# Daily Archives: November 9, 2011

## Bayes Theorem and the Monty Hall Problem

To learn Bayes Theorem, I am attempting to apply it to a probability problem that I “intuitively” “learned” the correct answer to: the Monty Hall problem.

The Monty Hall problem is basically choosing between three possible doors. One has a prize, and the other two doors have a goat. You can choose a door but before opening the door, the announcer says he will open a door that has a goat behind it. After opening one of the goat-doors, the announcer asks you whether you want to keep your door choice or choose the remaining door.

The way I learned it, you initially have a 1/3 chance of opening the correct door. Whatever door you choose, it will retain its 1/3 attribute. Once the announcer opens one of the doors, that door is taken out of the probability (its probability is effectively 1 at this point and is no longer uncertain). Because you chose the door at first when it was only 1/3 probability, it retains that 1/3 probability while the remaining unchosen door bumps up to 2/3 in order to make everything equal.

I was taught the math behind it in one of my undergrad compsci classes, but I forgot the process and simply invented a heuristic for remembering the correct answer. But I believe Bayes Theorem might help me remember the process.

Again, the simple form of Bayes Theorem is this: P(H | E) = P(E | H) * P(H) / P(E). The prior probability, P(H), of picking the correct door is .33. Our evidence is when the announcer opens the door… so what is the probability of the announcer picking a goat door? What is the probability of the announcer picking the goat door given that you picked the correct door? The announcer will pick a goat-door no matter what your selection is. Again, think about this wording: The announcer will pick a goat-door no matter what your selection is.

Isn’t that the definition of independence? Some hypothesis that is believed in no matter the evidence? This means that both P(E | H) [the probability of the announcer picking a goat door given that you picked the correct door] and P(E) [the probability of the announcer picking a goat door] are equivalent. That is, the announcer will pick a goat-door no matter what your actual selection is. And as I described two posts ago, when P(E | H) and P(E) are equivalent, this means that E exists independently of H. Also, the likelihood ratio is 0 decibles – it doesn’t scream in one direction or the other, it’s just dead silence.

So when E is independent of H, P(H | E) = P(H). So if P(H) is .33, then P(H | E) is also .33. But here’s a concept that I haven’t talked about, but is pretty obvious. P(H) + P(~H) has to equal 1.00. So this means that P(~H) is .66. It was .66 before the announcer picked the goat door, and because of independence, it is still .66. It’s just that now the .66 applies to one door instead of being spread across two doors.

So when faced with the Monty Hall problem, Bayes Theorem says that you should bet 66 cents out of a dollar on the non-selected door having the prize.

Comments Off on Bayes Theorem and the Monty Hall Problem

Posted by on November 9, 2011 in Bayes

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