A 52-card deck is thoroughly shuffled and you are dealt a hand of 13 cards.
(a) If you have one ace, what is the probability that you have a second ace?
(b) If you have the ace of spades, what is the probability that you have a second ace?
[Math] Card probability question
combinatoricsprobability
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Best Answer
The number of hands with exactly one ace is $4{48\choose 12}$. The number of hands with no aces is ${48\choose 13}$. So the number of hands with at least one ace is ${52\choose 13}-{48\choose 13}$. The number of those with more than $1$ ace is: ${52\choose 13}-{48\choose 13}-4{48\choose 12}$.
So the first probability is: $$\frac{{52\choose 13}-{48\choose 13}-4{48\choose 12}}{{52\choose 13}-{48\choose 13}}$$
You can simplify this by multiplying the numerator and denominator by $\frac{39!13!}{48!}$ to get relatively small numbers. I got a probability of $\frac{11199}{28996}$, but I wouldn't trust my arithmetic.
There are $51 \choose 12$ hands with the ace of spaces. If those, there are $48\choose 12$ without another ace. I'll leave the rest to you.