HMMT February

Solution:
We first compute the probability that $f(1) = 1$. Note that $f(1) = 1$ if and only if $1$ is part of a cycle whose length divides $2025$.
We claim that for any given $k$, the probability that $1$ is in a cycle of length $k$ is $\frac{1}{9}$. Indeed, the probability that $f(1) \neq 1$ is $\frac{8}{9}$. Given this, there are $8$ possible values remaining for $f(f(1))$, so the probability that $f(f(1)) \neq 1$ is $\frac{7}{8}$, and so on. Finally, there are $10 - k$ possible values remaining for $f^{k}(1)$, so the probability that $f^{k}(1) = 1$ given all previous assumptions is $\frac{1}{10 - k}$. Thus, the probability that $1$ is in a cycle of length $k$ is
$$\frac{8}{9} \cdot \frac{7}{8} \cdot \frac{10 - k}{11 - k} \cdot \frac{1}{10 - k} = \frac{1}{9}.$$
Hence, the probability that $f^{2025}(1) = 1$ is the probability that $1$ is in a cycle of length $1$, $3$, $5$, or $9$, which is $\frac{4}{9}$.
If $f(1) \neq 1$, then $f(1)$ is equally likely to be any of $2$ through $9$ by symmetry, averaging $5.5$.
Therefore, the expected value of $f(1)$ is
$$\frac{4}{9} \cdot 1 + \frac{5}{9} \cdot 5.5 = \left\lceil \frac{7}{2} \right\rceil.$$