The answer I have seen is unclear, which is
$P_i$ = probability that $x$ is in $a = \frac{1}{n}$ cycle of length $i$
$n\sum_{i=1}^n \frac{1}{i}·P_i = $ expected number of cycles = $\sum_{i=1}^n \frac{1}{i}$
I have no idea about how to get the answer, could you give me a detailed solution? It would be very helpful for my final exam. Thanks!
The answer may depend on the nature of randomness, i. e., on the way the graph is selected. If self-loop is allowed and the graph is selected uniformly among all such labeled digraphs on $n$ vertices, then this graph is equivalent to a random permutation. It is easy to prove that expected number of cycles in the random permutation of $n$ elements is corresponding harmonic number $H_n = \sum_{i = 1}^n \frac{1}{i}$.
If self-loop is not allowed and the graph is selected uniformly among all such labeled digraphs on $n$ vertices, then this graph is equivalent to a random derangement. The expected number of cycles of length $k$ in derangement of $n \ge 2$ elements is $$\frac{1}{k}\cdot \frac{1 / 0! - 1 / 1! + 1 / 2! + \cdots + (-1)^{n - k} / (n - k)!}{1 / 0! - 1 / 1! + 1 / 2! + \cdots + (-1)^n / n!} = \frac{1}{k}\cdot \frac{\sum_{i = 0}^{n - k}(-1)^i / i!}{\sum_{i = 0}^{n}(-1)^i / i!} = \frac{1}{k}\cdot \frac{!(n - k) / (n - k)!}{!n / n!},$$ where $!n$ is the number of derangements. So the expected number of cycles of all lengths is $$\sum_{k = 2}^n\frac{1}{k}\cdot \frac{!(n - k) / (n - k)!}{!n / n!} = \frac{n!}{!n} \cdot \sum_{k = 2}^n \frac{!(n - k)}{k \cdot (n - k)!}.$$
If the graph is uniformly selected among unlabeled digraphs, the problems becomes equivalent to computing expected number $k$ of summands in equation $n = a_1 + a_2 + \dots + a_k$, where $1 \le a_1 \le a_2 \le \ldots \le a_k$, when such equation is selected uniformly among all such possible equations. This problem is not so easy to get a closed formula, however both number of such representations of $n$ and the total number of summands in all representations can be computed using dynamic programming. (If self-loop is not allowed, then $a_1 \ge 2$.)
If the graph is selected non-uniformly, then we need to consider a certain probability distribution, which is not given.
