[jukofyork]

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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)


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There are 2 fewer cards to select from after the first player gets his (they aren't independant events) and there are two possible orderings for the players to get these cards, so wouldn't it be:

Probability of P1(AA) & P2(AK) =
= (6/1326)*(8/1225) + (16/1326)*(3/1225)

Probability of P1(AA) & P2(KQ)
= (6/1326)*(16/1225) + (16/1326)*(6/1225)

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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.

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That seems to make sense though, but could it also be effected by the fact that it's 8 ways out of 1326 and 6 ways out of 1225, etc?

Juk [img]/images/graemlins/smile.gif[/img]