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

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?

• The two ranks are picked using the same $13\choose2$ coefficient. – Empy2 Mar 16 '15 at 11:07
• @bodokaiser Note that $$\binom{13}{2} = \frac{1}{2}\binom{13}{1}\binom{12}{1}$$ – N. F. Taussig Mar 16 '15 at 12:04