Conditional expectation of first roll conditionally on sum of two rolls I have to find $E(X\mid Y)(y)$ where $X$ is the value of the first roll and $Y$ is the sum of the two dice.   I know that

$$E(X|Y)(y) = \sum_x{xP(X\mid Y)}=\frac{\sum_xxP(X=x, Y=y)}{P(Y=y)},$$

but this doesn't really get me anywhere.  From that summation, I get

$$E(X|Y)(y) = \frac{1\cdot P(1, Y=y)}{P(Y=y)} + \frac{2\cdot P(2, Y=y)}{P(Y=y)} + \dots + \frac{6\cdot P(6, Y=y)}{P(Y=y)}.$$

Here's where I'm stumped.  Can someone please answer this question for me.

Thanks!

EDIT:  To be clear, I need to find the expected value of the first roll given the sum of the two dice.

I get E(X∣Y=y)=y/2 for y≥2. Is that correct?
Use symmetry: 


$(X,Y)$ and $(Y-X,Y)$ have the same distribution hence $$E(X\mid Y)=E(Y-X\mid Y).$$ 
$X+(Y-X)=Y$  hence $$E(X\mid Y)+E(Y-X\mid Y)=E(Y\mid Y)=Y.$$


Thus, $$E(X\mid Y)=\frac12Y.$$
See this.  Additionally/alternatively, Bayes' theorem would let you write $P(X \mid Y)$ in terms of $P(Y \mid X)$, which may make more sense to you.