In Game Theory, why is it never correct to choose a strategy that isn't a best response to some belief about what the other player will do?

Question

In this lecture, it is stated, "the more general lesson, do not choose a strategy that is never a best response to any belief." A soccer example is used to conclude that the kicker should not kick to the middle.

Because at no point along the x-axis is the blue line greater than or equal to both the green line and the red line, the theory says that we can discard the blue line as a viable strategy.

This seems to be taken as obvious, but to me it is not. What if, instead, we were dealing with the following sort of situation?

Here the blue strategy (kicking to the middle, in the example) is slightly worse than the best response to each belief (perhaps even more infinitesimally so than I've drawn it), but much better than picking the worst response to almost all beliefs. What is the intuition for discarding the never-best-response blue strategy even in such a case? It seems to me that blue should be the best strategy overall in my modified example.

Answer 1

First a comment that may deepen your understanding of the subject: in this situation the $M$ strategy's blue graph has to be a line, it cannot be like the one represented in your image.

The expected payoff is a linear function of the belief $p$ of the player: $$ EU(M, p) = p \cdot U(M,l) + (1-p) \cdot U(M,r). $$ Thus as a function of $p$, it can be plotted as a line, one that connects the points $(0,U(M,r))$ and $(1,U(M,l))$.

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Now to your question: if for all beliefs a strategy $s$ is not a best response, then it can be shown that a (probably mixed) strategy exists that strictly dominates $s$. In your example this would be the strategy where the player plays $L$, $R$ with 50% probability each. If you would plot the expected payoff of this strategy it would be a horizontal line that crosses the point where the red and green lines intersect. As a result, this mixed strategy would always yield an expected payoff that is slightly higher than $M$'s.