What are the odds of me dealing two cards in a row and having them be a pair? Here is the problem.  I deal two cards face up on the table.  What are the odds they a pair?  Simple?  I thought so too.
Solution 1: Correct?
When considering the first card, we know that doesn't matter.  There is zero chance of it being a pair!  The second card is very valuable.  We know that there are only 3 cards left in that deck that actually match that first card.  There are 51 cards in the deck in total.  This leaves 0/52 + 3/51.
Solution 2: Correct?
Let's consider every single possible combination of cards we can get.  (The deck will all be divided to four to make our lives easier.)  We can get AA | 11 | 12 | 13 | 14 | 15 | 16 ...  (There were intentionally two pairs in the front.)  WE know that there are 169 possibilities  (52 / 4 choose 2).  NOW, we know, we see that there are 13 pairs to this entire deck.  (One pair for each 'row'.  )  That leaves 13 pairs / 169 cards.
Solution 1: ~5%
Solution 2: ~7
Here is another example (to clear up confusion) with a much simpler deck.
We have a deck of sick cards.  (1, 2, 3, 1, 2, 3).  Now, let's try solution one.
Solution 1:
0/6 + 1/5.  (1/5 because there is only one card left to actually match those other first card.  There are also only five cards left in the deck because the first one ask use.)
Solution 2:
Every combination
11
12
13
11
12
13
21
22
23
21
22
23
31
32
33
31
32
33
11
12
13
11
12
13
21
22
23
21
22
23
31
32
33
31
32
33
We see 12 pairs out of 36 possible combinations.  This is 12/36.
Solution 1: 20
Solution 2: 33
Someone please give the rights answer and explain why this is wrong>?
 A: Solution 1 is correct.
Solution 2 is incorrect because your denominator is wrong. You are assuming that a pair (e.g., 22) is as probable as a non-pair (e.g., 23), which is not the case. If you consider the suits of the cards (♠♥♦♣), there are 32 distinct ways to obtain each non-pair (assuming the order matters), but only 12 ways to obtain each pair.
It's the same in your simplified example. Each pair should appear twice, not four times, so there are 6 pairs out of 30 possibilities, same as the first solution.
A: +1 Doctor Milt's answer. A convenient way to look at these problems is use the binomial coefficient ${n \choose k}$, read as "n choose k". We choose out of 52 cards 2 of them, so the total number of possible hands is ${52 \choose 2}$, from those hands we want a pair (2) that is made out of 4 cards, i.e. ${4 \choose 2}$, and they are 13 ranks we can use either one, i.e. ${13 \choose 1}$. This brings us our probability to have a pair in the first two cards as $p = \frac{{4 \choose 2}{13 \choose 1}}{{52 \choose 2}}$ $=$ $ \frac{6 \times 13}{1326} $ $=$ $ 0.0588235294...$ which equates to $\frac{3}{51}$.
Finally, just to note, the odds of an event are computed as the ratio of the probability $p$ over one minus $p$; i.e. in this case the odds will be below $1$ (here $0.0526315789...$). The odds shown in the main post appear inverted.
