I was looking at this website http://www.cwu.edu/~glasbys/POKER.HTM and I read the explanation for how to calculate the probability of getting a full house. To me, the logic basically looked like you figure out the number of possible ranks and multiply by the number of ways to choose the cards from that given rank.
In other words, for a full house $P=$
$$\frac{{13\choose1}{4\choose3}{12\choose1}{4\choose2}}{52\choose5}$$
Following this logic, I tried to calculate the probability of getting two pair. My (incorrect) logic was that there are 13 possible ranks for the first pair and $4\choose2$ ways to choose two cards from that rank, 12 possible ranks for the second pair and $4\choose2$ ways to choose two cards from that rank, and 11 possible ranks for last card and $4\choose1$ ways to choose a card from that rank.
So I tried $P=$
$$\frac{{13\choose1}{4\choose2}{12\choose1}{4\choose2}{11\choose1}{4\choose1}}{52\choose5}$$
Obviously my solution was incorrect. I read explanation and the correct answer is $P=$
$$\frac{{13\choose2}{4\choose2}{4\choose2}{11\choose1}{4\choose1}}{52\choose5}$$
I'm still a bit fuzzy on where I went wrong though. Can anyone help me understand this problem a little better? Thank you very much for your help.
Best Answer
You have to choose the two card values you want as your pairs simultaneously. Remember--multiplying the numbers ${13\choose1}{4\choose2}{12\choose1}{4\choose2}{11\choose1}{4\choose1}$ assumes an $order$, i.e. you are counting, say, QQKK2 as different from KKQQ2.
This is why you have to do ${13\choose2}{4\choose2}{4\choose2}{11\choose1}{4\choose1}$. It makes the counting not sensitive to which pair you choose first.