= Solution
\b[There is exactly one such nontrivial game for every $n\ge1$: the <unanimity game> requiring every player.] Put
$$
\boxed{u_N(S)=
\begin{cases}
1,&S=N,\\
0,&S\ne N.
\end{cases}
\qquad\text{Number of games}=1.}
$$
It is a <simple cooperative game>, and permuting players preserves every <coalition> value. Thus it has the required <symmetry of players in a cooperative game>. It is also a <convex cooperative game>: if both $S,T$ are winning, all four terms in the <supermodular> inequality have value $1$; if exactly one is winning, their union is $N$ and their intersection loses, giving equality $1=1$; if neither wins, the right-hand side is zero and the inequality follows from nonnegativity. For $n=1$ the same definition gives the sole player value $1$.
To prove uniqueness, first establish that <convex simple games are unanimity games>. If $S,T$ are <winning coalitions>, the <supermodular> inequality gives
$$
2=v(S)+v(T)\le v(S\cup T)+v(S\cap T)\le2,
$$
so their intersection is also winning. The family of <winning coalitions> is finite and nonempty. Its intersection $K$ is therefore winning and nonempty, since the empty <coalition> loses. Every winning <coalition> contains $K$, and monotonicity makes every superset of $K$ winning. Hence $v=u_K$.
If $K$ were a proper nonempty <subset> of $N$, take $i\in K$ and $j\notin K$, and let $T=N\setminus\{i,j\}$. Then $T\cup\{i\}=N\setminus\{j\}$ contains $K$ and wins, whereas $T\cup\{j\}=N\setminus\{i\}$ loses. This violates <symmetry of players in a cooperative game>, which requires those two <coalition> values to agree. Thus $K=N$, proving the claimed count. The “nontrivial” convention excludes the identically zero game; if that degenerate game were also called simple, it would supply one additional symmetric convex example.
Back to article page
