Number of ways to pick at least three of a kind in 5-card poker - what's wrong with C(49,2) for the last two cards?

Question

In finding the number of ways to get a 5-card poker hand that contains at least three of a kind, what's wrong with the following $$n=\binom{13}{1}\binom{4}{3} \binom{49}{2}$$ So, we have 13 numbers to choose from, and for each of those there are 4 cards for which we choose 3. Finally, we don't care about the other two cards because we care only about the probability to get at least three of a kind, so we pick 2 out of the remaining 49 cards. Using this and a total of $$N = \binom{52}{2}$$ ways to pick 5 cards, we get probability $$P=\frac{n}{N}=0.0235294\approx2.353\text{ %}$$ However, poker probabilities Wikipedia page says that the probability to get three of a kind or better is 2.87%, so what is going on here ?

Answer 1

Wikipedia gives the probability "$3$ of a kind or better" which includes straights, flushes, full house, and $4$ of a kind.

Your calculation allows for $3$ or $4$ of a kind, or full house, but does not capture straights or flushes.