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.



Difference between two largest degrees
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](https://en.wikipedia.org/wiki/Hoeffding%27s_inequality#General_case_of_bounded_random_variables), 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. 


