Talk:Occupancy theorem

The example might be wrong - but the theorem is right —Preceding unsigned comment added by Derek dreery (talk • contribs)

The example is correct. It just lacks explanation. [15 Choose 12] possible combinations. If we assume each player has at least one card then there are [(15-4)Choose(12-4)] = [11 Choose 8] ways of doing this. Hence the probability is as stated. —Preceding unsigned comment added by 137.205.77.199 (talk) 17:34, 23 May 2008 (UTC)

Distinguish the buckets
it would be worthwhile to explicity say that the buckets are distinguishable.

Tashiro (talk) 19:35, 22 September 2012 (UTC)