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Monty Hall problem

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Monty Hall problem

The Monty Hall problem is a brain teaser, in the form of a probability puzzle, based nominally on the American television game show Let's Make a Deal and named after its original host, Monty Hall. The problem was originally posed in a letter by Steve Selvin to the American Statistician in 1975. It became famous as a question from reader Craig F. Whitaker's letter quoted in (and solved by) Marilyn vos Savant's "Ask Marilyn" column in Parade magazine in 1990:

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? Savant's response was that the contestant should switch to the other door. By the standard assumptions, the switching strategy has a ⁠2/3⁠ probability of winning the car, while the strategy of keeping the initial choice has only a ⁠1/3⁠ probability. When the player first makes their choice, there is a ⁠2/3⁠ chance that the car is behind one of the doors not chosen. This probability does not change after the host reveals a goat behind one of the unchosen doors. When the host provides information about the two unchosen doors (revealing that one of them does not have the car behind it), the ⁠2/3⁠ chance of the car being behind one of the unchosen doors rests on the unchosen and unrevealed door, as opposed to the ⁠1/3⁠ chance of the car being behind the door the contestant chose initially. The given probabilities depend on specific assumptions about how the host and contestant choose their doors. An important insight is that, with these standard conditions, there is more information about doors 2 and 3 than was available at the beginning of the game when door 1 was chosen by the player: the host's action adds value to the door not eliminated, but not to the one chosen by the contestant originally. Another insight is that switching doors is a different action from choosing between the two remaining doors at random, as the former action uses the previous information and the latter does not. Other possible behaviors of the host than the one described can reveal different additional information, or none at all, leading to different probabilities. In her response, Savant states:

Suppose there are a million doors, and you pick door #1. Then the host, who knows what's behind the doors and will always avoid the one with the prize, opens them all except door #777,777. You'd switch to that door pretty fast, wouldn't you? Many readers of Savant's column refused to believe switching is beneficial and rejected her explanation. After the problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine, most of them calling Savant wrong. Even when given explanations, simulations, and formal mathematical proofs, many people still did not accept that switching is the best strategy. Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation demonstrating Savant's predicted result. The problem is a paradox of the veridical type, because the solution is so counterintuitive it can seem absurd but is nevertheless demonstrably true. The Monty Hall problem is mathematically related closely to the earlier three prisoners problem and to the much older Bertrand's box paradox.

