A tournament is a directed graph in which every pair of vertices has exactly one directed edge between them—for example, here are two tournaments on the vertices {1,2,3}:
(1,2,3) is a Hamiltonian path, since it visits all the vertices exactly once, without repeating any edges, but (1,2,3,1) is not a valid Hamiltonian cycle, because the tournament contains the directed edge 1 → 3 and not 3 → 1. In the second tournament, (1,2,3,1) is a Hamiltonian cycle, as are (2,3,1,2) and (3,1,2,3); for this problem we’ll say that these are all different Hamiltonian cycles, since their start/end points are different.
Consider the following way of choosing a random tournament T on n vertices: independently for each (unordered) pair of vertices {i, j} ⊂ {1,...,n}, flip a coin and include the edge i → j in the graph if the outcome is heads, and the edge j → i if tails. What is the expected number of Hamiltonian paths in T? What is the expected number of Hamiltonian cycles?
To find the expected number of Hamiltonian paths, you need to find two things:
$N$, the total number of paths that could be Hamiltonian paths in the tournament. (So, the number of ways to put all $n$ vertices in order.)
$p$, the probability that a particular path does end up a Hamiltonian path. (The probability that all the arcs between the vertices have the correct direction.)
Then the expected number of Hamiltonian paths is just the product $Np$.
I'm not going to compute $N$ and $p$ for you, you will have to do that yourself.
To see why this happens, we can use linearity of expectation. The formal argument is this. We can write $X$, the number of Hamiltonian paths, as $X_1 + X_2 + \dots + X_N$, where $X_i$ is the indicator variable for the $i^{\text{th}}$ Hamiltonian path: it is $1$ if the $i^{\text{th}}$ sequence of vertices forms a Hamiltonian path, and $0$ if it doesn't. Then $$\mathbb E[X] = \mathbb E[X_1 + X_2 + \dots + X_N] = \mathbb E[X_1] + \mathbb E[X_2] + \dots + \mathbb E[X_N].$$ Meanwhile, the expected value of $\mathbb E[X_i]$ is $p$ for any $i$: with probability $p$, $X_i=1$, and otherwise $X_i=0$. Therefore $$\mathbb E[X] = \underbrace{p + p + \dots + p}_{N \text{ times}} = Np.$$
For Hamiltonian cycles, you will have the same value of $N$, since in either case, you have to put all $n$ vertices in order. However, the probability $p$ will be different, since you're asking for more of the tournament's edges to have the right orientation.