At least one ace in a poker hand

Question

You are dealt 5 cards from a well-shuffled standard 52-card deck. What is the probability that your hand contains at least one ace?

Accepted Answer

As with dice, the clean route is the complement: the probability of no aces. The twist is that cards are drawn without replacement, so the per-card probability shifts as the deck shrinks. There are 48 non-aces, and each successive draw must land among the non-aces remaining: P(	ext{no ace}) = \frac{48}{52} \cdot \frac{47}{51} \cdot \frac{46}{50} \cdot \frac{45}{49} \cdot \frac{44}{48} \approx 0.659. Equivalently, count hands: P(	ext{no ace}) = \dfrac{\binom{48}{5}}{\binom{52}{5}} = \dfrac{1{,}712{,}304}{2{,}598{,}960} \approx 0.659. Therefore P(	ext{at least one ace}) = 1 - 0.659 \approx 0…

At least one ace in a poker hand

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