The Monty Hall problem (sometimes called Monty Hall Dilemma or Monty Hall Paradox) is one of my favorite math problems involving probability. I first learned about it long ago, when I just started to learn more in depth about probability in college. And like many other students I was not immediately convinced and satisfied with the solution given by my lecturer at that time. The counter-intuitive nature of the solution to the problem is what intrigued me the most. Let me try to explain for those of you not familiar with this problem.

Supposed that in a game show there are three doors, behind one of which is a prize (let say a car). There’s nothing behind the other two doors. You choose one of the doors, say door no 1. The host of the game, who knows which door hides the car, opens another door, say no 3, which is empty. You are then given a choice again whether you want to stick to your first choice of door no 1 or whether you want to switch to door no 2. Now the question is, is it better for you to switch? In other words, do you have higher probability of getting the car by switching?

Note that in the original problem, the other two doors hide a goat each instead of being empty. I modified it a little in case some people consider a goat to be a better prize than a car. At first glance it seems simple enough that after the host has opened door no 3, there’s a 50-50 chance that the car is behind door no 1 or door no 2. And so it doesn’t matter whether you switch or not. But, is that really the case?

Well, let’s try to look at the problem from another angle. When you first chose door no 1, there was a 1/3 chance that that car was behind it. But it also means that there was a 2/3 chance that the car was not behind it. In other words, there’s a 2/3 chance that the car was behind door 2 or door 3. Now when the host opens door no 3, which is empty, the probability remains the same (because the host always open the empty door, since he knows what are behind the doors). So your chance of getting the car is actually higher if you switch.

There are other ways of understanding the problem, which I’m not going to explain all here. I list down some references below if you want to read and learn further about this problem. The mathematical proof of the solution involves conditional probability theory and can be shown using Bayes’ Theorem.

You can also try to understand the problems by listing out all the possible outcomes when you switch as follows:

• Outcome 1
  • The car: door no 1
  • Your first choice: door no 1
  • The host open door no 3
  • You switch, and you get nothing

• Outcome 2
  • The car: door no 2
  • Your first choice: door no 1
  • The host open door no 3
  • You switch, and you get the car

• Outcome 3
  • The car: door no 3
  • Your first choice: door no 1
  • The host open door no 2
  • You switch, and you get the car

From the above we can see that 2 out of 3, you will get the car when you switch. I hope by now you can understand the problem and the solution better. And if you still need further proof or explanation, you can try some the following resources:

• The Wikipedia entry on Monty Hall Problem – Quite a detailed explanation about the problem and the solutions, including the proof
• Monty Hall Dilemma – Another excellent article about the problem with some variations of the problem
• Monty Hall Simulation – An applet simulating the problem, here the other two doors hide a goat each
• Monty Hall Problem Cartoon at YouTube