Suppose we both roll two fair twenty sided die. Given that I rolled higher than you, what is the expected value of my roll?
I believe I could answer this question but I'm mainly wondering if there is a more efficient way to compute this answer rather than summing up the conditional probabilities of rolling each number n 1-20 and multiplying that by n?
Found the trick to solve it. Let $n$ be the max value of the die, and $(d_1,d_2)$ be a pair satisfying $d_2 > d_1$, ie. a favorable event, where the second element of the pair represents your roll, and the first represents theirs.
Let's represent the event $(d_1,d_2)$ where $d_2>d_1$ as the triple of differences: $$(v_1,v_2,v_3) = (d_1 - 0, \hspace{1mm} d_2-d_1, \hspace{1mm} (n+1)-d_2)$$
These three differences are uniformly distributed values between $0$ and $n+1$ (excluding $0$ and $n+1$), with the restriction that they sum to $n+1$.
We are looking for: $E(d_2) = E(v_1+v_2)$ . But $$E(v_1 + v_2) = E(v_2 + v_3) = E(v_3 + v_1)= $$ $$= E(\frac{(v_1+v_2) + (v_2+v_3) + (v_1 + v_3) }{3}) =$$ $$=E(\frac{2}{3}(v_1 + v_2 + v_3)) = $$ $$= E(\frac{2}{3}(n+1)) = \frac{2}{3}(n+1)$$
Summary of proof idea: