# Probability for the sum of two dice

I am trying to do the following exercise:

Imagine rolling two dice and summing the results up. (So we can get values between $$2$$ and $$12$$). What sum has the highest probability? What if we have $$n \in \mathbb{N}$$ dice?

My approach: In case of the two dice, the most simple approach would be to consider every single case. But this gets way too time-consuming for more than 2 dice. So here is the other approach I have been thinking about:

Let $$A_k$$ be the event that the sum equals $$k$$, where we know that $$2 \leq k \leq 12$$. Thus a successful outcome $$\omega$$ would be of the form $$(i,k-i)$$ for $$i \in \{1,...,6\}$$.

I think the next step would be to find a formula for $$|A_k|$$ depending on $$k$$. But I don't know how to find it.

• I don't think that the approach taken by the original poster will lead anywhere, as the resulting computations will increase in complexity, each time that the number of dice, $~n,~$ increase. For example, the probability of getting a $~10~$ on three dice is $$\sum_{k = 4}^9 \left[ ~f(k) \times \frac{1}{6} ~\right].$$ Here, the factor of $~\dfrac{1}{6}~$ reflects the probability that the third die specifically shows $~(10 - k),~$ while $~f(k) = \dfrac{k-1}{36} ~: ~k \leq 7,~$ and $~f(k) = \dfrac{13 - k}{36} ~: ~k > 7.$ Commented Aug 14 at 0:53
• It is quick and easy to say which sum has the highest probability for any number of dice. It is much more tedious to figure out what the probability of that sum is. Which question are you trying to ask, exactly? Commented Aug 14 at 0:56
• @DavidK I want to figure out the probability of that sum.
– NTc5
Commented Aug 14 at 1:00
• Then say so in the question. Commented Aug 14 at 2:50

A nice way to see how many ways you can sum a number, is through the notion of a convolution, which essentially you can see as making one variable go up, and the other down.

For example, to get the number of ways you can sum up to six with two dice, let's call that $$A_6^2$$, you can see it as pairs $$(1,5),(2,4),(3,3),(4,2),(5,1)$$. The magic of this is that if you know $$A^n_k$$, that is, the number of ways you can count up to $$k$$ using $$n$$ dice, then you can pair $$1$$ and $$A^n_k$$, that is, assuming the result of the first die is $$1$$, and checking how many ways the other dice can add up to $$5$$ to compensate.

Try to work out the numbers, and you will find easy-ish ways to compute what you want for $$n+1$$ dice, if you already have the results for $$n$$ dice, and proceed by induction.

$$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\iverson}[1]{\left[\left[\,{#1}\,\right]\right]} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$ $$\bbx{\begin{array}{c}\mbox{Albeit my answer seems weird at first sight, it illustrates} \\ \mbox{the general procedure for}\ \underline{any\ number}\ \mbox{of dices !!!.} \end{array}}$$

Hereafter, $$\ds{\on{P}_{S}}$$ is the probability of draw a sum of $$\ds{S = 2,3,\ldots,12}$$ with two $$6$$-side dices. Therefore, \begin{align} \color{#44f}{\left.\rule{0pt}{5mm}\on{P}_{S}\right\vert_{S\ \in\ \mathbb{N}_{\scriptscriptstyle \ 2\ \leq\ 12}}} & \sr{\rm def.}{\equiv} \color{#44f}{\sum_{a = 1}^{6}{1 \over 6} \sum_{b = 1}^{6}{1 \over 6} \bracks{z^{S}}z^{a + b}} \\[5mm] & = {1 \over 36}\bracks{z^{S}} \pars{\sum_{a = 1}^{6}z\, {z^{6} - 1 \over z - 1}}^{2} \\[5mm] & = {1 \over 36}\bracks{z^{S - 2}}\pars{1 -2z^{6} + z^{12}}\pars{1 - z}^{-2} \\[5mm] & = {1 \over 36}\left\{% \bracks{S \geq 2}{-2 \choose S - 2} \pars{-1}^{S- 2}\right. \\[2mm] & \rule{1.5cm}{0pt}-2\bracks{S \geq 8}{-2 \choose S - 8} \\[2mm] & \rule{1.5cm}{0pt}\left.+\bracks{S \geq 14}{-2 \choose S - 14}\right\} \\[5mm] & = {1 \over 36} \left\{\rule{0pt}{5mm}\bracks{S \geq 2}\pars{S - 1} - 2\bracks{S \geq 8}\pars{S - 7}\right. \\[2mm] & \left.\phantom{AAA\,\,}+ \bracks{S \geq 14}\pars{S - 13}\rule{0pt}{5mm}\right\} \\[5mm] & = \color{#44f}{\left\{\begin{array}{ccl} \ds{{S - 1 \over 36}} & \mbox{if} & \ds{2 \leq S \leq 7} \\[2mm] \ds{{13 - S \over 36}} & \mbox{if} & \ds{8 \leq S \leq 12} \end{array}\right.} \end{align}

You can obtain the distribution by iteratively computing "running sums". Assume that you know the distribution for $$n-1$$ dice, $$N_{n-1}^s$$ giving the number of ways to obtain the sums $$s$$ with $$n-1$$ dice (nonzero for $$s\in [n-1,6(n-1)]$$).

To obtain $$N_n^s$$, it suffices to consider all ways to obtain $$s-1$$ plus a $$1$$, or $$s-2$$ plus a $$2$$..., so $$N_n^s=\sum_{k=1}^6 N_{n-1}^{s-k}.$$

For a single dice, the distribution is $$1,1,1,1,1,1$$ over $$[1,6]$$. Then by running sum, for two dice, $$1,2,3,4,5,6,5,4,3,2,1$$ over $$[2,12]$$. Three dice, $$1,3,6,10,15,21,25,27,27,25,21,15,10,6,3,1$$

Here are more values obtained by an Excel sheet.

After normalization to get the probabilities, we see those plots. The curves tend to a Gaussian. If you are familiar with them, you can recognize B-spline weighting functions.

The expressions are not easy to derive. They are piecewise polynomial of degree $$n$$, in $$n$$ intervals of length $$6$$, with a common point.