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?


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.


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.

  • $\begingroup$ 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
  • 3
    $\begingroup$ @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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