# Games for which a "convexity argument" work?

A few days ago I was reading a text which introduced a simple cooperative game, played by Alice and Bob, whose analysis began by showing a pure strategy was the best.

They did this by first imagining that they employed a mixed strategy and pointed out that definitionally this is just a weighted-average of pure strategies. Thus, the win-probability of the mixed strategy is just some weighted average of the win-probabilities of the pure strategies. Since it's an average, there must be some pure strategy whose win-probability was at least as large as this average so, without loss of generality, we can assume the optimal strategy for the game is a pure one.

Upon first inspection, I felt this argument was extremely wide-reaching but after speaking with some friends I now think that it may apply to any cooperative game with finite moves. This is because the set of pure strategies can be seen as a set of points in a kind of "Game space" or parameter space of one's choices and then the set of mixed-strategies will be the convex hull of these points and then the payoff, being a linear combination of the deterministic payoffs, is a linear function of the parameters within the convex set. Such a function is maximised at a vertex of the this set and a vertex, we already said, is a deterministic strategy. For this reason (and a mysterious connection to centres of gravity of convex objects), I call this a "convexity argument".

Are these the only games for which this argument is valid? Are there more or are there actually fewer (i.e. my above argument assumes something about a general cooperative game which isn't true)?

• It could be related to a class of games called supermodular games. Commented Apr 24, 2020 at 9:50

Why is this important? Well, in some games no pure strategy equilibrium exists. Consider the matching pennies game. Both players can play Up or Down. Payoffs for player 1 are $$\pi_1(u,u)=\pi_1(d,d)=1$$ and $$\pi_1(u,d)=\pi_1(d,u)=0$$. Payoffs for player 2 are $$\pi_2(u,u)=\pi_2(d,d)=0$$ and $$\pi_2(u,d)=\pi_2(d,u)=1$$.
In this game, there is no pure strategy equilibrium. The only Nash equilibrium is in mixed strategies where both players play $$u$$ and $$d$$ with probability 0.5 each. Hence, by definition of a Nash equilibrium, both players have a best response that is in mixed strategies. This does not invalidate your claim above: There is also another best response for each player which is a pure strategy, but a pure strategy cannot be part of an equilibrium in this game, since this would make players too predictable.