Probability of a 2 pair in poker What is wrong with my method of find the probability of a 2 pair?
For the first card we have 52 choices, one this is picked( call it A) we have 3 choices(different suite than A but this same value) for the second card, ( lets call this card B). Now we have AB, to we have to throw away 2 more cards so we don't have 3 of a kind or a four of a kind. Therefore one our third card we have 48 choices and on our fourth card we have 3 choices again similar to above.  On our last card we only have 44 choices left. So the probablity of getting a 2 pair should be (52*3*48*3*44)/(52 choose 5).
But this isn't correct. 

Note: I edited this to "2 pair"; the original version was wrong.
 A: It sounds as though you are trying to calculate a two-pair given your posted attempt.  A two-pair in poker is defined to be made up of two doubles and one singleton (e.g. $J\heartsuit, J\spadesuit, 5\spadesuit, 2\diamondsuit, 5\clubsuit$) (note that order is unimportant, we may always think to rearrange cards from highest to lowest and from $\spadesuit,\heartsuit,\diamondsuit,\clubsuit$)
To solve for a two pair, set it up as a multiplication principle question:


*

*Pick the two numbers which are used in the pairs: $\binom{13}{2}$ (note: you must pick these simultaneously to avoid overcounting)

*Pick the two suits used in the larger of the two pairs: $\binom{4}{2}$ (note: you must pick these simultaneously to avoid overcounting, but since we have already picked the numbers, we can now tell which is larger)

*Pick the two suits used for the smaller of the two pairs: $\binom{4}{2}$

*Pick the number for the singleton: $\binom{11}{1}$

*Pick the suit for the singleton: $\binom{4}{1}$


Thus the number of two-pair hands is: $\binom{13}{2}\binom{4}{2}\binom{4}{2}\binom{11}{1}\binom{4}{1}$.
Divide this by $\binom{52}{5}$ to get the probability.
A: You have misunderstood what a straight is. You are insisting that the second card be the same value as the first, but a straight has no cards of the same value.  Your process seems to be describing a calculation of the probability of two pair, because you want the first two cards to match in rank, the second two to match and the fifth not to match.  This is still wrong, because the matching cards could be drawn at any two of the five positions.  You have counted drawing AA227, but not A27A2.  If you multiply the numerator by the number of permutations of two pair (counting the paired cards as identical and counting the swap of pairs as not a change) you are there.
