Right answer, wrong explanation, probability of grids? 
Two unit squares are selected at random without replacement from an $n\times n$ grid of unit squares. Find the least positive integer $n$ such that the probability that the two selected squares are horizontally or vertically adjacent is less than $\frac{1}{2015}$.

Here is my reasoning, I believe it is incorrect, but please help me out why.

I fixed a square $x$. To get an adjacent square, there are $4$ options (above, below, right, left) to get an adjacent square. And there are $n^2 - 1$ squares to choose from (since $x$ has been chosen). I set up: $$\frac{4}{n^2 - 1} < \frac{1}{2015}$$

In the end I got $90$, which is correct, but I think the method is incorrect? 
 A: The reason your argument doesn't work is that the corner and edge squares of the grid are NOT adjacent to $4$ squares each. Edge squares are adjacent to $3$, and corners are adjacent to $2$.
Instead, a good strategy would be: (1) count the number of PAIRS of squares that are adjacent (horizontally or vertically); (2) count the number of total pairs of squares; and then divide (1) by (2).
If you do this you get
$$
\frac{\underbrace{(n-1) \cdot n}_{\text{horizontally adjacent}} + \underbrace{n \cdot (n-1)}_{\text{vertically adjacent}}}{n^2 \choose 2}
= \frac{4 n (n-1)}{n^2 (n^2 - 1)}
= \frac{4}{n(n+1)}
$$
Then you solve
$$
\frac{4}{n(n+1)} < \frac{1}{2015} \iff n(n+1) > 8060 \iff n \ge \boxed{90}.
$$
A: There are $(n-1)n$ ways for your chosen squares to be adjacent horizontally, and the same number of ways to be adjacent vertically.
There are $n^2(n^2-1)/2$ ways to choose the two squares.
Taking $n=89$ gives a probability of $1/2002.5$ and $n=90$ gives $1/2047.5$.
So, you did get the correct answer.
The argument you made works only for squares away from the edges.  If you're on a non-corner edge, then you only have three choices.  If you're on one of the four corners, you only have two.
So, you're overestimating the probability.  But does it ever change the answer?  As it turns out, yes.  Had the probability been $1/2026$ instead of $1/2015$ the correct answer would still have been $90$, but your approximation would have told you the answer was $91$.
So, yes, under different circumstances, your method would have given the wrong answer.
A: If we exclude $\frac 12$ the corner squares and $\frac14$ of the border squares from the count, 
say the whole of the left column, our count will be correct, and the Pr equation will be:
$$\frac {4n(n-1)}{n^2(n^2-1)} = \frac{4}{n(n+1)}< \frac{1}{2015}$$ 
