Picking two cards of the same suit

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So, I am having a problem with this in that the method I use gives two completely separate answers.

Two cards are selected from a deck of $52$ playing cards. What is the probability they constitute a pair (that is, that they are of the same denomination)?

So, for the first method I reason this.

The first card picked has a $13/52$ chance of being in some suit. The second card picked has probability $12/51$ of being in the same suit.

So... The probability should be $(13/52)(12/52) = 3/52$.

The other method is by combinatorics.

I have $52 \cdot 51$ one ways of creating a pair of cards. But I have $13 \cdot 12$ different ways of creating a pair of the same suit. Now to me, the logical thing to do is to multiply this number by $4$, because I would have to count each valid pair from each suit.

This would give me

$$\frac{4 \cdot 12 \cdot 13}{52 \cdot 51}$$

What's wrong with the reasoning on the second one?

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The second is correct.

The first should be amended as follows. The first card has a chance of 1 of being in some suit, and the second should be 12/51 to be of the same suit, so the total is 12/51...

UPDATE

If the suit is fixed ahead of time, i.e. you need the probability to draw 2 hearts, not any 2 of the same suit, then first probability is 13/52 and second is 12/51, so indeed the total is $$\frac{13}{52} \times \frac{12}{51} = \frac{3}{51} = \frac{1}{17} \approx 5.88\%$$