Probability of getting exactly one pair of aces, knowing there's one ace in the draw?

Question

We draw 5 cards in a 52 cards game. What is the probability of getting exactly one pair of aces, knowing there's one ace in the draw ?

I know that the answer is $\frac{\binom{4}{2} \binom{48}{3}}{\binom{52}{5}-\binom{48}{5}}$, but I don't get why $\frac{\binom{3}{1}\binom{48}{3}}{\binom{51}{4}}$ is not the good answer.

I am following the same reasonning as for the question: "What is the probability of getting exactly one pair of aces?", i.e $\frac{\binom{4}{2}\binom{48}{3}}{\binom{52}{5}}$.

I can't get why there it is not the right answer.

Answer 1

It looks like you assume when the guaranteed ace is gonna appear - this is not the same. It can be drawn at any of the five draws.

Answer 2

The problem would be better phrased if it said that there is at least one ace in the draw.

We are conditioning on the fact that there is at least one ace in the draw. This is our sample space.

Since $\binom{48}{5}$ of the $\binom{52}{5}$ possible five-card hands have no aces, there are $$\binom{52}{5} - \binom{48}{5}$$ five-card hands with at least one ace in the draw.

The number of ways of selecting exactly one pair of aces is $$\binom{4}{2}\binom{48}{3}$$ since we must select two of the four aces and three of the other $48$ cards in the deck.

Hence, the probability that you receive exactly one pair of aces given that you are dealt at least one ace is $$\frac{\dbinom{4}{2}\dbinom{48}{3}}{\dbinom{52}{5} - \dbinom{48}{5}}$$

In your attempt, you assumed that you already had an ace and were being dealt four additional cards. What you calculated is the probability that you will select exactly one ace when dealt four of the $51$ cards that remain in the deck after you have been handed one of the aces in the deck. However, you are being dealt five cards. You are not being handed an ace, then given four additional cards.