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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)?

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

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I imagine when you say "a pure strategy was best", you mean it was best given some strategy profile of all other players, i.e., it was a best response.

What you are saying would seem to be generally true when talking about best responses: There is always a pure strategy that is a best response. However, this does not mean that a mixed strategy cannot be a best response as well. If at least two pure strategies give the highest payoff given the strategies of the other players, then mixing over these strategies gives the same payoff.

In game theory we are usually interested in equilibrium, that is, that not only one player best responds to others for a given strategy profile of the others, but that all players play mutual best responses.

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.

So I think for equilibrium analysis your observation is of somewhat limited value. When you consider mixed strategy equilibria, your goal is to find a strategy profile so that all players are indifferent between their pure strategies, so they can mix these strategies without reducing their payoff. In other words, your observation does not guarantee that an equilibrium in pure strategies exists, only that a best response in pure strategies exists.

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