A Dodecahedral Game of Craps: What Is the Probability of Winning?

 

Introduction

Craps is usually played with two six-sided dice, but what happens when we replace them with two twelve-sided dice?

At first glance, the rules of this modified game seem straightforward. A player rolls two regular dodecahedra, with each face numbered from 1 to 12. Certain sums produce an immediate win or loss, while every other sum establishes a “point” that remains relevant until it is rolled again—or until the losing sum 13 appears.

What makes the problem interesting is that the different sums are not equally likely (just like in the original game). Some sums can be obtained in many different ways, while others are much rarer. Once a point has been established, the game also enters a kind of race: will the player roll that particular sum first, or will 13 appear first?

The question, therefore, is not simply a matter of counting the possible outcomes of one throw. We have to combine the distribution of the sums with the probabilities of eventually hitting the point before 13.

This is a nice example of how a seemingly elementary dice game can turn into a problem involving conditional probability and an infinite sequence of possible throws. Can you solve this problem as stated below?

A Dodecahedral Game of Craps

A modified version of the dice game craps is played with two regular (i.e., perfectly symmetrical) dodecahedra. Each die has its sides numbered from \(1\) to \(12\) so that after each throw of the dice, the sum of the numbers on the top two surfaces of the dice would range from \(2\) to \(24\). If a player gets the sum \(13\) or \(23\) on his first throw (a natural), he wins. If he gets \(2\), \(3\), or \(24\) on his first throw (craps), he loses. If he gets any other sum (his point), he must throw the dice again. On this or any subsequent throw, the player loses if he gets the sum \(13\) and wins if he gets his point, but must throw both dice again if any other sum occurs. The player continues until he either wins or loses. To the nearest whole percent, what is the probability at the start of any game that a dice thrower will win?\(^{1}\)

There is a total of \(12\times 12=144\) possible outcomes. Let \(s\) be a given sum of the two numbers on the top two surfaces of the dice. Define the function \(D:\{ 2,3,...,24 \}\to \{ 1,2,...,12 \}\) by the rule: For any given sum \(s\),

$$D(s) = \begin{cases}
s-1 & \text{if } \: 2\le s\le 13 \\
25-s & \text{if } \: 13\le s\le 24 \\

\end{cases}$$

$$=\text{The number of ways to obtain}\:s$$

Why then, you may ask, does this function give the correct number of ways in which a sum \(s\) can be obtained from some dice throw? To answer this question, remember that we wish to count the number of integer pairs \(\left( n,m \right)\in \{ 1,2,3,...,12 \}^{2}\) with 

$$n+m=s, \:\:\:1\le n\le 12,\:\:\: 1\le m\le 12,\:\:\: 2 \le s \le 24$$

If \(s\le 13\), then both \(n\) and \(m\) may be at most \(s-1\) , which is \(12\) at most (the upper bound comes from \(m=s-n\ge1 \Rightarrow s-1\ge n \)). In this case, there is no restriction, and the number of solutions is simply \(s-1\) (pick \(n=1,2,3,...,s-1\)). For example, let \(s=10\), which can be attained in 9 different ways: \((1,9), (2,8), ..., (9,1)\). We will call this our naive count \(N(s)\). For \(s\gt 13\), we must subtract the number of invalid solutions in which either \(m\ge 13\) or \(n\ge 13\) but not both (if \(m\ge 13\) and \(n\ge 13\), then \(m+n=s\ge 26\), which is impossible!) from our naive count. If \(n \ge 13\) then \(n\) may be equal to \(13,14,15, ..., s-1\) because \(13\le n\le s-1\). In other words, the upper bound constraint still applies in addition to the new lower bound. Consequently, the number of possible solutions in this case is \((s-1)-13+1=s-13\) (We count both endpoints of the interval). Similarly, if \(m \ge 13\), the number of possible solutions is simply \(s-13\). Then, we subtract these invalid solutions from \(N(s)\) to get the expression for all \(s \ge 13\),

