Every person plays with another in a tournament of $n$ people. It is called $k$-fair if for any set of $k$ players, someone beats everyone in thee set. Show that for all positive integers $k$ there exist $n$ and some tournament with $n$ people that is $k$-fair 
My approach: Note $E(d^{+}(v_i))=\frac{n-1}{2}$, and there exist a graph such that the outdegree of every vertex is $\frac{n-1}{2}$ for all $n$ odd. There are $\binom{\frac{n-1}{2}}{k}$ $k$-tuples that are beaten by any vertex, and $\binom{n}{k}$ k tuples in total. Note for $n$ large, $n\binom{\frac{n-1}{2}}{k}>\binom{n}{k}$
So there doesn’t exist a k-tuple such that cannot be beaten by one person. Is this a correct solution?

No. The problem with your solution is that a given $k$-set (not $k$-tuple, because order doesn't matter) could be beaten by more than one vertex. So just because you've found more than $\binom nk$ instances of a $k$-set being beaten by a vertex, doesn't mean that every $k$-set is part of one of those instances.
A $k$-set could be beaten by up to $n-k$ vertices, so you'd need $n \binom{\frac{n-1}{2}}{k}$ to exceed $(n-k)(\binom nk-1) + 1$, which isn't going to happen for any $n$.

Instead, consider a random tournament. Find the probability that a $k$-set is not beaten by any vertex, and from there, find the expected number of $k$-set with this property.
If the expected number of $k$-sets that are not beaten by any vertex is less than $1$, then it's possible for that number to be $0$, which would produce a tournament in which every $k$-set is beaten by some vertex.
