 Solution

Continue with $0<h<d$ and set $r=d/h$. Part (e) requires an interior count to satisfy $s\le r\le s+1$ with $s\ge k$; the all-deer endpoint $s=n$ is allowed when $r\ge n$. If $r$ is nonintegral and $k\le r<n$, only $s=\lfloor r\rfloor$ works. If $r=m$ is an integer with $k+1\le m\le n$, both $s=m-1$ and $s=m$ work. At $r=k$ only $s=k$ remains, while $r>n$ permits only $s=n$. When $r<k$, there is no positive equilibrium count.

Thus there are at most two positive counts, and exactly one precisely whenThis includes $k=n$, when the excluded set is empty. It counts numbers of deer hunters, not the many choices of which hunters participate. The argument is the threshold stag-hunt equilibrium count.

The original paper states only $h<d$ explicitly. Its assertion needs the usual strict positivity of the hare reward. If zero is included, take $n=6,k=4,d=1,h=0$. Counts $s=1$ and $s=2$ both give unsuccessful hunters zero, and one additional deer hunter still cannot reach the threshold; all unilateral alternatives also pay zero. The count $s=6$ is an equilibrium as well. These are three positive equilibrium counts, contradicting an unrestricted reading. More generally, for $h=0<d$, the positive counts are $1,\dots ,k-2$ and $n$.