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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.

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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 $k\gt n$.

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  • $\begingroup$ That's a good point, Barry. I want to condition it on $X>0$. Please see my edit. $\endgroup$
    – Paul S.
    Nov 7 '13 at 20:31
  • $\begingroup$ (Responding to your edit) You're right. I've asked a new question here instead. $\endgroup$
    – Paul S.
    Nov 7 '13 at 20:46
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$X$ is binomially distributed with parameters $n$ and $p=1/k$. A succes if ball number $1$ is thrown, a failure otherwise.

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