Probability of getting pair on a poker hand I am doing probability course again, and I have this problem:
" If it is assumed that all $\binom {52}{5}$ poker hands are equally likely,what is the probability of being dealt one pair?". Now, my logic gives me expression: $$\frac {\binom {50}{3} \cdot 13 \cdot \binom {4}{2}} {\binom {52}{5}}$$
My logic is following: After having my pair, I can choose any 3 cards from 50 which are left. There is $\binom {4}{2}$ to choose the pair from 4 cards, and 13 types of cards to choose pair of. Now, my answer gives me $\approx 58.8$% probability of getting a pair, but my textbook suggests that the answer is $\approx 42.26$%. I don't really see any other way to look at this. Where am I wrong?
 A: Be aware that after choosing one pair it is not "allowed" anymore to choose another pair (presupposed that you want exactly one pair), and also it is not allowed anymore to choose cards that make the pair a triple or a quadruple. 
These extra conditions are not taken into account in your logic. 
A: drhab's comment above is exactly right. The correct probability is equal to: $$\frac {13 \cdot \binom42 \cdot \frac{48 \cdot 44 \cdot 40}{6}}{\binom{52}{5}} \approx 0.4226$$
The $\frac{48 \cdot 44 \cdot 40}{6}$ part is number of ways to choose remaining 3 cards without making another pair and without making a three or four of kind by accident. We have 48 cards to choose from, because 2 is already taken and we can't take 2 more, because that would make a three or four of kind. After taking another card, this removes 4 to take to prevent making two pairs. This ends up us with ${48 \cdot 44 \cdot 40}$. Since order of cards doesn't matter, we divide by number of permutations of those 3 cards: $3 \cdot 2 \cdot 1 = 6$, ending up with correct answer.
