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# probability : poker problem

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probability : poker problem [#permalink]  20 Mar 2009, 10:45
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]  20 Mar 2009, 13:18
Expert's post
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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Re: probability : poker problem [#permalink]  20 Mar 2009, 13:44
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, 13:44
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