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Thread: poker probability

  1. #1
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    poker probability

    In order to find the probability of dealing a full house in a 5 card poker hand, you use the folowing:
    (13-choose-1)x(4-choose-3)x(12-choose-1)x(4-choose-2) / (52-choose-5)

    But in order to find the probability of dealing 2 pairs, you use the this:
    (13-choose-2)x(4-choose-2)x(4-choose-2)x(11-choose-1)(4-choose-1) / (52-choose-5)

    What I don't understand is why doesn't the 2 pairs probability method work the same way as the full house method (choosing 1 from 13 then 1 from 12 rather than choosing 2 from the 13 at the start) and work as follows:
    (13-choose-1)x(4-choose-2)x(12-choose-1)x(4-choose-2)x(11-choose-1)(4-choose-1) / (52-choose-5)
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  2. #2
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    Re: poker probability

    Quote Originally Posted by Phoebert View Post
    What I don't understand is why doesn't the 2 pairs probability method work the same way as the full house method (choosing 1 from 13 then 1 from 12 rather than choosing 2 from the 13 at the start) and work as follows:
    (13-choose-1)x(4-choose-2)x(12-choose-1)x(4-choose-2)x(11-choose-1)(4-choose-1) / (52-choose-5)
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  3. #3
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    Re: poker probability

    Hello, Phoebert!

    I assume you understand the purpose of each of the factors.

    There is a subtle difference in the two methods you mentioned.


    In order to find the probability of dealing a Full House in a 5-card poker hand,

    you use the following: .$\displaystyle \dfrac{{13\choose1}{4\choose3}{12\choose1}{4 \choose2}} {{52\choose5}} $

    Choose one of the 13 values for the Triple: $\displaystyle {13\choose1}$ ways.
    Choose 3 of the 4 cards of that value: $\displaystyle {4\choose3}$ ways.
    Choose one of the other 12 values for the Pair: $\displaystyle {12\choose1}$ ways.
    Choose 2 of the 4 cards of that value: $\displaystyle {4\choose2}$ ways.

    There are: .$\displaystyle {13\choose1}{4\choose3}{12\choose1}{4\choose2}$ ways to get a Full House.
    . . and divide by the number of 5-card hands: $\displaystyle {52\choose5}$
    . . to get the probability.




    But in order to find the probability of dealing Two Pairs,

    you use the this: .$\displaystyle \dfrac{{13\choose2}{4\choose2}{4\choose2}{11 \choose1}{4\choose1}}{{52\choose5}}$

    Choose 2 of the 13 values for the Two Pairs: $\displaystyle {13\choose2}$ ways.
    Choose 2 of the 4 cards of one value: $\displaystyle {4\choose2}$ ways.
    Choose 2 of the 4 cards of the other value: $\displaystyle {4\choose2}$ ways.
    Choose 1 of the other 11 values: $\displaystyle {11\choose1}$ way.
    Choose 1 of the 4 cards of that value: $\displaystyle {4\choose1}$ way.

    There are: .$\displaystyle {13\choose2}{4\choose2}{4\choose2}{11\choose1} {4\choose1}$ ways to get two Two Pairs.
    . . divide by the number of 5-cards hands: $\displaystyle {52\choose5}$
    . . to get the probability.




    What I don't understand is why doesn't the 2 pairs probability method work the same way as the Full House method
    (choosing 1 from 13 then 1 from 12, rather than choosing 2 from the 13 at the start) and work as follows:

    . . $\displaystyle \frac{{13\choose1}{4\choose2}{12\choose1}{4\choose 2}{11\choose1}{4\choose1}}{{52\choose5}}$

    Your method produces a number twice as large as necessary.
    Here's the reason why.

    The original method chooses 2 values from the 13 available values.
    There are: $\displaystyle {13\choose2} \:=\:78$ possible choices for values.

    We can list them if we like:

    . . $\displaystyle \begin{array}{c}A2\;A3\;A4\;A5\;\cdots\;AQ\;AK \\ 23\;24\;25\;\cdots\;2Q\;2K \\ 34\;35\;\cdots\;3Q\;3K \\ \vdots \\ JQ\;JK \\ QK \end{array}$


    Your method has: $\displaystyle {13\choose1}{12\choose1} \:=\:156$ possible choices for values.

    Your list looks like this:

    . . $\displaystyle \begin{array}{c}A2\,A3\,A4\,A5\,\cdots\,AQ\,AK \\ 2A\;23\;24\;25\;\cdots\;2Q\;2K \\ 3A\;32\;34\;35\;\cdots\;3Q\;3K \\ \vdots \\ K\!A\,K\!2\,K\!3\,K\!4\,\cdots\,K\!J\,K\!Q \end{array}$


    Your list considers $\displaystyle A2$ to be different from $\displaystyle 2A$.

    That is "two Aces and two 2's" is not the same as "two 2's and two Aces."

    And we know better . . .
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  4. #4
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    Re: poker probability

    Thanks so much. I understand now
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