$$N(s)-2(s-13)=(s-1)-2s+26=25-s$$

And that is why the function \(D\) indeed gives the correct count for the number of ways in which a particular sum \(s\) is made. Now, to determine the overall probability of winning this game, we must add up all the probabilities that contribute to it, i.e., all the ways a given player may win. As stated in the problem, if the player gets the sum \(13\) or \(23\) on his first roll, he wins. Then,

$$\mathbb{P}\left( \text{A Natural} \right)=\frac{D(13)+D(23)}{144}=\frac{12+2}{144}=\frac{14}{144}=\frac{7}{72}$$

Another way to win is by setting a point first (any other sum has occurred on his first roll except \(2\), \(3\), or \(24\)). The probability of setting a point on his first roll is therefore

$$\frac{D(s)}{144}, \text{where} \:s\in \{ 4,5,...,12,14,...,22 \}$$

If a point has been set, the player continues to roll until either a subsequent roll matches his point and wins or gets the sum \(13\) and loses. Let \(p=\text{P}(\text{roll s})\)(his point), \(q=\text{P}(\text{roll 13})\) (lose), and \(r=1-p-q\) (other sums). After a point has been set, each roll is either decisive, with probabilities of winning or losing, \(p\) and \(q\) respectively, or non-decisive, with probability \(r\) (In which case, he must throw the dice again). For a player to win at all, the first decisive roll must be his point \(s\), which can happen on the very next roll, having a probability of \(p\), after a non-decisive roll, having a probability of \(r\cdot p\), after two non-decisive rolls, having a probability of \(r^{2}\cdot p\), and so forth. Hence,

$$\mathbb{P}\left( \text{Eventually Win} \right)=p+rp+r^{2}p+r^{3}p+...=p\sum_{k=0}^{\infty }r^{k}=\frac{p}{1-r}$$

Then, since \(r=1-p-q\), by substitution,

$$\frac{p}{1-(1-p-q)}=\frac{p}{p+q}=\frac{\text{P}(\text{roll s})}{\text{P}(\text{roll s})+\text{P}(\text{roll 13})}=\frac{\frac{D(s)}{144}}{\frac{D(s)}{144}+\frac{D(13)}{144}}=\frac{D(s)}{D(s)+12} $$

By the general product rule of probability (the conjunction of setting a point first and then winning once that point is matched on any subsequent roll), the total contribution from all possible points is 

$$\sum_{s\in \text{points}}^{}\frac{D(s)}{144}\cdot\frac{D(s)}{D(s)+12}=\sum_{s\in \text{points}}^{}\frac{D(s)^{2}}{144\left( D(s)+12 \right)}$$

Consequently, by the sum rule of probability for independent events, the overall probability of winning is the sum of all the aforementioned contributions, that is,

$$\mathbb{P}\left( \text{Win} \right)=\frac{7}{72}+\sum_{s\in \{ 4,5,...,12,14,...,22 \}}^{}\frac{D(s)^{2}}{144\left( D(s)+12 \right)}$$

What follows is pure arithmetic. Although the test maker's instructions state that we are only allowed to use pocket calculators—which isn't a significant issue—it’s worth noting that it's no longer \(1985\), and what's crucial is the mathematical reasoning, not the computation itself, in my judgment. So, to speed up the process, we will use Excel to compute the final result.

s represents the points. Note that the rightmost column shows the contribution of each point to the overall probability. 

Thus,

$$\mathbb{P}\left( \text{Win} \right)=\frac{7}{72}+\sum_{s\in \{ 4,5,...,12,14,...,22 \}}^{}\frac{D(s)^{2}}{144\left( D(s)+12 \right)}\approx 0.437680591
$$

Which, rounded to the nearest whole percent, is \(44\%\). I don't know about you, but I wouldn't play this game!

                                                          

References:

1.  Adapted from Hoeflin, R. K. (1985, April). Problem 37 in The Mega Test. Omni, 7(4), p. 129-132.