What is the probability of drawing two cards from the same deck of cards that have the same suite

It is cyclic so: ...Jack, Q, K, A, 2...

The first card you pick doesn't matter. You have a 100% chance of drawing a card that fits the requirements. There are now 51 cards in the deck. If the card was the n of some suit, the next card must be (n+1) or (n-1) of that same suit. Therefore there are two cards in the 51 that will satisfy the requirements. So, the P of drawing two consecutive cards of the same suit is 2/51

I wasn't given an answer, I was simply told I was wrong. Where is my error?

Want to improve this question? Add details and clarify the problem by editing this post.

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?