<p>That is a fine proof. Though you probably want to show your last statement "The other inequalities don't require <span class="math-container">$\epsilon_n$</span> to be restricted". It is fairly obvious, but won't hurt to be super clear.</p>

<p>Perhaps a simpler proof would be that any convex combination of the two strategies already found will also work. That is <span class="math-container">$\lambda(0,1/2,1/2)+(1-\lambda)(0,1/3,2/3)$</span> also strictly dominates <span class="math-container">$L$</span> for any <span class="math-container">$\lambda\in(0,1)$</span>. So, in fact, there are not only countably many strategies that strictly dominate <span class="math-container">$L$</span> but uncountably many of them. </p>

<p>(Idea of the proof:) Notice that this holds because strict dominance is given by a set of inequalities (3 in this example) so if you have two strategies that satisfy those inequalities, any convex combination of these strategies will necessarily also satisfy the inequalities since expected utility is linear in probabilities. I'll leave the formal proof for you to write, but hopefully the logic is clear.</p>

CC BY-SA 4.0
2020-05-08T16:41:47.153Z
2020-05-08T16:41:47.153Z
Regio
<p><a href="https://i.sstatic.net/xmRX7.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/xmRX7.png" alt="enter image description here"></a></p>

<p>My attempt: </p>

<p><span class="math-container">$(0, 1/3, 2/3)$</span> works as a strategy that strictly dominated <span class="math-container">$L$</span>. </p>

<p>To prove that there are infinitely many of them:</p>

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

<p>I chose <span class="math-container">$n&gt; 10$</span> since <span class="math-container">$2&lt;2.5-5\epsilon_n \Leftrightarrow 0,1&gt; \epsilon_n$</span> and <span class="math-container">$1/n&lt;0,1$</span> for <span class="math-container">$n&gt;10$</span>. The other inequalities don't require <span class="math-container">$\epsilon_n$</span> to be restricted. </p>

<p>Would this be a correct way to prove this?</p>

CC BY-SA 4.0
2020-05-08T16:41:06.797Z
Sir, you are posting a [lot of questions](https://economics.stackexchange.com/users/21632/the-poor-jew?tab=questions) in a short time period. Perhaps try to work them out a bit longer yourself.
Giskard
CC BY-SA 4.0
2020-05-08T16:41:38.373Z
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
CC BY-SA 4.0
2020-05-08T16:42:40.927Z
Lastly, questions of the "Is this correct" type are a poor fit for the SE format.
Giskard
CC BY-SA 4.0
2020-05-08T16:50:11.233Z
@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
Slim Shady
<game-theory>
Proving there are infinitely many strictly dominant strategies
Slim Shady
# Proving there are infinitely many strictly dominant strategies

(Asked by: username_0 on 2020-05-08)

[![enter image description here](https://i.sstatic.net/xmRX7.png)](https://i.sstatic.net/xmRX7.png)

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?

**Tags:** game-theory

> **Comments:**
> **username_1**: Sir, you are posting a [lot of questions](https://economics.stackexchange.com/users/21632/the-poor-jew?tab=questions) in a short time period. Perhaps try to work them out a bit longer yourself.  
> **username_1**: 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.  
> **username_1**: Lastly, questions of the "Is this correct" type are a poor fit for the SE format.  
> **username_0**: @username_1 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  

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## 1 answers

### Answer by username_2 on 2020-05-08. Score: 1 | Accepted answer

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.

(Idea of the proof:) Notice that this holds because strict dominance is given by a set of inequalities (3 in this example) so if you have two strategies that satisfy those inequalities, any convex combination of these strategies will necessarily also satisfy the inequalities since expected utility is linear in probabilities. I'll leave the formal proof for you to write, but hopefully the logic is clear.


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