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Math Help - Poker Probability (3 of a kind)

  1. #1
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    Question Poker Probability (3 of a kind)

    I'm having trouble trying to figure some of this problem out. I'll provide what I know under the question below.

    What is the probability that a five card poker hand will contain three of a kind? (Note: This should not include a full house. That is three of one kind and two of another)

    |S| = C(52,5) = 2,598,960
    |T| = C(13,1) x C(4,3) x C(48,1) x C(44,1) = 109,824

    I know that i'm having trouble with |T|. I've heard that all I have to do is multiply it by 1/2 but I do not know why. Someone please explain throughly. Thanks!

    Final Answer will be |S|/|T| = 88/4165
    Last edited by dhardin07; November 16th 2009 at 12:37 PM. Reason: typo's
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  2. #2
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    Hello dhardin07

    Welcome to Math Help Forum!
    Quote Originally Posted by dhardin07 View Post
    I'm having trouble trying to figure some of this problem out. I'll provide what I know under the question below.

    What is the probability that a five card poker hand will contain three of a kind? (Note: This should not include a full house. That is three of one kind and two of another)

    |S| = C(52,5) = 2,598,960
    |T| = C(13,1) x C(4,3) x C(48,1) x C(44,1) = 109,824

    I know that i'm having trouble with |T|. I've heard that all I have to do is multiply it by 1/2 but I do not know why. Someone please explain throughly. Thanks!

    Final Answer will be |S|/|T| = 88/4165
    What you have heard is correct: you do need to divide by 2!. The reason is that in your working so far, you have selected the final two cards in order. Therefore every possible pair will occur 2! times.

    If you divide your expression for |T| by 2, then:
    \frac{|T|}{2}= 54912

    \Rightarrow \frac{|T|}{|S|}=\frac{88}{4165}
    An alternative method for selecting the final two cards is initially to ignore the restriction: select 2 from 48; and then subtract from this the number of ways of selecting two of a kind (so that's selecting 1 from 12 and then 2 from 4). This gives:
    \binom{48}{2}-\left(\binom{12}{1}\times\binom{4}{2}\right)=1056=  \frac{\binom{48}{1}\times\binom{44}{1}}{2!}
    Grandad
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