Tables

· Paradox › Simple solutions
Goat
Goat
Behind door 1
Goat
Behind door 2
Goat
Behind door 3
Car
Result if staying at door #1
Wins goat
Result if switching to the door offered
Wins car
Goat
Goat
Behind door 1
Goat
Behind door 2
Car
Behind door 3
Goat
Result if staying at door #1
Wins goat
Result if switching to the door offered
Wins car
Car
Car
Behind door 1
Car
Behind door 2
Goat
Behind door 3
Goat
Result if staying at door #1
Wins car
Result if switching to the door offered
Wins goat
Behind door 1
Behind door 2
Behind door 3
Result if staying at door
Result if switching to the door offered
Goat
Goat
Car
Wins goat
Wins car
Goat
Car
Goat
Wins goat
Wins car
Car
Goat
Goat
Wins car
Wins goat
· Solutions using conditional probability and other solutions › Conditional probability by direct calculation
Player initially picks Door 1, 300 repetitions
Player initially picks Door 1, 300 repetitions
Car hidden behind Door 3(on average, 100 cases out of 300)
Player initially picks Door 1, 300 repetitions
Host must open Door 2 (100 cases)
Host must open Door 2 (100 cases)
Car hidden behind Door 3(on average, 100 cases out of 300)
Host must open Door 2 (100 cases)
Car hidden behind Door 1(on average, 100 cases out of 300)
Host randomly opens Door 2(on average, 50 cases)
Car hidden behind Door 1(on average, 100 cases out of 300)
Host randomly opens Door 3(on average, 50 cases)
Car hidden behind Door 2(on average, 100 cases out of 300)
Host must open Door 3 (100 cases)
Probability ⁠1/3⁠(100 out of 300)
Probability ⁠1/3⁠(100 out of 300)
Car hidden behind Door 3(on average, 100 cases out of 300)
Probability ⁠1/3⁠(100 out of 300)
Car hidden behind Door 1(on average, 100 cases out of 300)
Probability ⁠1/6⁠(50 out of 300)
Car hidden behind Door 1(on average, 100 cases out of 300)
Probability ⁠1/6⁠(50 out of 300)
Car hidden behind Door 2(on average, 100 cases out of 300)
Probability ⁠1/3⁠(100 out of 300)
Switching wins
Switching wins
Car hidden behind Door 3(on average, 100 cases out of 300)
Switching wins
Car hidden behind Door 1(on average, 100 cases out of 300)
Switching loses
Car hidden behind Door 1(on average, 100 cases out of 300)
Switching loses
Car hidden behind Door 2(on average, 100 cases out of 300)
Switching wins
On those occasions when the host opens Door 2,switching wins twice as often as staying (100 cases versus 50).
On those occasions when the host opens Door 2,switching wins twice as often as staying (100 cases versus 50).
Car hidden behind Door 3(on average, 100 cases out of 300)
On those occasions when the host opens Door 2,switching wins twice as often as staying (100 cases versus 50).
Car hidden behind Door 1(on average, 100 cases out of 300)
On those occasions when the host opens Door 3,switching wins twice as often as staying (100 cases versus 50).
Car hidden behind Door 3(on average, 100 cases out of 300)
Car hidden behind Door 1(on average, 100 cases out of 300)
Car hidden behind Door 2(on average, 100 cases out of 300)
Player initially picks Door 1, 300 repetitions
Host must open Door 2 (100 cases)
Host randomly opens Door 2(on average, 50 cases)
Host randomly opens Door 3(on average, 50 cases)
Host must open Door 3 (100 cases)
Probability ⁠1/3⁠(100 out of 300)
Probability ⁠1/6⁠(50 out of 300)
Probability ⁠1/6⁠(50 out of 300)
Probability ⁠1/3⁠(100 out of 300)
Switching wins
Switching loses
Switching loses
Switching wins
On those occasions when the host opens Door 2,switching wins twice as often as staying (100 cases versus 50).
On those occasions when the host opens Door 3,switching wins twice as often as staying (100 cases versus 50).
Possible host behaviors in unspecified problem · Variants › Other host behaviors
The host acts as noted in the specific version of the problem.
The host acts as noted in the specific version of the problem.
Host behavior
The host acts as noted in the specific version of the problem.
Result
Switching wins the car two-thirds of the time. (Specific case of the generalized form below with p = q = ⁠1/2⁠)
The host always reveals a goat and always offers a switch. If and only if he has a choice, he chooses the leftmost goat with probability p (which may depend on the player's initial choice) and the rightmost door with probability q = 1 − p.
The host always reveals a goat and always offers a switch. If and only if he has a choice, he chooses the leftmost goat with probability p (which may depend on the player's initial choice) and the rightmost door with probability q = 1 − p.
Host behavior
The host always reveals a goat and always offers a switch. If and only if he has a choice, he chooses the leftmost goat with probability p (which may depend on the player's initial choice) and the rightmost door with probability q = 1 − p.
Result
If the host opens the rightmost (P = ⁠1/3⁠ + ⁠q/3⁠) door, switching wins with probability ⁠1/1+q⁠. Vice versa, if the host opens the leftmost door, switching wins with probability ⁠1/1+p⁠. Always switching is the sum of these: (⁠1/3⁠ + ⁠q/3⁠) / (1+q) + (⁠1/3⁠ + ⁠p/3⁠) / (1+p) = ⁠1/3⁠ + ⁠1/3⁠ = ⁠2/3⁠.
"Monty from Hell": The host offers the option to switch only when the player's initial choice is the winning door.
"Monty from Hell": The host offers the option to switch only when the player's initial choice is the winning door.
Host behavior
"Monty from Hell": The host offers the option to switch only when the player's initial choice is the winning door.
Result
Switching always yields a goat.
"Mind-reading Monty": The host offers the option to switch in case the guest is determined to stay anyway or in case the guest will switch to a goat.
"Mind-reading Monty": The host offers the option to switch in case the guest is determined to stay anyway or in case the guest will switch to a goat.
Host behavior
"Mind-reading Monty": The host offers the option to switch in case the guest is determined to stay anyway or in case the guest will switch to a goat.
Result
Switching always yields a goat.
"Angelic Monty": The host offers the option to switch only when the player has chosen incorrectly.
"Angelic Monty": The host offers the option to switch only when the player has chosen incorrectly.
Host behavior
"Angelic Monty": The host offers the option to switch only when the player has chosen incorrectly.
Result
Switching always wins the car.
"Monty Fall" or "Ignorant Monty": The host does not know what lies behind the doors, and opens one at random that happens not to reveal the car.
"Monty Fall" or "Ignorant Monty": The host does not know what lies behind the doors, and opens one at random that happens not to reveal the car.
Host behavior
"Monty Fall" or "Ignorant Monty": The host does not know what lies behind the doors, and opens one at random that happens not to reveal the car.
Result
Switching wins the car half of the time.
The host knows what lies behind the doors, and (before the player's choice) chooses at random which goat to reveal. He offers the option to switch only when the player's choice happens to differ from his.
The host knows what lies behind the doors, and (before the player's choice) chooses at random which goat to reveal. He offers the option to switch only when the player's choice happens to differ from his.
Host behavior
The host knows what lies behind the doors, and (before the player's choice) chooses at random which goat to reveal. He offers the option to switch only when the player's choice happens to differ from his.
