# What is the probability of the sum of four dice being 22?

## Question

Four fair six-sided dice are rolled. The probability that the sum of the results being $22$ is $$\frac{X}{1296}.$$ What is the value of $X$?

## My Approach

I simplified it to the equation of the form:

$x_{1}+x_{2}+x_{3}+x_{4}=22, 1\,\,\leq x_{i} \,\,\leq 6,\,\,1\,\,\leq i \,\,\leq 4$

Solving this equation results in:

$x_{1}+x_{2}+x_{3}+x_{4}=22$

I removed restriction of $x_{i} \geq 1$ first as follows-:

$\Rightarrow x_{1}^{‘}+1+x_{2}^{‘}+1+x_{3}^{‘}+1+x_{4}^{‘}+1=22$

$\Rightarrow x_{1}^{‘}+x_{2}^{‘}+x_{3}^{‘}+x_{4}^{‘}=18$

$\Rightarrow \binom{18+4-1}{18}=1330$

Now i removed restriction for $x_{i} \leq 6$ , by calculating the number of bad cases and then subtracting it from $1330$:

calculating bad combination i.e $x_{i} \geq 7$

$\Rightarrow x_{1}^{‘}+x_{2}^{‘}+x_{3}^{‘}+x_{4}^{‘}=18$

We can distribute $7$ to $2$ of $x_{1}^{‘},x_{2}^{‘},x_{3}^{‘},x_{4}^{‘}$ i.e$\binom{4}{2}$

We can distribute $7$ to $1$ of $x_{1}^{‘},x_{2}^{‘},x_{3}^{‘},x_{4}^{‘}$ i.e$\binom{4}{1}$ and then among all others .

i.e

$$\binom{4}{1} \binom{14}{11}$$

Therefore, the number of bad combinations equals $$\binom{4}{1} \binom{14}{11} – \binom{4}{2}$$

Therefore, the solution should be:

$$1330-\left( \binom{4}{1} \binom{14}{11} – \binom{4}{2}\right)$$

However, I am getting a negative value. What am I doing wrong?

EDIT

I am asking for my approach, because if the question is for a larger number of dice and if the sum is higher, then predicting the value of dice will not work.

#### Solutions Collecting From Web of "What is the probability of the sum of four dice being 22?"

There aren’t too many to just count.

Permutations of $6+6+6+4$: $\binom41=4$

Permutations of $6+6+5+5$: $\binom42=6$

These are the only options, so your numerator must be $4+6=10$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

The number of configurations that satisfies “the sum is $\ds{22}$” is given by:

\begin{align}
X & =
\sum_{d_{1} = 1}^{6}\sum_{d_{2} = 1}^{6}\sum_{d_{3} = 1}^{6}\sum_{d_{4} = 1}^{6}
\bracks{z^{22}}z^{d_{1} + d_{2} + d_{3} + d_{4}} =
\bracks{z^{22}}\pars{\sum_{d = 1}^{6}z^{d}}^{4} =
\bracks{z^{22}}\pars{z\,{z^{6} – 1 \over z – 1}}^{4}
\\[5mm] & =
\bracks{z^{18}}{1 – 4z^{6} + 6z^{12} – 4z^{18} + z^{24}\over \pars{1 – z}^{4}}
\\[5mm] & =
\bracks{z^{18}}\pars{1 – z}^{-4} – 4\bracks{z^{12}}\pars{1 – z}^{-4} + 6\bracks{z^{6}}\pars{1 – z}^{-4} – 4\bracks{z^{0}}\pars{1 – z}^{-4}
\\[5mm] & =
{-4 \choose 18}\pars{-1}^{18} – 4{-4 \choose 12}\pars{-1}^{12} +
6{-4 \choose 6}\pars{-1}^{6} – 4 =
{21 \choose 18} – 4{15 \choose 12} + 6{9 \choose 6} – 4
\\[5mm] & =
1330 -4 \times 455 + 6 \times 84 – 4 = \bbx{10}
\end{align}

The bad combinations criteria is atleast one $x_i \geq 7$.

The number of bad combinations when:

1. One of $x_i$ is forced to be greater than or equal to $7$ is $\binom{4}{1}\binom{12 + 4 – 1}{12} = 1820$.

2. Two of $x_i$’s are forced to be greater than or equal to $7$ is $\binom{4}{2}\binom{6+4-1}{6} = 504$.

3. Three of $x_i$’s are forced to be greater than or equal to $7$ is $\binom{4}{3}\binom{0+4-1}{0} = 4$.

4. Four of $x_i$’s are forced to be greater than or equal to $7$ is $0$.

So, total bad combinations $= 1820 – 504 + 4 – 0 = 1320$

I used $n(\cup_{i=1}^{4}A_i) = \sum_{i=1}^{4}n(A_i) – \sum_{i,j, i\neq j}n(A_i\cap A_j) + \ldots$

So, possible combinations $= 1330 – 1320 = 10$.

