Monty Hall Problem is an interesting problem, and is a great example to understand Bayes Rule.

What is Monty Hall Problem? Here is a description from Wikipedia.

Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?

Maybe someone answer it’s 0.5, because, there are two doors left, behind which one is car, and the other one is goat. At first glance, the reason is very convincing, so as to you have no idea. To be honest, my answer at the first is 0.5. But after several minutes I realize my answer is wrong.

Let us check the question, “Is it to your advantage to switch your choice”, it ask will you choose No. 1 door still, or switch to No. 2 door? In other words, the question is which door you choose will give you more probability to win? It’s probability problem actually, so we need to treat this problem with probability thinking.

Just imagine, there is a Monty Hall Problem game to play in your mobile phone, you always pick the first door. And you play it for many times, and record the doors combination and result for each time. Play once is a event, and this event is independent, I will describe one record with “G(pick) C G(open)”, “G” is for goat, “C” is for car, and “pick” is marking the door you pick, and “open” is indicating the door host open. After many times you play the game, to find the rounds in which host open the No.3 door, and count how many times car in No. 2 door and No. 3 door in these rounds. If you play the game enough times, I think you can caculate to find which probability is larger?

For example, I list several rounds here:

1. C(pick)   G         G(open)
2. C(pick)   G(open)   G
3. G(pick)   C         G(open)
4. G(pick)   C         G(open)
5. G(pick)   G(open)   C
6. G(pick)   G(open)   C

Why I list these rounds for example? Because these are all the cases when you pick No. 1 door, and the probability of each case is same, so each cse I list twice. But the special point is that round 1 and 2, since in case “CGG”, after you pick door No. 1, the host may open door No. 2 or No. 3, and the probability is equal, so round 1 open door No. 2, and round 2 open door No. 3.

Now, you can see which round host open door No. 3, round 1, 3 and 4. In round 1, door No. 1 has car, In round 3 and 4, door No. 2 has car. We can realize that the probability of door No. 1 and door No. 2 has car is different, 1/3 for No. 1, 2/3 for No. 2. The reason is the probobility of host open the door No. 3 is different, if car is behind No. 1, the probility is 1/2, and 1 for if car is behind No. 2. This probability of host open the door No. 3 is called likelihood in Bayes Rule. Bayes Rule can caculate this problem perfectly.

Suppose

Let $j = 1, j = 3$, the probability of car is behind door No. 1 is $P(C_1 H_{13})$ .

Obviously, if we switch to choose door No. 2, we have more probability to win the game.