I recently read Leyton/Brown's Essentials of Game Theory and I keep coming back to this (Sec. 2.2):
When the support of a best response $s∗$ includes two or more actions, the agent must be indifferent among them—otherwise, the agent would prefer to reduce the probability of playing at least one of the actions to zero. But thus any mixture of these actions must also be a best response, not only the particular mixture in $s∗$. Similarly, if there are two pure strategies that are individually best responses, any mixture of the two is necessarily also a best response.
This is followed by an example on the next page of using this observation as the basis of computing a mixed-strategy Nash equilibrium.
I get it, but I don't get it "in my gut", which is troubling. It just seems to sound highly counter-intuitive that having a "best response" implies that we are thus "indifferent" in our response. I've tried to make this feel natural, but I haven't succeeded.
Does anyone have a better explanation/analogy/picture by which this can be made memorable and intuitive?
Sure. Let's make a simple example. What if you and I both, on count of three, put down a penny on the table. If the faces match, I pay you one dollar, and if they don't, you pay me one dollar. Moreover, let's say you know I am randomizing 50-50 heads vs. tails. Given I am doing that: $$U_{\textrm{You}}(\textrm{Heads}) = \underbrace{1\cdot (0.5)}_{\textrm{Payoff if I end up heads}} -\underbrace{1\cdot (0.5)}_{\textrm{Payoff if I end up tails}} = 0$$
$$U_{\textrm{You}}(\textrm{Tails}) = \underbrace{1\cdot (0.5)}_{\textrm{Payoff if I end up tails}} -\underbrace{1\cdot (0.5)}_{\textrm{Payoff if I end up heads}} = 0$$
But moreover, any mixture for you also gets you zero expected profit!