Playing Yahtzee

Yahtzee is a popular game played with five dice. The game is similar to the card game Poker – in both games, one is trying to achieve desirable patterns in the dice faces or cards, and some types of patterns are similar in the two games. In this section, we describe some of the dice patterns in the first roll in Yahtzee and consider the problem of determining the chances of several of the patterns.

Outcomes of one roll of five dice

When a player rolls five dice in the game Yahtzee, the most valuable result is when all of the five dice show the same number such as

2, 2, 2, 2, 2.

This is called a “Yahtzee” and the player scores 50 points with this pattern. A second valuable pattern is a “four-of-a-kind” where you observe one number appearing four times, such as

3, 4, 3, 3, 3.

The following table gives all of the possible patterns when you roll five dice in Yahtzee. When you play the game, some of these patterns are worth a particular number of points and these points are given in the right column.

Pattern	Sample of pattern	Point value
Yahtzee	4, 4, 4, 4, 4	50
Four of a kind	6, 6, 6, 4, 6
Large straight	2, 6, 4, 5, 3	40
Small straight	4, 2, 1, 3, 2	30
Full house	5, 1, 1, 5, 1	25
Three of a kind	2, 2, 3, 4, 2
Two pair	6, 3, 3, 6, 2
One pair	4, 3, 4, 1, 5
Nothing	1, 3, 2, 5, 6
TOTAL

Total number of outcomes

As in the case of two dice, it is useful to distinguish the five dice when we count outcomes. We can represent an outcome by placing a value of individual die rolls (1 through 6) in the six slots.

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So two possible outcomes are

2, 3, 4, 5, 5 and 3, 2, 4, 5, 5

Each die has 6 possibilities and so, applying the multiplication rule, the total number of outcomes in the rolls of five dice is

$6 \times 6\times 6\times 6\times 6 = 7776.$

Since all of the outcomes are equally likely, we assign a probability of 1/7776 to each outcome.

Probability of a Yahtzee

We can represent the Yahtzee roll as the outcome

x, x, x, x, x

where x denotes an arbitrary roll of one die. There are six possible choices for x, and so the number of possible Yahtzees is 6.

Since each outcome has probability 1/7776, the probability of a Yahtzee is

Prob(Yahtzee) = 6/7776.

Probability of four of a kind

In the pattern “four of a kind”, we want to have one number appear four times and a second number appear once. In other words, we are interested in counting outcomes of the form

x, x, x, x, y

where the four x’s and the single y can be in different orders.

To apply the multiplication rule, we think of writing down a possible “four of a kind” in three steps.

Step 1: We first choose the number for x (the number that appears four times).

Step 2: We next choose the number for the singleton y.

Step 3: We mix up the orders of the four x’s and the one y.

For example, we choose the outcome 5, 5, 5, 3, 5 by (1) choosing 5 to be the number that appears four times, (2) choosing 3 as the number that appears once, and then arranging the digits 5, 5, 5, 5, 3 to get 5, 5, 5, 3, 5.

We next count the number of ways of doing each of the three steps.

Step 1: There are 6 ways of choosing x.

Step 2: Once x has been chosen, there are 5 ways of choosing the value for y.

Step 3: Last, once x and y have been selected, there are $_{5}C_{4}=5$ = 5 ways of mixing up the x’s and y’s.

To find the number of four-of-a-kinds, we use the multiplication rule using the number of ways of doing each of the three steps:

Number of ways = $6 \times 5 \times 5 = 150.$ 6×5×5=150.

The corresponding probability of four-of-a-kind is

Prob(four-of-a-kind) = 150/7776.

PRACTICE: Yahtzee

1. (Probability of a two pair.) We follow the same basic method as described above in counting the number of two-pairs. Represent an outcome by the sequence

x, x, y, y, z

where x and y denote the numbers that will each occur twice, and z is the number that appears one.

(a) How many ways are there for choosing the numbers x and y? (Note that since both x and y each appear twice, it is incorrect to say that the number of ways of choosing x and y is 6 $\times$ 5 = 30.)

(b) How many ways are there for choosing z, the identity of the number that appears once?

(c) The number of ways of mixing up the orders of x, x, y, y, z is

$\frac{5!}{2!2!1!}$ 5!2!2!1!

(This is a generalization of the earlier counting rule of arrangements of two different types.) Combine this result with the results of parts (a) and (b) to find the number of two-pairs.

(d) Find the probability of a two-pair.

2. (Probability of a large straight.) A large straight is observing five consecutive numbers in our roll. We write down a large straight in two steps:

Step 1: We write down the possible five numbers in the large straight.

Step 2: We arrange the numbers.

(a) How many ways can you do step 1? In other words, how many possible choices are there for the five numbers that make up the straight?

(b) How many ways can you arrange the five numbers in the straight?

(c) Combining the two steps, how many ways can you have a large straight?

(d) What is the probability of a large straight?