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Thread: 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; Nov 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 $\displaystyle 2!$. The reason is that in your working so far, you have selected the final two cards in order. Therefore every possible pair will occur $\displaystyle 2!$ times.

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

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