Re: Prop bet and a math ?
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find the probability of not hitting one of the three cards and subtract from one.
1-[(43/52)*(42/51)*(41/50)] = .441 = 44.1%
in other words, more likely that a Q, 7 or 2 will come on flop than not.
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Maybe I'm being stupid here, but aren't there a total of 12 Q's, 7's, and 2's in the deck. That means that 40/52 are not any of these on the first draw, and we need a 40/52*39/51*38/50 shot to come in for none of these cards to come up = 44.7% that none of the cards will come up = 55.3% that one of them will. You want to bet for one of these cards to come.
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