The question asks what the probability is of being dealt 2 pairs in a 5 card poker hand (from a shuffled deck of 52 cards).

So you need 1 pair and another numerically different pair (otherwise it will be four of a kind) and a final card which matches neither pair (otherwise it will be a full house).

The answer is approx. 0.0475 which can be found in plenty of places as the answer to this question with a web search and is explained here: https://math.stackexchange.com/a/1770961/484087

I got the answer wrong by exactly a factor of 2, i.e. 0.095. Clearly I have over-counted but I can not see what is wrong with the logic I used. Can someone explain where I went wrong please? Here's what I did:

Num. of combinations of 2 things and 2 things and 1 thing from 5 things: $\displaystyle \frac{5!}{2!·2!·1!}=\frac{120}{4}=30$

1st card can be anything: $\frac{52}{52}$

2nd card must match the 1st: $\frac{3}{51}$

3rd card must not match 1st or 2nd: $\frac{48}{50}$

4th card must match the 3rd: $\frac{3}{49}$

5th card must not match 1st, 2nd, 3rd, or 4th: $\frac{44}{48}$

$\displaystyle 30·\frac{52}{52}·\frac{3}{51}·\frac{48}{50}·\frac{ 3}{49}·\frac{44}{48}=\frac{396}{4165}≈0.095078$