0
$\begingroup$

(Politics game) Consider a population of voters uniformly distributed along the ideological spectrum from left $(x = 0)$ to right $(x = 1)$.

Each of the candidates for a single office simultaneously chooses a campaign platform (i.e. a point on the line between $x = 0$ and $x = 1$).

The voters observe the candidates’ choices and then each voter votes for the candidate whose platform is closest to the voter’s position on the spectrum.

If there are two candidates and they choose platforms $x_1 =0.3$ and $x_2 = 0.6$, for example, then all voters to the left of $0.45$ vote for candidate $1$ and all those to the right vote for candidate $2$, and $2$ wins with $55$ percent of the vote.

Suppose that the candidates care only about being elected. If there are two candidates what is the pure strategy Nash equilibrium?

If there are three candidates exhibit a pure strategy Nash equilibrium.

$\textbf{Hint:}$ Assume that any two candidates that choose the same platform equally split the votes cast of that platform, and that ties among the leading vote-getters are resolved by coin flips.

What is the solution of the above game?

$\endgroup$
6
  • $\begingroup$ Hi, meta is not for asking questions, meta is for site discussion $\endgroup$
    – 1muflon1
    Mar 27 at 17:05
  • $\begingroup$ @1muflon1 sorry...do you want me to delete it? $\endgroup$ Mar 27 at 17:07
  • $\begingroup$ I migrated it, next time be more careful $\endgroup$
    – 1muflon1
    Mar 27 at 17:10
  • $\begingroup$ 1. What "given numbers"? 2. The game is not well-defined, you write "Suppose that the candidates care only about being elected." What exactly are the election rules? What happens if no one gets a majority? 3. This assumption is to make the game more well-defined. You have to specify the number of votes/election chances for every situation. You did this yourself when writing "If $x_1=x_2$ then the population of voters in evenly split among the candidates." $\endgroup$
    – Giskard
    Mar 27 at 17:35
  • $\begingroup$ @Giskard to avoid all such complexities...I re-edited the question and I believe everything is nicely defined. $\endgroup$ Mar 27 at 17:43

1 Answer 1

1
$\begingroup$

The two player version is the spatial competition variation of Hotelling's "Main Street"/"line city" model.

The three player version is slightly different, because instead of maximizing votes you try to get the most votes of the three players. The goal is close enough that the logic applied in the traditional model will show you that there is no pure strategy Nash-equilibrium in this case.


There are two properties of an equilibrium to consider:

  1. Is it possible that a left-most (or right-most) candidate is alone in his spectrum point in equilibrium?

  2. (For the three-player case) is it possible that three players are in the same spectrum point in equilibrium?

You can answer each point with a little thinking/experimentation, and these two properties are enough to completely characterize equilibra in the two, three player versions. (A little additional thinking will also get you to four players, and after that infinity is not too far off.)


If instead of thinking, you'd rather read:

A brief (perhaps too brief) answer for the two player version is given here.

There are also several papers on this topic, e.g.;

"The Principle of Minimum Differentiation Reconsidered: Some New Developments in the Theory of Spatial Competition" by Eaton and Lipsey (1975)

"Hotelling's "Main Street" with More Than Two Competitors" by Economides (1993).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.