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I thought about an example game to use to illustrate what "Pareto-optimal" means, and I can't think of an outcome of Shotgun game (rock, paper, scissors) played by three players that would be Pareto-optiomal.

I was just wondering whether I'm not mistaking?

PS. I saw this question: Nash Equilibria for zero-sum games (Rock Paper Scissors) but it's about the overall optimal strategy, not Pareto-optimal.

The scoring is like this:

  • When you draw against another player, you get 0 points.

  • When you win against another player, you get 1 point.

  • When you loose against another player, you get -1 point.

  • The score is a sum of two outcomes of your match against two other players. Example: player A has scissors, player B has scissors and player C has rock, then players A and B have each 0 (from draw) - 1 (defeated by the rock), i.e. both have -1. Player C has 2 (since that player defeated both A and B and both victories give this player a +1).

Oh... but now that I've described it, I think that every such game would be Pareto-optimal, because improving one's scoring will necessary worsen the scoring of someone else. Ouch, sorry :(

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What are the rules? –  Arnaud Jun 2 '13 at 22:24
    
@Arnaud oh, I'll update the question now. –  wvxvw Jun 3 '13 at 6:02
    
Your question and your answer are both interesting ! –  Arnaud Jun 3 '13 at 7:23
    
@wvxvw: Why not either write up your answer and accept it (so that people can tell it has been answered), or delete it? –  robjohn Jun 3 '13 at 15:43
    
This was flagged as belonging on Cross-Validated, but it really does not seem suited to that site. There is no game theory site, so it should stay here (besides, it has been answered). –  robjohn Jun 3 '13 at 16:53
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It appears that I misapprehended the conditions I put for the problem, but after more thought it becomes apparent that, in fact any shotgun game is Pareto-optimal because whenever you improve a score of one of the players, you necessarily worsen the score of some other player.

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