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My question is a variant of the balls-in-bins problem (similar to the coupon collector problem), where I'm interested in how many of the $n$ bins will be empty after throwing $k$ balls.

As far as I understand, this normally equals $\mathbf{E}[X] = n\left(\frac{n-1}{n}\right)^k$. Where I get lost is when I try and extend this problem as follows:

Rather than $n$ bins that can hold any number of balls, let us say that $m$ of the bins are no longer available after receiving one ball (i.e. without replacement). The other $n-m$ bins continue to act as normal. If $m=n$, then of course $X=n-k$ bins will be empty after throwing $k$ balls.

For other values of $m$ though, how does this affect the number of empty bins?

Let me know if I need to clarify anything. This is my first time posting! Thank you for your help :)

Sam
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  • I don't have an answer, but if you google "balls into bins with finite capacity" (no quotes), you get helpful links like: https://math.stackexchange.com/questions/665178/how-many-ways-to-distribute-k-indistinguishable-balls-over-m-of-n-distingu https://math.stackexchange.com/questions/707996/number-of-ways-to-distribute-indistinguishable-balls-into-distinguishable-boxes https://math.stackexchange.com/questions/213755 -- I realize your problem is different since only SOME bins have finite capacity, but generating functions may be the way to go here. –  Sep 26 '17 at 15:49

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