*Originally posted by uzless*

**you have four sets of numbers from 1 to 16 for a total of 64 numbers.
**

You put them all in a hat.

12 people draw 5 numbers each out of the hat.

What are the odds that at least one person will not draw a number from 1-4?

Just to clarify, it sounds like there are 16 available top picks, call them 1A-4A, 1B-4B, 1C-4C and 1D-4D. I think the easiest way to solve this is to work backwards and find the probability that all 12 players will get at least one top pick, and then subtract this probability from 1 to find the answer.

First off, there are 64c60 ways (i.e. "64 choose 60" or 64!/(60!*4!) ways) to choose which picks will get assigned, and then (60c5)*(55c5)*...(10c5)(5c5) ways to assign those picks to the players 5 at a time, so there are (64c60)*(60c5)*(55c5)*...(10c5)*(5c5) assignments in total.

Now, in order to ensure all 12 players receive a top pick we choose 12 out of the 16 top picks and assign one to each player randomly. Then, of the remaining 52 choices we choose 48 and assign them 4 at a time. The total number of ways to do this is (16c12)*12!*(52c4)*(48c4)*(44c4)*...(8c4)*(4c4). The 12! comes from the random assignment of the 12 top picks, because (12c1)*(11c1)*...(2c1)*(1c1) = 12!.

The overall probability that each player receives at least one top pick is just the ratio of these:

P = (16c12)*12!*(52c48)*(48c4)*(44c4)*...(8c4)*(4c4) / (64c60)*(60c5)*(55c5)*...(10c5)*(5c5) = 13.5%

The probability that at least one player does not receive any top picks is simply 1-P, which is approximately 86.5%.