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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?

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  • $\begingroup$ 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. $\endgroup$ Sep 27, 2023 at 19:06
  • $\begingroup$ 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 $p = 1,2$. $\endgroup$
    – user807606
    Sep 27, 2023 at 19:32
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    $\begingroup$ 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. $\endgroup$ Sep 27, 2023 at 19:40

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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:

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