. I would leave the K & W formulation until after the simple explanation. This leaves the question intentionally more ambiguous which gives greater freedom for a simple explanation. I would also want to add some pictures or diagrams (like the one below only with pretty pictures) for the simple explanation and a discussion of why people get the answer wrong. The first section should be easy to understand and convincing. Martin Hogbin (talk) 12:37, 2 January 2010 (UTC)[reply]

What is the purpose of this 'Construction'. Is it to rewrite the MHP Article from the ground up? I would rather see us start with a copy of the existing article, and make changes to it, so that all additions, deletions and changes have an audit trail.

I think the combining doors solution with images should remain in the Simple solution section. Glkanter (talk) 13:13, 2 January 2010 (UTC)[reply]

Is this a part of the mediation process? Has there been unanimous agreement to work with this mediator? What does that mean that we are each committing to? Glkanter (talk) 13:15, 2 January 2010 (UTC)[reply]

>>>Does your solution refer to the above formulation of the MHP??Nijdam (talk) 11:14, 17 December 2009 (UTC)[reply]

>>>I guess it does not, as in the above formulation the player always chooses door 1. The question answered is probably quite close to that which you have asked Jeff about and is this reasonable interpretation of Whitaker's original question:

>>>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, and the host, who knows what's behind the doors, opens another door, which has a goat. He then says to you, "Do you want to pick the remaining door ?" Is it to your advantage to switch your choice?. Martin Hogbin (talk) 12:50, 17 December 2009 (UTC)[reply]

Then call x the number of the door chosen, and h the number of the opened door. Give your solution as a function of x and h. Nijdam (talk) 14:50, 20 December 2009 (UTC)[reply]

>>>But I do not care to specify which door was chosen by the host or the player. Any problem can be made artificially conditional, for instance: ' There are three numbered doors with a car placed behind door 1. A player picks a door randomly what is the probability of picking the car'. Do we need to consider three separate cases, that the player picks door 1, door 2, door 3? Martin Hogbin (talk) 15:18, 20 December 2009 (UTC)[reply]

That's why I say: call the chosen door x etc. So you don't have to specify it. But anyhow one of the three doors has been chosen and another opened and shown to the player. In your example: yes, you need to consider 3 seperate cases, or reason that the choice 1 gives the same answer as the other choices, or, call your choice x etc. Nijdam (talk) 17:08, 20 December 2009 (UTC)[reply]

Or put it another way: suppose you are the player. Tell me which door you have chosen and which one you see opened. Nijdam (talk) 17:11, 20 December 2009 (UTC)[reply]

What the player sees or knows is not important. We are to answer the problem only on the information given in the question. The door opened by the host is not specified. Martin Hogbin (talk) 18:04, 20 December 2009 (UTC)[reply]

We are to see the problem through the eyes of the player. The host offers her the opportunity to change her choice. And, as far as I know, almost everyone (except you and ...) put themselves in the position of the player and consider the door opened. Actually we also have the same information: door 1 (or another) chosen and door 3 (or another) opened. Nijdam (talk) 13:56, 21 December 2009 (UTC)[reply]

I will respond on the arguments page, discussion here is making the page untidy. Martin Hogbin (talk) 21:32, 21 December 2009 (UTC)[reply]

Nijdam, I am interested in pursuing the question of exactly when and why a probability problem becomes conditional but I think that here is not the best place. I suggest the arguments page. Martin Hogbin (talk) 21:41, 20 December 2009 (UTC)[reply]

>>>Nijdam, let me try to turn this problem round so that we might work together towards a solution. I, and I think many others, would like to have the simple solution shown below as the first solution in the article, because it is the solution most likely to be understood by the vast majority of our readers. The question I would ask you is how best to achieve this whist still remaining mathematically correct. What question do think the solution below is the answer to and can this question justifiable be called a formulation of the MHP? Martin Hogbin (talk) 21:30, 20 December 2009 (UTC)[reply]

The table below shows the possible outcomes for three equally likely choices of door. It can be seen that, of the three possibilities, the player who switches wins two and the player who sticks wins one.

