Page 469 - Discrete Mathematics and Its Applications

448 7 / Discrete Probability

Poker, and other card games, are growing in popularity. To win at these games it helps to
know the probability of different hands. We can find the probability of specific hands that arise
in card games using the techniques developed so far. A deck of cards contains 52 cards. There
are 13 different kinds of cards, with four cards of each kind. (Among the terms commonly used
instead of “kind” are “rank,” “face value,” “denomination,” and “value.”) These kinds are twos,
threes, fours, fives, sixes, sevens, eights, nines, tens, jacks, queens, kings, and aces. There are
also four suits: spades, clubs, hearts, and diamonds, each containing 13 cards, with one card of
each kind in a suit. In many poker games, a hand consists of five cards.

EXAMPLE 5 Find the probability that a hand of five cards in poker contains four cards of one kind.
Solution: By the product rule, the number of hands of five cards with four cards of one kind
is the product of the number of ways to pick one kind, the number of ways to pick the four of
this kind out of the four in the deck of this kind, and the number of ways to pick the fifth card.
This is

C(13, 1)C(4, 4)C(48, 1).

By Example 11 in Section 6.3 there are C(52, 5) different hands of five cards. Hence, the
probability that a hand contains four cards of one kind is

C(13, 1)C(4, 4)C(48, 1) 13 · 1 · 48
= ≈ 0.00024. ▲
C(52, 5) 2,598,960

EXAMPLE 6 What is the probability that a poker hand contains a full house, that is, three of one kind and
two of another kind?
Solution: By the product rule, the number of hands containing a full house is the product of the
number of ways to pick two kinds in order, the number of ways to pick three out of four for
the first kind, and the number of ways to pick two out of four for the second kind. (Note that
the order of the two kinds matters, because, for instance, three queens and two aces is different
from three aces and two queens.) We see that the number of hands containing a full house is

P(13, 2)C(4, 3)C(4, 2) = 13 · 12 · 4 · 6 = 3744.
Because there are C(52, 5) = 2,598,960 poker hands, the probability of a full house is

3744
≈ 0.0014. ▲
2,598,960

EXAMPLE 7 What is the probability that the numbers 11, 4, 17, 39, and 23 are drawn in that order from a bin
containing 50 balls labeled with the numbers 1, 2,..., 50 if (a) the ball selected is not returned
to the bin before the next ball is selected and (b) the ball selected is returned to the bin before
the next ball is selected?

Solution: (a) By the product rule, there are 50 · 49 · 48 · 47 · 46 = 254,251,200 ways to select
the balls because each time a ball is drawn there is one fewer ball to choose from. Consequently,
the probability that 11, 4, 17, 39, and 23 are drawn in that order is 1/254,251,200. This is an
example of sampling without replacement.
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(b) By the product rule, there are 50 = 312,500,000 ways to select the balls because there are
50 possible balls to choose from each time a ball is drawn. Consequently, the probability that
11, 4, 17, 39, and 23 are drawn in that order is 1/312,500,000. This is an example of sampling
with replacement. ▲