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Let's consider two pairs in a 52 cards deck of poker where every person gets five cards.

My idea to approach this problem is to take following steps:

  1. First pair
    • There are ${4 \choose 2}$ combinations getting two cards of the same rank
    • There are ${13 \choose 1}$ combinations of having a specific rank out of a suit
  2. Second pair
    • There are still ${4 \choose 2}$ combinations to get two cards of the same rank
    • However as one card per suit is gone we only have ${12 \choose 1}$ for each combination out of a suit
  3. Any card
    • There are ${4 \choose 1}$ combinations getting one card out of the same rank
    • There are ${11 \choose 1}$ combinations to getting one card out of a suit

This would yield in:

$$P(TP) = \frac{{4 \choose 2}{13 \choose 1} \cdot {4 \choose 2}{12 \choose 1} \cdot {4 \choose 1}{11 \choose 1}}{{52 \choose 5}}$$

According to wikipedia the correct probability would be calculated as:

$$P(TP) = \frac{{13 \choose 2}{4 \choose 2}{4 \choose 2} \cdot {4 \choose 1}{11 \choose 1}}{{52 \choose 5}}$$

What is the mistake in my model and how could I think of the one provided in wikipedia?

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Realize that in your way to select pairs first you are fixing a rank say 1 and then in second part another rank say 2. also the possibility that in first rank was 2 and second rank was 1 is counting different in your way counting but actually its same So for choice of pairs you have to divide it by 2 and then your answer will match automatically.

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Your number is twice wikipedia's number. That is because you counted two Jacks, two kings, as well as two kings,two jacks. So you counted each hand twice.

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  • $\begingroup$ Is it possible to use four binomial coefficient to show this? I am confused that they only use three though these are separate steps. $\endgroup$ – bodokaiser Mar 16 '15 at 10:54
  • $\begingroup$ The two ranks are picked using the same $13\choose2$ coefficient. $\endgroup$ – Empy2 Mar 16 '15 at 11:07
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    $\begingroup$ @bodokaiser Note that $$\binom{13}{2} = \frac{1}{2}\binom{13}{1}\binom{12}{1}$$ $\endgroup$ – N. F. Taussig Mar 16 '15 at 12:04

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