Probability of $2$ runners both finishing in any of the top $3$ positions

Question

Given a race of $n$ runners, what is the probability of $2$ chosen runners both finishing in any of the top $3$ positions of a race.

I think the probability is $6(n-2)/n!$, but it seems intuitively wrong.

Answer 1

If $n= 2$ the answer is trivially $1$. If $n>2$, the two runners need to be set in one of these three permutations: The two runners are first and second, first and third, or second and third. That's $3$ permutations, and we multiply by $2$ because we can switch the two runners for each permutation. So we get $6$ permutation for the two runners.

There are left $(n-2)$ runners which can be set in $(n-2)!$ different permutations. So overall we have $6(n-2)!$ valid permutations. Without any limitations, all the permutations are $n!$. Thus, the answer is $\frac{6(n-2)!}{n!}$.

I hope it is clear enough.