# I have 2 tosses of a die, $X$ is the $\min$ and $Y$ is the $\max$

If we let $X_1$ be the toss of the first die, and $X_2$ be the toss of the second die, and as stated in the title, $X=\min(X_1,X_2)$ and $Y=\max(X_1,X_2)$. I'm asked to find $E(Y|X=x)$ and I know that it's $$\sum_{y=1}^6yP(Y=y|X=x)$$ But I'm just a little stuck on the conditional probability part. I know that when $x=1$ the expected value is $\frac{41}{11}$ but I don't understand how that is. I keep getting $\frac{41}{36}$, because

\begin{align}&\sum_{y=1}^6yP(Y=y|X=1)\\ &\qquad=1\left(\frac{1}{36}\right)+2\left(\frac{2}{36}\right)+3\left(\frac{2}{36}\right)+4\left(\frac{2}{36}\right)+5\left(\frac{2}{36}\right)+6\left(\frac{2}{36}\right)\\ &\qquad=\frac{41}{36}\end{align}

Where did I go wrong?

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Hint: Think of the $6\times 6$ array with entries $(i,j), 1 \leq i, j, \leq 6$. The event $\{X=1\}$ is comprised of the $11$ entries of the top row and leftmost column, and in these $Y$ takes on values $1, 2, 3, 4, 5, 6$ how many times? and so the expected value can be obtained from these. You have the $41$ already, the above explains why the denominator is $11$, not $36$.
Fix $x \in \{1,...,6\}$. The possible ordered pairs are $(6,x),...,(x+1,x),(x,x),(x,x+1),...,(x,6)$. A quick check shows that there are $13-2x$ pairs, and the sum of the maximum of all pairs is $(x+7)(6-x)+x=42-x^2$ (add the first and last, the second and second last, all the way into $(x+1,x)$, then add the remaining $(x,x)$).
Hence $E(Y|X=x) = \frac{42-x^2}{13-2x}$. Setting $x=1$ gives $E(Y|X=1) = \frac{41}{11}$