HMMT February, 2022

Solution:

Let $n=10$ and call two numbers close if there is at most one number between them and a circular permutation focused if only $n$ is greater than all numbers close to it. Let $A_{n}$ be the number of focused circular permutations of $\{1,2, \ldots, n\}$.
If $n \geq 5$, then there are 2 cases: $n-1$ is either one or two positions from $n$. If $n-1$ is one position from $n$, it is either on its left or right. In this case, one can check a permutation is focused if and only if removing $n$ yields a focused permutation, so there are $2A_{n-1}$ permutations in this case. If $n-1$ is two positions from $n$, there are $n-2$ choices for $k$, the element that lies between $n$ and $n-1$. One can show that this permutation is focused if and only if removing both $n$ and $k$ and relabeling the numbers yields a focused permutation, so there are $2(n-2)A_{n-2}$ permutations in this case. Thus, we have $A_{n}=2A_{n-1}+2(n-2)A_{n-2}$.
If we let $p_{n}=A_{n}/(n-1)!$ be the probability that a random circular permutation is focused, then this becomes
$$p_{n}=\frac{2p_{n-1}+2p_{n-2}}{n-1}$$
Since $p_{3}=p_{4}=1$, we may now use this recursion to calculate
$$p_{5}=1,\quad p_{6}=\frac{4}{5},\quad p_{7}=\frac{3}{5},\quad p_{8}=\frac{2}{5},\quad p_{9}=\frac{1}{4},\quad p_{10}=\frac{13}{90}.$$