General Monty Hall Problem and Information Theory

Problem: There are three doors and only there is one door behind which there is a prize.You first select a door, then the host of the game will open one door behind which there is no prize (he can't open the door you had selected). Then you are given the choice to switch the door or not. This is the typical Monty Hall problem people talk about. The general form is that n doors with only one door behind which there is a prize. The host of the game will open m doors (m<n-1) behind which there are no prize. Then you are given the choice to switch to another door.

Solution: The most most important observation about this problem is that "probability" is actually not about "randomness", but about information. How much information we have will influence the probability".  Now let's first look at the basic form of the Monty Hall problem.

• assume you chose door A, the probability that you had made the correct choice, P(A),  is 1/3. Assume the host opened the door C (the case for door B is the same), let us denote this action as O. The probability, P(A|O), which means the probability that you had made the correct choice given the host opened the door C, is what we are interested in. Actually P(A|O) = P(A) = 1/3, since unless you move the prize, there is no way you can change the odds of your original choice.
• Besides, we can systematically calculate P(A|O). According to Bayes's theorem, P(A|O) = P(O|A)*P(A)/P(O). We know P(A) = 1/3. P(O|A) should equal 1/2, since if A is the right answer, B and C are both empty doors. The the probability for the host to open door C is simply 1/2. Then we need to calculate P(O). 
• P(O) = P(O|A)*P(A) + P(O|B)*P(B) + P(O|C)*P(C). It is easy to know, P(O|C) = 0, also  P(O|A)*P(A) =1/6. We have P(O|B) = 1. The reason is that we had already chosen A, but B is the right answer, so the only choice for the host is C. Therefore, P(O) = 1/6+1*1/3 = 1/2. Then we have P(A|O) = 1/3.
• Then the probability to win if we switch is P(B|O) = P(O|B)*P(B)/P(O) = 1*1/3 / (1/2) =2/3, so we need to switch!

Now let's try to tackle the general problem.

• The probability to win if we stick to the original choice (assume it is A again), P(A) = 1/n, also we have P(A|O) = 1/n.
•  The probability to win if we switch, P(S|O) = P(We switch to the correct door | Our original choice is wrong) = (1/(n-m-1))*(1-1/n) = ((n-1) / (n-m-1))*1/n). It is easy to know P(S|O)  > P(A|O), so we still need to switch.
• Thinking in the way of information theory, after host reveal some empty doors, we are given more information. These information will change the probability distribution!

See Also: Monty hall and Bayesian probability theory,  The Monty Hall problem -- over easy