Who's Counting: The Monty Hall Problem

A New Question and a More General Question

If this is all clear, here's a variation of the problem to test your understanding of the probability involved. (The answer is at the end of this column.) Let's say that "Let's Make a Deal" were to attempt a comeback. The producers, fearing the audience would be small if the game were exactly the same, devise a variant game in which the contestant is presented with 10 doors. Again behind one of them is a car, behind the others booby prizes. After the contestant picks a door, Monty (or his avatar) opens just seven of the remaining nine unopened doors, but is careful never to open the door hiding the car. There are now three unopened doors -- the one that the contestant originally picked and two others. Which strategy works best, switching to one of the other two unopened doors or sticking with the original pick? Furthermore, what is the probability of winning by following these two strategies?

One last question: Can you think of any real-world situations -- crime mysteries, world politics, administrative deceptions -- which might be modeled on some close variant of the Monty Hall problem (and not simply by Bayes' theorem)? That is, are there situations in which the "contestant," say a reporter, must choose among various alternatives and the "host," say an official, knows the true answer, but is evasive about it and instead answers a question different from the one the contestant asks?

Answer: The chance the prize is behind the door originally chosen by the contestant is 1/10 and remains 1/10. The chance it's behind one of the nine other unopened doors is 9/10. Since the host opens seven of these nine other unopened doors, the 9/10 probability that it's behind one of them is divided between two of these nine doors. So the contestant should switch to one of these two. Doing so raises his probability of winning from 1/10 to one-half of 9/10 or 45 percent.


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