Poker, two aces in five cards

Question

Poker hand: $5$ cards are dealt from a deck of $52$. What is the probability of receiving exactly $2$ aces? (Hypergeometric: $N=52$, $K=4$, $n=5$, $k=2$.)

Solution

The cards are dealt without replacement, so the model is hypergeometric: $P(X=2) = \frac{\binom{4}{2}\binom{48}{3}}{\binom{52}{5}} = \frac{6\cdot 17296}{2\,598\,960} = \frac{103\,776}{2\,598\,960} \approx \boxed{0{,}0399}.$ About $4\%$: receiving two aces in a hand is a fairly rare event.

Links

Topics: Probability distributions
Concepts: Hypergeometric distribution
Methods: Hypergeometric distribution
Skills: Combinatorial calculation · Probability calculation
Exercise type: Probability problem