This is a problem of choosing the best position
In Game Theory this problem is known as the Median Voter Theorem.
Suppose we have two candidates $s_1$ and $s_2$
Utilities
So there can be the following scenarios
Case 1: $s_1 < s_2$
Case 2: $s_1 = s_2 < 0.5$
Case 3: $s_1 = s_2 = 0.5$
But there is no Nash Equilibria for three candidates
Consider this
We suppose (without loss of generality) that
Utility functions
$s_a < s_b < s_c$ | $\left\{\begin{matrix}
u^{(1)}_a(s_a, s_b, s_c) = \cfrac{s_a + s_b}{2} \\ u^{(1)}_b(s_a, s_b, s_c) = 1 - \cfrac{s_b + s_c}{2} \\ u^{(1)}_c(s_a, s_b, s_c) = \cfrac{s_b + s_c}{2} - \cfrac{s_a + s_b}{2} \\ \end{matrix}\right.$ |
$a$ may deviate: $u_a(s_a + \epsilon, s_b, s_c) > u_a(s_a, s_b, s_c)$ | |
$s_a < s_b = s_c$ | $\left\{\begin{matrix}
u^{(2)}_a(s_a, s_b, s_c) = \cfrac{s_a + s_b}{2} \\ u^{(2)}_b(s_a, s_b, s_c) = \cfrac{1 - \cfrac{s_a + s_b}{2}}{2} \\ u^{(2)}_c(s_a, s_b, s_c) = u^{(2)}_b(s_a, s_b, s_c) \\ \end{matrix}\right.$ |
$a$ may deviate: $u_a(s_a + \epsilon, s_b, s_c) > u_a(s_a, s_b, s_c)$ | |
$s_a = s_b < s_c$ | $\left\{\begin{matrix}
u^{(3)}_a(s_a, s_b, s_c) = \cfrac{s_b + s_c}{2 \cdot 2} \\ u^{(3)}_b(s_a, s_b, s_c) = u^{(3)}_a(s_a, s_b, s_c) \\ u^{(3)}_c(s_a, s_b, s_c) = 1 - \cfrac{s_b + s_c}{2} \\ \end{matrix}\right.$ |
$c$ may deviate: $u_c(s_a, s_b, s_c - \epsilon) > u_c(s_a, s_b, s_c)$ | |
$s_a = s_b = s_c \ne 0.5$ | $\left\{\begin{matrix}
u^{(4)}_a(s_a, s_b, s_c) = \cfrac{1}{3} \\ u^{(4)}_b(s_a, s_b, s_c) = \cfrac{1}{3} \\ u^{(4)}_c(s_a, s_b, s_c) = \cfrac{1}{3} \\ \end{matrix}\right.$ |
$a$ may deviate: $u_a(s_a + \epsilon, s_b, s_c) > u_a(s_a, s_b, s_c)$ | |
$s_a = s_b = s_c = 0.5$ | $\left\{\begin{matrix}
u^{(4)}_a(s_a, s_b, s_c) = \cfrac{1}{3} \\ u^{(4)}_b(s_a, s_b, s_c) = \cfrac{1}{3} \\ u^{(4)}_c(s_a, s_b, s_c) = \cfrac{1}{3} \\ \end{matrix}\right.$ |
$a$ may deviate: $u_a(s_a + \epsilon, s_b, s_c) > u_a(s_a, s_b, s_c)$ |
So in all cases there is somebody who wants to deviate:
This is the allocation problem: