Shared out-neighbor of maximum-degree vertices in fully connected digraph? Given any tournament $G = (V,E)$ (so $\forall i,j \in V$, either $(i,j) \in E$ or $(j,i) \in E$), let $X \subset V$ be the vertices with maximal out-degree, and let $Y = V \setminus X$.
If $|Y| > 0$, is it necessarily true that there must exist $y \in Y$ such that $\forall x\in X,\;\;(x,y) \in E $ ?
I think a simple argument by average degrees can prove this for $|Y| \leq 2$, but I'm not sure how to generalize.
 A: This is only guaranteed when $|X| \le 2$ or $|Y|=1$;

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*When $Y = \{y\}$, deleting $Y$ leaves a nearly-regular tournament where all outdegrees differ by at most $1$; the vertices with higher degree are the ones that $y$ had edges to. If $|X|$ is odd, then the only such tournament is regular, which means that $y$ must have had outdegree $0$. If $|X|=2k$, the only such tournament is one with $k$ vertices of degree $k-1$ and $k$ vertices of degree $k$, which means $y$ must also have had degree $k$; this is a contradiction, because then $Y$ would also be a max-degree vertex.

*When $X = \{x\}$ and $|Y|>0$, $x$ can't have maximum outdegree and also have outdegree $0$, so there must be some $y \in Y$ with an edge $(x,y)$.

*When $X = \{x_1, x_2\}$, if there is no vertex $y \in Y$ with edges $(x_1, y)$ and $(x_2,y)$, that means $\deg^+(x_1) + \deg^+(x_2) \le |Y|+1$, counting the edge $(x_1, x_2)$ or its reverse. So the average outdegree of vertices in $X$ is at most $\frac{n-1}{2}$ (in an $n$-vertex tournament). But this is the average outdegree of all vertices, so we must get a regular tournament, contradicting $|Y|>0$.

Despite what you claim, such a vertex $y \in Y$ might not exist when $|Y|=2$. Here is an example for $|Y|=2$ with the vertices labeled according to their outdegree: the vertices in $Y$ are highlighted, and for each vertex $y \in Y$, I've also highlighted the edge that shows that not all vertices $x \in X$ have the edge $(x,y)$:

It is tedious to cook up examples for other pairs of $|X|$ and $|Y|$, but I think it's likely that all of the examples exist that I haven't ruled out above.
