Threshold stag-hunt equilibrium count ($k\le s<n,d/(s+1)\le h\le d/s$)

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 $s\ge k$, and is zero otherwise. Assume $0<h<d$ and $2\le k\le n$. 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 $d/n\ge h$.

Writing $r=d/h$, positive equilibrium counts lie at $s\le r\le s+1$ with $s\ge k$, or at $s=n$ when $r\ge n$. There are at most two. Exactly one exists when $r\ge k$ and $r\notin \{k+1,\dots ,n\}$. 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 $1\le s\le k-2$ as well as $n$.