# Adjusting payoffs has unintuitive effect on optimal strategies in bimatrix games

Consider the game Rock-Paper-Scissors. If we award a win with $$1$$, a loss with $$-1$$ and a draw with $$0$$, we get the following bimatrix game (with rewards ordered as row player, then column player):

R P S
R $$0,0$$ $$-1,1$$ $$1,-1$$
P $$1,-1$$ $$0,0$$ $$-1,1$$
S $$-1,1$$ $$1,-1$$ $$0,0$$

Obviously this has a mixed strategy Nash equilibrium of $$(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}), (\frac{1}{3}, \frac{1}{3}, \frac{1}{3})$$, i.e. both players choose at random, and both players have an expected reward of exactly $$0$$.

Suppose I wanted the row player to favor Rock. I might give it a higher reward (2 instead of 1) if it wins with Rock, like so:

R P S
R $$0,0$$ $$-1,1$$ $$2,-1$$
P $$1,-1$$ $$0,0$$ $$-1,1$$
S $$-1,1$$ $$1,-1$$ $$0,0$$

However the effect this has on the Nash equilibrium is completely opposite my intentions: the row player's strategy in equilibrium is still $$(\frac{1}{3}, \frac{1}{3}, \frac{1}{3})$$, but now the column player's strategy in equilibrium $$(\frac{4}{12}, \frac{5}{12}, \frac{3}{12})$$. The row player's expected reward goes up to $$\frac{1}{12}$$ and the column player's reward stays at $$0$$.

Mathematically, I know this is the result I'm getting, but it seems completely unintuitive to me. Not in the least because the column player's reward didn't actually change: if the column player has the mixed strategy $$(\frac{1}{3}, \frac{1}{3}, \frac{1}{3})$$, its reward is also $$0$$. I get that this would allow the row player to increase its reward drastically (by always picking Rock, for example), which in turn would incentivize the column player to pick Paper more often. But the end result is that the column player "changes" its strategy purely to "prevent" the row player from gaining more reward. This feels contradictory with the fact that in a non-zero-sum game, both players play to maximize their own reward, not to minimize the reward of the other player(s).

Intuitively, I would say there could also be an equilibrium where both players have a positive expected reward, the row player because they pick Rock slightly more often, and the column player because they pick Paper slightly more often. Why isn't there?

In fact, as I'm writing this I've thought of something else. If you were to play a meta-game where both players pick their strategy as their first move and receive their expected reward with that strategy as their reward in the meta-game, this possibility for cooperation does appear:

$$(\frac{1}{3}, \frac{1}{3}, \frac{1}{3})$$ $$(\frac{4}{12}, \frac{5}{12}, \frac{3}{12})$$
$$(\frac{1}{3}, \frac{1}{3}, \frac{1}{3})$$ $$\frac{1}{9}, 0$$ $$\frac{1}{12}, 0$$
$$(1, 0, 0)$$ $$\frac{1}{3},0$$ $$\frac{1}{12},\frac{1}{6}$$

Of course there are many mixed strategies missing from this meta-game, which explain why $$(1,0,0), (\frac{4}{12}, \frac{5}{12}, \frac{3}{12})$$ isn't an equilibrium. But if players can cooperative, shouldn't there be an equilibrium where they do?