The Monty Hall problem is a probability puzzle loosely based on the American television game show *Let’s Make a Deal* and named after the show’s original host, Monty Hall. The problem, also called the Monty Hall paradox, is a *veridical paradox* because the result appears odd but is demonstrably true. The Monty Hall problem, in its usual interpretation, is mathematically equivalent to the earlier Three Prisoners problem, and both bear some similarity to the much older Bertrand’s box paradox.

The problem was originally posed in a letter by Steve Selvin to the *American Statistician* in 1975. A well-known statement of the problem was published in 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?

Vos Savant’s response was that the contestant should always switch to the other door. If the car is initially equally likely to be behind each door, a player who picks Door 1 and doesn’t switch has a 1 in 3 chance of winning the car while a player who picks Door 1 and does switch has a 2 in 3 chance. Consequently, contestants who switch double their chances of winning the car.

Many readers refused to believe that switching is beneficial. After the Monty Hall problem appeared in *Parade*, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine claiming that vos Savant was wrong. (Tierney 1991) Even when given explanations, simulations, and formal mathematical proofs, many people still do not accept that switching is the best strategy.

The Monty Hall problem has attracted academic interest because the result is surprising and the problem is interesting to formulate. Furthermore, variations of the Monty Hall problem are made by changing the implied assumptions, and the variations can have drastically different consequences. For example, if Monty only offered the contestant a chance to switch when the contestant had initially chosen the car, then the contestant should never switch. Variations of the Monty Hall problem are given below.

Extended problem description

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 (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, often but not always meaning by that, at random with equal probabilities. The resulting set of assumptions gives what is called “the standard problem” by many sources (Barbeau 2000:87). According to Krauss and Wang (2003:10), even if these assumptions are not explicitly stated, people generally assume them to be the case. A fully unambiguous, mathematically explicit version of the standard problem is:

Suppose you’re on a game show and you’re given the choice of three doors [and will win what is behind the chosen door]. Behind one door is a car; behind the others, goats [unwanted booby prizes]. 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 [uniformly] at random. 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

## Solutions

### A simple way to demonstrate that a switching strategy really does win two out of three times on the average is to simulate the game with playing cards (Gardner 1959b; vos Savant 1996:8). Three cards from an ordinary deck are used to represent the three doors; one ‘special’ card such as the Ace of Spades should represent the door with the car, and ordinary cards, such as the two red twos, represent the goat doors.

The simulation, using the following procedure, can be repeated several times to simulate multiple rounds of the game. One card is dealt face-down at random to the ‘player’, to represent the door the player picks initially. Then, looking at the remaining two cards, at least one of which must be a red two, the ‘host’ discards a red two. If the card remaining in the host’s hand is the Ace of Spades, this is recorded as a round where the player would have won by switching; if the host is holding a red two, the round is recorded as one where staying would have won.

By the law of large numbers, this experiment is likely to approximate the probability of winning, and running the experiment over enough rounds should not only verify that the player *does* win by switching two times out of three, but show why. After one card has been dealt to the player, it is *already determined* whether switching will win the round for the player; and two times out of three the Ace of Spades is in the host’s hand.

If this is not convincing, the simulation can be done with the entire deck, dealing one card to the player and keeping the other 51 (Gardner 1959b; Adams 1990). In this variant the Ace of Spades goes to the host 51 times out of 52, and stays with the host no matter how many *non*-Ace cards are discarded.

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.

### Vos Savant’s solution

The solution presented by vos Savant in *Parade* shows the three possible arrangements of one car and two goats behind three doors and the result of switching or staying after initially picking Door 1 in each case:

Door 1 | Door 2 | Door 3 | result if switching | result if staying |
---|---|---|---|---|

Car | Goat | Goat | Goat |
Car |

Goat | Car | Goat | Car |
Goat |

Goat | Goat | Car | Car |
Goat |

A player who stays with the initial choice wins in only one out of three of these equally likely possibilities, while a player who switches wins in two out of three. The probability of winning by staying with the initial choice is therefore 1/3, while the probability of winning by switching is 2/3.

Nice! I get it. Must admit took a while.

If you liked that this will blow your mind, if it doesn’t bore you to death in the meantime.

http://deskarati.com/2011/05/19/non-transitive-dice/