Excerpted from The American Statistician, August 1975, Vol. 29, No. 3
On the Mony Hall Problem
I have received a number of letters commenting on my "Letters to the Editor" in The American Statistician of February, 1975, entitled "A Problem in Probability." Several correspondents claim my answer is incorrect. The basis to my solution is that Monty Hall knows which box contains the keys and when he can open either of two boxes without exposing the keys, he chooses between them at random. An alternative solution to enumerating the mutually exclusive and equally likely outcomes is as follows:
A = event that keys are contained in box B
B = event that contestant chooses box B
C = event that Monty Hall opens box A
Then
P(keys in box B | contestant selects B and Monty opens A)
| = P(A | BC) | = P(ABC)/P(BC) =P(C | AB)P(AB)/P(C | B)P(B) =P(C | AB)P(B | A)P(A)/P(C | B)P(B) =(1/2)(1/3)(1/3)(1/2)(1/3) 1/3 |
If the contestant trades his box B for the unopened box on the table, his probability of winning the card is 2/3.
D.L. Ferguson presented a generalization of this problem for the case of n boxes, in which Monty Hall opens p boxes. In this situation, the probability the contestant wins when he switches boxes is (n-1)/[n(n-p-1)].
Benjamin King pointed out the critical assumpitons about Monty Hall's behavior that are necessary to solve the problem, and emphasized that "the prior distribution is not the only part of the probabilistic side of a decision problem that is subjective."
Monty Hall wrote and expressed that he was not "a student of statistics problems" but "the big hole in your argument is that once the first box is seen to be emprty, the contestant cannot exchange his box." He continues to say, "Oh and incedentally, after one [box] is seen to be empty, his chances are no longer 50/50 but remain what they were in the first place, one out of three. It just seems to the contestant that one box having been eliminated, he stands a better chance. Not so." I could not have said it better myself.
Steve Selvin
School of Public Health
University of California
Berkley, CA
Monte's Posterior
A response to Steve Selvin's "A Problem in Probability"
Selvin's note solved an interesting problem concerning the well-known TV show "Let's Make a Deal." However, his solution via an enumeration approach is confirmed by a more straight-forward procedure as follows. Consider the following events and probabilities:
P(A = Key box in hand) = 1/3
P(B = Key box on table) = 2/3
P(C = Monte opens empty box on table) = 1
Then, revising probabilities in the light on Monte's action yields:
Hence, the probability of the remaining unopened box on the table containing the keys is 2/3, the contestant should switch.
Joseph G. Van Matre
School of Business
University of Alabama in Birmingham
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olecartalkfan
Your proof of this is faulty. The fact that you learn which curtain has one of the zonks behind it does not increase your chance of winning BY SWITCHING. Your odds of winning once you are aware of where one of the zonks is is 50%, but it stays 50% whether you switch or not.
Kevin Strom
An easier way to understand the Monty Hall puzzler is this: Imagine you are Monty Hall, about to reveal one of the zonk prizes. There are only two possibilities: 1) The contestant has already picked the big prize, and you are just messing with him to trick him into switching. The probability of this is one third, simply because that's the probability of the contestant initially picking the big prize box. 2) The contestant has picked a zonk, and you are showing him the other zonk (since obviously you can't show him the big prize). The probability of this is two thirds. There are no other possible scenarios. Simple. All the best and love your show, Kevin.