#1
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Combinatorics messing with me...
OK, this is a rather simple calculation that for some reason I keep getting the wrong number.
Game is Razz, and you are dealt 3 to the wheel. What is the probability that you will complete the wheel after all 7 cards are out? Total remaining 4 card combinations = combin(49,4) Total ways to form a combination of the 2 cards I need, unpaired = 4*4 = 16 The other 2 cards, given that a 4 and 5 represent the other 2 = combin(47,2) This should account for situations like 4445, which is a sucess holding A23. Probability = 16*combin(47,2)/combin(49,4) = 8.16% The MCU Poker Chart lists this as 7.15% Where am I messing this up? |
#2
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Re: Combinatorics messing with me...
You are overcounting the situations where you receive more
than one of the two key cards. Of the C(49,4)=211876 combinations, you get these number of combinations for: 2 key cards: 4 x 4 x C(49-2x4,2) = 16C(41,2) = 13120 3 key cards (pair): 2xC(4,2)x4x41 = 2x6x4x41 = 1968 "two pairs" of key cards: C(4,2)xC(4,2) = 6x6 = 36 "trips" with key cards: 2xC(4,3)x4 = 2x4x4 = 32 total: 15156 probability ~ 0.07153240575 |
#3
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Re: Combinatorics messing with me...
[ QUOTE ]
You are overcounting the situations where you receive more than one of the two key cards. Of the C(49,4)=211876 combinations, you get these number of combinations for: 2 key cards: 4 x 4 x C(49-2x4,2) = 16C(41,2) = 13120 3 key cards (pair): 2xC(4,2)x4x41 = 2x6x4x41 = 1968 "two pairs" of key cards: C(4,2)xC(4,2) = 6x6 = 36 "trips" with key cards: 2xC(4,3)x4 = 2x4x4 = 32 total: 15156 probability ~ 0.07153240575 [/ QUOTE ] Thanks man... this had be pissing me off. I understand it now. Thanks for the help. |
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