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Old 12-15-2004, 01:46 PM
Dave H. Dave H. is offline
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Default Probability of AT LEAST a four flush

Question 1: Are my first 2 assumptions correct given a suited pocket?

Assumption 1: Probability of EXACTLY a four flush on the flop = (C(11,2) * 39) / C(50,3) = .1094

Assumption 2: Probability of EXACTLY a five flush on the flop = C(11,3)/C(50,3) = .0028

Question 2: Do I just sum the above (.1094 + .0028) to arrive at the probability of AT LEAST a four flush on the flop? If not, how would that be computed please?
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Old 12-15-2004, 01:53 PM
TRBNGR TRBNGR is offline
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Default Re: Probability of AT LEAST a four flush

The answer to question 2 is Yes.
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Old 12-15-2004, 02:37 PM
Lost Wages Lost Wages is offline
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Default Re: Probability of AT LEAST a four flush

Probability of EXACTLY a four flush on the flop = (C(11,2) * 39) / C(50,3) = .1094

Correct

Probability of EXACTLY a five flush on the flop = C(11,3)/C(50,3) = .0028

You have the formula correct but you made a mistake on the math, it should be .0084

Question 2: Do I just sum the above (.1094 + .0028) to arrive at the probability of AT LEAST a four flush on the flop? If not, how would that be computed please?

That's fine. (.1094 + .0084) = .1178

Lost Wages
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