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If we have two actions $a_1$ and $a_2$ that in an equilibrium mixed strategy should be mixed $70\%/30\%$ with both having an expected value of say $10$. What happens to the expected values of the actions if we $100\%$ of the time do action $a_1$?

My thinking was that the expected value of the action we always do $a_1$ will go down?

Edit: Practical example. In a hand of poker, we calculate the optimal strategy, which is a mixed strategy between betting our hand for size $A$ $70\%$ of the time and size $B$ $30\%$ of the time. The expected value of these two bets is equal since it is a mixed strategy. My question then is by deviating from this optimal strategy, what would happen to the expected values. My intuition was that if we start betting our hand for size $A$ $100\%$ of the time instead of $70\%$, then the expected value of this action would go down? But maybe it stays the same, that's what I'm asking.

Thanks a lot

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  • $\begingroup$ Remember that expectation is a linear operator, so if you have $\mathbb{E}[X]$ and $\mathbb{E}[Y]$ then $$ \mathbb{E}[aX + bY] = a\mathbb{E}[X] + b \mathbb{E}[Y] $$ for any constants $a,b$. $\endgroup$
    – Matti P.
    Commented Nov 10, 2020 at 12:16
  • $\begingroup$ But anyway, there's something I don't quite understand about your question. You say that both actions have the same expected value, equal to 10. So do you mean that $$ \mathbb{E}[A_1] = 10 \qquad \text{and} \qquad \mathbb{E}[A_2] = 10 $$ or $$ \mathbb{E}[0.7A_1 + 0.3 A_2] = 10 $$ ? $\endgroup$
    – Matti P.
    Commented Nov 10, 2020 at 12:18
  • $\begingroup$ Yes i mean since in mixed strategies all actions have the same expected value but should be performed different amount of the times $\endgroup$
    – wuannnn
    Commented Nov 10, 2020 at 12:18
  • $\begingroup$ If you have different random variables with the same expected value (10), then a linear combinations of those (so that the factors add to 1) also has an expected value of 10. That's due to the linear nature of expected value. $\endgroup$
    – Matti P.
    Commented Nov 10, 2020 at 12:19
  • $\begingroup$ Ok, so doing action a1 100% instead of 70% of the time as suggested by the equilibrium it will still have an expected value of 10? $\endgroup$
    – wuannnn
    Commented Nov 10, 2020 at 12:21

1 Answer 1

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If you mix actions, it means that you are indifferent between them so, given the other player does not change his strategy, you might as well play just one of them $100\%$ of the time and get the same payoff. However, if the other player responds, then there might be multiple scenarios and it also depends whether you consider a new equilibrium or just the payoff of the new strategy profile.

For example, in the "matching pennies" game, if you play pure then the other player best responds and your payoff is lower. But in such a game, there is no pure NE, and you can best respond to his best response and so on.

On the other hand, in a game such as "Battle of the Sexes", if you play purely and the other player best responds, you both arrive at one of the pure equilibria of the game, where the payoff is strictly higher than the mixed one.

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