Re: What\'s the probability that the flop missed everyone?
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The probability of not holding it is 1-(6/C(50,2)). The probability for 2 palyers not to hold a specific pair will be 2(1-(6/C(50,2)).
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I'm not sure why you think you can add here but it's not correct. Do you think its twice as likely that two players won't hold a specific pair? Shouldn't it be less likely?
1-(6/C(50,2)) is the probability that a player doesn't hold a specific pocket pair, the question is how often do two players not have any PP and not pair the flop.
Unless someone can show me a mistake, I'm pretty sure my way is correct and necessary. Look at the first way I did it if you want to see my method.
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