# Monty Hall and conditional probabilities [duplicate]

I am a bit confused. I thought after Monty opens the door with the goat, conditional probabilities kick in and our sample space shrinks down to 2 doors instead of 3, and since they were all equally likely, the two doors now have probabilities of 1/2. But apparently, we are not supposed to use conditional probability here?

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## marked as duplicate by Alex BeckerApr 4 '14 at 0:05

This question was marked as an exact duplicate of an existing question.

Yes, we are supposed to do conditional probabilities but the doors are not equally likely because the door that was opened did not have the prize and also the door that was open was not the initially chosen.

The easiest way to see this is consider two strategies: S) always switch the door and N) never switch the door.

If you choose $S$ then you lose only if your original door had the prize, so you win $\frac{2}{3}$ of the time.

If you choose $N$ you only win $\frac{1}{3}$ of the time.

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Because you chose a door before Monty opened one means that he did not have an equal probability to chose either unopened door. If he opened a door before you chose, then you would have a 50% chance.

Consider this problem:

Your neighbor has 2 children. You learn that he has a son, Joe. What is the probability that Joe’s sibling is a brother?

The possible children the neigher has is $\{BB, BG, GB, GG\}$. Since you know he has a son, this is reduced to $\{BB, BG, GB\}$.

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