Probability, Part 5, The Monty Hall Problem

In the 1960’s (so I’m told), a new game show on TV appeared called Let’s Make A Deal. The show was hosted by Monty Hall. In this show, a contestant was shown three doors on the stage and was asked to choose one. One door had a great prize like a car. The other two had less desirable prizes like a goat. Well right away, if you’ve been diligent reading my posts, you know that the contestant has a one-third chance of winning.

However, Monty would then open one of the doors not picked an expose a less desirable prize. Monty would then ask the contestant if she would like to switch. The question is, should she? Does it make a difference or is the probability still one-third either way? Well let’s see the two probability trees for each strategy: stay or switch.

First the “stay” strategy. Let’s assume that the car is behind door 1 so we can label the branches appropriately. The second level branches have probability 0 or 1, because once you choose a door, the strategy of staying or switching determines exactly what you will end up with:

So if you look at all the scenarios (branches) that end up with a car and add those probabilities together, you get 1/3 as expected. Please see my previous post regarding probabilities trees if needed.

Now let’s create a tree where we switch doors after Monty shows what’s behind one of the “loser” doors:

Now add up the probabilities that end up with a car and you get 2/3! You double your chances of winning if you switch. You see, Monty adds more information to the problem by exposing one of the loser doors and you take advantage of this by switching doors. Because you have 2/3 chance of initially picking a goat, if you do pick a goat and Monty exposes the other door with a goat, you will have no choice but to end up with the car if you switch.