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I'm going to try to rewrite my question in a better way:

I have a set of $N$ boxes, and one of those boxes is filled. I sample the boxes with uniform probability and without replacement until I find the filled box. What is the mean and variance for the number of empty boxes I've opened? For example, if $N = 2$, and I open two boxes to find the full box, I've opened one empty box.

My guess is that I need to open $\mu = \frac{N}{2}$ boxes to find the full box. Thus, I need to open $\frac{N}{2}-1$ empty boxes prior to opening the full box. However, because each trial depends on the next, I don't know how to calculate the variance?

I suppose I'd like something like the negative binomial distribution, but where we're sampling without replacement?


Update: Byron Schmuland answers the first part of my question here: Expectation of number of trials before success in an urn problem without replacement

We just need to recast my empty boxes as "red" balls and my full box as a "blue" ball, and ask how many red balls we need to draw before we draw a "blue" ball. However, how might we calculate the variance?

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  • $\begingroup$ I would not "use the hypergeometric distribution", rather I would note that the number of empty boxes one must open before finding the filled box is uniformly distributed on 0..N. $\endgroup$ – Did May 5 '13 at 20:29
  • $\begingroup$ Mean: look better. Variance: try. $\endgroup$ – Did May 5 '13 at 21:02
  • $\begingroup$ If $X$ is uniformly distributed on $\{0,1,\ldots,N\}$, what is $E[X]$, say, when $N=2$? Is it $\frac{N+1}2-1$? $\endgroup$ – Did May 5 '13 at 21:34
  • $\begingroup$ Quote: the number of empty boxes I need to open. There are $N$ empty boxes hence $X=N+1$ is impossible. $\endgroup$ – Did May 5 '13 at 21:38
  • $\begingroup$ Why do you erase your comments after I answered them? Why do you modify your post after it was discussed in the comments? Such a heavy (and silent) rewriting is not how the site is supposed to function. $\endgroup$ – Did May 6 '13 at 5:07
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Each box is equally likely to be the filled box, so the number of empty boxes opened before you open the fill box is uniformly distributed from $0$ to $N-1$. Thus the expected number of empty boxes opened is $(N-1)/2$, and the variance is

$$ \frac1N\sum_{k=0}^{N-1}k^2-\left(\frac{N-1}2\right)^2=\frac16\frac{(N-1)N(2N-1)}N-\left(\frac{N-1}2\right)^2=\frac{N^2-1}{12}\;. $$

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  • $\begingroup$ The variance should be the same for the total number of boxes opened, right? $\endgroup$ – Jack May 5 '13 at 22:38
  • $\begingroup$ @Jack: Yes, only the mean is shifted by $1$ in that case. $\endgroup$ – joriki May 5 '13 at 22:42
  • $\begingroup$ Sorry, one more question if you have the time - how might the mean and variance change if we had multiple filled boxes in the population? $\endgroup$ – Jack May 5 '13 at 22:49
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    $\begingroup$ @Jack: On your first comment: No, I don't think there's much of a connection there. On your second comment: That's slightly more complicated; I suggest to ask that in a question of its own. $\endgroup$ – joriki May 5 '13 at 22:51

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