Conditional expectation of the number of phone calls There are twenty individuals numbered $1,2,...,20$.Each individual chooses 10 others from this group in a random fashion,independently of the choices of the others, and makes one phone call to each of the $10$.
Let $X$ be the number of calls handled(incoming as well as outgoing) by Individual 1.Let $Y$ be the number of calls between Individual 1 and 2. Find $E(X|Y=1)$.
My approach:
See that, $X=10+Y_1$ where $Y_1:$ number of phone calls incoming to $1$.
Thus, $E(X)=10+E(Y_1)$. See that $Y_1 \sim Bin(19,\frac{1}{19})$. Thus, $E(X)=10+\frac{19}{19}=11$. Also, $Y=  \left\{
\begin{array}{ll}
      0  \ \text{with probability} \frac{100}{361} \\
      1  \ \text{with probability} \frac{180}{361} \\
      2  \ \text{with probability} \frac{81}{361} \\
\end{array} 
\right. $
Now,given $Y=1$,
One case can be that individual 1 has 2 on his caller list but 2 does not call him.
And the other is that individual 1 does not call 2 but 2 calls 1. Let $X_3,X_4+1$ be the number of calls received by $1$ in the two situations respectively.
So,$E(X|Y=1)=\frac{[E(10+X_3)]\frac{90}{361}+[10+1+E(X_4)]\frac{90}{361}}{\frac{180}{361}}$
Now, here $X_3 \sim Bin(18,\frac{1}{18})$ and $X_4 \sim Bin(18,\frac{1}{18})$.
So, we plug in the expectations to get a value.
But is this approach correct?
Anyone having a better idea to calculate the conditional expectation?
 A: $X$ denotes the number of calls handled (incoming as well as out-going) by Individual $1$. Further, it is certain that the number of outgoing calls for $1$ will always be $10$.
Consider now the expected number of incoming calls handled by $1$. If we, for a while, forget about individual $2$. Then the expected number of calls from $18$ remaining individuals to individual 1 will be equal to $18.\dfrac{10}{19}$ (from the Binomial Distribution with parameters $n = 18$ and $p = 10/19$).
Further it is given that there is only $1$ call between individual $1$ and individual $2$. Now, this call can be either made from $1 \to 2 \text{ or }2 \to 1$.
$P(1\to2\mid Y=1) = P(2\to1\mid Y=1) = \dfrac{\frac{90}{361}}{\frac{180}{361}} = \dfrac{1}{2}$. That is, with probability $\frac{1}{2}$, the call has already been counted in the $10$ calls made by one and with probability $\frac{1}{2}$, the call yet remains to be counted. Then including this information with the above quanitites, we get
$E(X\mid Y=1) = 10 + \dfrac{1}{2} + 18.\dfrac{10}{19}$ 

Note: The answer is almost same as that provided in the previous answer, only that he missed $10$ outgoing calls. I had to write this
  answer since I wasn't able to comment on his.

A: I think the following argument works and computes the expectation more easily:
Let $X'$ be the number of calls handled by individual $2$. By symmetry, $E(X|Y=1)=E(X'|Y=1)$, and by linearity of expectation, $E(X+X'|Y=1)=2E(X|Y=1)$, so we can answer the question by solving a simpler problem: What is the expected number of calls involving either individual $1$ or individual $2$, given that there was exactly one call between them?
Aside from the known call between them, all other calls handled by $1$ or $2$ were placed by one of the other $18$ individuals. For one of these other individuals, say individual $x$, the expected number of calls to $1$ is $10/19$ and the expected number of calls to $2$ is also $10/19$, because their calls are independent of the known condition, which doesn't involve them.
Therefore the number of calls involving either $1$ or $2$ is $1+18(2\cdot{10\over19})$, and the desired expected value is half that, or ${1\over2}\cdot\left(1+18(2\cdot{10\over19})\right)$.
