The Monte Hall 'Make a Deal' Problem

One of the most controversial recreational problems is the Monte Hall Problem. The idea stems from the original game show entitle 'Let's Make A Deal.' The problem appeared first in an article by Steve Selvin in the journal American Statistician.

The problem gained attention when it became the object of a Marilyn Vos Savant column in Parade magazine where it was stated in the following form:

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 is 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 says that the probability of obtaining the car improves if you switch. Many people find this answer runs counter to intuition. This answer like all answers depends on the assumptions that are made especially in regard to the statement

the host, who knows what is behind the doors,...

If we assume that the strategy of the host is to minimize the number of cars given away, then he will not always show the contestant a goat, but the common assumption in the problem is that the host will always show a door that has a goat behind it.

Let us assume that the host does show a goat to the contestant no matter which door is originally chosen. The probability that the car was originally chosen is 1/3rd. The probability that the contestant did not chose the car is 2/3rds, so it follows that the contestant will obtain the car 2/3rds of the time by switching after the host shows a goat.

It is easier to see this relation by considering the question when the number of doors is large. For instance, suppose there are 100 doors. We further assume you pick one door at random and then the host opens 98 doors with goats in them. Obviously, the probability is 1/100 that you originally picked the right door and the probability that the car is behind the remaining door is 99/100.

But suppose the host wishes to minimize the likelihood that the contestant will get the car. What strategy ought he to pursue ?