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Probability of poker hands

  1. Dec 31, 2005 #1
    I'm trying to figure out the probabilities for various 5 card poker hands, and I'm not having a very good time of it.
    Consider "two pairs".

    For the 1st pair, there are [itex]\binom{13}{1}[/itex] choices for denomination. Out of that, there are [itex]\binom{4}{2}[/itex] choices for suit.

    For the second pair, there are [itex]\binom{12}{1}[/itex] choices for denomination (not 13, otherwise it'd be "4 of a kind"). And again, there are [itex]\binom{4}{2}[/itex] choices for suit.

    For the last card, there are [itex]\binom{11}{1}[/itex] choices for denomination and [itex]\binom{4}{1}[/itex] choices for suit.

    There's a grand total of [itex]\binom{52}{5}[/itex] things that can happen for 5 draws from 52 cards. So the probability for 2 pair should be:


    which works out to be .095. A book I have says that two pair should be .048.

    As another example, consider 3 of a kind.

    There are [itex]\binom{13}{1}[/itex] choices for denomination of the "3 of a kind", and [itex]\binom{4}{3}[/itex] choices for suit.

    Now there are 2 cards left, and 12 denominations. So there are [itex]\binom{12}{2}[/itex] ways of choosing a denomination for the two single cards, and each card can have any of [itex]\binom{4}{1}[/itex] suits. So the probability for 3 of a kind should be:


    which works out to be .0079. The book says .021.

    At this point, I'm ready to admit there's a huge conceptual flaw in my head. There's something fundamental about counting that I'm just not "getting". Can someone please help clear my conceptual misunderstanding about counting?

    Many thanks!
    Last edited by a moderator: Dec 31, 2005
  2. jcsd
  3. Dec 31, 2005 #2


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    The way you went about finding it out for 2 pair, you made the order of the pairs matter, i.e. you counted two 5's and two 9's as a separate hand from two 9's and two 5's. A hand of two pairs consists of 3 denominations, so do 13C3 first. Then, of those three denominations, two of them are paired, so do 3C2. Finally, choose the suits.
  4. Dec 31, 2005 #3


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    On the second one you can break it down to:
    You have to divide by two because there's two ways to order the two single cards (similar to the first one in this aspect).
    Last edited: Dec 31, 2005
  5. Dec 31, 2005 #4


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    For your expression, I get something closer to 0.00528.., which is roughly a fourth of the books answer. This missing factor of 4 comes in your last two cards, there are [itex]\binom{12}{2}[/itex] choices for the denominations as you say, but [itex]\binom{4}{1}[/itex] choices for each of these two cards suits, so you need another [itex]\binom{4}{1}[/itex] factor.
  6. Dec 31, 2005 #5


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    Often times, counting just requires a good way to describe the objects being counted.

    Let's consider how you tried to do two pair, but rephrase it from this perspective:

    You have described a two pair as:
    (1) A choice for the denomination of the first pair
    (2) A choice of the two suits in the first pair
    (3) A choice of the denomination for the second pair
    (4) A choice of the two suits in the second pair
    (5) A choice of the denomination for the last card.
    (6) A choice of the suit for the last card.

    Now you ask yourself:
    (A) Can every two-pair be described this way?
    (B) Is every such description a different hand?

    The answer to (A) is, of course, yes. The answer to (B) is no, so this is no good! However, it can be salvaged by asking yourself how many times is each hand counted? In this case, every two-pair is counted exactly twice, so you simply have to divide the result by 2.

    Of course, you could come up with a slightly different description that counts each two-pair exactly once.
  7. Dec 31, 2005 #6


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  8. Jan 1, 2006 #7
    Is math a tautology?

    Poincare':The syllogism can teach us nothing essentially new,
    and, if everything is to spring from the principle of identity, everything
    should be capable of being reduced to it. Shall we then admit that the
    enunciations of all those theorems which fill so many volumes are nothing
    but devious ways of saying A is A?"

    Not the case in poker hands because we model this from the real world.
    Last edited: Jan 1, 2006
  9. May 16, 2011 #8
    This thread seems to be so old. But the topic is useful to those poker hopefuls who want to win. This may help a bit. After all, theories exist first before application.
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