In order for the sum to equal 22, either three dice equal $6$ and one equals $4$, or two dice equal $6$ and two dice equal $5$. The number of valid outcomes thus equals:

$${4 \choose 1} + {4 \choose 2} = 4 + 6 = 10$$

As such, the probability of the four dice having a sum of $22$ equals:

$$\frac{{4 \choose 1} + {4 \choose 2}}{6^4} = \frac{10}{1296} = \frac{5}{648} \approx 0.00772$$

I am going to show you how you can reduce the given problem a simpler one by using symmetry.

You wish to find the number of solutions of the equation
$$x_1 + x_2 + x_3 + x_4 = 22 \tag{1}$$
in positive integers not exceeding $6$. Since $x_k$, $1 \leq k \leq 4$, is a positive integer satisfying $x_k \leq 6$, then $y_k = 7 – x_k$ is also a positive integer not exceeding $6$. Substituting $7 – y_k$ for $x_k$, $1 \leq k \leq 4$, in equation 1 yields
\begin{align*}
7 – y_1 + 7 – y_2 + 7 – y_3 + 7 – y_4 & = 22\\
-y_1 – y_2 – y_3 – y_4 & = -6\\
y_1 + y_2 + y_3 + y_4 & = 6 \tag{2}
\end{align*}
Equation 2 is an equation in the positive integers when the given restrictions are imposed. A particular solution of equation corresponds to placing an addition sign in three of the five spaces between successive ones in a row of six ones. For instance,
$$1 + 1 1 + 1 + 1 1$$
corresponds to the solution $y_1 = 1$, $y_2 = 2$, $y_3 = 1$, and $y_4 = 2$ (or $x_1 = 6$, $x_2 = 5$, $x_3 = 6$, $x_4 = 5$) while
$$1 1 1 + 1 + 1 + 1$$
corresponds to the solution $y_1 = 3$, $y_2 = y_3 = y_4 = 1$ (or $x_1 = 4$, $x_2 = x_3 = x_4 = 6$). Thus, the number of solutions of equation 2 in the positive integers is the number of ways of selecting which three of the five spaces between successive ones in a row of six ones will be filled by addition signs, which is
$$\binom{5}{3} = 10$$

Note: If you prefer to work in the nonnegative integers, then we wish to find the number of solutions of the equation
$$x_1′ + x_2′ + x_3′ + x_4′ = 18 \tag{1’}$$
subject to the restrictions that $x_k’ = x_k – 1 \leq 6 – 1 = 5$ for $1 \leq k \leq 4$. Since $x_k’$, $1 \leq k \leq 5$, is a nonnegative integer not exceeding $5$, so is $y_k’ = 5 – x_k’$. Substituting $5 – y_k’$ for $x_k’$ in equation 1′ and simplifying yields
$$y_1′ + y_2′ + y_3′ + y_4′ = 2 \tag{2’}$$
which is an equation in the nonnegative integers. A particular solution corresponds to the insertion of three addition signs in a row of two ones. For instance,
$$+ + + 1 1$$
corresponds to the solution $y_1 = y_2 = y_3 = 0$, $y_4 = 2$ (or $x_1 = x_2 = x_3 = 6$, $x_4 = 4$), while
$$1 + 1 + +$$
corresponds to the solution $y_1 = y_2 = 1$, $y_3 = y_4 = 0$ (or $x_1 = x_2 = 5$, $x_3 = x_4 = 6$). Thus, the number of solutions of equation 2 in the nonnegative integers is
$$\binom{2 + 3}{3} = \binom{5}{3}$$
since we must choose which three of the five positions (two ones and three addition signs) will be filled with addition signs.

