There are 6 doors in the game. Behind two (different) doors chosen at random is a prize. Alice chose one of the doors, marked x. Moderator Bob randomly opens one of the other doors that does not contain a prize.
- Let's mark with the event that there is a prize behind door x. Calculate Pr[].
In the following sections, let ≠ be one of the unopened doors. Let be the event that behind door y there is a prize. Let be the event that the moderator opens door y.
- Calculate Pr[A∩∩̅].
- Calculate Pr[A].
- Calculate Pr[|A̅].
I have a homework exercise that involves a variation of Monty Hall. In section 1, I think the answer is 1/3 because she didn't change her choice, but my friends say the answer is 2/5 because the moderator opened one door without a prize, so it increases the probability of Alice's choice.
Regarding section 2, this is the section where I got stuck and I can't understand what I should take into account. I do know that I need to calculate the probability that the moderator opened door y, behind which there is no prize and behind door x there is a prize.
I would appreciate help first in section 2, thanks
Before moving to question 2, first a comment regarding question 1. Please note that question 1 just asks about the unconditional or a priori probability $P(X)$, which is just $\frac{1}{3}$ ... you and your friends are already considering the fact that Bob opens a door, but that would be the conditional probability $P(X|A)$. Now, as it so happens, since Bob is certain to open an empty door after your initial pick, the conditional probability P(X|A) would still be $\frac{1}{3}$, and so you happen to have the right answer... but you didn't think about quite correctly, since question 1 was just asking you for the unconditional/initial/a priori probability of $P(X)$
For question 2: note that event $A$ implies event $Y'$: Given the constraints on this game, if Bob opens door $y$, then that automatically means that $y$ does not contain the prize. This means that $P(A \cap X \cap Y')$ is simply the same as $P(X \cap A)$. Moreover, to work out that probability, it may be a little easier to assume that you picked door $x=1$, and that Bob opens up door $y=2$. That is, you need to ask yourself: Given that you picked door $1$, what is the probability that (event $X$) that door $1$ has the prize (given that it could have been elsewhere of course) and that (event $A$) Bob opens up door $y=2$ (given that he could have opened up some other door)? Also, ask yourself: are $X$ and $A$ independent or not? If they are independent, you can use $P(X \cap A) = P(X) * P(A)$. But if not, you need to use $P(X \cap A) = P(X) * P(A|X)$. In fact, you can always use the latter formula ... and in this case that might actually be easier to work with anyway!