Conditional Expectation of dice rolls A 6-sided die is rolled for 10 times.  The die rolls are mutually independent.  Each die uniformly likely to result in a digit from 1,2, 3, 4, 5, and 6.  Let X denote the sum of the first 6 rolls and Y denote the sum of the last six rolls.
Find E(X|Y).
I have found E(X), E(Y), Var(X), Var(Y), Cov(X,Y), and Corr(X,Y).
I wish I had input, but am relatively new to statistics and don't know where to start on this problem. Thanks in advance. 
 A: You need to juggle around a bit with the random variables a bit, then you can solve this without going through all the casses one by one.
First lets split up a bit. Let $U$ denote the sum of the first 4 throws, $V$ the sum of the 5th and 6th throw, and $W$ the sum of the last 4 throws. We then have $X=U+V$ and $Y=V+W$. We also have that $U, V, W$ are independent of each other, and also $U$ being independent of $Y$. We thus already know $E(U|Y)=E(U)=4 \cdot 3.5 = 14$.
Now we can separate
$$ E(X|Y=k) = E(U+V|Y=k) = E(U|Y=k) + E(V|Y=k) = 14+E(V|Y=k)$$.
So we are left to find $E(V|Y=k)$. We can get this by splitting up further: Define $V'$ to be sum of the 7th and 8th throw and $V''$ the sum of the 9th and 10th throw. We then have $Y = V+V'+V''$ and can calculate
$$ E(V|Y=k)+E(V'|Y=k)+E(V''|Y=k)=E(Y|Y=k)=k $$
as we know that if $Y=k$ then certainly $Y=k$.
Now, for symmetry reasons we see that
$$ E(V|Y=k)=E(V'|Y=k)=E(V''|Y=k)$$ 
and thus  $E(V|Y=k)=k/3$. Combining the above we get for every $k$
$$ E(X|Y=k) = 14+E(V|Y=k) = 14 + k/3 $$
and thus
$$ E(X|Y) = 14 + \frac{1}{3}Y.$$
A: Hint: let $X_i$ be the outcome of tossing die $i=1,2, \ldots , 12$.
Then
\begin{align}
E(X|Y) & =  E\left( \sum_{i=1}^6 X_i \Big\vert \sum_{i=5}^{10} X_i\right) \\
& =  E\left( X_1+X_2+X_3+X_4 \Big\vert \sum_{i=4}^{10} X_i\right)+E\left( X_5+X_6 \Big\vert \sum_{i=5}^{10} X_i\right)\\
& =  4E(X_i) +E\left( X_5+X_6 \Big\vert \sum_{i=5}^{10} X_i\right) \end{align}
A: Let $X_i$ denote the value of the $i$-th roll. Then
\begin{eqnarray*}
E(X_1+\cdots+X_6|X_5+\cdots+X_{10})&=&4 E(X_1)+E(X_5+X_6|X_5+\dots+X_{10})\\
&=&4\times7/2+2\times \frac{X_5+\cdots+X_{10}}{6}\\
&=& 14+\frac{X_5+\cdots+X_{10}}{3},
\end{eqnarray*}
where independence was used in the first step and symmetry in the second. 
