I'm trying to estimate some probability in a random tournament, and I know that what I have is false, as it leads to contradicting results published some 40 years ago. But I don't know where the reasoning is faulty exactly - any help would be appreciated.
A tournament $T = (V, A)$ is a directed graph in which for each distinct $x, y \in V$, exactly one of $(x, y)$ or $(y, x)$ is in $A$. We say $T$ is balanced if $|V|$ is odd and each vertex has outdegree precisely $(n - 1)/2$ (and thus indegree $(n - 1)/2$).
Suppose that $T$ is a random balanced tournament. That is, $T$ is chosen randomly and uniformly among the set of all balanced tournaments on $|V|$ vertices.
Fix $S \subset V$ with $|S| = k$. For some vertex $v \notin S$, denote by $D_v$ the event that $v$ dominates $S$. That is, that $(v, s)$ is an arc for each $s \in S$. Denote by $\overline{D_v}$ the opposite, i.e. some $s \in S$ has $(s, v) \in A$.
What is $Pr[D_v]$ for a fixed vertex $v$ ?
I say each $(n - 1)/2$ subset of $V \setminus \{v\}$ has an equal chance of being the neighbourhood of $v$, and so $Pr[D_v] = \frac{{{n - 1} \choose {(n - 1)/2 - k}}}{{{n - 1} \choose {(n - 1)/2}}}$.
Now I want to know (or bound) the probability that no $v$ dominate $S$.
So I say \begin{align} Pr[\bigcap_{v \in V \setminus S} \overline{D_v}] &= Pr[\overline{D_{v_1}}]Pr[\overline{D_{v_2}} | \overline{D_{v_1}}] \ldots Pr[\overline{D_{v_{n - k}}} | \overline{D_{v_1}} \cap \ldots \cap \overline{D_{v_{n - k - 1}}}] \\ &\leq Pr[\overline{D_v}]^{n - k} \\ &= \left(1 - \frac{{{n - 1} \choose {(n - 1)/2 - k}}}{{{n - 1} \choose {(n - 1)/2}}}\right)^{n - k} \end{align}
Here I'm assuming that $Pr[\overline{D_{v_2}} | \overline{D_{v_1}}] \leq Pr[\overline{D_{v_2}}]$. I have intuitive reasons to believe it's true, though it might not. I'm interested in any opinion on that matter.
Revised answer after your edit: I'm not sure where you're getting the numerator ${n-1 \choose (n-1)/2-k}$ in your expression for $P[D_v]$. It looks like you're trying to count the number of ways to choose the $(n-1)/2 - k$ remaining out-neighbors for $v$, since we're assuming that $v$ is adjacent to each of the $k$ vertices in $S$. But you're making this choice from a set of $(n-1)-k$ remaining vertices, not $n-1$. So, I think the numerator should instead be ${n-1-k \choose (n-1)/2 - k}$.
Taking $n=5$ and $k=1$, your original expression says that $P[D_v] = {4 \choose 1}/{4 \choose 2} = 2/3$, but the corrected expression gives $P[D_v] = {3 \choose 1}/{4 \choose 2} = 1/2$, as intuition would suggest.