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June 21st, 2011, 07:06 PM  #1 
Joined: Jun 2011 Posts: 2 Thanks: 0  Combinations (nCr)
Context: The problem asks for the probability that you'll get two pairs in a poker hand. I'm going to omit (11C1 * 4C1) as it isn't relevant to my question. Why is this: 13C1 * 4C2 * 12C1 * 4C2 Not equal to this: 13C2 * 4C2 * 4C2 I must be misunderstanding this concept somewhere. It would appear to me that if you're choosing the same amount of cards in 4CX for multiple groups then you MUST select the groups of four together in the same 13CX combination. This is obvious once you browse the Wikipedia page on poker probability. A full house is (13C1 * 4C3 * 12C1 * 4C2), but (13C2 * 4C3 * 4C2) yields a different answer and is incorrect. It's easy enough to do the algebra. Just cancel out factorials and look at what is left in both cases to see how they are (obviously) different from each other. However, conceptually my brain keeps telling me these two things should be equal. So, why aren't they? 
June 22nd, 2011, 07:19 AM  #2  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 2,776 Thanks: 133  Re: Combinations (nCr) Hello, nates1984! Quote:
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June 22nd, 2011, 02:27 PM  #3 
Joined: Jun 2011 Posts: 2 Thanks: 0  Re: Combinations (nCr)
Thanks for the reply. That made everything very clear. I'm also glad to see this forum has latex tags, I'll make sure to use tex if I post any more questions or offer any answers.


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