The game of poker dice and the multinomial theorem

This post presents an application of the multinomial theorem and the multinomial coefficients to the game of poker dice. See the previous post The multinomial theorem for a discussion of the multinomial theorem and multinomial coefficients.

The game of poker dice is different from the standard poker game. Instead of drawing five cards from a deck of 52 cards, five fair dice are rolled. The resulting five scores from the dice form a poker dice hand. The possible hands are ranked as follows:

  • Five of a kind: one score occurs five times (e.g. 2,2,2,2,2).
  • Four of a kind: two distinct scores appear, one score occurs four times and the other score occurs one time (e.g. 2,2,2,2,5).
  • Full house: two distinct scores appear, one score occurs three times and the other score occurs two times (e.g. 3,3,3,1,1).
  • Three of a kind: three distinct scores appear, one score occurs three times, the other two scores occur one time each (e.g. 2,2,2,3,6).
  • Two pair: three distinct scores appear, two of the scores occur two times each and the other score occurs once (e.g. 4,4,1,1,5).
  • One pair: four distinct scores appear, one score occurs two times and the other three scores occur one time each (e.g. 5,5,1,2,4).
  • High card: five distinct scores occur (e.g. 2,3,5,4,1).

 
In rolling 5 dice, there are 6^5=7776 ordered outcomes. For example, assuming that the five dice are rolled one at a time, 2,3,2,6,2 indicates outcome that the first die results in a two and the second die results in a three and so on (this is a three of a kind). To find the probability of a three of a kind, we simply divide the number of ways such hands can occur by 7776. We use the multinomial coefficients to obtain the number of outcomes for each type of poker dice hands.

Rolling k dice (or rolling a die k times) can also be regarded as the occupancy problem of assigning k balls into 6 cells. As will be shown below, the problem of computing the probabilities of poker dice hands is seen through the lens of the occupancy problem of randonly placing 5 balls into 6 cells. For example, five of a kind is equivalent to all five balls being placed in different cells. Three of a kind is equivalent to three of the balls being in one cell, and the other two balls in two other different cells. For discussion on the occupancy problems in this blog, see the following posts:

In placing 5 balls into 6 cells, we use 6-tuples to indicate how many balls are in each of the 6 cells. For example, (0,3,1,0,0,1) denotes a three of a kind hand of 2,2,2,3,6 (the score of two appearing three times, the score of three appearing one time and the score of six appearing one time). Note that the 6 coordinates represent the six scores of a die (six cells) and the sum of the coordinates is 5. The 6-tuple of (3,1,1,0,0,0) is also a three of a kind hand, representing the outcome that the score of one appearing three times, the score of two appearing one time and the score of three appearing one time. We use the multinomial coefficients to determine how many of the 7776 ordered outcomes correspond to a 6-tuple such as (3,1,1,0,0,0). With respect to the occupancy problem, such 6-tuples are called occupancy number sets.

The Multinomial Theorem
For any positive integer n and any positive integer k, we have the following polynomial expansion:

    \displaystyle \biggl(x_1+x_2+ \cdots +x_n\biggr)^k=\sum \limits_{a_1+ \cdots +a_n=k} \frac{k!}{a_1! \ a_2! \cdots \ a_n!} \ \ x_1^{a_1} x_2^{a_2} \cdots x_n^{a_n}

Remark
In addition to being a formula for polynomial expansion, there are two important interpretations of the multinomial theorem and multinomial coefficients. One is that of determining the number of ordered strings that can be formed using a set of alphabets. For example, with one m, four i's, four s's and two p's, there are \displaystyle \frac{11!}{1! \ 4! \ 4! \ 2!}=\text{34,650} possible 11-letter strings that can be formed, of which mississippi is one specific example.

Another interpretation is that of partitioning a set of distinct objects into several groupings where objects in each grouping are indistinguishable. For example, in a group of 11 candidates, how many ways can we form four committees such that the Committee 1 has only one member, Committee 2 has four members, Committee 3 has four members and Committee 4 has two members (assuming that each person can only serve in one committee)? The answer, as in above example, is \text{35,650}.

