What is wrong with ${13 \choose 1}{4 \choose 2} \cdot {12 \choose 1}{4 \choose 2}$ as combinations for two pair in poker?

Question

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?

Answer 1

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.

Answer 2

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.