[_D&L_]

Pretend we have two players that play only two hands:

AK or AA.

First player: AK - 16 combos, AA 6 combos, total 22
First Players chance of playing AK is 16/22
First Players chance of playing AA is 6/22

Now assign the second player:

If player 1 was assigned AK (p = 16/22), then
Player 2: (Baysian updating)
Plays AK with 9/12 probability
Plays AA with 3/12 probability
If player 1 was assigned AA (p = 6/22), then
Player 2: (Baysian updating)
Plays AK with 8/9 probability
Plays AA with 1/9 probability

Now what is Player 2's chance of playing AA
(16/22)*(3/12) + (6/22)*(1/9) = 7/33

Now the problem: Player 1 and Player 2 had the same distributions. But 7/33 != 6/22. The problem is that assigning PLAYER 1 his distribution frist, gives player 2 an unequal chance of getting AA. The player most likely to be dealt pocket pairs, is the one where all the aces are in deck when he gets a card on his distribution. Notice that Player 2's chance of being distributed AA is less than Player 1, regardless of whether Player 1 was dealt AK or AA. (6/22 >> 3/12 or 1/9)

A correct algorithm, would at the very least, have to give both players (with equal distributions) an equal chance of getting AA. Thus, dealing everyone random hands, and discarding until everyone gets a hand on their distribution is an unbiased, but albeit extremely slow method of getting the right answer.