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[[File:Monty open door.svg|thumb|In search of a new car, the player picks a door, say 1. The game host then opens one of the other doors, say 3, to reveal a goat and offers to let the player pick door 2 instead of door 1.]]
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{{quote|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 he knows 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?|[[#refWhitaker1990|Whitaker 1990]]}}
Certain aspects of the host's behavior are not specified in Marilyn vos Savant's wording of the problem. For example, it is not clear if the host considers the position of the prize in deciding whether to open a particular door or is required to open a door under all circumstances ([[#refMueserandGranberg1999|Mueser and Granberg 1999]]). Almost all sources make the additional assumptions that the car is initially equally likely to be behind each door, that the host must open a door showing a goat, and that he must make the offer to switch. Many sources add to this the assumption that the host chooses at random which door to open if both hide goats. The resulting set of assumptions gives what is called "the standard problem" by many sources ([[#refBarbeau2000|Barbeau 2000:87]]). According to Krauss and Wang ([[#refKraussandWang2003|2003:10]]), even if these assumptions are not explicitly stated, people generally assume them to be the case.
==Solution==
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As [[Keith Devlin]] says ([[#refDevlin2003|Devlin 2003]]), "By opening his door, Monty is saying to the contestant 'There are two doors you did not choose, and the probability that the prize is behind one of them is 2/3. I'll help you by using my knowledge of where the prize is to open one of those two doors to show you that it does not hide the prize. You can now take advantage of this additional information. Your choice of door A has a chance of 1 in 3 of being the winner. I have not changed that. But by eliminating door C, I have shown you that the probability that door B hides the prize is 2 in 3.{{' "}}
==Aids to understanding==
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Another simulation, suggested by vos Savant, employs the "host" hiding a penny, representing the car, under one of three cups, representing the doors; or hiding a pea under one of three shells.
===Increasing the number of doors===
That switching has a probability of 2/3 of winning the car runs counter to many people's intuition. If there are two doors left, then why isn't each door 1/2? The intuition may be aided by generalizing the problem to have a large number of doors so that the player's initial choice has a small chance of winning.
It may be easier to appreciate the solution by considering the same problem with 1,000,000 doors instead of just three ([[#refvosSavant1990|vos Savant 1990]]). In this case there are 999,999 doors with goats behind them and one door with a prize. The player picks a door. His initial probability of winning is 1 out of 1,000,000. The game host then opens 999,998 of the other doors revealing 999,998 goats. (Imagine the host starting with the first door and going down a line of 1,000,000 doors, opening each one, skipping over only the player's door and one other door.) The host then offers the player the chance to switch to the only other unopened door. On average, in 999,999 out of 1,000,000 times the other door will contain the prize, as 999,999 out of 1,000,000 times the player first picked a door with a goat. A rational player should switch. Intuitively speaking, the player should ask how likely is it, that given a million doors, he or she managed to pick the right one. The example can be used to show how the likelihood of success by switching is equal to (1 minus the likelihood of picking correctly the first time) for any given number of doors. The chance that the player's door is correct hasn't changed. It is important to remember, however, that this is based on the assumption that the host knows where the prize is and must not open a door that contains that prize, randomly selecting which other door to leave closed if the contestant manages to select the prize door initially.
To extend the above, it's as if Monty gives you the chance to keep your one door, or open all 999,999 of the other doors, of which he kindly opens 999,998 for you, leaving, deliberately, the one with the prize. Clearly, one would choose to open the other 999,999 doors rather than keep the one.
This example can also be used to illustrate the opposite situation in which the host does ''not'' know where the prize is and opens doors randomly. There is a 999,999/1,000,000 probability that the contestant selects wrong initially, and the prize is behind one of the other doors. If the host goes about randomly opening doors not knowing where the prize is, the probability is likely that the host will reveal the prize before two doors are left (the contestant's choice and one other) to switch between. If the host happens to not reveal the car, then both of the remaining doors have an equal probability of containing a car. This is analogous to the game play on another game show, ''[[Deal or No Deal]]''; in that game, the contestant chooses a numbered [[briefcase]] and then randomly opens the other cases one at a time.
Stibel et al. ([[#refStibeletal2008|2008]]) propose working memory demand is taxed during the Monty Hall problem and that this forces people to "collapse" their choices into two equally probable options. They report that when increasing the number of options to over 7 choices (7 doors) people tend to switch more often; however most still incorrectly judge the probability of success at 50/50.
==Sources of confusion==
When first presented with the Monty Hall problem an overwhelming majority of people assume that each door has an equal probability and conclude that switching does not matter ([[#refMueserandGranberg1999|Mueser and Granberg, 1999]]). Out of 228 subjects in one study, only 13% chose to switch ([[#refGranbergandBrown1995|Granberg and Brown, 1995:713]]). In her book ''The Power of Logical Thinking'', vos Savant ([[#refvosSavant1996|1996:15]]) quotes cognitive psychologist [[Massimo Piattelli-Palmarini]] as saying "... no other statistical puzzle comes so close to fooling all the people all the time" and "that even Nobel physicists systematically give the wrong answer, and that they ''insist'' on it, and they are ready to berate in print those who propose the right answer." Interestingly, pigeons make mistakes and learn from mistakes, and experiments, [[#refHerbransonandSchroeder2010|Herbranson and Schroeder, 2010]], show that they rapidly learn to always switch, unlike humans.
