The Monty Hall Problem

(From 8/13/06 on my old blog)

Suppose Monty Hall offers you the choice of one of three doors. You get to keep whatever is behind the door that you select. However, only one of the doors has something good behind it – a new car! Behind the other two doors are goats.

After selecting one of the doors, Monty, who knows what is behind each door, reveals what is behind one of the doors that you did not choose. The door he reveals has a goat behind it. Now there are two doors left, the one you selected and the door that Monty did not show you. He then offers you the opportunity to switch doors. Based on probability, should you switch?

You might think that that since there are two doors left and one of them has the car and the other has a goat that it’s a 50-50 chance so it doesn’t matter if you switch or not. You would be wrong.

It is actually better to switch. There is a 2/3 probability that the unselected door has the new car.

Consider. There are three possibilities with your selection in caps:
CAR goat goat
car GOAT goat
car goat GOAT

Notice that in only one of the three cases you select the good door. So, there’s a 1/3 probability of selecting the good door. Since Monty knows which door is good and which ones are bad, it doesn’t matter which one he reveals when there are two bad doors left. Let’s reveal (eliminate) a bad door that you didn’t originally select in each of the cases. That leaves these possibilities:
CAR goat (goat)
car GOAT (goat)
car (goat) GOAT

We started with three cases and there are still three cases. In two of the three cases, switching results in a good door. This problem has been argued over and over again over the last number of years. Here is a link to a page with another good explanation.