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Write $k=2m+1$, where $m\ge 2$. For three distinct sites $a<b<c$, the left, middle and right firms receive respectivelyThese follow by counting the two midpoint boundaries, including split ties. Therefore, after two firms occupy $\ell <r$, the final entrant compares $2(\ell -1)$ at $\ell -1$, $2(k-r)$ at $r+1$, and $r-\ell$ at every available interior site. Omit an unavailable region and retain every maximizing tie. This is the final entrant payoff in a discrete location game.

First suppose $C$ opens at the central site $m+1$. If $S$ chooses $s\le m$, the final entrant's best possible payoffs on the left, right and interior are at most $2m-2$, exactly $2m$, and at most $m$, respectively. Hence $N$'s unique best response is $m+2$. The resulting payoff to $S$ is $m+s$, uniquely maximized on that side at $s=m$. Reflection gives the other optimal choice $s=m+2$, followed by $N=m$. Under either choice, $C$ receives only $2$. Any mixture of the tied optimal second-mover choices has the same consequence.

Now let $C$ deviate to $c=m$. If $S$ chooses $s<m$, the unique final best response is $m+1$, yielding $S$ at most $2m-2$. If $S$ chooses $m+1$, the unique final best response is $m+2$, yielding $S$ only two. Both classes will be strictly inferior to an available alternative, independently of how later ties are resolved:

Accordingly, in every subgame perfect equilibrium continuation after $C=m$, the second mover must choose a site $s\ge m+2$. This remains true with mixed strategies: every excluded choice pays strictly less than the displayed guaranteed alternative, under every possible continuation tie policy.

For any such $s$, every legal final site gives $C$ at least three. If $N$ lies left of $m$, then $C$ is the middle firm and receives $s-N\ge 3$. If $N$ lies right of $m$, then $C$ is the left firm and receives $m+\min (s,N)-1\ge 2m\ge 4$. Thus the deviation gives $C$ at least $3$, against every subgame-perfect continuation, whereas the central opening gives it $2$.

In this original example, a final-mover tie produces the line segment for $S=6$: its points are expected payoff pairs from a mixed strategy over two final best responses. The second mover can independently secure eight by choosing $S=7$, so only the segment to the right of that guarantee can support its optimal choice. Mixing over a tied final response cannot make the first mover's payoff fall below three.

A subgame perfect equilibrium must prescribe optimal continuation play even after the first mover's off-path deviation. Since the deviation is strictly profitable under every such prescription, no subgame perfect equilibrium can have $C$ open centrally. The argument does not select favourable ties or assume that off-path players cooperate with the deviator.