Please, between this line and the next, only propositions to alter the article
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.
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.
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).
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:
The strategy of the host is reflected by:
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:
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)
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)
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 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?
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.
(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)
Please, between this line and the next, only propositions to alter the article
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.
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.
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).
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:
The strategy of the host is reflected by:
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:
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)
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)
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 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?
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.
(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)