[cbloom]

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Imagine if player #1 only plays AA and player #2 only plays AK and KQ. If you don't permute the order then the fact that player #1 has taken away two of the aces every time before player #2 gets his selection would mean that he would end up with KQ far more often than he should (as there will always only be 2 aces left to choose from). On the other hand, if you permute the player order then 50% of the time player #2 gets to choose when there are 4 aces still available and 50% of the time he has just 2 aces left to choose from.


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You guys are chasing ghosts.

You deal Player 1 random cards. If he doesnt have AA he folds.
You deal Player 2 cards excluding the ones already dealt.
This correctly gives Player 2 an equal frequency of cards.

Now, if you want to consider the cases where Player 1 *didn't fold* and Player 2 also *doesn't fold* then you are look at the probably of

P1 Dealt AA AND Player 2 dealt AK or KQ

Probably of P1(AA) & P2(AK)
= (6/1326) * (8/1326)

Probably of P1(AA) & P2(KQ)
= (6/1326) * (16/1326)

Now, if you're trying to just assign hands, you have to count the multiplicity. It doesn't matter what order you do it.

P1 then P2 :

Give P1 AA - 6 ways
Give P2 AK - 8 ways
or Give P2 KQ - 16 ways

P1(AA) and P2(AK) = 6*8 = 48 ways
P1(AA) and P2(KQ) = 6*16 = 96 ways

P2 then P1 :

P2 AK - 16 ways
then give P1 AA - 3 ways

or P2 KQ - 16 ways
then give P1 AA - 6 ways

the product of ways is the same in either order. There's no need to try different permutations.