Expected number of empty bins with clumpy balls? I have $b$ bins and a reservoir of $n$ balls. The balls have a tendency to "clump", that is, when I try to grab one from the reservoir, the number of balls removed is uniformly random on $[1, m]$, where $m$ is the number of balls remaining in the reservoir for that draw. The "clump" is added to one of the $b$ bins, the bin selected from all bins with equal probability. When the reservoir is depleted, how many empty bins are expected? I don't even know where to start... Edit: I asked a neighbor (an actuary) this question, found this on a note on my windshield this morning, seems to match the answer by Quasi : $\frac{b\prod _ {z=\frac{b-1}{b}}^{\frac{b-1}{b}+n-1} z}{n!}$

For positive integers $n,b$, let $f(n,b)$ be the expected number of empty bins at the end of the process.

I'm not sure if there's a closed form for $f(n,b)$, but here's a recursion, implemented in Maple, to compute $f(n,b)$ for given (not too large) values of $n,b$, . . .

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Shown above is a sample calculation for $n=10,\;b = 6$, which shows that $$f(10,6) = \frac{751583152441}{208971104256} \approx 3.596588893$$ For $n=10000,\;b=100$, the recursion fails (too many levels of recursion for my version of Maple), but a simulation gives the approximate result $$f(10000,100) \approx 90.67$$