The game show problem

by John Stapel

Here’s a notorious example of dealing with the unseen that other people see.

“You are at a game show. There are three doors. One door hides a car and two doors each hide a goat(*). Presumably you want the car. The game show host as you to pick one door without opening. After opening the door, the game show host selects a door and opens it showing a goat. He now asks you if you want to stick with your choice of door or whether you want to change it. Should you change it?”

This problem causes endless confusion because its answer depends on how it is phrased.

(*) Sometimes the car is the road to freedom and the goats are tigers. I’m not sure what is worse. Being nibbled at by a goat or being eaten by a tiger.

Let’s look at the unseen.

There are currently three possible outcomes: OOX, OXO, and XOO.

Without loss of generality, we can assume that you always pick the first door. At this point, your chance is 1 in 3 or 1/3 as one of them has the X at the first position.

The game show host now opens one of the other doors. There are four ways he can do it: O=X, OX=, XO=, and X=O. If the notation is not obvious at this point, then X is the car, O is a goat, and = is an open door.

If he intentionally selects a door with a goat behind it, there are, however, only three ways he can do it. Here’s why. If your door currently has a goat behind it, he has to reveal the other goat. The leaves a car behind the other choice. If your door currently has a car behind it, he can reveal any of the two goats. Hence there are 3 outcomes and 2 of them leads to the car if you change your choice. Hence, you should elect to change your choice. The point of confusion lies in the fact that if your door is currently hiding a car, then the host’s choice of door does not change the conditional probability; he could choose either one.

Now suppose the game show host does not intentionally pick the goat. Then there are 6 outcomes: O=X, OO=, OX=, O=O, XO=, X=O, some of which reveals the car. Now assume that the outcome shows a goat that was randomly selected by the host. The following outcomes then apply. O=X, OX=, XO=, X=0. You will note that with your choice of the first door, the probability is 2/4 or 50%.

Just for fun, this can be extended to a game show with four doors. In that case: OOOX, OOXO, OXOO, and XOOO.

A selecting host will create O=OX and OO=X out of the first. O=XO and OOX= out of the second. OX=O and OXO= out of the third and X=OO, XO=O, and XOO= out of the fourth.

It is however still only important what kind of choices he’s giving you. There are only four effective choices! One choice from each of the four above.

OO=X, OOX=, OXO=, and XOO=

In three out of four cases, you will have 1/2 chance if you change. This gives you a combined probability of 3/4*1/2=3/8 if you change your choice. If you stick with your original choice, you remain at 1/4.

Originally posted 2008-11-20 07:27:38.