# Can we ignore certain strategies when looking for mixed-strategy Nash equilibria?

I run into this problem in my attempt to answer another question on math.stackexchange

We have a 2-player zero-sum game with an infinite set of strategies.

Imagine that we have one class of infinite strategies, let's call them $H_b$, where $b \in [0,1]$. You can view $b$ as the parameter that tweaks the general class to a specific strategy. Imagine also that we have another class of infinite strategies, call them $R_a$ with $a \in [0,1]$. $R_a$ strategies are "inferior" to $H_b$ strategies. More specifically, if we denote as $G(x,y)$ the gain of strategy $x$ over strategy $y$, then strategies $R$ are inferior to $H$ in the sense that $G(H_b, R_a) > 0, \space \forall a \in [0,1], \forall b < \frac{2}{3}$. In other words, any strategy in the first two thirds of strategy class $H$, wins all of the strategies in strategy class $R$.

My question is: Should I include strategies $R$ when I am looking for a mixed-strategy equilibrium, or can I safely ignore them?

EDIT: I know that if a strategy is strictly dominated we can safely ignore it when looking for a mixed strategy equilibrium. But strategies $R$ are not strictly dominated (if I understand the term correctly). Strictly dominated would mean that some strategy Y provides higher gains than $R_a$ for all possible opponents. This does not happen in our case. Let's choose any strategy $H_b, b \le 2/3$ to be our potential strictly dominant strategy over all strategies $R$. We can always find strategies $H_c, c> 2/3$ and $R_a$ such that $G(H_b,H_c)<0$ but $G(R_a,H_c)>0$. In other words, there exists an opponent, namely strategy $H_c$, that $R_a$ can perform better than $H_b$. For example, let's examine if strategy $H_{0.5}$ is strictly dominant over $R$. We can find that strategy $H_{0.8}$ wins over $H_{0.5}$, but $H_{0.8}$ loses over $R_{0.4}$. So, for opponent $H_{0.8}$, $R_{0.4}$ is better than $H_{0.5}$, and thus (by definition) not strictly dominated by $H_{0.5}$.

To complete the proof that no strategy in $H$ can dominate over all of $R$, let's consider strategies $H_b, b > 2/3$ to be our potential strictly dominant strategy. We can easily show that this cannot be, as we can find strategy $R_a$ such that $G(H_b, R_a) < 0$ and we know that $G(R_a, R_a) = 0$.

Even if strategies $R$ are not strictly dominated, can we use/exploit the "inferiority" of strategies $R$ when looking for a mixed strategy equilibrium?

I think there is a way out of having to consider strategies $R$ (even if theory says we should include them). If we find the mixed strategy equilibrium including only strategies $H$ then we can check if the found mixed strategy is better than every $R$ strategy. If it is, then we are clear. Is my reasoning correct?

EDIT 2: I found something that slightly simplifies the problem, but does not change my crux of my question. I found that strategy $R_1$ strictly dominates over all other strategies $R$. First I noted that $G(R_b, R_a) = \frac{b-a}{2}$, so $b=1$ gives you the best possible result if we restrict our choices in $R$ strategies. Furthermore, if we also consider $H$ strategies we can show that: $$G(R_1, H_b) > G(R_a, H_b), \forall b \in [0,1], \forall a \in [0,1)$$ So we can discard all other $R$ strategies and only keep $R_1$. The main question remains though, because $R_1$ is not dominated by any strategy $H$.

Assume I find a mixed strategy considering only strategies $H$ and then find the mixed strategy's gain against $R_1$. If this gain is positive, can I now ignore $R_1$?

• If I told you that strictly dominated actions were never played with positive probability in mixed nash equilibria (which is true), would that answer your question? Sep 28, 2016 at 20:42
• @Shane, this is the crux of my question and confusion. As far as I understand, strategies $R_a$ are not strictly dominated. Strictly dominated would mean that some $H_b$ strategy would provide higher gains than $R_a$ for all possible opponents. This does not happen in my case. For example, strategy $H_{0.8}$ wins over $H_{0.5}$, and strategy $H_{0.5}$ wins over $R_{0.1}$ but $R_{0.1}$ wins over $H_{0.8}$. I'll edit my question to make this part clearer. Sep 29, 2016 at 2:53
• You are comparing strategies of two different players. I understand that in your original question the game is symmetric, but here it seems that the question implicitly assumes the game is symmetric (and zero sum) without stating it. Sep 29, 2016 at 18:35
• For infinite games, I do not know if it works, but my guess is what you are looking for is elimination of never best responses. See, Theorem 11 in homepages.cwi.nl/~apt/stra/ch4.pdf Sep 29, 2016 at 18:38
• @SergioParreiras, thanks for the suggestions. I edited the question to make it clearer, and in the process I also discovered a fact the simplifies the problem a little. But my main question stands. Thank you for the reference. Yes, I think that $R_1$ can be named as "never best response", but I am not sure the theorem holds because the game is not finite. How about the other proposal I make (to exclude and then test)? Does it make sense? Sep 30, 2016 at 7:28