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

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. 
