Suppose we have $n\geq 4$ players $p_1,\cdots,p_n$ of a game between two teams: team $A$ and team $B$ (disjoint sets, each with two or more players, so that $|A|+|B|=n$). Assume that each player $p_i$ has a skill value $a_i$ (a positive real number, but we can also assume it's a positive integer).

Team $A$ wins if and only if $\sum_{p_i\in A} a_i > \sum_{p_i\in B} a_i$ (hence, team $B$ loses; if the two sums are equal we have a tie). Assume that no player has a skill value greater than the sum of the values of the remaining players.

For integer $k$, $\frac n2 \leq k \leq n-2$, let $w_k$ be the number of different $k$-sets $A$ (sets of cardinality $k$) that we can form so that $A$ wins or ties.

For example, let $n=5$, $(a_1,\cdots,a_5)=(1,3,6,7,10)$. Then $w_3=8$, since out of the $\binom 5 3=10$ different ways to choose the players of team $A$, only the teams $\{p_1,p_2,p_3 \}$ and $\{p_1,p_2,p_4 \}$ will not win.

(In general, if $l_k$ is the number of $k$-sets $A$ that lose, then $l_k+w_k=\binom nk$, and $w_{n-k}=l_k$, which justifies the interval considered for $k$'s).

I have two questions:

1- The highest number of winning (or tieing) $k$-sets is clearly $\binom nk$, which can be achieved with $n$ players of equal skill value. Intuitively, the smallest number of winning (or tieing) $k$-sets seems to be $\binom {n-1}{k-1}$, which is achieved with one player with great value and all others with very small value (again, for $k$ such that $\frac n2 \leq k \leq n-2$). For example, with $n=5$, $k=3$ as before, 6 winning triples can be formed with $(a_1,\cdots,a_5)=(1,2,3,4,9)$, obtained with the best player $p_5$ and any combination of 2 players from the remaining 4. However, I was not able to give a rigorous proof of this.

2- Suppose it is true that $\binom {n-1}{k-1} \leq w_k \leq \binom nk$. Given an integer $m$ in this interval, does there exist an $n$-tuple of skill values so that $w_k=m$?

Suppose we have $n\geq 4$ players $p_1,\cdots,p_n$ of a game between two teams: team $A$ and team $B$ (disjoint sets, each with two or more players, so that $|A|+|B|=n$). Assume that each player $p_i$ has a skill value $a_i$ (a positive real number, but we can also assume it's a positive integer).

Team $A$ wins if and only if $\sum_{p_i\in A} a_i > \sum_{p_i\in B} a_i$ (hence, team $B$ loses; if the two sums are equal we have a tie). Assume that no player has a skill value greater than the sum of the values of the remaining players.

For integer $k$, $\frac n2 \leq k \leq n-2$, let $w_k$ be the number of different $k$-sets $A$ (sets of cardinality $k$) that we can form so that $A$ wins or ties.

For example, let $n=5$, $(a_1,\cdots,a_5)=(1,3,6,7,10)$. Then $w_3=8$, since out of the $\binom 5 3=10$ different ways to choose 3 players to be in team $A$, only the triples $\{p_1,p_2,p_3 \}$ and $\{p_1,p_2,p_4 \}$ will not win.

(In general, if $l_k$ is the number of $k$-sets $A$ that lose, then $l_k+w_k=\binom nk$, and $w_{n-k}=l_k$, which justifies the interval considered for $k$'s).

I have two questions:

1- The highest number of winning (or tieing) $k$-sets is clearly $\binom nk$, which can be achieved with $n$ players of equal skill value. Intuitively, the smallest number of winning (or tieing) $k$-sets seems to be $\binom {n-1}{k-1}$, which is achieved with one player with great value and all others with very small value (again, for $k$ such that $\frac n2 \leq k \leq n-2$). For example, with $n=5$, $k=3$ as before, 6 winning triples can be formed with $(a_1,\cdots,a_5)=(1,2,3,4,9)$, obtained with the best player $p_5$ and any combination of 2 players from the remaining 4. However, I was not able to give a rigorous proof of this.

2- Suppose it is true that $\binom {n-1}{k-1} \leq w_k \leq \binom nk$. Given an integer $m$ in this interval, does there exist an $n$-tuple of skill values so that $w_k=m$?