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Solution for #13

Remember that the host knows where the prize is. When you pick a door, there's a 66.7% chance you're wrong. If you're wrong, the host will always open the one door left that doesn't contain the prize. So if you were wrong (66.7% chance), you're better off switching to the door that the host leaves closed.

What you are really being asked here is whether the odds are better with your first choice, or with both of the other two. All the host really does is prove that it isn't in one of the remaining two -- but we knew that anyway, because there's only one prize.

Another way to look at it is this: consider that there are 1000 doors. You pick one. Before you open it, the host opens 998 of the 999 remaining doors. There's a 0.1% chance you guessed correctly -- if you guessed wrong (99.9% chance), then you know the prize is in the one door the host left shut. So you're better off changing your guess.

Still not convinced? This is Brain Food's most contested answer, but it is correct! Let's break it down a little more explicitly:

There are nine possible scenarios:

1. Prize is behind door A. You pick door A.
2. Prize is behind door A. You pick door B.
3. Prize is behind door A. You pick door C.
4. Prize is behind door B. You pick door A.
5. Prize is behind door B. You pick door B.
6. Prize is behind door B. You pick door C.
7. Prize is behind door C. You pick door A.
8. Prize is behind door C. You pick door B.
9. Prize is behind door C. You pick door C.

Now the host, who knows which door the prize is behind, deliberately chooses a door that the prize is not behind. For the three cases where you chose correctly in the beginning, the host has the choice of which other door to open. It doesn't matter which one he picks. But for the six cases where you chose wrong, he deliberately chooses the wrong guess of the remaining two choices, revealing it to be a wrong guess. It all breaks down as follows:

1. Prize is behind door A. You pick door A. Host reveals door B or C as empty.
2. Prize is behind door A. You pick door B. Host reveals door C as empty.
3. Prize is behind door A. You pick door C. Host reveals door B as empty.
4. Prize is behind door B. You pick door A. Host reveals door C as empty.
5. Prize is behind door B. You pick door B. Host reveals door A or C as empty.
6. Prize is behind door B. You pick door C. Host reveals door A as empty.
7. Prize is behind door C. You pick door A. Host reveals door B as empty.
8. Prize is behind door C. You pick door B. Host reveals door A as empty.
9. Prize is behind door C. You pick door C. Host reveals door A or B as empty.

(We've had several complaints that the above listing is incorrect because scenarios 1, 5, and 9 should be broken out. In fact, it is incorrect to break them out, because what's important here is that each listed scenario have an equal probability of occurring, not that each scenario be broken down to irreducible terms.)

Now we come to the point where we must decide whether to stick to the original guess or to switch:

1. Prize is behind door A. You pick door A. Host reveals door B or C as empty. Switching loses.
2. Prize is behind door A. You pick door B. Host reveals door C as empty. Switching wins.
3. Prize is behind door A. You pick door C. Host reveals door B as empty. Switching wins.
4. Prize is behind door B. You pick door A. Host reveals door C as empty. Switching wins.
5. Prize is behind door B. You pick door B. Host reveals door A or C as empty. Switching loses.
6. Prize is behind door B. You pick door C. Host reveals door A as empty. Switching wins.
7. Prize is behind door C. You pick door A. Host reveals door B as empty. Switching wins.
8. Prize is behind door C. You pick door B. Host reveals door A as empty. Switching wins.
9. Prize is behind door C. You pick door C. Host reveals door A or B as empty. Switching loses.

Final conclusion: switching wins six out of nine times, which is equal to two thirds of the time, or about 66.7%.

Still not convinced? Check out Wikipedia's page on the Monty Hall Paradox.