Suppose there are two political candidates, each trying to receive as many votes as possible. There are 10 political positions they could choose, from #1 to #10. The voters are evenly distributed across these 10; 10% at #1, 10% at #2, and so on. The candidates must simultaneously, independently choose a position, and the voters will vote for the candidate whose position is closest to their own (if there is a tie, the votes will split evenly).
The median voter theorem says that the candidates should choose the median, i.e. #5 or #6. This is fairly easy to demonstrate with the game theory technique of deleting dominated strategies. #1 is always worse than #2, and can be eliminated. If #1 is never chosen, then #2 is always worse than #3, and can be eliminated. Then #3 is always worse than #4, and so on - the outermost positions are eliminated until only #5 and #6 remain.
So much for two candidates. What happens with three? #1 and #10 are weakly dominated, but that is the only simple conclusion. There are 10^3 = 1000 possible outcomes (although the first half and the second half are symmetric), and I'm not sure how to make this problem manageable. Out of desperation I once entered all 1000 outcomes into Gambit and got an equilibrium which does work. It suggests there might be a way to eliminate #1, #2, #9 and #10. But it is asymmetric; since it is a symmetric game, shouldn't there be a symmetric equilibrium? How can it be identified?
Gambit Equilibrium:
First candidate
Choose 5: 56.58%
Choose 7: 43.28%
Choose 6: 0.14%
Second candidate
Choose 6: 41.00%
Choose 3: 22.19%
Choose 4: 19.65%
Choose 8: 17.15%
Third candidate
Choose 7: 40.49%
Choose 4: 35.40%
Choose 5: 16.04%
Choose 3: 8.07%
Payoffs (average percentage of votes)
First candidate: 34.19
Second candidate: 33.37
Third candidate: 32.43
~~Critical Update~~
I found the unique symmetric Nash equilibrium for three players!
The Game's Payoffs
For the sake of clarity, I will offer a few examples to illustrate how the payoffs are set up. I have considered the game in which one's payoff is proportional to the percentage of votes they get; I have not considered the game in which one's payoff is proportional to their chances of winning the election, which is also worth considering, and easier, and must differ in some respects.
The payoffs must always have the same sum, such that if every player chooses the same strategy, their individual payoffs will be equal to 1/3 of the sum.
If you choose, say, position 4, and your opponents both choose position 5, your payoff will be equal to the four out of ten positions you won, i.e., 2/5, and your opponents will split the remaining six positions, each receiving 3/10.
If you choose 1 (which I do not advise, but the illustration will suffice), and one opponent chooses 3, and the other chooses 7, we will find that certain positions will be subject to a two-way split. Namely, position 2 will be split between you and one opponent, and position 5 will be split between your opponents. In this case, each receives an equal share in that position, and thus we can calculate our payoff as one position out of ten plus half a position, or 3/20. Your opponents will receive 3/10 and 11/20, respectively.
Finally, if you choose position 4 and both your opponents choose position 8, position 5 will be subject to a three-way split. In this case, your payoff will be five out of ten positions, plus one-third of a position - in total, 1/2 + 1/30 = 8/15. (Your opponent's individual payoffs are 7/30.)
Previous Efforts
Unlike the similar game for two players, this game is much too complicated to analyse by hand. The only way forward that I have found is to undertake the laborious task of describing it in full to a computer program. As mentioned above, I originally chose Gambit, a game theory tool which does a very able job of finding Nash equilibria for any game whose payoff matrix can be given in full. It does have its limits, but I expected that it could give me at least some equilibrium. When I finished the undertaking of 3000 manual calculations to describe the payoff matrix, I asked for the quantal response equilibrium (QRE) and it gave me an approximation to the asymmetric equilibrium given above.
I made sure to verify by independent means that this was indeed an equilibrium, but I still don't know why it would give me an asymmetric equilibrium for a symmetric game. How does it distinguish between the players, or between the symmetric strategies 5 and 6, 4 and 7, 3 and 8? Maybe I am overlooking something about the nature of quantal response equilibrium; or, perhaps more likely, I made a small mistake somewhere in my 3000 calculations, which was not, however, enough to result in a measurably different equilibrium.
New Methods
To obtain the symmetric equilibrium, I used Mathematica to specify a system of equations. I was able to make two assumptions which simplified the process to 500 calculations: In the symmetric equilibrium, (1) every player employs the same strategy, and (2) the probability of playing a strategy is equal to the probability of playing its pair; for example, position 4 and position 7 must be played with equal probability.
Once each strategy's payoffs have been defined in terms of the opponent's probabilities of playing each strategy, Mathematica can (provided the answer to the query is not stupendously difficult) give the conditions under which one strategy is better than another, or the two are equal. Since the nature of mixed strategy Nash equilibria is being indifferent between the strategies employed in the mixture, this is invaluable. For example, if it is found that there is no situation in which two strategies can be considered equal, they cannot both be part of the equilibrium.
By constructing these equations, I was able to determine the following for the case in which the opponents must play symmetrical strategies (I wanted to consider the whole game, but this proved too much for even Mathematica to handle (at least, with the way I framed the question)):
Position 1 (and 10) is strictly dominated by every other strategy.
Position 2 (and 9) is strictly dominated by 4/7 and 5/6, and weakly dominated by 3/8 (they are equal against 1/10, which is itself dominated).
There is no dominance relation between positions 3/8, 4/7 and 5/6.
Therefore, the symmetric equilibrium, just like the asymmetric equilibrium, must be some mixture between positions 3 to 8.
But this much was expected. It is time to reveal the unique answer Mathematica gave.
The Symmetric Equilibrium
Positions 3 and 8 are each played with the following probability:
$\frac{1}{780} (277 - \frac{15421}{(2987531 - 5850 \sqrt{153645})^{\frac{1}{3}}} - (2987531 - 5850 \sqrt{153645})^{\frac{1}{3}})$
$0.018340620407844103926308393051783287215179749928538061332...$
Positions 4 and 7 are each played with the following probability:
$-\frac{25}{156} + \frac{1}{312} (1118944 - 2808 \sqrt{153645})^{\frac{1}{3}} + \frac{1}{156} (139868 + 351 \sqrt{153645})^{\frac{1}{3}}$
$0.342260202632860679061642533008242882389659507825233421522...$
Positions 5 and 6 are each played with the following probability:
$\frac{1}{390} (119 - \frac{6431}{2340 \sqrt{153645} - 758503)^{\frac{1}{3}}} + (2340 \sqrt{153645} - 758503)^{\frac{1}{3}})$
$0.139399176959295217012049073939973830395160742246228517145...$
And the average payoffs are, necessarily, 1/3 each.
It is possible to apply the same method to solve the four-player game. However, that would involve 5000 manual calculations - $5*10^{n-1}$.