$[1-x,1]$      $rx$        $x$     $ 1/2 -x $

We now solve $\Expct[p(s_1)] = \Expct[p(s_2)] $ for $x$, namely, $$(1-r)\cdot x + rx \cdot \frac x2 + rx \cdot x + r(1-2x) \cdot x = \frac12 - x~.$$ This yields $rx^2 - 4x + 1 = 0$, $$rx^2 - 4x + 1 = 0 \text{,}$$ and recalling that $0<x<1$ we obtain $x=(2-\sqrt{4-r})/r$. $$x=\frac{2-\sqrt{4-r}}{r} \text{.}$$

By the definition of $x$, this means that for every $ r $, when $\loc_1=\loc_2=(2-\sqrt{4-r})/r$ and $\loc_3=\loc_4=1-(2-\sqrt{4-r})/r$(See Fig. \[fig:Hgame-4p-Disconnections\]), each server $s_i$ receives the same expected payoff: $\Expct[p(s_i)]=1/2 - \loc $. We claim that this is a Nash equilibrium, since no server may increase his payoff by moving to another location. This follows from Observation \[obs:disconnect\]. As a motivating exercise, in Appendix \[app:nash4\] we explicitly show that no improving move exists. The detailed calculation is deferred to the full paper.

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

Moreover, this Nash equilibrium is unique due to the following considerations. First, by Observation \[obs:disconnect\], the peripheral servers must be paired with their neighbors. Second, the peripheral servers must have hinterlands of the same length, otherwise one would take the hinterland of the other. Third, every two paired servers must have the same 