442 CHAPTER 7 Counting and Combinatorics
Basic questions about hands for card games include how many hands of a certain kind
exist and what percentage of the total number of hands are of a certain kind.
Example 5. How many different poker hands are there?
Solution. This answer is just the number of ways of choosing five cards from the 52-card
deck:
52!
C(52, 5) = = 2,598,960
47! 5!
Example 6. What percentage of poker hands contain four of a kind? A hand that contains
four of a kind has four cards with the same value. For example, a hand with four Queens
and a fifth card having any other value is of this type.
Solution. The total number of poker hands was determined in Example 5. It remains to
be determined how many different poker hands contain four of a kind.
Since there are 13 different card values in a deck and four of a kind consists of four
cards of the same value, there are 13 possibilities for these four cards. It remains to be
determined in how many ways the hand can be completed. Since four of a kind uses four
of the 52 cards in the deck, there are 48 cards remaining from which the last card in the
hand can be chosen. Use these two observations and the Multiplication Principle to find
the total number of such hands:
(# Hands with four of a kind) = (# Possibilities for four of a kind)- (# Ways to complete the hand)
= 13 .C(48, 1)
= 624
The percentage of such hands is 624/2,598,960 ; 0.02%. U
Example 7. How many poker hands contain exactly one pair and three other cards, no
two of which have the same value?
Solution. As an example, there are four Jacks. Any two of these can be chosen for thepair, so there are six pairs of Jacks. There are C(4, 2) = 6 pairs for any card value. Since
there are 13 values for cards and we want to choose one of these values to represent two
of the cards in the poker hand, there are C(13, 1) • C(4, 2) ways to choose two cards with
the same value. The remaining three cards have values different from the value of the pair,
and at the same time, no two of the three have the same value. Since there are 12 values
for cards not in the pair, there are C(12, 3) ways to choose the values for the three other
cards. After choosing the three values for the other cards, the suit of each of these cards
can be chosen in C(4, 1) ways, since there are four cards from different suits in the deck
with each value:(# Hands with 1 pair and 3 unmatched cards) = (# Pairs) • (# Ways to complete the hand)
= C(13, 1) C(4, 2) C(12, 3) C(4, 1)3
= 1,098,240
The percentage of such hands out of the total number of hands possible is 1,098,240/
2,598,960 • 42%. U