TaintedRogue
08-09-2005, 09:21 PM
My estimate is that in a 10 player game, 50% of the players will have a pocket pair. Anyone know how to compute this?
Thanks,Ken
AaronBrown
08-09-2005, 10:23 PM
That's too high.
There are 52*48*47*46/(4*3*2*1) = 270,725 possible Omaha hands.
13*6*48*44/2 = 82,368 of them contain exactly one pocket pair (not trips or quads, not two pair). There are 13 ranks, each one can select two cards in 6 different ways, there are 48 cards that don't match and 44 that don't match either the pair or the third card, divided by 2 because the last two cards can come in either order. That's 30%.
If you add the two pairs, trips and quads, it doesn't amount to even 2% more hands.
With ten hands the numbers are slightly different, but not much. About 3 of them can be expected to have pocket pairs.
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