In a random tournament with $n$ nodes, for any pairs of nodes $a$ and $b$, we create either the edge $a\rightarrow b$ or the edge $b\rightarrow a$ with 50-50 probability.
What is the expected length of the longest chain (meaning transitive subtournament)? If there is no nice closed formula, are there good upper bounds?
EDIT: this question is under improvement, in great part thanks to the comment section
I am terrible at probabilities (it is my weakest area in maths) but I will give it a shot; I might inspire someone.
I will try to explicitly calculate the probability that the graph has a chain of length $k$ and then we can apply the definition of expected value to compute that.
Fix $k \leq n$, $n$ being the number of vertices of the graph (let us call it $G$).
There are $\frac{n!}{(n-k)!}$ ways of creating an ordered tuple of length $k$ with distinct vertices. The probability of that tuple representing the chain $a_1 \rightarrow a_2 \rightarrow \cdots \rightarrow a_k$ is precisely $\frac1{2^{k-1}}$.
This could mean
$$C_k = P(\text{graph has chain of length }k) = \frac{n!}{(n-k)!}\frac1{2^{k-1}}$$
I believe that
$$p_k = P(G\text{ has biggest chain of length }k) = P(G \text{ has chain of length }k) - P(G \text{ has chain of length } k+1)$$
Hence
$$p_k = C_k - C_{k+1}$$
and now you would only have to use the definition of expected value. If it were, we'd have
$$\mathbb{E}[\text{size of longest chain}] = \sum_{k=1}^n k\cdot\left(C_k - C_{k+1}\right)$$
The summation is
$$C_1 - C_2 + 2C_2 - 2C_3 + 3C_3 - 3C_4 + \cdots + (n-1)C_{n-1} - (n-1)C_{n} + nC_n = \sum_{k=1}^n C_k$$
Thus we need only to be able to compute $C_k$. The current formula can get $> 1$ for values of $(n, k)$. If you are able to compute $C_k$ correctly, the answer follows from this. Good luck! I will give it more thought as well.