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  #1  
Old 11-25-2007, 02:51 PM
jr4284 jr4284 is offline
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Location: Folding to turn raises
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Default 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?
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  #2  
Old 11-25-2007, 05:11 PM
bigpooch bigpooch is offline
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Default 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
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  #3  
Old 11-25-2007, 05:26 PM
jr4284 jr4284 is offline
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Default 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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