If the chosen positions are $(a, b)$ and $(c, d)$, then we are looking for the probability that $\{a, c\} \cap \{b, d\} = \emptyset$. Case 1: If $a \neq c$, then $b, d$ both have $n - 2$ choices. The total number of choices is $n(n-1)(n - 2)^2$. Case 2: If $a = c$, then $b, d$ must be different. There are $n(n - 1)(n - 2)$ choices. Thus the probability is equal to $\frac{n(n - 1)(n - 2)^2 + n(n - 1)(n - 2)}{n^2(n^2 - 1)} = \frac{(n - 1)(n - 2)}{n(n + 1)}$.

Although the question can be reduced to a counting problem, I think that it's a very good exercise. In particular , we see that, from $n = 7$, the probability that $A^2=0$ is $>1/2$.