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?