I was thinking about this situation:
Suppose there are $n$ boxes. In each box we randomly throw one of the balls numbered $1,2,\ldots,k$, independently of other boxes. Let $X$ be the number of boxes with ball number $1$. What is $E[X]$, and what is $E[\dfrac{1}{X}\mid X>0]$?
It seems that there should be "roughly" $n/k$ boxes with ball number $1$, so $E[X]=n/k$, and $E[\dfrac{1}{X}\mid X>0]=k/n$. But is it right?
Edit: I've asked a new question here to make it direct to the point.
You're right about $E[X]$ but wrong about $E[1/X]$. Since $X$ can take the value $0$ with nonzero probability, $E[1/X]$ is undefined.
Added after OP's edit: Even if you condition on $X$ being positive, $k/n$ can't be the right answer in general. Since $X$ is the number of boxes with ball number 1, it's a whole number and consequently $1/X\le1$ if $X\gt0$, which means the expected value can't exceed $1$. But $k/n\gt1$ if $k\gt n$.
$X$ is binomially distributed with parameters $n$ and $p=1/k$. A succes if ball number $1$ is thrown, a failure otherwise.
