Common problems

1 Problems about counting

Probability that a five-card hand contains

a standard 52-card deck with four suits (Clubs, Diamonds, Hearts, and Spades) and thirteen ranks (2,..., 10, jack, Queen, King, and Ace)

#ways of selecting 5 cards from 52 cards: \(\binom{52}{5}\)

  1. the ace of diamonds

    #ways that the ace of diamonds was selected in 5 cards (i.e., select four other cards from the remaining 51 cards): \(1 \times \binom{51}{4}\)

    \(P = \frac{1 \times \binom{51}{4}}{\binom{52}{5}}\)

  2. at least an ace

    Which is easier to calculate the compensate - counting #ways of no ace: \(\binom{48}{5}\)

    \(P = 1 - \frac{\binom{48}{5}}{\binom{52}{5}}\)

  3. at least a diamond

    #ways of no diamond: \(\binom{39}{5}\)

    \(P = 1 - \frac{\binom{39}{5}}{\binom{52}{5}}\)

  4. the probability that two cards drawn from a standard deck without replacement have the same rank

    #ways of selecting two cards: \(\binom{52}{2}\)

    #ways of selecting two cards in the same rank: \(\binom{13}{1}\binom{4}{2}\)

    \(P = \frac{\binom{13}{1}\binom{4}{2}}{\binom{52}{2}}\)