expected payoff, otherwise one would take the half-market of the other. It follows that the configuration above is the only Nash equilibrium.

Applying the considerations above to the game with *five servers* we find that the only equilibrium is when $s_1$ and $s_2$ are paired at $\loc_1=\loc_2=x$, $s_3$ is isolated at $\loc_3=1/2$, and $s_4$ and $s_5$ are paired at $\loc_4=\loc_5=1-x$, such that the distance between $s_3$ and each peripheral pair is exactly twice $x- r \cdot x^2/2$. (See Fig. \[fig:Hgame-5p-Disconnections\].)

Hence $x$ satisfies $1/2 - x = 2 \cdot (x - r x^2/2)$, which yields $r x^2 -3x + 1=0$, and recalling that $0<x<1$, we obtain $x=(3-\sqrt{9-4r})/(2r)$. Hence $x$ satisfies $$\frac12 - x = 2 \cdot \left( x- \frac {r \cdot x^2}2 \right)~,$$ which yields $$r x^2 -3x + 1=0~,$$ and recalling that $0<x<1$, we obtain $$x=\frac{3-\sqrt{9-4r}}{2r}~.$$

![Equilibrium with 5 servers ($x=\frac{3-\sqrt{9-4r}}{2r}$).[]{data-label="fig:Hgame-5p-Disconnections"}](figs/Hgame-5p-Disconnections){width="\textwidth"}

We next consider the general $n$-server game $H_{{\mbox{\small \sc lf}}}(n,r,[0,1])$. By Obs. \[obs:disconnect\], for any given configuration of servers, adding link failures to the game only affects the expected payoff gained from the hinterlands. Namely, a hinterland of length $x$ shrinks by $r x^2/2$ in expectation, while every other half-market retains its length in expectation. This leads to 