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Math Help - Two-Pair probability

  1. #1
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    Two-Pair probability

    The question is: What is the probability of drawing a two-pair in a 5 card set from a normal 52-card deck? That is, QQJJ2 or 337AA, for example.

    I have my answer, but for some reason it is exactly double the correct answer. I haven't been able to figure out why. Here's my solution:

    First you draw an arbitrary card. To pair it, the probability is 3/51. Next, you want to draw a different card; there are 48 such cards out of 50, so the probability is 48/50. You seek its pair, which has probability 3/49. Your final card must be different from both the originals (otherwise we'd have a full house), so it has probability 44/48.

    That yields approx. 0.00316. There are precisely 5! / 2! * 2! * 1! orderings of two pair hands = 30. Thus, 30 * 0.00316 = 0.09507. But the correct answer is half that. Any pointers in the right direction would be awesome!
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  2. #2
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    Re: Two-Pair probability

    Quote Originally Posted by arcketer View Post
    The question is: What is the probability of drawing a two-pair in a 5 card set from a normal 52-card deck? That is, QQJJ2 or 337AA, for example.

    I have my answer, but for some reason it is exactly double the correct answer. I haven't been able to figure out why. Here's my solution:

    First you draw an arbitrary card. To pair it, the probability is 3/51. Next, you want to draw a different card; there are 48 such cards out of 50, so the probability is 48/50. You seek its pair, which has probability 3/49. Your final card must be different from both the originals (otherwise we'd have a full house), so it has probability 44/48.

    That yields approx. 0.00316. There are precisely 5! / 2! * 2! * 1! orderings of two pair hands = 30. Thus, 30 * 0.00316 = 0.09507. But the correct answer is half that. Any pointers in the right direction would be awesome!
    \dfrac{\binom{13}{2}\binom{4}{2}\binom{4}{2}(44)}{  \binom{52}{5}} does that give the correct answer?

    If so WHY? If not WHY NOT?
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  3. #3
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    Re: Two-Pair probability

    Quote Originally Posted by Plato View Post
    \binom{13}{2}\binom{4}{2}\binom{4}{2}(44) does that give the correct answer?

    If so WHY? If not WHY NOT?

    LOL!
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  4. #4
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    Re: Two-Pair probability

    Hello, arcketer!

    What is the probability of drawing Two Pair in a poker hand from a 52-card deck?

    I have my answer, but for some reason it is exactly double the correct answer.
    I haven't been able to figure out why. Here's my solution:

    First you draw an arbitrary card.
    To pair it, the probability is 3/51.

    Next, you want to draw a different card.
    There are 48 such cards out of 50, so the probability is 48/50.
    You seek its pair, which has probability 3/49.

    Your final card must be different from both the originals
    (otherwise we'd have a full house), so it has probability 44/48.

    That yields approx. 0.00316.
    There are precisely 5!/(2!*2!*1!) orderings of two pair hands = 30.
    Thus, 30 * 0.00316 = 0.09507.
    But the correct answer is half that.

    Your reasoning is correct, but there is a subtle omission.


    For the first card, there are 52 choices.. . P(\text{any}) = \tfrac{52}{52}
    Suppose it is a Jack.

    The second must be another Jack. . P(\text{another Jack}) = \tfrac{3}{51}

    The third card must not be a Jack: . P(\text{another value}) = \tfrac{48}{50}
    Suppose it is a 7.

    The fourth card must another 7. . P(\text{another 7}) = \tfrac{3}{49}

    The fifth card must not be a J or 7. . P(\text{other}) = \tfrac{44}{48}

    All this is correct.


    You had ordered the five cards, so to un-order them,
    . . you multiplied by: {5\choose2,2,1} = 30 . . . . correct

    But this last factor is the number of ordered partitions.

    It considers \{JJ|77|3\} to be different from \{77|JJ|3\}


    And that is where the "times two" comes from.

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  5. #5
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    Re: Two-Pair probability

    Edit: NVM!
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  6. #6
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    Re: Two-Pair probability

    Quote Originally Posted by arcketer View Post
    Edit: NVM!
    Please explain yourself! Why "never mind"?
    That is unfair to anyone reading this thread.
    You may not need to post here again if you choose not to answer my question.
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  7. #7
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    Re: Two-Pair probability

    I'm not finished with the problem. I had typed out a response to you, but after submission read a more detailed response from Soroban, so I quickly edited away what I wrote. I haven't had a chance to look over his response yet as I just got home, but I will!
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  8. #8
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    Re: Two-Pair probability

    Ahh I see my mistake now, Soroban. So the general formula that I was using gives ordered partitions. Is there any such a formula for unordered partitions? It seems simple but I don't want to make a similar mistake on a similar problem later. Thank you for the help!
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