# Different payouts of pure strategies in mixed strategies

I have a question with mixed strategies. The question is as follows, if we're in a strategy profile that is a Nash equilibrium and a player is playing a mixed strategy, can the pure strategies that are part of such a mixed strategy give different payoffs?

My first impression is no. If, for example, we consider a game such that the mixed strategy assigns $$\mu$$ to pure strategy $$A$$ and $$1-\mu$$ to pure strategy $$B$$ and if $$A$$ gives higher payouts than $$B$$, then another mixed strategy that gives probability 1 to playing $$A$$ and 0 playing $$B$$ would give higher payoffs for said player, contrary to it being a Nash equilibrium.

I don't know if I'm right or if I have a mistake, I would really appreciate your comments or if you have a better way of arguing it or have a counterexample.

Update: For example, in matching pennies, we are in an equilibrium when we play the strategy (0.5,0.5) for each player and yet the payoffs of the associated pennies are different. I don't know if this serves to exemplify that it is possible.

• Yes, I think you should generally not delete answered questions. If they are poor questions they are supposed to be closed/deleted by the community. Apr 11 at 16:45
• Done, an apology, on other stack exchange sites they have sometimes closed me, they ask like that.
– Haus
Apr 11 at 16:46

The point is that expected payoffs are linear ina a player's own strategy: $$\sum_{(s_1,\ldots,s_n)}\sigma_1(s_1)\cdots\sigma_n(s_n)~u_i(s_1,\ldots,s_n)=\sum_{s_i}\sigma_i(s_i)\sum_{(s_1,\ldots,s_{i-1},s_{i+1},\ldots,s_n)}u_i(s_1,\ldots,s_n).$$ So the player's expected payoff will be a weighted average of the payoffs of the pure strategies, and it can only be a best response if it mixes between pure best responses.