# Proving there are infinitely many strictly dominant strategies My attempt:

$$(0, 1/3, 2/3)$$ works as a strategy that strictly dominated $$L$$.

To prove that there are infinitely many of them:

Let $$A_n=(0,1/2+\epsilon_n,1/2-\epsilon_n)$$ where $$\epsilon_n=1/n$$ for $$n=1,2,3,...$$. We can see that for $$n > 10$$, $$A_n$$ strictly dominates L. So, since $$n$$ diverges, there are infinitely many $$n$$ for which $$A_n$$ is a strictly dominant strategy.

I chose $$n> 10$$ since $$2<2.5-5\epsilon_n \Leftrightarrow 0,1> \epsilon_n$$ and $$1/n<0,1$$ for $$n>10$$. The other inequalities don't require $$\epsilon_n$$ to be restricted.

Would this be a correct way to prove this?

• Sir, you are posting a lot of questions in a short time period. Perhaps try to work them out a bit longer yourself. – Giskard May 8 at 16:41
• Also, you are posting very specific questions. These types of questions are unlikely to be of help to future visitors, which is the primary goal of SE sites. – Giskard May 8 at 16:41
• Lastly, questions of the "Is this correct" type are a poor fit for the SE format. – Giskard May 8 at 16:42
• @Giskard sorry, coming from Math SE there's a special tag named "Solution-verification", and there such questions are normal. (At least I assume so since no one told me otherwise). Also, I work on these questions quite a bit, and if I get stuck or need feedback, considering there are no answers in my textbook for these, this is the only site I can refer to – The Poor Jew May 8 at 16:50

That is a fine proof. Though you probably want to show your last statement "The other inequalities don't require $$\epsilon_n$$ to be restricted". It is fairly obvious, but won't hurt to be super clear.
Perhaps a simpler proof would be that any convex combination of the two strategies already found will also work. That is $$\lambda(0,1/2,1/2)+(1-\lambda)(0,1/3,2/3)$$ also strictly dominates $$L$$ for any $$\lambda\in(0,1)$$. So, in fact, there are not only countably many strategies that strictly dominate $$L$$ but uncountably many of them.