Back

Solution: 2018 Fall Final - 12

Author: Michiel Smid

Question

In a standard deck of 52 cards, each card has a suit and a rank. There are four suits (spades ♠, hearts ♡, clubs ♣, and diamonds ♢), and 13 ranks (Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, and King).
Assume you get a uniformly random hand consisting of 5 cards. What is the probability that the 5 cards in this hand are all of the same suit?
(a)
$4 \cdot \left. {13 \choose 5} \middle/ {52 \choose 5} \right.$
(b)
$\left. {13 \choose 5} \middle/ {52 \choose 5} \right.$
(c)
$4 \cdot \left. {52 \choose 5} \middle/ {13 \choose 5} \right.$
(d)
$\left. {52 \choose 5} \middle/ {13 \choose 5} \right.$

Solution

So there are 13 (spades $ spadesuit $ cards, 13 hearts $ heartsuit $ cards, 13 clubs $ clubsuit $ cards, and 13 diamonds $ diamondsuit $) cards

  • Let S be all possible hands of 5 cards: $ |S| = \binom{52}{5} $
  • Let A be the event that the 5 cards are all of the same suit
    First, we choose 1 of the 4 suits for the hand: $ \binom{4}{1} = 4 $
    Then from any of these 13 card suits, we choose 5 cards: $ \binom{13}{5} $
    $ |A| = 4 \cdot \binom{13}{5} $
    $ Pr(A) = \frac{ 4 \cdot \binom{13}{5} }{ \binom{52}{5} } $