Expected winning amounts for $2$ players with different number of sided dice

A simpler way of getting the same answer is as follows (again I assume $m>n$ are the two sizes of dice, with yours smaller).

• If your opponent rolls higher than $n$, (probability $\frac{m-n}{m}$), you lose the amount he rolls. This is uniform between $n+1$ and $m$, so on average it is $\frac{n+1+m}{2}$. So this gives a contribution of $-\frac{(m-n)(n+m+1)}{2m}=-\frac{m^2-n^2+m-n}{2m}$.

• If you both roll different numbers which are $n$ or below (probability $\frac{n-1}{m}$) then it is totally symmetrical so your expected win/loss is $0$.

• If you both roll the same number (probability $\frac{1}{m}$), then you lose however much that number is; on average it is $\frac{1+n}{2}$. So this contributes $-\frac{1+n}{2m}$.

Your total expectation is therefore $-\frac{m^2-n^2+m+1}{2m}$, which is the same as Math1000's result.

(This also gives you a quick way to get the answer if you have the larger die, but still pay your opponent when the rolls are equal: just make the first term positive before adding them. Here $m$ is now the number of sides on your die since we assume $m>n$.)

More generally assume that your die has $n$ sides and your oppponent's die has $m$ sides. Let $X$, $Y$ be the score obtained by you and your opponent, respectively. Then your winnings are $W := X\mathsf 1_{\{X>Y\}} - Y\mathsf 1_{\{X\leqslant Y\}}$, and so for each $k\in\{2,3,\ldots n\}$ we have $$ \mathbb P(W = k) = \mathbb P(X\mathsf 1_{\{X>Y\}} - Y\mathsf 1_{\{X\leqslant Y\}} = k) = \mathbb P(X=k)\mathbb P(k>Y) =\begin{cases} \frac 1n\cdot\frac{k-1}m = \frac{k-1}{nm},& k\leqslant m\\ \frac1n\cdot\frac mm = \frac1n ,& k>m\\ \end{cases} $$ Similarly, for $k\in\{-1,-2,\ldots,-m\}$ we have $$ \mathbb P(W=k) = \mathbb P(Y = |k|)\mathbb P(X\leqslant |k|) = \begin{cases} \frac1m\cdot\frac{|k|}n = \frac{|k|}{mn},& |k|\leqslant n\\ \frac1m\cdot\frac nn = \frac1m,& |k|>n. \end{cases} $$ Hence for $m>n$ (as in the original problem), we have \begin{align} \mathbb E[W] &= \sum_{k\in\mathbb Z}k\cdot \mathbb P(W=k)\\ &= \sum_{k=2}^n k\cdot\frac{k-1}{nm} - \left(\sum_{k=1}^{n} k\cdot \frac{k}{mn}+\sum_{k=n+1}^{m} k\cdot\frac1m\right)\\ &= \frac{(n-1) (n+1)}{3 m} - \left(\frac{(n+1) (2 n+1)}{6 m} + \frac{(m-n) (m+n+1)}{2 m} \right)\\ &= \frac{(n+1)(n-1)-m(m+1)}{2 m}. \end{align} For $n=20$, $m=30$ we have $\mathbb E[W] = -\frac{177}{20}$.