Given that you have exactly one pair, what is the probability you have two aces?

Question

You are drawing 5 cards from a 52 card deck in a game of Poker.

Here's what I've got so far, but I'm a bit stuck on how to proceed.

Let $S = \{h \in 2^D : |h| = 5\}$ and $P(h) = \frac{1}{52 \choose 5}$ for all $h \in S.$
Let $A$ be the event of getting two aces.
Let $B$ be the event of getting a pair.

I want to find $P(A|B) = \frac{P(A \cap B)}{P(B)}$.

I know that $P(A) = \frac{2162}{54145}$ and $P(B) = \frac{352}{833}$.

It seems I always get stuck trying to find $P(A \cap B)$. Is there only one element in $P(A \cap B)$, which would be the hand that contains two aces? How can I proceed with this problem?

Answer 1

No computation is needed. Given we have exactly $1$ pair, by symmetry that pair is equally likely to be a pair of $2$'s, of $3$'s, and so on. So the probability is $1/13$.