Counting pair in a poker hand - counting confusion

Question

I want to be good enough to do math competitions, and one of the main topics in them is always combinatorics. So I'm going through a combinatorics book, "A Walk through Combinatorics."

The confusion I have is with the following problem:

Find the prob that a poker hand has a pair( and nothing better than a pair).

My attempt: We know that a poker hand has five cards from a deck of 52 cards. Each card is equally likely so the total combinations is: $${52} \choose 5$$

The number of hands that contain only a pair and nothing better than a pair are: $${{13} \choose {4}} * {{4} \choose {2}}* {{4} \choose {1}}^3$$

There are 13 cards of each card. We choose 4 the 13 values. Then from one of the groups we want a pair, and from the remaining group we want a single.

So, the probability I got was:

$$\frac {{{13} \choose {4}} * {{4} \choose {2}}* {{4} \choose {1}}^3}{{52} \choose 5}$$.

However, according to the answers:

The correct probability should be : $\frac {{{13} \choose {1}} *{{12} \choose {3}} * {{4} \choose {2}}* {{4} \choose {1}}^3} {{52} \choose 5}$

Where is my logic flawed? Where am I going wrong? Am I double counting somewhere? I think my answer produces a smaller probability so am I undercounting?

Answer 1

The number of hands containing only a pair can be so calculated

1) Select the pair: AA, 22,...,KK and you have $13$ choices

2) select the suits of the pair: you have $\binom{4}{2}=6$ choices

3) select the three single cards: $\binom{12}{3}$ choices

4) for each single triplet you have to choice the suits (the suits can be repeated) : $4^3$ choices

....multiply all and get the number of choices containing ONLY one pair

$$13\times\binom{4}{2}\times\binom{12}{3}\times 4^3=1,098,240$$

Answer 2

The difference is your ${13 \choose 4}$ versus their ${13 \choose 1} {12 \choose 3}$ (your ${4 \choose 2}$ and ${4 \choose 1}$ factors are correct). You're choosing the four of 13 ranks in the hand, but not specifying which rank has the pair. They choose the rank for the pair with ${13 \choose 1}$ and then the three other ranks with ${12 \choose 3}$.