Refer to Example 1. For each poker holding below, (1) find

Chapter 9, Problem 11E

(choose chapter or problem)

Problem 11E

Refer to Example 1. For each poker holding below, (1) find the number of five-card poker hands with that holding; (2) find the probability that a randomly chosen set of five cards has that holding.

a. royal flush

b. straight flush

c. four of a kind

d. full house

e. flush

f. straight

g. three of a kind

h. one pair

i. neither a repeated denomination nor five of the same suit nor five adjacent denominations

Example 1

Poker Hand Problems

The game of poker is played with an ordinary deck of cards (see Example 2). Various five-card holdings are given special names, and certain holdings beat certain other holdings. The named holdings are listed from highest to lowest below.

Royal flush: 10, J, Q, K, A of the same suit

Straight flush: five adjacent denominations of the same suit but not a royal flush—aces can be high or low, so A, 2, 3, 4, 5 of the same suit is a straight flush.

Four of a kind: four cards of one denomination—the fifth card can be any other in the deck

Full house: three cards of one denomination, two cards of another denomination

Flush: five cards of the same suit but not a straight or a royal flush

Straight: five cards of adjacent denominations but not all of the same suit—aces can be high or low

Three of a kind: three cards of the same denomination and two other cards of different denominations

Two pairs: two cards of one denomination, two cards of a second denomination, and a fifth card of a third denomination

One pair: two cards of one denomination and three other cards all of different denominations

No pairs: all cards of different denominations but not a straight, or straight flush, or flush, or royal flush

a. How many five-card poker hands contain two pairs?

b. If a five-card hand is dealt at random from an ordinary deck of cards, what is the probability that the hand contains two pairs?

Solution

a. Consider forming a hand with two pairs as a four-step process:

Step 1: Choose the two denominations for the pairs.

Step 2: Choose two cards from the smaller denomination.

Step 3: Choose two cards from the larger denomination.

Step 4: Choose one card from those remaining.

The number of ways to perform step 1 is  because there are 13 denominations in all. The number of ways to perform steps 2 and 3 is  because there are four cards of each denomination, one in each suit. The number of ways to perform step 4 is  because the fifth card is chosen from the eleven denominations not included in the pair and there are four cards of each denomination. Thus

b. The total number of five-card hands from an ordinary deck of cards is  =2,598,960. Thus if all hands are equally likely, the probability of obtaining a hand with two pairs is

Example 2

Probabilities for a Deck of Cards

An ordinary deck of cards contains 52 cards divided into four suits. The red suits are diamonds  and hearts  and the black suits are clubs  and spades . Each suit contains 13 cards of the following denominations: 2, 3, 4, 5, 6, 7, 8, 9, 10, J (jack), Q (queen), K (king), and A (ace). The cards J, Q, and K are called face cards. Mathematician Persi Diaconis, working with David Aldous in 1986 and Dave Bayer in 1992, showed that seven shuffles are needed to “thoroughly mix up” the cards in an ordinary deck. In 2000 mathematician Nick Trefethen, working with his father, Lloyd Trefethen, a mechanical engineer, used a somewhat different definition of “thoroughly mix up” to show that six shuffles will nearly always suffice. Imagine that the cards in a deck have become—by some method—so thoroughly mixed up that if you spread them out face down and pick one at random, you are as likely to get any one card as any other.

a. What is the sample space of outcomes?

b. What is the event that the chosen card is a black face card?

c. What is the probability that the chosen card is a black face card?

Solution

a. The outcomes in the sample space S are the 52 cards in the deck.

b. Let E be the event that a black face card is chosen. The outcomes in E are the jack, queen, and king of clubs and the jack, queen, and king of spades. Symbolically,

c. By part (b), N(E) = 6, and according to the description of the situation, all 52 outcomes in the sample space are equally likely. Therefore, by the equally likely probability formula, the probability that the chosen card is a black face card is