Application of Statistical Mechanics and Density of States to Rock Paper Scissors

I recently read an article from a leading poker theoretician who hypothesised that a strategy having lower exploitability or smaller distance to Nash Equilibrium would statistically mean that the player gain edge vs a player with larger Nash distance. His argument was very much in the way of entropy saying that in a 5% exploitable strategy the system would tend to it’s most probable state where the more accurate player would gain EV. I decided to dive deeper into the idea and apply counting techniques from the density of states in statistical mechanics to either prove or disprove his theory.

Instead of poker I am going to try to solve this problem for the simpler problem of rock, papers scissors. For this rock paper scissors game I will limit strategies to having only integer frequencies. This is I am confident is a reasonable simplification for the problem and this assumption is further motivated by the final solution. To be specific a strategy of 30,30,40 is allowed but 30.5, 30.5, 39 is not. This does mean 1/3, 1/3, 1/3 is not allowed however this is not a particularly relevant point when we talk about strategies a distance from Nash. Playing 1/3,1/3,1/3 cannot possibly be +EV against any other strategy is a good indicator that being closer to equilibrium does not actually increase our chances of winning. We could also make the further argument that deviating epsilon from Nash could not possibly make more than delta vs any other strategy .

I will briefly define what I mean by exploitability just in case it is not the common usage. A strategy of 33, 33, 34 has an exploitability of 1%. If it is 34% rock then optimal exploit is 100% paper which wins .34 and splits .33=.165 , This would yield .505 of the pot which I will define to be 1% exploitability. This ensures that if a player gains 100% of the pot the opponent’s strategy was 100% exploitable. Therefore exploitability is defined in terms of percentage of the original pot share that the exploiting player has gained. In this case the player was originally entitled to .5 of the pot but now is entitled to .505 pot corresponding to 1%.

One can in fact convince themselves that 33, 33 ,34 is the only possibly strategy with 1% exploitability. This corresponds to 3 possible combinations however one can assume without loss of generality that rock is player 34%. We can now take a look at 100% exploitable strategies. This is 3 combinations of 100, 0, 0. In the case of 100 rock then Player A (34,33,33) is entitled to .34(50) + (.33)(100)+(.33)(0) = .50. In the case of 100 paper then Player A is entitled to (.34)(0)+(.33)(50)+(.33)(100)=49.5 and in the case of 100 scissors then Player A is entitled to (.34)(100)+(.33)(0)+(.33)(50)=50.5. Hence one can see if each of these strategies is taken 1/3 of the time the players break even despite player A having a much less exploitable strategy. One may take this as a special case and so I have also calculated values for 5% exploitability.

In this case there are 2 possible options for 5% exploitable strategies. 35, 35, 30 with 3 combinations and 36, 33, 31 with 6 combinations. One can again assume Player A plays rock 34% of the time and calculate his EV against each of these 9 strategies. This does indeed give player A 50% pot share.

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One can see these solutions have a highly symmetric structure. If one strategy gains x EV another strategy will gain -x EV. Poker is of course a less symmetric game than rock paper scissors but I would be surprised if this result no longer holds. Given that the solution to the problem gives rise to a fairly obvious symmetry it is quite clear that 50% pot share will be common to any sort of exploitability ratio even for non integer strategic frequencies. Therefore the system does not evolve into being favourable for one player but rather the symmetries of the game constrain both player to an even pot share providing both players remain informationally neutral.