The fourth and fifth cards cannot be the same as each other nor the first three cards. There are 12 values left, and we need to pick 2: B(12,2) = 12! /2! (12-2)! = 12! /2! 10! = (12×11)/2 = 132/2 = 66 combinations of value. But there are 4 suits for each of the fourth and fifth cards. So there are 66×4×4 = 1,056 ways to choose the fourth and fifth cards.
That makes for (13×4)x(66×4×4) = 54,912 ways to chose a five-card hand that produces three of a kind.
Another way to do this, which is easier to understand, is as follows:
The fourth card in the hand may not be the same as the first three. So we have only (13-1)x4 = 48 cards left. Our total so far is 13×4 x 48.
The fifth card should not be the same as first three nor the fourth, or the latter case will be a full house. (This had me stumped for over an hour.) So there are only (13-2)x4 = 44 cards to choose from for the fifth card.
That suggest that there are 48×44 = 2,112 ways to pick the fourth and fifth cards, but that would be true if order mattered. Since it doesn’t, we have to divide by 2! = 2, producing 2,112/2 = 1,056, as before.
Thus, the total number of 5-card trip hands is 13×4 x 1,056 = 54,912 =~ 1/50 odds.
Let’s calculate the number of one-pair hands before two-pair. It’s done similarly to trips (three of a kind). The first card can be any of 13 values. Each value has 4 representative cards, but we just want unique combinations of 2 cards. Pull out the old binomial formula, and …