Nash Equilibria in Median Voter Theorem for More Than 2 Candidates

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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




game-theory nash-equilibrium




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    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




    game-theory nash-equilibrium




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      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




      game-theory nash-equilibrium




      share|cite|improve this question











      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





      game-theory nash-equilibrium





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