My Math Forum Combinations (nCr)

 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
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Re: Combinations (nCr)

Hello, nates1984!

Quote:
 $\text{The problem asks for the probability that you'll get Two Pairs in a poker hand.}$ $\text{Why is this: }\:\left(_{13}C_1\right)\left(_4C_2\right)\left(_{ 12}C_1\right)\left(_4C_2\right) \:\text{ not equal to this: }\:\left(_{13}C_2\right)\left(_4C_2\right)\left(_4 C_2\right)$ $\text{I must be misunderstanding this concept somewhere.}$

$\text{Consider the second expression: }\:\left(_{13}C_2\right)\left(_4C_2\right)\left(_4 C_2\right)$

$\text{There are 13 values for the cards.}$
[color=beige]. . [/color]$\text{We choose 2 of them: }\:_{13}C_2\text{ choices.}$

$\text{They could be, for example, }7's\text{ and }K's.$
[color=beige]. . [/color]$\text{The order does not matter.}$

$\text{There are: }\:_4C_2\text{ ways to choose the two 7's.}$
$\text{There are: }\:_4C_2\text{ ways to chooise the two K's.}$

$\text{Hence, there are: }\:\left(_{13}C_2\right)\left(_4C_2\right)\left(_4 C_2\right) \text{ ways to get Two Pairs.}$

$\text{Now consider the first expression: }\:\left(_{13}C_1\right)\left(_4C_2\right)\left(_{ 12}C_1\right)\left(_4C_2\right)$

$\text{Case 1}$

$\text{Choose the value of the first pair: }\:_{13}C_1\text{ choices.}$
[color=beige]. . [/color]$\text{Suppose it is a 7.}$
$\text{Then there are: }\:_4C_2\text{ ways to get two 7's.}$

$\text{Choose the value of the second pair: }\:_{12}C_1\text{ choices.}$
[color=beige]. . [/color]$\text{Suppose it is a K.}$
$\text{Then there are: }\:_4C_2\text{ ways to get two K's.}$

$\text{Hence, there are: }\:\left(_{13}C_1}\right)(_4C_2)\left(_{12}C_1\rig ht)(_4C_2) \text{ ways to get two 7's and two K's.}\;\;[1]$

$\text{Case 2}$

$\text{Choose the value of the first pair: }\:_{13}C_1\text{ choices.}$
[color=beige]. . [/color]$\text{Suppose it is a K.}$
$\text{Then there are: }\:_4C_2\text{ ways to get two K's.}$

$\text{Choose the value of the second pair: }\:_{12}C_1\text{ choices.}$
[color=beige]. . [/color]$\text{Suppose it is a 7.}$
$\text{Then there are: }\:_4C_2\text{ ways to get two 7's.}$

$\text{Hence, there are: }\:\left(_{13}C_1\right)(_4C_2)\left(_{12}C_1\righ t)(_4C_2) \text{ ways to get two K's and two 7's.}\;\;[2]$

$\text{But [1] and [2] represent the }same\;hand.\;\text{ It has been counted twice!}$
[color=beige]. . [/color]$\text{The first expression is twice as large as it should be.}$

$\text{The second expression is the correct one: }\:(_{13}C_2)(_4C_2)(_4C_2)$

$\text{But we are not done . . .}$

$\text{Suppose the Two Pairs are 7's and K's.}$
$\text{The deck contains: four 7's, four K's, and forty-four Others.}$

$\text{We have the two 7's and two K's. \;The }fifth\text{ card must be one of the Others.}$
[color=beige]. . [/color]$\text{And there are: }\:_{44}C_1\text{ choices.}$

$\text{Therefore, there are: }\:(_{13}C_2)(_4C_2)(_4C_2)(_{44}C_1) \text{ ways to get Two Pairs.}$

 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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