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Aces in a poker hand

ProbabilityHard

A hand of $5$ cards is dealt from a well-shuffled standard deck of $52$ cards, which contains $4$ aces. Given that the hand contains at least $1$ ace, what is the probability that it contains at least $2$ aces? Give a fraction in lowest terms.

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Hint

Count hands with no aces and hands with exactly $1$ ace, and subtract from all hands.

Solution

The number of all hands is $(5 52) = 2598960$ , and the number with no aces is $(5 48) = 1712304$ .
So the number of hands with at least $1$ ace is $2598960 - 1712304 = 886656$ .
The number of hands with exactly $1$ ace is $4 \times (4 48) = 4 \times 194580 = 778320$ .
So the number of hands with at least $2$ aces is $886656 - 778320 = 108336$ .
The conditional probability is $\frac{108336}{886656}$ , which reduces to $\frac{2257}{18472}$ .

Answer: $\frac{2257}{18472}$

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