Excellent! Thanks very much. (For my own peace of mind, I suppose I'll have to figure out how that relates to the off-the-cuff inverse factorial heuristic...) --Michael Kleber On 8/26/05, Edwin Clark <eclark@math.usf.edu> wrote:
On Thu, 25 Aug 2005, Michael Kleber wrote:
Suppose you place N things in N buckets. How many do expect the fullest bucket to contain?
My colleague Stephen Suen says:
I believe I have a reference for it.
Kolchin, Sevastyanov and Chistyakov, Random Allocations, Wiley, 1978.
It is known that fullest urn has ln(n)/ln(ln(n)) balls (with probability tending to 1 as n goes to infinity).
I think the exact answer is
(1/n^n) ((n-1)n^n - \sum_{k\ge1} \sum_{i_1,...,i_n} n!/(i_1! ... i_n!) )
where the second sum is over all n-tuples (i_1,...,i_n) such that (1) \sum_j i_j = n, and (2) i_j <= k.
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