Many Dice

We denote $P(X = x)$ as the probability that $X = x$ and $\mathbb{E}[X]$ as the expected value of $X$ .

Let $X$ be the sum of Edgar's rolls and let $X_i$ , $1 \leq i \leq n$ be the value of the $i$ th die. We see that $X = X_1 + X_2 + ... + X_n$ . Observe that from linearity of expectation, $\mathbb{E}[X] = \mathbb{E} \left [ \sum_{i = 1}^n X_i \right ] = \sum_{i = 1}^n \mathbb{E} [X_i] = \frac{n(p+1)}{2}$ since the expected value of a single one of Edgar's die is $\frac{p+1}{2}$ .

Now, observe that if Edgar rolls a sum of $i$ , then Alan must roll $i+1, i+2, ...$ or $m$ . Thus, the desired probability is $\sum_{i = p}^{np} P(X = i) \frac{m-i}{m}$ . We can simplify this in the following manner: $\begin{aligned} \sum_{i = p}^{np} P(X = i) \frac{m-i}{m} & = \sum_{i = p}^{np} P(X = i) - \frac{1}{m} \sum_{i = p}^{np} iP(X = i) \\ & = 1 - \frac{1}{m} \mathbb{E}[X] = \boxed{1 - \frac{n(p+1)}{2m}} \end{aligned}$ as desired.

Hack: The answer must work with $n = p = m = 1$ , in which the answer is 0. The only option that gives this answer is $\boxed{1 - \frac{n(p+1)}{2m}}$ .

Next time, maybe actually use concrete numbers so you can't bash your way through like this?