The conditional probability table below shows how cases, in all of which the player initially chooses door 1, would be split up, on average, according to the location of the car and the choice of door to open by the host. Conditional probability by direct calculation[ edit ] Tree showing the probability of every possible outcome if the player initially picks Door 1 By definition, the conditional probability of winning by switching given the contestant initially picks door 1 and the host opens door 3 is the probability for the event "car is behind door 2 and host opens door 3" divided by the probability for "host opens door 3".

Most statements Its not whether you win or the problem, notably the one in Parade Magazine, do not match the rules of the actual game show Krauss and Wang, Then I simply lift up an empty shell from the remaining other two.

However, the probability of winning by always switching is a logically distinct concept from the probability of winning by switching given that the player has picked door 1 and the host has opened door 3. One analysis for one question, another analysis for the other question. In particular, vos Savant defended herself vigorously.

Another way to understand the solution is to consider the two original unchosen doors together Adams ; Devlin; Williams ; Stibel et al.

After the player picks a door, the host opensof the remaining doors. Confusion and criticism[ edit ] Sources of confusion[ edit ] 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 Mueser and Granberg, For this variation, the two questions yield different answers.

What is the probability of winning the car given the player has picked door 1 and the host has opened door 3?

After the host reveals a goat, you now have a one-in-two chance of being correct. There, the possibility exists that the show master plays evil by opening other doors only if a door with a goat was initially chosen.

University of Florida vos Savant a Vos Savant wrote in her first column on the Monty Hall problem that the player should switch vos Savant a. Your choice of door A has a chance of 1 in 3 of being the winner. 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.

The solutions in this section consider just those cases in which the player picked door 1 and the host opened door 3. But, these two probabilities are the same.

These are the only cases where the host opens door 3, so if the player has picked door 1 and the host opens door 3 the car is twice as likely to be behind door 2. Whether you change your selection or not, the odds are the same.

On average, intimes out of 1, the remaining door will contain the prize. The fact that these are different can be shown by varying the problem so that these two probabilities have different numeric values.

The assertion therefore needs to be justified; without justification being given, the solution is at best incomplete. The discussion was replayed in other venues e.

A show master playing evil half of the times modifies the winning chances in case one is offered to switch to "equal probability". You can now take advantage of this additional information. In this situation, the following two questions have different answers: An intuitive explanation is that, if the contestant initially picks a goat 2 of 3 doorsthe contestant will win the car by switching because the other goat can no longer be picked, whereas if the contestant initially picks the car 1 of 3 doorsthe contestant will not win the car by switching Carltonconcluding remarks.

As Cecil Adams puts it Adams"Monty is saying in effect: What is the probability of winning the car by always switching? Moreover, the host is certainly going to open a different door, so opening a door which door unspecified does not change this.

Among these sources are several that explicitly criticize the popularly presented "simple" solutions, saying these solutions are "correct but It is based on the deeply rooted intuition that revealing information that is already known does not affect probabilities.

The problem continues to attract the attention of cognitive psychologists.

The fact that the host subsequently reveals a goat in one of the unchosen doors changes nothing about the initial probability. Behrends concludes that "One must consider the matter with care to see that both analyses are correct"; which is not to say that they are the same.

Later in their response to Hogbin and Nijdamthey did agree that it was natural to suppose that the host chooses a door to open completely at random, when he does have a choice, and hence that the conditional probability of winning by switching i.

But the answer to the second question is now different:TheINQUIRER publishes daily news, reviews on the latest gadgets and devices, and INQdepth articles for tech buffs and hobbyists. The Monty Hall problem is a brain teaser, in the form of a probability puzzle, loosely based on the American television game show Let's Make a Deal and named after its original host, Monty mi-centre.com problem was originally posed (and solved) in a letter by Steve Selvin to the American Statistician in (Selvin a), (Selvin b).It .

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