Question

find the probability that a hand of bridge will contain at least one ace

Answer 1

Answer on Question#39805 – Math - Other

There are $\binom{48}{13}$ hands with no ace, because we've got to choose 13 cards from 48 (52 - 4 aces = 48). So there must be $\binom{52}{13} - \binom{48}{13}$ hands with at least one. Thus the proportion that contain at least one ace is

$$P = \frac {\binom {5 2} {1 3} - \binom {4 8} {1 3}}{\binom {5 2} {1 3}} = 1 - \frac {\binom {4 8} {1 3}}{\binom {5 2} {1 3}} = 1 - \frac {\frac {4 8 !}{3 5 ! \cdot 1 3 !}}{\frac {5 2 !}{3 9 ! \cdot 1 3 !}} = 1 - \frac {4 8 \cdot 4 7 \cdot \ldots \cdot 3 6}{5 2 \cdot 5 1 \cdot \ldots \cdot 4 0} = 1 - \frac {3 9 \cdot 3 8 \cdot 3 7 \cdot 3 6}{5 2 \cdot 5 1 \cdot 5 0 \cdot 4 9} \approx 0. 7$$

Answer: 0.7.