Probability of a three-pairs in 6-card poker Suppose out of a deck of 52 regular playing cards, you are dealt a hand of 6 cards. What is the probability of you getting a three-pair hand? That is, three faces that each appear twice. I've worked out $${\binom{13}{1}\binom{4}{2}\cdot\binom{12}{1}\binom{4}{2}\cdot\binom{11}{1}\binom{4}{2}\over\binom{52}{6}}\approx0.01821$$ as the answer, since you're picking three faces and any two suits. Is this correct, or did I miss something?

It's a good thought, but you're overcounting. For example, your numbers would suggest that the following are different: (1) a pair of aces, then a pair of twos, then a pair of sixes and (2) a pair of sixes, then a pair of aces, then a pair of twos. Due to this overcounting, your answer is precisely $6$ times as large as it should be, since you're counting each ordered arrangement of the three pairs differently, and there are $3!=6$ ways to arrange $3$ distinct objects in order.

Rather, to determine the faces, we use instead $\binom{13}{3}$ to prevent such overcounting (since we're choosing $3$ of the $13$ possible face values). Then we use $\binom{4}{2}$ for each face (since we're choosing $2$ of the available $4$ suits for each face), so we have: $$\frac{\binom{13}3\cdot\binom{4}{2}^3}{\binom{52}{6}}$$