26 Aug
2005
26 Aug
'05
2:51 p.m.
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.