(Hotelling’s voting model) Consider a population of voters uniformly distributed along the ideological spectrum from left (x = 0) to right (x = 1). There are two candidates i = 1,2 for a single ofﬁce and they choose their ideological positions $x_i$ $\in$ [0,1] simultaneously. Voters observe the choices of the candidates and then each voter votes for the candidate whose position is closest to the voter’s ideological position. If the candidates choose $x_1$ = .3 and $x_2$ = .6, for instance, then any voter to the left of 0.45 votes for candidate 1 and anyone to the right of 0.45 votes for candidate 2. Therefore, candidate 1 gets 45% of votes and 2 gets 55% so the winner is candidate 2. If both candidates choose the same position, the votes are split equally, and the winner is chosen by ﬂipping a fair coin. Suppose that the candidates care only about their chance of winning, and are not at all interested in ideology
Suppose there are three candidates. Show that in any pure strategy Nash equilibrium: (i) there is a player who wins for sure, (ii) the winner’s position is the most left or the most right among the three candidates, (iii) if the winner is in the most left, then the positions of both losers are strictly to the right of 1/2, and the position of the winner must be strictly between 0 and 1/2. (iv) For any x $\in$ (0,1/2), construct a pure strategy Nash equilibrium in which the winner’s position is x .
Sorry, since it's game theory, I wasn't sure where to post this, but it is very brainteaser-ish to me and I thought the math people might be able to take a stab at it