length of $s_1$’s hinterland is $L(s_1)=\loc_1-f$. Otherwise, $L(s_1)=\loc_1$. The expected length is therefore $$L_{{\mbox{\small \sc lf}}}(s_1) =
     \Expct[L(s_1)] = (1-r)\cdot\loc_1
                    +r\cdot\left[\int_{0}^{\loc_1}(\loc_1-f)df+\int_{\loc_1}^{1}\loc_1 df\right]
                    =\loc_1-\frac{r\cdot\loc_1^2}2~.$$

-   **Calculating $\mathbf{R_{{\mbox{\small \sc lf}}}(s_1)}$:** if no failure occurs, or if the line is disconnected at location $0<f<\loc_1$ or $\loc_2<f<1$, then the length of the half-market is $R(s_1)=(\loc_2-\loc_1)/2$. If, on the other hand, an edge is disconnected at $\loc_1<f<\loc_2$, then $R(s_1)=f-\loc_1$. Therefore, $$R_{{\mbox{\small \sc lf}}}(s_1) =
        \Expct[R(s_1)]=
        (1-r)\cdot\frac{\loc_2-\loc_1}2+r\cdot\frac{\int_{\loc_1}^{\loc_2}(f-\loc_1)df}{\loc_2-\loc_1}=
        \frac{\loc_2-\loc_1}2~.$$

![Calculating the expected payoff in the two-player game with link failures.[]{data-label="fig:Hgame-2p"}](figs/Hgame-2p-Disconnections){width="\textwidth"}

It follows that the payoff of $s_1$ is $$p_{{\mbox{\small \sc lf}}}(s_1)=L_{{\mbox{\small \sc lf}}}(s_1)+R_{{\mbox{\small \sc lf}}}(s_1)=
    \loc_1-\frac{r\cdot\loc_1^2}2+\frac{\loc_2-\loc_1}2=\frac{\loc_2+\loc_1}2-\frac{r\cdot\loc_1^2}2~.$$ This function attains its maximum at $\loc_1=1/(2r)$. That is, as long as $s_1$ remains on the left of $s_2$, $s_1$ would move to $1/(2r)$. If $1/2 \leq \loc_2 < 1/(2r)$, then $s_1$’s best response would be pairing with $s_2$.

Note that $1/(2r)>1/2$ for every $0<r<1$ and thus if $\loc_1=1/(2r)$ and $\loc_2>1/(2r)$, then $s_2$ would prefer to move to the left of $s_1$. It follows that $\loc_2<1/(2r)$. By symmetry, it also holds that $\loc_1>1-1/(2r)$. Hence, condition \[p1\] holds, and as in the basic Hotelling game, the only equilibrium is when both servers are paired at the center of the line, i.e., $\loc_1=\loc_2=1/2$. 