The Three Doors Problem

Chandler Gregg

4/10/13

Math 7 – 13S – Professor Winkler

The Three Doors Problem

In the movie ’21,’ a talented MIT student, Ben, learns to count cards from a crooked professor and a few other students in order to pay for his impending medical school tuition. In one of the scenes, the professor poses a problem that I like to call the ‘three doors problem.’ The problem goes like something this:

You are a contestant on a game show trying to win a brand new car. The host of the game show presents you with three doors, behind two of which are sheep and behind the other is the car. You choose a door. The host goes to a door you did not choose and opens it, revealing a sheep. The host then asks you if you would like to switch doors.

What gives you statistically the best chance of winning the car: staying with your first choice or switching? What is the probability of winning the car under the optimal choice?

Ben, one of many in a large lecture class, answers the problem correctly, gaining the attention of the professor and the eventual opportunity to count cards.

The answer to the first question is yes and to the second, 66.7% (2/3), but the answer is not obvious. Most people assume that under the conditions of a game show or any other game of chance, a host would not provide a chance of winning greater than 50/50. In fact, most of the time the probability of winning on a game show is significantly less than 50%. When people try to solve this problem, they believe that staying with the door they initially chose or switching is a 50/50 tradeoff; they believe that when the game show host opens one door that contains a sheep, it simply reduces the problem to just two doors. However, these assumptions are incorrect.

The key distinction between 50% and 66.7% is when the choice is made relative to the events of the game show. If the game show host opened the door before the contestant chose a door, it would be a simple 50/50 problem. In the three doors problem, however, the contestant has already made a choice and, by revealing a sheep behind an opened door, the host doubles the contestant’s probability of winning a prize.

Here is a way to think through it graphically:

As you can see, there are two possibilities out of three of winning a car from switching, while there is only one possibility out of three of winning a car from staying.

Another way to think of this problem is to avoid choosing the car at first and then switching. Consider that there is only a 33% percent that your first choice is the car. It appears easier (66% chance) to initially choose a sheep and then switch, because once you choose a sheep and switch, you are guaranteed the car. In essence, this reverses the challenge of the problem and makes it seem twice as easy.

One loose end with this problem arises when considering the switching strategy for other games of chance. Let’s say you are a contestant in a similar game show. The game show asks you to choose between the same three doors and gives you a chance to switch after your first choice. Should you do it? Applying the logic from the first problem, you might consider it. However, the host does nothing to aid your chances of winning, so this strategy is useless. No matter how many times you switch your choice, each time you have the same probability of choosing the prize door. The key distinction with the three doors problem is the door reveal.

The reason I like this problem is because it is easily solvable when considered from a graphical or visual approach, but in the heat of the moment, a game show contestant or puzzle solver would likely proceed illogically. It is a mathematical and logical problem, broadening the appeal of the problem. What about the grading criterion for puzzles that we considered in class? This problem shines on its visual component (the question is easy to picture and my favorite solution is graphical) and its elegant but at the same time paradoxical solution. All things considered, I would say this problem is an A+.