Consider an $N$-sided die. I roll it twice. I want to find the expectation of the first dice roll given that it is greater than the second dice roll.
So if we let $X$ and $Y$ denote the random variable value for the first and second rolls, respectively. We want to find $E[X | X > Y]$. Conditioning on $X > Y$ means we only consider the following ordered pairs $(2,1), (3,1), (3,2), (4,1), (4,2), (4,3)$ and so on. For an $N$-sided die there are $1 + 2 + \ldots + N-1 = \frac{N(N-1)}{2}$ of these ordered pairs.
So
$$ E[X | X > Y] = \frac{2}{N(N-1)} (1 * 2 + 2 * 3 + 3 * 4 + \ldots + (N-1)*N) $$
So this leads in to a series. I am wondering if there's an easier way to solve this without having to solve this series? I feel like there should be based on the nice structure of this problem