The Monty Hall problem

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Pigeon
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The Monty Hall problem

Post by Pigeon » Tue Sep 18, 2012 2:02 pm

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 problem was originally posed in a letter by Steve Selvin to the American Statistician in 1975. One published 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 [but the door is not opened], 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, because the host has removed an incorrect option from the unchosen doors, so 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. Even when given explanations, simulations, and formal mathematical proofs, many people still do not accept that switching is the best strategy. According to the French magazine Le Figaro Magazine, Paul Erdős, the most prolifically published mathematician of modern times, found it very hard to believe the result of his own calculations, and admitted he was not fully convinced until a simple computer simulation confirmed the predicted result.

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.

An intuitive explanation is to reason that a player whose strategy is to switch loses if and only if the player initially picks the car, which happens with probability 1/3, so switching must win with probability 2/3.

Simply put, if the contestant picks a goat (to which two of the three doors lead) they will win the car by switching as the other goat can no longer be picked, while if the contestant picks the car (to which one door leads) they will not win the car by switching. So, if you switch, you win the car if you originally picked a goat and you won't if you picked the car, and as you have a 2 in 3 chance of originally picking a goat you have a 2 in 3 chance of winning by switching.

Another way to understand the solution is to consider the two original unchosen doors together. Instead of one door being opened and shown to be a losing door, an equivalent action is to combine the two unchosen doors into one since the player cannot choose the opened door.


Stibel et al. proposed 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.

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Royal
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Re: The Monty Hall problem

Post by Royal » Tue Sep 18, 2012 8:07 pm

Isn't this calculation based on the assumption that the Host wants the contestant to win? (Which is why he looks at other doors and provides the optional choice- the door that doesn't have a goat.) If he's truly objective in the matters, say he's a robot, it's always 1/3 chance.

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Pigeon
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Re: The Monty Hall problem

Post by Pigeon » Tue Sep 18, 2012 8:37 pm

Monty said he can influence his game.

But in a purer sense, it is as stated. Looks odd but true. Think about the 7 doors.

Disclaimer: Psychics excluded.

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Pigeon
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Re: The Monty Hall problem

Post by Pigeon » Mon Oct 08, 2012 8:56 pm

Here is a web page that evaluates the Monty Hall problem

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