Result
Switching wins the car half of the time.
The host opens a door and makes the offer to switch 100% of the time if the contestant initially picked the car, and 50% the time otherwise.
The host opens a door and makes the offer to switch 100% of the time if the contestant initially picked the car, and 50% the time otherwise.
Host behavior
The host opens a door and makes the offer to switch 100% of the time if the contestant initially picked the car, and 50% the time otherwise.
Result
Switching wins ⁠1/2⁠ the time at the Nash equilibrium.
Four-stage two-player game-theoretic. The player is playing against the show organizers (TV station) which includes the host. First stage: organizers choose a door (choice kept secret from player). Second stage: player makes a preliminary choice of door. Third stage: host opens a door. Fourth stage: player makes a final choice. The player wants to win the car, the TV station wants to keep it. This is a zero-sum two-person game. By von Neumann's theorem from game theory, if we allow both parties fully randomized strategies there exists a minimax solution or Nash equilibrium.
Four-stage two-player game-theoretic. The player is playing against the show organizers (TV station) which includes the host. First stage: organizers choose a door (choice kept secret from player). Second stage: player makes a preliminary choice of door. Third stage: host opens a door. Fourth stage: player makes a final choice. The player wants to win the car, the TV station wants to keep it. This is a zero-sum two-person game. By von Neumann's theorem from game theory, if we allow both parties fully randomized strategies there exists a minimax solution or Nash equilibrium.
Host behavior
Four-stage two-player game-theoretic. The player is playing against the show organizers (TV station) which includes the host. First stage: organizers choose a door (choice kept secret from player). Second stage: player makes a preliminary choice of door. Third stage: host opens a door. Fourth stage: player makes a final choice. The player wants to win the car, the TV station wants to keep it. This is a zero-sum two-person game. By von Neumann's theorem from game theory, if we allow both parties fully randomized strategies there exists a minimax solution or Nash equilibrium.
Result
Minimax solution (Nash equilibrium): car is first hidden uniformly at random and host later chooses uniform random door to open without revealing the car and different from player's door; player first chooses uniform random door and later always switches to other closed door. With his strategy, the player has a win-chance of at least ⁠2/3⁠, however the TV station plays; with the TV station's strategy, the TV station will lose with probability at most ⁠2/3⁠, however the player plays. The fact that these two strategies match (at least ⁠2/3⁠, at most ⁠2/3⁠) proves that they form the minimax solution.
As previous, but now host has option not to open a door at all.
As previous, but now host has option not to open a door at all.
Host behavior
As previous, but now host has option not to open a door at all.
Result
Minimax solution (Nash equilibrium): car is first hidden uniformly at random and host later never opens a door; player first chooses a door uniformly at random and later never switches. Player's strategy guarantees a win-chance of at least ⁠1/3⁠. TV station's strategy guarantees a lose-chance of at most ⁠1/3⁠.
Deal or No Deal case: the host asks the player to open a door, then offers a switch in case the car has not been revealed.
Deal or No Deal case: the host asks the player to open a door, then offers a switch in case the car has not been revealed.
Host behavior
Deal or No Deal case: the host asks the player to open a door, then offers a switch in case the car has not been revealed.
Result
Switching wins the car half of the time.
Host behavior
Result
The host acts as noted in the specific version of the problem.
Switching wins the car two-thirds of the time. (Specific case of the generalized form below with p = q = ⁠1/2⁠)
The host always reveals a goat and always offers a switch. If and only if he has a choice, he chooses the leftmost goat with probability p (which may depend on the player's initial choice) and the rightmost door with probability q = 1 − p.
If the host opens the rightmost (P = ⁠1/3⁠ + ⁠q/3⁠) door, switching wins with probability ⁠1/1+q⁠. Vice versa, if the host opens the leftmost door, switching wins with probability ⁠1/1+p⁠. Always switching is the sum of these: (⁠1/3⁠ + ⁠q/3⁠) / (1+q) + (⁠1/3⁠ + ⁠p/3⁠) / (1+p) = ⁠1/3⁠ + ⁠1/3⁠ = ⁠2/3⁠.
"Monty from Hell": The host offers the option to switch only when the player's initial choice is the winning door.
Switching always yields a goat.
"Mind-reading Monty": The host offers the option to switch in case the guest is determined to stay anyway or in case the guest will switch to a goat.
Switching always yields a goat.
"Angelic Monty": The host offers the option to switch only when the player has chosen incorrectly.
Switching always wins the car.
"Monty Fall" or "Ignorant Monty": The host does not know what lies behind the doors, and opens one at random that happens not to reveal the car.
Switching wins the car half of the time.
The host knows what lies behind the doors, and (before the player's choice) chooses at random which goat to reveal. He offers the option to switch only when the player's choice happens to differ from his.
Switching wins the car half of the time.
The host opens a door and makes the offer to switch 100% of the time if the contestant initially picked the car, and 50% the time otherwise.
Switching wins ⁠1/2⁠ the time at the Nash equilibrium.
Four-stage two-player game-theoretic. The player is playing against the show organizers (TV station) which includes the host. First stage: organizers choose a door (choice kept secret from player). Second stage: player makes a preliminary choice of door. Third stage: host opens a door. Fourth stage: player makes a final choice. The player wants to win the car, the TV station wants to keep it. This is a zero-sum two-person game. By von Neumann's theorem from game theory, if we allow both parties fully randomized strategies there exists a minimax solution or Nash equilibrium.
Minimax solution (Nash equilibrium): car is first hidden uniformly at random and host later chooses uniform random door to open without revealing the car and different from player's door; player first chooses uniform random door and later always switches to other closed door. With his strategy, the player has a win-chance of at least ⁠2/3⁠, however the TV station plays; with the TV station's strategy, the TV station will lose with probability at most ⁠2/3⁠, however the player plays. The fact that these two strategies match (at least ⁠2/3⁠, at most ⁠2/3⁠) proves that they form the minimax solution.
As previous, but now host has option not to open a door at all.
Minimax solution (Nash equilibrium): car is first hidden uniformly at random and host later never opens a door; player first chooses a door uniformly at random and later never switches. Player's strategy guarantees a win-chance of at least ⁠1/3⁠. TV station's strategy guarantees a lose-chance of at most ⁠1/3⁠.
Deal or No Deal case: the host asks the player to open a door, then offers a switch in case the car has not been revealed.
Switching wins the car half of the time.

