Solving the Monty Hall Problem
The premise of the Monty Hall problem is simple. You are on a game show, and the host shows you three doors labeled A, B, and C. One door has a car hidden behind it, whereas the other two has a goat each. Obviously, you want to pick the one with the car. The host initially tells you to pick one door, which you do. After you’ve chosen, the host then opens one of the other two doors, and this opened door will never have the car behind it. And now you are posed with one last question: do you want to switch doors, or do you want to stay with your original choice?
At first, the answer seems trivial. It shouldn’t matter right? After all, no matter what door you pick, there’s always a 1/3 chance that you will pick the car, and 2/3 chance that you will pick a goat. However, we may need to think again. Let’s use Baye’s rule to work out the math behind this sticky situation.
Let A, B, and C denote the event that the car is behind A, B, or C, respectively. Then, let AA, BB, CC denote the door that the host opens.
So what happens is we choose door A? What is the probability that we will win the car P[A|BB] if the host opens door B? (or C, it doesn’t matter)
P[A|BB] = P[A]*P[BB|A]/P[BB] = (1/3)*(1/2)/(1/2) = 1/3
P[A] is 1/3 because the car can be behind any of the three doors. P[BB|A] is 1/2 because the host can pick either door B or C, because the car is behind A as specified. P[BB] = 1/2 because there are only two options for the host.
This makes sense because there is a 1/3 chance that we picked the car in the first place. Now what happens if we switch to C if the host opened B? (or B if the host opened C)
P[C|BB] = P[C]*P[BB|C]/P[BB] = (1/3)*(1)/(1/2) = 1/2
Last before, P[C] = 1/3 because the car can be behind any of the three doors. However, P[BB|C] = 1 because the host cannot pick C because C has the car. Therefore, he has an 100% chance of opening B. P[BB] is 1/2 because there are two doors to pick from.
Consequently, we notice that the player has a 1/2 chance of choosing the right door; however, if we switch, our chances increase to 2/3. This is because when the host opens the door without the car, you have a better chance to find the car than you did before. By staying, you have a 1/3 chance of getting the car (unchanged), but if you switch, the odds increase to 1-(1/3) = 2/3. And voila, we’ve solved the Monty Hall problem using Baye’s rule.
References
http://www.cut-the-knot.org/Probability/BayesTheorem.shtml#SickChildSolution