Here is the problem:
A and B play a game. Each entity is equally likely to win, and the first to $3$ rounds wins. If $B$ wins the second round and A wins the game, what is the probability that B wins the first game?
The problem is quite simple with casework, but it would be nice if I could generalize beyond casework. It seems like this is a conditional probability problem with two conditions: that A wins the game and B wins the second round.
Normally conditionally probability for two events $X$ and $Y$ is written as $P(X | Y) = \frac{P(X \cap Y)}{P(Y)}$, and in this case "A wins the game and B wins the second round" is $Y$. It seems too that this is a conditional probability statement as well, but I'm not sure how to calculate it. A winning the game has a 1/2 chance of happening, but what about B winning the second round? After we figure out this probability, we can apply the same method to finding the probability that B wins the first round given that A wins the game and B wins the second round, but I have tried and can't get very far in generalizing beyond the numbers in the original problem.
$$P(\text{B wins first round} \mid \text{A wins whole game and B wins 2nd round}) = \frac{P(\text{B wins first round and A wins whole game and B wins 2nd round})}{P(\text{A wins game and B wins 2nd round})}$$
Compute the numerator and denominator separately.
For the numerator, the only sequence of wins allowed is "BBAAA" (if B wins the first two rounds, then the only way A can win the whole game is by winning the next 3 games).
For the denominator, there are several valid sequences: ABAA, ABABA, ABBAA, BBAAA.