Number of possibility of getting at least a pair of poker cards The question: randomly drawing a hand (5 cards) from a deck of 52 poker cards, what is the number of possibilities of getting at least one pair in the 5 cards? A pair is two cards within the same denomination.
My intuition is to use (the total number of possibility for a poker hand) - (number of poker hand without pairs at all), which leads to 
$\dbinom{52}{5} - \dbinom{13}{5}\times 4^5 = 1,281,072$
and I see that a similar question has been asked here.
On the other hand, I read about another solution that reasoned as the following:
Considering the one pair that is within the 5 cards: its number of possible denomination is $\dbinom{13}{1} $ and the pair could have $\dbinom{4}{2}$ possible colors; then the other three cards have $\dbinom{50}{3} $. 
$\dbinom{13}{1}\times\dbinom{4}{2}\times\dbinom{50}{3} = 1,528,800$
I suspect that the second solution is wrong, but I don't see how it's counting wrong.
Thanks.
 A: The second formula overcounts the hands with at least one pair. For it multiple counts the the $2$ pairs hands, the $3$ of a kind hands, the $4$ of a kind hands, and the full house hands. 
For example, the $4$ Kings and $7$ of diamonds hand is counted $\binom{4}{2}=6$ times. For the process counts as different the hand that has the King of spades and King of hearts, counted in the $\binom{13}{1}\binom{4}{2}$ part, and the other two Kings (counted in the $\binom{50}{3}$ part, and the hand that has the King of spades and the King of diamonds (counted in the $\binom{13}{1}\binom{4}{2}$ part), and the other two Kings (counted in the $\binom{50}{3}$ part).
Remark: The intuition behind your second formula probably is that you find the number of ways of getting "the" pair, and multiply by the number of ways of getting the remaining cards. However, in the cases of overcount listed above, there is no such thing as "the" pair. The process, with $\binom{13}{1}\binom{4}{2}\binom{12}{3}\binom{4}{1}^3$ works perfectly well to count the hands that have precisely $1$ pair.
A: The first case is right.
The second is wrong in 5 ways:
1. You would use (52 1) rather than (13 1) - the first card in the pair can be anything
2. You would use (3 1) rather than (4 2) for the second card in the pair.
3. You would divide these by two, to get all possible pairs.
4. You would multiply this by (51 3) to get up to the kind of approach you're talking about.
This would give 1,624,350 (http://www.wolframalpha.com/input/?i=133+%2851+Combination+3%29)
However,
5. this approach is still wrong, because there will be pairs in the (51 3) which are then going to get counted twice.
