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Message: [QUOTE="Dorkus Malorkus, post: 434649, member: 6020"] 2 other players (I'm assuming the three players left includes you, if only to avoid having to extend the following working even further), they hold 4 'mystery' cards. There are four 'gutshot' cards to choose from remaining deck of 46 (52, less 4 board cards, less your two cards) Let's call the cards in villain's hands A,B,C, and D, and assume we're working on a completely 'random hand' basis. [B]P(A is a 6) = 4/46 = 0.087[/B] [B]P(A is not a 6) = 1-0.087 = 0.913 [/B]| = "given that", for those who aren't acquainted with stat notation. :) P(B is a 6 and A is not a 6) = 0.913*(4/45) = 0.913*0.089 = 0.081 P(B is a 6 | A is not a 6) = P(B is a 6 and A is not a 6) / P(A is not a 6) P(B is a 6 | A is not a 6) = 0.081 / 0.913 [B]P(B is a 6 | A is not a 6) = 0.089[/B] P(A and B are not 6s) = 0.913*(1-0.089) = 0.831 P(C is a 6 and A and B are not 6s) = 0.831*(4/44) = 0.831*0.091 = 0.076 P(C is a 6 | A and B are not 6s) = P(C is a 6 and A and B are not 6s) / P (A and B are not 6s) P(C is a 6 | A and B are not 6s) = 0.076 / 0.831 [B]P(C is a 6 | A and B are not 6s) = 0.092[/B] P(A, B, and C are not 6s) = 0.831*(1-0.092) = 0.755 P(D is a 6 and A, B, and C are not 6s) = 0.755*(4/43) = 0.755*0.093 = 0.070 P(D is a 6 | A, B, and C are not 6s) = P(D is a 6 and A, B, and C are not 6s) / P(A, B, and C are not 6s) P(D is a 6 | A, B, and C are not 6s) = 0.070 / 0.755 [B]P(D is a 6 | A, B, and C are not 6s) = 0.093[/B] P(A, B, C, or D are a 6) = P(A is a 6) + P(B is a 6 | A is not a 6) + P(C is a 6 | A and B are not 6s) + P(D is a 6 | A, B, and C are not 6s) P(A, B, C, or D are a 6) = 0.087 + 0.089 + 0.092 + 0.093 [B]P(A, B, C, or D are a 6) = 0.361 = 36%[/B] This doesn't account for situations where more than one of A,B,C,or D are a 6, but the probability of that is low and won't affect the outcome much. I've just realised after working all this I could have accounted for it by working Bombjack's way, but I'm too lazy to go through it again. I wouldn't think it would change the result more than a couple of percent though. I might have gone wrong somewhere (it has been a few years since I did my maths course, after all!), but the result is around about what I expected, so perhaps not. :o [/QUOTE]

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