Solving a Three-Dice Problem Using Two Approaches

Last updated on Feb 26, 2023 4 min read

Introduction

In his book, Do Dice Play God, Ian Stewart wrote

For three dice, Cardano solved a long-standing conundrum. Gamblers had long known from experience that when throwing three dice, a total of 10 is more likely than 9. This puzzled them …

In this note, I will focus my attention to the so-called 3-dice problem: We roll three dice simultaneously; what’s the probability that we get a total $t$ , where $t = 3, 4, \dots, 18$ ?

The probability that we get a total $t$ is

$\frac{C (t)}{6^{3}} = \frac{C (t)}{216},$

where $C (t)$ is the count of throws with total being equal to $t$ . For getting $C (t)$ , I will next show a generating function approach and a brute force approach (i.e. an R function).

Generating function approach

We define

$\begin{array}{ccl} g (x) & = & (x + x^{2} + \dots + x^{6})^{3} \\ = & x^{3} (1 + x + \dots + x^{5})^{3} . \end{array}$

The desired number

C (t)

is nothing but the coefficient of

x^{t}

g (x)

, or

C (t)

is equal to the coefficient of

x^{t - 3}

$\begin{array}{ccl} g_{1} (x) & = & (1 + x + \dots + x^{5})^{3} \\ = & {(\frac{1 - x^{6}}{1 - x})}^{3} \\ = & (1 - x^{6})^{3} (1 - x)^{- 3} . \end{array}$

We know that

$(1 - x^{6})^{3} = 1 - 3 x^{6} + 3 x^{12} - x^{18},$

and

$\begin{array}{ccl} (1 - x)^{- 3} & = & \sum_{i = 0}^{+ \infty} (- 1)^{i} (\binom{- 3}{i}) x^{i} \\ = & \sum_{i = 0}^{+ \infty} (\binom{3 + i - 1}{i}) x^{i} \\ = & \sum_{i = 0}^{+ \infty} (\binom{i + 2}{2}) x^{i} . \end{array}$

Letting

$u = t - 3,$

thus,

for $0 \leq u \leq 5$ ,

$C (t) = (\binom{u + 2}{2});$

for $6 \leq u \leq 11$ ,

$\begin{array}{ccl} C (t) & = & (\binom{u + 2}{2}) - 3 (\binom{u - 6 + 2}{2}) \\ = & (\binom{u + 2}{2}) - 3 (\binom{u - 4}{2}); \end{array}$

for $12 \leq u \leq 15$

$\begin{array}{ccl} C (t) & = & (\binom{u + 2}{2}) - 3 (\binom{u - 4}{2}) + 3 (\binom{u - 12 + 2}{2}) \\ = & (\binom{u + 2}{2}) - 3 (\binom{u - 4}{2}) + 3 (\binom{u - 10}{2}) . \end{array}$

Replacing

u

with

t - 3

in the above, we have

for $3 \leq t \leq 8$ ,

$C (t) = (\binom{t - 1}{2});$

for $9 \leq t \leq 14$ ,

$C (t) = (\binom{t - 1}{2}) - 3 (\binom{t - 7}{2});$

for $15 \leq t \leq 18$ ,

$C (t) = (\binom{t - 1}{2}) - 3 (\binom{t - 7}{2}) + 3 (\binom{t - 13}{2}) .$

Remark: In the above we have used the following well known result:

$\begin{array}{ccl} (1 - x)^{- n} & = & \sum_{i = 0}^{+ \infty} (- 1)^{i} (\binom{- n}{i}) x^{i} \\ = & \sum_{i = 0}^{+ \infty} (\binom{n + i - 1}{i}) x^{i} \\ = & \sum_{i = 0}^{+ \infty} (\binom{n + i - 1}{n - 1}) x^{i} . \end{array}$

Brute force approach

n_dice_problem <- function(n = 3)
{x <- rep(list(1:6), n) 
 y <- as.matrix(expand.grid(x))
 table(rowSums(y))
}

(n_dice_problem(n = 3))

## 
##  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 
##  1  3  6 10 15 21 25 27 27 25 21 15 10  6  3  1

(n_dice_problem(n = 4))

## 
##   4   5   6   7   8   9  10  11  12  13  14  15  16  17  18  19  20  21  22  23 
##   1   4  10  20  35  56  80 104 125 140 146 140 125 104  80  56  35  20  10   4 
##  24 
##   1

Generalizing 3-dice problem to n-dice problem

An obvious generalization of the 3-dice problem is an $n$ -dice problem, where $n$ can be any a positive integer. Let $C^{(n)} (t)$ denote the count of throws with the total of $n$ dice being equal to $t$ . Since we already knew how to get $C^{(3)} (t)$ (generating-function approach), it is easy to have

$C^{(n)} (t) = \sum_{i = 0}^{⌊ (t - n) / 6 ⌋} (- 1)^{i} (\binom{n}{i}) (\binom{t - 6 i - 1}{n - 1}),$

where

t = n, n + 1, \dots, 6 n

In the brute-approach section, we already have R function n_dice_problem(), but because of computer memory limit the value of $n$ cannot be larger than 10. We have the following R code, in which the value of $n$ can be large, say at least as large as 100.

library(gmp)
c_n_t  <- function(t, n) {
  a <- floor((t - n) / 6)
  i <- as.bigz(0:a)
  temp <- (-1)^i * chooseZ(n, i) * chooseZ(t - 6*i - 1, n - 1)
  sum(temp)
}
c_n_t(300, 100)

## Big Integer ('bigz') :
## [1] 207241756759032546720209656767808329086730171654102671945216674007548108775

Acknowledgment: I thank Professor Martin Maechler of ETH Zurich for pointing me to gmp package.

Reference

[1] Stewart, I. Do Dice Play God? pp. 29-30.