Let $X$ be the value of the first die and $Y$ the sum of the two dice. Find $E(X / Y)$ Consider an experiment of rolling two dice. Let $X$ be the value of the first die and $Y$ the sum of the two dice. Find $E(X / Y)$, ie, obtain the value of $E(x/y) (y)$ for all $y$
Good evening, I could solve the problem. I have no idea how to start. You could give me some tips to solve it, please. Or some bibliographic references with similar exercises?
 A: Assuming you mean $\mathrm{E}(X|Y)$, then since the distribution of each die is identical and independent, if we are given that their sum is $y$, the expected value of each die would be the same: $y/2$.
If you did mean $\mathrm{E}(X/Y)$, then in light of the preceding, this would be $\frac12$.

Explicit Calculation of $\boldsymbol{\mathrm{E}(X|Y)}$
Given that the sum of the dice is $y$, the possibilities for $x$ are from $y-6$ to $6$ with equal probability of each. That is, the probability that the first die is $x\in\mathbb{Z}$ is
$$
\frac1{13-y}[y-6\le x\le6]
$$
where the brackets are Iverson brackets.
Thus, the expected value would be
$$
\begin{align}
\sum_{x=y-6}^6\frac{x}{13-y}
&=\frac{\frac126(6+1)-\frac12(y-7)(y-6)}{13-y}\\
&=\frac{21-21+\frac{13}2y-\frac12y^2}{13-y}\\[9pt]
&=\frac y2
\end{align}
$$
This nicely agrees with the symmetry argument given above.
A: We show how to handle the problem for one value of $y$, say $y=9$.
Given that $Y=9$, $X$ takes on values $3$ to $6$ with equal probabilities. Thus $$E(X|Y=9)=\frac{3+4+5+6}{4}.$$  
One value of $y$ done, $10$ more to do. 
Remark: The symmetry argument of robjohn is much better.