Most statements of the problem, notably the one in ''Parade Magazine'', do not match the rules of the actual game show ([[#refKraussandWang2003|Krauss and Wang, 2003:9]]), and do not fully specify the host's behavior or that the car's location is randomly selected ([[#refGranbergandBrown1995|Granberg and Brown, 1995:712]]). Krauss and Wang ([[#refKraussandWang2003|2003:10]]) conjecture that people make the standard assumptions even if they are not explicitly stated. Although these issues are mathematically significant, even when controlling for these factors nearly all people still think each of the two unopened doors has an equal probability and conclude switching does not matter ([[#refMueserandGranberg1999|Mueser and Granberg, 1999]]). This "equal probability" assumption is a deeply rooted intuition ([[#refFalk1992|Falk 1992:202]]). People strongly tend to think probability is evenly distributed across as many unknowns as are present, whether it is or not ([[#refFoxandLevav2004|Fox and Levav, 2004:637]]).
In addition to the "equal probability" intuition, a competing and deeply rooted intuition is that ''revealing information that is already known does not affect probabilities''. Although this is a true statement, it is not true that just knowing the host can open one of the two unchosen doors to show a goat necessarily means that opening a specific door cannot affect the probability that the car is behind the initially-chosen door. If the car is initially placed behind the doors with equal probability and the host chooses uniformly at random between doors hiding a goat (as is the case in the standard interpretation) this probability indeed remains unchanged, but if the host can choose non-randomly between such doors then the specific door that the host opens reveals additional information. The host can always open a door revealing a goat ''and'' (in the standard interpretation of the problem) the probability that the car is behind the initially-chosen door does not change, but it is ''not because'' of the former that the latter is true. Solutions based on the assertion that the host's actions cannot affect the probability that the car is behind the initially-chosen door are very persuasive, but lead to the correct answer only if the problem is completely symmetrical with respect to both the initial car placement and how the host chooses between two goats ([[Monty Hall problem#refFalk1992|Falk 1992:207,213]]).
==Other solutions==
Another approach showing switching wins with probability 2/3 is to determine the [[conditional probability]] the car is behind Door 2 given that the player has initially picked Door 1 and the host has opened Door 3 ({{Harvnb|Selvin|1975b}}; [[#refMorganetal1991|Morgan et al. 1991]]; [[#refGrinsteadandSnell2006|Grinstead and Snell 2006:137]]). Referring to the [[decision tree]] as shown to the right ([[#refChun1991|Chun 1991]]) or the equivalent figure below, and considering only cases where the host opens Door 3 after the player picks Door 1, switching loses in a case with probability 1/6 but wins in a case with probability 1/3. The conditional probability the car is behind Door 2 is therefore 2/3 = 1/3 / (1/6 + 1/3), while the conditional probability the car is behind Door 1 is only 1/3 = 1/6 / (1/6 + 1/3).
Many other solutions, all showing the player should switch, have been published including solutions using [[game theory]] (see "[[#Game theory approach|Game theory approach]]", below) and formal solutions using [[Bayes Theorem]] (see "[[#Mathematical formulation|Mathematical formulation]]", below).
<br clear=all/>
{| class="wikitable" style="margin:auto; text-align: center;" width="90%"
|-
! width="33%" | Car hidden behind Door 3
! colspan=2 width="33%" | Car hidden behind Door 1
! width="33%" | Car hidden behind Door 2
|-
! colspan=4 | Player initially picks Door 1
|-
| [[File:Monty-RightCar.svg|150px|Player has picked Door 1 and the car is behind Door 3]]
| [[File:Monty-MiddleCar.svg|150px|Player has picked Door 1 and the car is behind Door 2]]
|-
| Host must open Door 2
| colspan=2 | Host randomly opens either goat door
| Host must open Door 3
|-
| bgcolor=#cccccc | [[File:Monty-RightCarSwitch.svg|177px|Host must open Door 2 if the player picks Door 1 and the car is behind Door 3]]
| width=16% bgcolor=#cccccc | [[File:Monty-LeftCarSwitch2.svg|88px|Host opens Door 2 half the time if the player picks Door 1 and the car is behind it]]
| width=16% | [[File:Monty-LeftCarSwitch1.svg|88px|Host opens Door 3 half the time if the player picks Door 1 and the car is behind it]]
| [[File:Monty-MiddleCarSwitch.svg|177px|Host must open Door 3 if the player picks Door 1 and the car is behind Door 2]]
|-
| bgcolor=#cccccc | Probability 1/3
| bgcolor=#cccccc | Probability 1/6
| Probability 1/6
| Probability 1/3
|-
| bgcolor=#cccccc | Switching wins
| bgcolor=#cccccc | Switching loses
| Switching loses
| Switching wins
|-
| colspan=2 bgcolor=#cccccc | If the host has opened Door 3, these cases have not happened
| colspan=2 | If the host has opened Door 3, switching wins twice as often as staying
|}
===Alternative derivations===
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* [http://www.nytimes.com/2008/04/08/science/08monty.html The Monty Hall Problem] at The New York Times (simulation)
* [http://dl.dropbox.com/u/548740/www/DoorKeeperGame/index.html The Door Keeper Game] (simulation)
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