Example 1
In a random poker dice hand, what is the probability of obtaining a 4 two times, a 3 one time, a 5 one time and a 6 one time? Note that this is a specific example of the poker dice hand of one pair.

We consider the 6-tuple of (0,0,1,2,1,1). We are trying to partition 5 scores into four subgroups, one group having two identical scores of 4, one group with a score of 3, one group with a score of 5 and one group with a score of 6. Thus consider the following multinomial coefficient:

    \displaystyle \frac{5!}{1! \ 2! \ 1! \ 1!}=60

So out of \text{7,776} possible hands, 60 of them satisfiy the condition that a 4 appearing two times, a 3 appearing one time, a 5 appearing one time and a 6 appearing one time. The probability is:

    \displaystyle \frac{60}{7776}=0.0077160494

Example 2
What is the probability that one score appears two times, three other scores appear one time each in a random poker dice hand?

Here, we need to count all the possible poker dice hands of one pair. Both (0,0,1,2,1,1) and (2,1,1,1,0,0) are examples of one pair. In essense, we need to count all the occupancy number sets such that among the 6 coordinates (cells), one of the cells is a 2 and three of the cells are 1. To this end, we apply the multinomial theorem twice, one time on the five rolls of dice and one time one the 6 cells.

Consider the occupancy number set (2,1,1,1,0,0). Note that the multinomial coefficient is 60 as in Example 1 (the first application of the multinomial thoerem). Now look at the 6 coordinates of the occupancy number set (2,1,1,1,0,0). We wish to partition these 6 coordinates into three groupings, one with one 2, one with three 1's and one with two 0's. The following is the multinomial coefficient (the second application of the multinomial theorem):

    \displaystyle \frac{6!}{1! \ 3! \ 2!}=60

Thus the number of possible poker dice hands of one pair is: 60 \times 60=\text{3,600} and for a random poker dice hand, the probability that it is a one pair is:

    \displaystyle \frac{3600}{7776}=0.462962963

Remark
Example 2 provides the algorithm for computing the remaining poker dice hand probabilities. The key is to apply the multinomial coefficients twice, one time on a representative occupancy number set, the second time on the six cells (the six faces of a die in this case). Then the number of poker dice hands in question is the product of the two multinomial coefficients.

Example 3
What is the probability that a random poker dice hand is three of a kind?

Consider the occupancy number set of (3,1,1,0,0,0). The associated multinomial coefficient for the five rolls of dice is:

    \displaystyle \frac{5!}{3! \ 1! \ 1!}=20

Now partition the six cells into three groupings (one 3, two 1's, three 0's):

    \displaystyle \frac{6!}{1! \ 2! \ 3!}=60

Thus the probability that a random poker hand is three of a kind is:

    \displaystyle \frac{20 \times 60}{7776}=0.1543209877

Summary
The following are the probabilities of poker dice hands.

    \displaystyle P(\text{five of a kind})=\frac{6}{7776}

    \displaystyle P(\text{four of a kind})=\frac{150}{7776}

    \displaystyle P(\text{full house})=\frac{300}{7776}

    \displaystyle P(\text{three of a kind})=\frac{1200}{7776}

    \displaystyle P(\text{two pair})=\frac{1800}{7776}

    \displaystyle P(\text{one pair})=\frac{3600}{7776}

    \displaystyle P(\text{high card})=\frac{720}{7776}

________________________________________________________________________
\copyright  \ \text{2010 - 2015 by Dan Ma} (Revised March 28, 2015)

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One thought on “The game of poker dice and the multinomial theorem

  1. I now understand so much better how one calculates these poker dice probabilities! You saved me before the test. I wish my lecturer explained this to us! ps. any books you recommend for a first year? I don’t like the way things are explained in “A First Course in Probability” by Sheldon Ross..

    Thanks again!

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