Probability of a five-card poker hand containing 2 pairs

Question

I am trying to solve this question in my own way. I tried counting like this- First card can be chosen in 52 ways, second in 3 ways because it has to be of same kind as first, third in 48 ways, fourth in 3 ways and fifth in 44 ways. I got the answer as : (52*3*48*3*44)/5! as the favorable events. I was trying to solve it like the way it is solved here Can anyone help me get it correct because I am not getting the correct answer this way

Answer 1

I think you're dividing by 5! because that's how many times you're overcounting each combination. However this isn't the right amount. Each hand $\{A_1, A_2, B_1, B_2, C\}$ is counted in $2 \times 2 \times 2 = 8$ different ways by your method since the first pair can be in any order, the second pair can be in any order and the order of the pairs can be swapped. This gives $$ \frac{52 \times 3 \times 48 \times 3 \times 44}{2 \times 2 \times 2} = 123552$$ hands which are 2 pairs. This agrees with the frequency on wikipedia: https://en.wikipedia.org/wiki/Poker_probability#Frequency_of_5-card_poker_hands

Answer 2

If you start with the pairs, you're going to run into trouble since at some point you have to divide by multiple $2$s to account for swapping cards within pairs, as well as swapping the pairs. Rather, you should start with the kicker, and also fix one property at a time and not one card at a time:

• $13$ ways to select the rank of the kicker, $4$ choices for its suit
• $\frac{12×11}2$ ways to select the ranks of the two pairs
• $\frac{4×3}2×\frac{4×3}2$ ways to select the suits of each pair