Difference between two largest degrees Consider a uniform random tournament with $n$ vertices. (Between any two vertices $x,y$, with probability $0.5$ draw an edge from $x$ to $y$; otherwise draw an edge from $y$ to $x$.) Let $S$ be the set of all out-degrees. Let $s_1$ be the largest element of $S$, and $s_2$ the next largest. (If $S$ is a singleton, let $s_2=s_1$.) 
Let $c\in (0,1)$ be a constant. What is $\lim_{n\rightarrow\infty}\text{Pr}[s_1-s_2<cn]$?
My guess is that the limit should go to $1$, that is, the two largest out-degrees are close to each other compared to the size of the tournament.
 A: You guess is correct, assuming that by $x_1$ and $x_2$ you meant $s_1$ and $s_2$. 
Indeed, the probability in question is $1-p_n$, where 
\begin{equation}
 p_n:=P(\exists i\in[n]\ D_i-\max_{j\in[n]\setminus\{i\}}D_j\ge cn), 
\end{equation}
where $[n]:=\{1,\dots,n\}$ and $D_i$ is the out-degree of the $i$th vertex. We can write 
\begin{equation}
 D_i=\sum_{j\in[n]}D_{ij},
\end{equation}
where 
\begin{equation}
 D_{ij}:=I\{X_{ij}=1\},
\end{equation}
$I\{\cdot\}$ is the indicator function, and $(X_{ij})$ is a random $n\times n$ skew-symmetric matrix whose above-diagonal entries are independent Rademacher random variables (r.v.'s), with $P(X_{ij}=\pm1)=1/2$ if $1\le i<j\le n$. 
We need to show that $p_n\to0$ (as $n\to\infty$), for each $c\in(0,1)$. In fact, 
\begin{equation}
 p_n\le nP(D_1-\max_{j\in[n]\setminus\{1\}}D_j\ge cn)
 \le nP(D_1-D_2\ge cn). 
\end{equation}
Next, 
\begin{equation}
 D_1-D_2=D_{12}-D_{21}+\sum_{j=3}^n Y_j,
\end{equation}
where $Y_j:=D_{1j}-D_{2j}$, so that the $Y_j$'s are iid zero-mean r.v.'s, with $|Y_j|\le1$. Also, $D_{12}-D_{21}\le1$. So, by (say) Hoeffding's inequality, for all large enough $n$, 
\begin{equation}
 p_n\le nP(D_1-D_2\ge cn)\le nP\Big(\sum_{j=3}^n Y_j\ge cn-1\Big)\le ne^{-(cn-1)^2/(2(n-2))}\to0, 
\end{equation}
as desired. 
A: The outdegree of every fixed vertex is distributed binomialy with the mean $n/2$ and the variance $n/4$. Hence, the probability that the outdegree deviates from $n/2$ by $cn/3$ at least is extremely small, and by the union bound so is the probability that there is at least one vertex with the outdegree deviating from $n/2$ by $cn/3$. This means that with probability $1-o(1)$, the outdegrees of all vertices are in the range $(n/2-cn/3,n/2+cn/3)$, confirming your conjecture.
