Re: how do i do this 2-7 tripple draw math
I would approach it as how many ways are there to make each hand that you would consider favorable (16 ways to make 47 etc, will vary based on any dead cards). Then divide each of those possibilities by the total possible 2 card combos ((52-5)*(52-5-1)/2) for the example you mentioned. Then finaly add up all individual two card combinations. You can also look at how many ways to draw 1 good card, and then one you consider bad, or how many ways to catch two bricks (1 - all good 2 card and good 1 draws should equal all bad two card draws)
I think it's similar to looking at hold'em, (52*51)/2 = 1326 possible hands, 6 ways to make AA etc.
Anyone else confirm my way or know of an easier way?
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