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 126.96.36.199 (talk) 17:34, 23 May 2008 (UTC)
Distinguish the buckets
it would be worthwhile to explicity say that the buckets are distinguishable.