Re: math question
i'll give it a shot.
These calculations assume the first player already holds the pocket aces.
The number of regular flushes (not including straight flushes) that do not include the 2 aces in the first players hand = 2*combin(13,5) + 2*combin(12,5) - 36 = 4122
This is any 5 cards from the 13 in the 2 suits not represented by the aces and any 5 cards from the 12 remaining cards in the aces' suits, minus the 36 straight flushes that do not contain the 2 aces.
The number of possible 5-card hands the 2nd player could hold = combin(50,5)= 2118760
The odds of his flopping the flush is therefore = 4122/2118760 = 0.001945478
Since each hand is independent, the individual probabilities can be multiplied to get the probability of their happening concurrently.
(0.001945478)^3
small enough to not worry much about its happening even once, much less 3 times in a row.
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