You choose a goat		You choose a goat		You choose a car
The host opens a door to reveal a goat		The host opens a door to reveal a goat		The host opens a door to reveal a goat
You Stick	You Swap	You Stick	You Swap	You Stick	You Swap
You get a Goat	You get a Car	You get a Goat	You get a Car	You get a Car	You get a Goat

If you always swap, you 'always' get the opposite (goat or car) to what you would get if you always stick. Thus if you have a 1/3 chance of initially choosing the door with the car, by always switching you have a 2/3 chance of finishing up with the car.

>>> I think that the footnote, which I have moved out of the table, is totally irrelevant to the Simple Solution to the Simple (unconditional) MHP: "We take it that it is unimportant which of the two possible doors that would reveal a goat the host opens. In the case that the host makes this choice randomly it turns out that this is correct, but nevertheless the problem is strictly one of conditional probability (ref Morgan), the condition being the door that the host opens. This, together with the variation that the host is known to choose non-randomly, is discussed in more detail below." Gill110951 (talk) 13:53, 20 December 2009 (UTC)[reply]

>>> Gill we all need to make some effort to compromise here. Many editors insist that the problem must be conditional - a footnote would be a small price to pay for their approval. Martin Hogbin (talk) 14:26, 20 December 2009 (UTC)[reply]

>>>>I agree that the text of the footnote belongs somewhere around here; not just for the sake of compromise, but because I find the conditional problem also important and fascinating. I only propose to move it outside of the table, since its presence there obscures the simple argument that "always switching" gives you an unconditional probability of wining of 2/3, if it is true that the unconditional probability of first having picked the good door is 1/3. Gill110951 (talk) 19:12, 21 December 2009 (UTC)[reply]

Paper by Morgan et al.[edit]

>>>I am trying to write something here that we can all agree on. >>>Nijdam, do you accept this as a fair and reasonable description of the Morgan paper and its importance to the MHP?

>>>>This looks OK, except that I wouldn't call the Morgan et al. paper "important". It's a simple paper which appeared in a journal for people who teach basic statistics. It *is* a nice step in the development of ideas around the Monty Hall Problem. Gill110951 (talk) 19:22, 21 December 2009 (UTC) [I have removed 'important'] Martin Hogbin (talk) 11:05, 27 December 2009 (UTC)[reply]

In 1991 an paper concerning the Monty Hall problem was published by Morgan et al. This considered a more general version of the problem

They discus possible game rules and, in common with many others they consider mainly the game rules in which the host always offers the swap. They also take it that the car is initially positioned randomly and that the player's initial choice is random but they go on to consider the possible non-random actions of the host, which they parametise. They address the following formulation of Whitaker's original question, 'The player has chosen door 1, the host has then revealed a goat behind door 3, and the player is now offered the option to switch'.

Morgan confirm the well known result that if the host chooses any unchosen door randomly (including the possibility of revealing the car, it being assumed that if the host reveals the car the game is considered void or is replayed) the probability of the player winning by switching is 1/2. They continue to show that even if the host never reveals a car and is known to choose a door revealing a goat non-randomly (for example they may always choose door 3 where possible) the player cannot do worse by switching than sticking, pointing out that this problem is one of conditional probability. Finally they confirm that if the host chooses a door randomly, when he can do so without revealing the car, the probability of winning by switching, calculated by the methods of conditional probability, is 2/3, the same value as that calculated by solving an unconditional formulation of the problem. Morgan give as as an example of such an unconditional formulation, 'You will be offered the choice of three doors, and after you choose the host will open a different door, revealing a goat. What is the probability that you win if your strategy is to switch'.

Following Morgan's lead, many later sources treat the problem as one of conditional probability.

>>>Next we could have some of the current diagrams in which the choice of goat door is considered significant.

