# Dice conditional expectation

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?

• What is the expected value of my roll given that it is greater than yours? Please rephrase your question. As it is written it is not clear what you mean. Sep 27, 2023 at 19:06
• I'll denote with $X, Y$ the outcome of the two dices and assuming they are independent. I don't know whether there's an even ''simpler'' way, but computing by Bayes the probability of $P(X=k \mid X > Y) = \frac{P(X > Y \mid X =k)}{P(X > Y)} P(X=k) = \frac{k-1}{190}$ and then taking the expected value seems rather straightforward once one knows the result of the summations $\sum_{k=1}^n k^p$ for $p = 1,2$. Sep 27, 2023 at 19:32
• Of the $n^2$ pairs of possible results, $\frac{n(n-1)}{2}$ of these have your dice of greater value than mine. If you rolled $m$, there are $m-1$ possible cases where you are higher than me. So, we calculate the total sum of your values over the winning cases to be $(\sum_{m=1}^{n} m(m-1)) = \frac{n(n+1)(n-1)}{3}$ . Dividing by number winning cases, ie by $\frac{n(n-1)}{2}$, we get expectation of $\frac{2(n+1)}{3}$, where $n=20$. Sorry it's not clever. Sep 27, 2023 at 19:40

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 $$E(v_1 + v_2) = E(v_2 + v_3) = E(v_3 + v_1)=$$ $$= E(\frac{(v_1+v_2) + (v_2+v_3) + (v_1 + v_3) }{3}) =$$ $$=E(\frac{2}{3}(v_1 + v_2 + v_3)) =$$ $$= E(\frac{2}{3}(n+1)) = \frac{2}{3}(n+1)$$