It is the coefficient of $x^{22}$ in the expansion of the generating function $$(x^6+x^5+x^4+x^3+x^2+x^1)^4=\left(\frac{x(1-x^6)}{1-x}\right)^4$$ which is $$1\cdot{{x}^{24}}+4\cdot {{x}^{23}}+10\cdot {{x}^{22}}+20\cdot {{x}^{21}}+35\cdot {{x}^{20}}+56\cdot {{x}^{19}}+80\cdot {{x}^{18}}\\+104\cdot {{x}^{17}}+125\cdot {{x}^{16}}+140\cdot {{x}^{15}}+146\cdot {{x}^{14}}+140\cdot {{x}^{13}}+125\cdot {{x}^{12}}+104\cdot {{x}^{11}}\\+80\cdot {{x}^{10}}+56\cdot {{x}^{9}}+35\cdot {{x}^{8}}+20\cdot {{x}^{7}}+10\cdot {{x}^{6}}+4\cdot {{x}^{5}}+1\cdot{{x}^{4}}$$

so the answer is $10$.

This generalises easily to more dice or even those with different numbers of faces

This is a method similar to @GTonyJacobs and @adhg’s answers, but in this answer I use a slightly different way of thinking:

The maximum number possible is a set of six $4$’s $(6, 6, 6, 6)$, which totals $6 * 4 = 24$. To reach $22$, we need to take off $2$ in total from the set.

There are $2$ ways to do this:

– take off $1$ from $2$ numbers out of $4$.

– take off $2$ from $1$ number out of $4$.

Following this, the first method can be done in $4 \choose 2$ ways, while the second can be done in $4 \choose 1$ ways, which is $10$ ways in total.

Therefore, the probability of $4$ dice totalling $22$ is $\frac{10}{6^4}$, which is approximately $0.0077$.

On an intuitive level, simply count how many ways are there to get 22. There are 2 ways to do so:

(A) 6,6,6,4 (B) 6,6,5,5

Each way can have different permutation (example 6,6,6,4, 6,6,4,6 etc) so for the first way (A), you can have 4 ways to arrange the result because: 4!/3!=24/3=4

and for (B) there are 6 ways to arrange the result because 4!/(2!*2!)=6.

So in total you get 10 ways to get 22. So 10/1296

We have that
\eqalign{ & x_{\,1} + x_{\,2} + x_{\,3} + x_{\,4} = 22\quad \left| {\;1 \le x_{\,k} \le 6} \right.\quad \Rightarrow \cr & \Rightarrow \quad y_{\,1} + y_{\,2} + y_{\,3} + y_{\,4} = 18\quad \left| {\;0 \le y_{\,k} \le 5} \right. \cr}
and
\eqalign{ & {\rm N}{\rm .}\,{\rm of}\,{\rm solutions}\,{\rm without}\,{\rm upper}\,{\rm restriction} = \cr & = {\rm N}{\rm .}\,{\rm of}\,{\rm (strong)}\,4{\rm – elements}\,{\rm compositions}\,{\rm of}\,22 = \cr & = {\rm N}{\rm .}\,{\rm of}\,{\rm (weak)}\,4{\rm – elements}\,{\rm compositions}\,{\rm of}\,18 = \cr & = \left( \matrix{ 22 – 1 \cr 4 – 1 \cr} \right) = 1330 \cr}

Concerning bad compositions, we have that if you write for instance
\eqalign{ & \overline {x_{\,1} } + x_{\,2} + x_{\,3} + x_{\,4} = 22\quad \left| \matrix{ \;7 \le \overline {x_{\,1} } \hfill \cr \;1 \le x_{\,k} \hfill \cr} \right.\quad \times \;\left( \matrix{ 4 \cr 1 \cr} \right)\quad \Rightarrow \cr & \Rightarrow \quad x_{\,1} + x_{\,2} + x_{\,3} + x_{\,4} = 16\quad \left| {\;1 \le x_{\,k} } \right.\quad \times \;\left( \matrix{ 4 \cr 1 \cr} \right) = \cr & = 4\;\left( \matrix{ 16 – 1 \cr 4 – 1 \cr} \right) = 1820 \cr}
you get a number of bad compositions which is higher than the total number, and it is not the number
of compositions with a least one term $\ge 7$. That is clearly due to the fact that , when adding $6$ to $x_1$ that could be already $\ge 7$.

And if you write
$$\overline {x_{\,1} } + x_{\,2} + x_{\,3} + x_{\,4} = 22\quad \left| \matrix{ \;7 \le \overline {x_{\,1} } \hfill \cr \;1 \le x_{\,k} \le 6 \hfill \cr} \right.$$
you are back to the original problem, with one element less and one sum to apply.

So that is not the right approach to this problem.

$\ds{\bbox[#dfd,5px]{\ The\ general\ formula\ }}$
for getting a sum $s$ with $m$ dices each with $r$ faces is given
in this post.