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Server Time: 12/1/2008 5:25:46 PM PACIFIC |
 need help from probability EXPERT, dalessi, 22. Aug 2003 20:52 | ||
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| I know it is astronomical, and very difficult to figure out, but this happened in tourney I was in. What are the odds of 3 people flopping sets in a given hand, not assuming that they start with pocket pairs, but from scratch. So I guess it's odds 3 or more have pocket pairs multiplied by odds that 3 actually do flop sets. Incidentally I had 33, other hands were 55 and 22, and flop came 532. As you can imagine I got carried out on hand, lol. Can anyone tell me the true odds? Thanks | ||
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| Re: need help from probability EXPERT, Schuster, 22. Aug 2003 23:54 | ||
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| Assuming just 3 people dealt in with random starting hands, and assuming I did the math right, 8.413e(-10), or 1,1886,234,600 to 1. The odds go up as more people are dealt in, of course, and they go up considerable if all 3 people already start with pocket pairs. | ||
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| Re: need help from probability EXPERT, Eaglesfan1, 23. Aug 2003 01:32 | ||
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| Same thing happened to me. I had 22. Another guy had 55. Another had KK. Flop came K52, I busted out on this hand and it left me wondering about the strength of 22. | ||
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| Re: need help from probability EXPERT, George Rice, 23. Aug 2003 22:25 | ||
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| Most flops will qualify as they do not have a pair. 52x48x44/52x51x50 = .82823529 or 82.8 % If exactly three players see the flop then the odds they have the three matching pairs are (2x9/49x48) x (2x6/47x46) x (2x3/45x44) = 1.2872023x10**-7 Multiply the two together yields 1.0661064x10**-7 = 9,379,926.1:1 Or if you prefer, figure it the other way. The chances of exactly three players seeing the flop with three different pairs is (2x6x13/52x51) x (2x6x12/50x49) x (2x6x11/48x47) = 2.0229368x10**-4 = 4942.3:1 The chances of the flop hitting all three pairs is 2x2x2x6/46x45x44 = 5.2700922x10**-4 = 1896.5:1 Multiplying the two together yields the same 1.0661064x10**-7 = 9,379,926.1:1 Interesting how it's more unlikely to have the three different pairs dealt to exactly three players, than it is to flop three sets given the three pairs. | ||
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