References

  1. Selvin 1975a.
  2. Selvin 1975b.
  3. vos Savant 1990a.
  4. Tierney 1991.
  5. vos Savant 1991a.
  6. Vazsonyi 1999.
  7. Gardner 1959a.
  8. Gardner 1982.
  9. Mueser & Granberg 1999.
  10. Krauss & Wang 2003, p. 9.
  11. vos Savant 1990b.
  12. Carlton 2005 concluding remarks
  13. Carlton 2005.
  14. Adams 1990.
  15. Devlin 2003.
  16. Devlin 2005.
  17. Williams 2004.
  18. Stibel, Dror & Ben-Zeev 2008.
  19. vos Savant 2012.
  20. Granberg 2014.
  21. Granberg & Brown 1995.
  22. vos Savant 1996, p. 15.
  23. Herbranson & Schroeder 2010.
  24. VerBruggen 2015.
  25. Krauss & Wang 2003, p. 10.
  26. Falk 1992, p. 202.
  27. Fox & Levav 2004, p. 637.
  28. Kahneman, Knetsch & Thaler 1991.
  29. Samuelson & Zeckhauser 1988.
  30. Gilovich, Medvec & Chen 1995.
  31. Kaivanto, Kroll & Zabinski 2014.
  32. Morone & Fiore 2007.
  33. Enßlin & Westerkamp 2018.
  34. Rosenthal 2005a.
  35. Gillman 1992.
  36. Lucas, Rosenhouse & Schepler 2009.
  37. Eisenhauer 2001.
  38. Morgan et al. 1991.
  39. Volokh 2015.
  40. Gillman 1992, emphasis in the original.
  41. Seymann 1991.
  42. vos Savant 1991c.
  43. Rao 1992.
  44. Bell 1992.
  45. Hogbin & Nijdam 2010.
  46. Rosenhouse 2009.
  47. Behrends 2008.
  48. Falk 1992, pp. 207, 213.
  49. Grinstead & Snell 2006, pp. 137–138.
  50. Chun 1991.
  51. Gill 2002.
  52. Henze 2011.
  53. Rosenthal 2005b.
  54. Gill 2011a.
  55. Gardner 1959b.
  56. vos Savant 1996, p. 8.
  57. vos Savant 1996.
  58. Granberg 1996, p. 185.
  59. Granberg & Brown 1995, p. 712.
  60. Gill 2010.
  61. Gill 2011.
  62. Granberg 1996, p. 188.
  63. Flitney & Abbott 2002.
  64. D'Ariano et al. 2002.
  65. Barbeau 1993.
  66. Hall 1975.
  67. Nalebuff 1987.
  68. Martin 1993.
  69. vos Savant 1996, p. xv.
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