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# probability : poker problem

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20 Mar 2009, 11:45
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Find the probability of a full house (three of a kind and two of another kind), say three kings and two aces.
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Re: probability : poker problem [#permalink]

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20 Mar 2009, 14:18
We can choose the two types of cards we'll have in 13*12 ways. We then need to choose three cards of the first type from the four cards available, and two cards of the second type from the four cards available. The probability is thus:

13*12*4C3*4C2/52C5
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Senior Manager
Joined: 06 Jul 2007
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Kudos [?]: 51 [0], given: 0

Re: probability : poker problem [#permalink]

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20 Mar 2009, 14:44
IanStewart wrote:
We can choose the two types of cards we'll have in 13*12 ways. We then need to choose three cards of the first type from the four cards available, and two cards of the second type from the four cards available. The probability is thus:

13*12*4C3*4C2/52C5

thanks Ian, makes very much sense. I appreciate it.
Re: probability : poker problem   [#permalink] 20 Mar 2009, 14:44
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