Most likely sum to be thrown with $2n$ dice is $7n$, with $2n+1$ dice most likely sums $7n+3$, $7n+4$

combinatoricsdiceprobability

Here's a problem from my probability textbook:

Show that the most likely sum to be thrown with $2n$ dice is $7n$; and with $2n+1$ dice the most likely sums are $7n+3$ and $7n+4$.

Random (unrelated) thoughts:

  • I don't even know to begin with this problem. Intuitively it makes sense since when $n = 1$ and we have two dice the most likely sum to be thrown is $7$ since we have $(1, 6)$, $(2, 5)$, $(3, 4)$, $(4, 3)$, $(5, 2)$, $(6,1)$ giving us probability ${1\over{6}}$. But I am not sure how to bootstrap this up for higher $n$.

  • Can we use some sort of induction?

  • Stars and bars gets very ugly for$$x_1 + x_2 + \ldots + x_{2n} = k$$where $k$ varies from $2n$ to $12n$ since the inclusion-exclusion gets pretty messy.

  • A key observation in the $n=1$ case is that there is a bijection between $\ell$ and $7 – \ell$ where $\ell \in \{1,2,3,4,5,6\}$, so maybe we want to produce a bijection between $m$ and $7n – m$ where $m \in \{1, 2, \ldots, 7n – 2, 7n-1\}$ and show that it is "maximal" relative to other $k$ where $k$ varies from $2n$ to $12n$?

EDIT: Matías Vera in the comments suggests using something called convolution. But my textbook suggests this problem can be done with entirely elementary methods.

EDIT 2: Okay, I just realized that the answer is evident when considering the coefficients of the terms of the expansions$$(x + x^2 + x^3 + x^4 + x^5 + x^6)^{2n}, \quad (x + x^2 + x^3 + x^4 + x^5 + x^6)^{2n+1}.$$But that feels like a cheap solution, I'd like a way to solve the problem that's a bit more combinatorial/"explicit" and sheds more light on what's going on.

EDIT 3: The answer in the back of my book is as follows:

The number of ways in which a number $m$ can be thrown with $r$ dice is the coefficient of $x^m$ in the expansion of$$(x + x^2 + x^3 + x^4 + x^5 + x^6)^r.$$But by the process of continued multiplication when $x + x^2 + x^3 + x^4 + x^5 + x^6$ is multiplied by itself, and again by itself, any number of times, the coefficients of the terms from either end must be the same and the middle coefficient or coefficients must be the highest. With $2n$ dice the mean throw is $7n$, and with $2n+1$ dice the mean throws will be $7n+3$ and $7n+4$. These will therefore be the throws most likely to occur.

This is much simpler than Mike Earnest's answer (which Mike even himself admits is not even complete!), so presumably the answer in the back of my book is not correct. How specifically does it fall short?

Best Answer

Let $P(n,s)$ be the probability of obtaining a sum of $s$ with $n$ dice. We will prove something stronger: that

$P(n,s+1)< P(n,s)$ when $s< 7n/2$,${}^*$ and
$P(n,s+1)> P(n,s)$ when $s\ge 7n/2$.

${}^*$Exception: When $n$ is even and $s=(7n-1)/2$, we have equality: $P(n+1,s)=P(n,s)$.

That is, the probability mass function increases up to the middle, and decreases after the middle, or is unimodal. We will prove this by induction on $n$.

By conditioning on the roll of one of the dice, you get for all $n \ge 2$ that $$ P(n,s)=\frac16\Big[P(n-1,s-1)+P(n-1,s-2)+\dots+P(n-1,s-6)\Big] $$ Now, for short, let $\Delta P(n,s)=P(n,s+1)-P(n,s)$. Then $$ \Delta P(n,s)=\frac16\Big[\Delta P(n-1,s-1)+\dots+\Delta P(n-1,s-6)\Big]\tag{$\star$} $$ We need to show the LHS is positive for $s< 7n/2$, and negative for $s\ge 7n/2$.

  • If $(s-1)< 7(n-1)/2$, that is if $s<(7n-5)/2$, then the induction hypothesis gives us that all of the $\Delta$'s on the RHS of $(\star)$ are positive, so we conclude the LHS is positive.

  • Similarly, if $(s-6)\ge (7n-1)/2$, i.e. when $s\ge (7n+5)/2$, all of the $\Delta's$ are negative, so the LHS of $(\star)$ is negative.

  • All that remains is the set of $s$ for which $$ (7n-5)/2 \le s < (7n+5)/2 $$ These are the middle values of $s$. When $n$ is even, it is the middle $5$ values, and when $n$ is odd, it is the middle $6$ values. This is the trickiest case. What happens is that some of the $\Delta$'s will be positive, and some will be negative, but the symmetry of the problem implies some positives will cancel come negatives, and only the desired signs remain.

    For example, consider $n=3$ and $s=9$. We get $$ \Delta P(3, 9)=\frac16\Big[\Delta P(2,3)+\Delta P(2,4)+\color{blue}{\Delta P(2,5)}+\color{red}{\Delta P(2,6)}+\color{red}{\Delta P(2,7)}+\color{blue}{\Delta P(2,8)}\Big] $$ Since the $n=2$ distribution is symmetric about $7$, you can show the red $\Delta$'s are equal and opposite each other, and same for the blue $\Delta$'s. The remaining $\Delta$'s are positive, so $\Delta P(3,9)\ge 0$, as desired.

This is the intuition; writing this up in a way that rigorously covers of all of the details might be a huge pain.

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