# Nash equilibria for “presidential election” game

Suppose, in a country there are $$m$$ different social issues, positions on which are being indexed with numbers $$[-1; 1]$$, with radicals on the opposing ends and moderates in the center. In this country there are also $$n$$ candidates running for president. Before they run, they have to choose their position on all those social issues by placing themselves into some point of this $$[-1;1]^m$$ «political compass». Suppose they have chosen their positions $$a_1, … , a_m$$ (as vectors in $$\mathbb{R}^n$$). Then define $$Pref(x)$$ as the set of the preferable candidates for voters with position $$x$$ (the candidates, the Euclidean distance from whose position to $$x$$ is minimal). Suppose, the voter positions are uniformly distributed on $$[-1;1]^m$$ and that each voter in the position $$x$$ chooses one of the candidates from $$Pref(x)$$ with equal probability. (So the total percentage of votes received by $$k$$-th candidate is $$\int_{[-1;1]^m} \frac{I_{Pref(x)}(k)}{2^m|Pref(x)|}dx$$). After the votes are counted, the president is chosen at random with equal probability from those candidates, that received the maximal numbest of votes. The candidates do not have their own opinion on those $$m$$ issues and only want to win the election (thus their final payoff is the exact probability of them winning). Is there some sort of classification of Nash equilibria in this class of games (for different $$m$$-s and $$n$$-s)?

For $$n = 1$$, every position is a Nash equilibrium as the only candidate will win the election no matter what they do.

For $$n = 2$$ both candidates should take the extremely moderate position that is $$0$$. That is due to the «geometric» fact, that if the positions of two candidates are distinct, then the one who is closer to the center wins. So if $$a_1 \neq 0$$, then the second candidate can adopt the position $$\frac{a_1}{2}$$ to win the election for sure, and vice versa.

However, I do not know, how to deal with the general problem.

This question on MSE. There is also a very useful partial answer by @antkam, with analysis of the particular case when $$m = 1$$ and $$n$$ is even.