Calculating combinations of three of a kind in Texas Hold 'Em The question sounds quite simple - in a game of Texas Hold'Em (or simply out of ${52 \choose 7}$), how many combinations of cards are there, that would result in three of a kind being the highest hand?
The Wikipedia article about poker probabilities says, that the frequency of three of a kind in a 7-card poker game is 6,461,620.
I have tried to arrive at that solution by first calculating all possible three card combinations: ${4 \choose 3} 13 = 52$, and then multiplying that number by the amount of possible combinations of the other 4 cards, however quickly realised that 6,461,620 isn't divisible by 52 which throws my plan out of the window.
How do I arrive at the correct amount of possible combinations?
 A: First we count the number of hands with three of a kind, no four of a kind, and no other pair.  There are $13$ ways to pick the three of a kind, $4$ ways to pick the three cards of that kind, $48$ acceptable cards for the first odd card, then $44,40,36$ for the following ones, but we have to divide by $4!=24$ for the orders of picking the four odd cards.  That gives
$$13\cdot 4 \cdot 48 \cdot 44 \cdot 40 \cdot 36/24=6,589,440$$
hands with three of a kind and no full house or four of a kind.  Note this is not many more than the Wikipedia result.  We now need to deduct the number of hands that have a straight or flush.  
The flushes are reasonably easy.  Again there are $13\cdot 4$ ways to get the three of a kind.  Then there are $3$ ways to choose the suit of the flush and ${12 \choose 4}=495$ ways to select the other cards for a total of $77,220$ hands with three of a kind and a flush (which includes the straight flushes).  
For straights we will start with the straight.  The Wikipedia page shows (assuming that ace low straights count) there are $10,240$ ways to choose the five cards of a straight including straight flushes.  Then there are $5$ ways to pick which card will have the three of a kind and $3$ ways to pick the two other cards, but we have to divide by $3$ for which of the three of a kind was part of the original straight so $10,240\cdot 5 \cdot 3/3=51,200$ hands to deduct.  
We have deducted the straight flush hands twice, once for the straight and once for the flush.  We need to add them in once.  There are $40$ straight flushes, $5$ ways to pick the rank that has three of a kind, and $3$ ways to pick the missing suit for $600$
This winds up with $6,589,440-77,220-51,200+600=6,461,620$ hands with three of a kind and nothing higher, in agreement with the Wikipedia page.
