RPS Equilibrium Conundrum
Clearly, it’s absurd that paper beats rock, but if rock beat paper then the game would become pointless.
Suppose we changed the rules such that paper only scores 1/2 point against rock. A full victory (rock against scissors or scissors against paper) scores 1 point, and a loss scores -1 point. Draws score 0 points. What mixed strategy is best in this game?
I found this equilibrium: p(rock) = p(paper) = 2/5, and p(scissors) = 1/5. If the opponent plays this strategy, then anything we do has an expected utility of 0. If both players use this strategy, then neither player has an incentive to change, so it’s an equilibrium.
This result seems strange to me. The rule change makes paper worse, and yet in the resulting equilibrium, we increase the probability of throwing paper. Who wants to explain this?
To clarify, rock against paper scores -1/2 point, right? (Otherwise your equilibrium is not an equilibrium.)
Here’s my explanation: Va n mreb-fhz tnzr, vg’f cbvagyrff gb gel gb jva. Gur orfg lbh pna qb vf gb cerirag gur bccbarag sebz jvaavat. Va guvf pnfr gung vaibyirf znxvat nyy bs uvf pubvprf unir rdhny rkcrpgrq fpberf. (Bgurejvfr ur pna jva ol cvpxvat gur pubvpr jvgu gur uvturfg rkcrpgrq fpber.) Fvapr cncre vf jrnxre, lbh arrq gb cynl vg zber bsgra gb cerirag ebpx sebz orpbzvat gur orfg pubvpr sbe lbhe bccbarag. Naq lbh arrq gb cynl ebpx zber bsgra gb onynapr bhg gur rkgen cncre, bgurejvfr fpvffbef orpbzrf gur orfg pubvpr sbe lbhe bccbarag.
Comment by Wei Dai — January 20, 2010 @ 6:32 am
The simple answer is that paper is not worse against the strategy (2/5,2/5,1/5). It produces a payoff of 0, same as rock or scissors. Moreover, paper is better than either rock or scissors against other strategies, like (1/2,1/3,1/6).
Here’s an intuition for the result: In RPS, the payoff to a strategy is higher if your prey is more common, and lower if your predator is more common. For the payoffs you gave, rock is “better” and paper “worse” against the baseline of (1/3,1/3,1/3), so we might expect more rock and less paper. But more rock and less paper is a bad environment for scissors to live in, so they begin dying out. As scissors dies out, the environment begins to look friendlier for paper and less friendly for rock. The paper population recovers, rock begins to die out, and when they are brought back into balance scissors begins to make a comeback. We end up at (2/5,2/5,1/5).
Comment by Jonathan — January 20, 2010 @ 12:10 pm
The simple version:
You have to throw a lot of paper in order to make it worthwhile for your opponent to throw scissors once in a while – and if your opponent never throws scissors, you can’t get the full benefits of throwing your new improved Super-Rock.
Comment by Doug S. — February 14, 2010 @ 5:37 am
The simple answer is that paper is not worse against the strategy (2/5,2/5,1/5). It produces a payoff of 0, same as rock or scissors. Moreover, paper is better than either rock or scissors against other strategies, like (1/2,1/3,1/6).
Here’s an intuition for the result: In RPS, the payoff to a strategy is higher if your prey is more common, and lower if your predator is more common. For the payoffs you gave, rock is “better” and paper “worse” against the baseline of (1/3,1/3,1/3), so we might expect more rock and less paper. But more rock and less paper is a bad environment for scissors to live in, so they begin dying out. As scissors dies out, the environment begins to look friendlier for paper and less friendly for rock. The paper population recovers, rock begins to die out, and when they are brought back into balance scissors begins to make a comeback. We end up at (2/5,2/5,1/5).
Comment by Bruce — May 19, 2010 @ 11:26 am