= Threshold stag-hunt equilibrium count
{title2=$k\leq s<n,\quad d/(s+1)\leq h\leq d/s$}
Consider $n$ players choosing a safe action worth $h$ or a joint action whose total reward $d$ is shared equally among its $s$ participants if $s\geq k$, and is zero otherwise. Assume $0<h<d$ and $2\leq k\leq n$. The safe-action profile is a <Nash equilibrium>. For $0<s<n$, a profile is a <Nash equilibrium> exactly under the displayed conditions: a joint participant must prefer $d/s$ to $h$, and a safe participant must prefer $h$ to $d/(s+1)$. An unsuccessful participant would strictly prefer the safe action. The all-joint endpoint is an equilibrium exactly when $d/n\geq h$.
Writing $r=d/h$, positive equilibrium counts lie at $s\leq r\leq s+1$ with $s\geq k$, or at $s=n$ when $r\geq n$. There are at most two. Exactly one exists when $r\geq k$ and $r\notin\{k+1,\ldots,n\}$. An integer $r$ in that excluded set permits both $r-1$ and $r$. Strict positivity matters: when $h=0<d$, the positive equilibrium counts instead include every $1\leq s\leq k-2$ as well as $n$.
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