>>>>I notice the word "randomly" being used above when the writer means "uniformly at random". This is a major source of confusion in basic probability especially when non-experts try to use probability. For instance, not making the distinction arguably led to the conviction for serial murder of the Dutch nurse Lucia de Berk, http://en.wikipedia.org/wiki/Lucia_de_Berk , which now appears to have been a major miscarriage of justice. Gill110951 (talk) 19:22, 21 December 2009 (UTC)[reply]

>>>>Could you explain the difference please. Martin Hogbin (talk) 09:34, 22 December 2009 (UTC)[reply]

>There is much I don't like about this. Saying "an important paper" is POV: it is saying what the editor thinks is important. Morgan's is not a more general version of the problem, it is a different problem that is broader in its scope. And the difference between being "more general" and "broader" is that the broadening is done in areas that are not supposed to be part of the MHP (see my above edits). The game rules Morgan discusses were never intended by those who originated the problem (see my edit above). Nor are they considered as possible rules by many, if not most, of the problem's readers at any level of education or mathematical background. Plus, Morgan really only addresses one game rule - how to choose a door to open - because they can prove that that rule doesn't affect the answer "at least as good to switch." So considering Morgan doesn't add to the understanding of the MHP, as most people see it, in any way. In fact, it is makes it more complicated and harder to understand.

'More general version' means 'broader in scope, this is common usage of the term. Martin Hogbin (talk) 23:01, 22 December 2009 (UTC)[reply]

You misunderstand. The Morgan paper expands the problem in ways that were not in the original. Not in ways that were left out by omission, but in ways that were intentionally not supposed to be a part of the MHP. Remember "anything else is a different problem" ? It is the expressed opinion of the problem's originator that Morgan's analysis is "a different problem." JeffJor (talk) 14:27, 23 December 2009 (UTC)[reply]

>I think it is POV to include Morgan at all in the same article as though it addresses the same problem. There are two views in the literature about the MHP: those who solve it without host strategies whether or not they use conditional probability and those who consider host strategies and therefor must use conditional probability. These two views are not compatible with each other, but the difference has nothing to do with whether door numbers are important or not. It has to do with host strategies only. It is only of they are considered that it door numbers matter (see G&S, where they talk about "without loss of generality" assumptions that work if no strategy is considered). So whether or not anybody can find a handful of references that say the problem does require door numbers (and one of those used actually says the statement, by itself, doesn't require door numbers) is irrelevant. The difference in Morgan is the q, not the door numbers themselves.

>But supporting just one of these views as "the MHP" is also POV because BOTH have support in literature (although I personally can't fathom why). However, the main interest (I'd guess 95% at least) in the MHP is because of the veridical paradox which can be discussed and resolved in its entirety without resorting to discussion of host strategies. In fact, the bulk of the MHP's appearences in literature do just that. Doing so in the article is not contradicting literature, it is paralleling how there are two views in the literature. After that is complete, we can introduce the broader problem that includes host strategies. Bring in Morgan, describe how they parameterize the alternate strategies, and say that the parameterization requires treating door numbers differently.

That is exactly my point, both views should be represented in the article, the simple view first, then, for the few that are interested, the more complicated (obfuscated in my opinion) version of Morgan. My Morgan summary was intended to be a link between the two versions, that is to say it should come after the simple solution and explanations. It also has to be acceptable to the 'Morgan fans'. Martin Hogbin (talk) 23:01, 22 December 2009 (UTC)[reply]

> Gill, "randomly" does mean "uniformly random" unless another distribution is specified. The case you link to seems to be an example of improper handling of dependencies, not the underlying distributions themselves. JeffJor (talk) 18:31, 22 December 2009 (UTC)[reply]

(outindented) The main importance to me of the Morgan paper for this article is its clear analysis of the "standard" case, i.e. the case q=1/2. Their study of host strategies is interesting, but not specific for the Wikipedia reader. Nijdam (talk) 19:51, 23 December 2009 (UTC)[reply]