<p>Going by the definition of dominant strategy from <a href="https://web.mit.edu/14.12/www/02F_lecture3-602.pdf" rel="nofollow noreferrer">here</a>:</p>
<p>A strategy <span class="math-container">$s_i^∗$</span> strictly dominates <span class="math-container">$s_i$</span> iff <span class="math-container">$u_i(s^∗_i , s_{−i}) &gt; u_i(s_i, s_{−i})$</span>, <span class="math-container">$∀s_{−i} \in S_{−i}$</span>.</p>
<p>Here we have <span class="math-container">$S_{-me}=\{\alpha, \beta\}$</span> and <span class="math-container">$s_{me} \in \{\alpha, \beta\} = S_{me}$</span>, with</p>
<p><span class="math-container">$u_{me}(\alpha, \alpha) = 0 &gt; -1 = u_{me}(\beta, \alpha)$</span>,</p>
<p><span class="math-container">$u_{me}(\alpha, \beta) = 3 &gt; 1 = u_{me}(\beta, \beta)$</span></p>
<p>Combining the above two,</p>
<p><span class="math-container">$u_{me}(\alpha , s_{−me}) &gt; u_{me}(s_{me}, s_{−me})$</span>, <span class="math-container">$∀s_{−me} \in S_{−me}$</span> and <span class="math-container">$s_{me} \in S_{me} - \{\alpha\}$</span></p>
<p><span class="math-container">$\implies s^∗_{me} = \alpha$</span>, i.e., <span class="math-container">$\alpha$</span> is the dominant strategy for me.</p>

CC BY-SA 4.0
2023-09-21T09:13:02.933Z
2023-09-21T09:23:32.063Z
2023-09-21T09:23:32.063Z
Sandipan Dey
<blockquote>
<p>if my pair plays Beta, I'm better off playing Beta than Alpha.</p>
</blockquote>
<p>In case of <span class="math-container">$(\beta,\beta)$</span>, you get 1, but in case of <span class="math-container">$(\alpha,\beta)$</span>, meaning you play <span class="math-container">$\alpha$</span> against pair's <span class="math-container">$\beta$</span>, you get 3, and <span class="math-container">$$3&gt;1.$$</span></p>

CC BY-SA 4.0
2023-08-07T19:45:15.870Z
Oh lort, I misspoke. Let's see... if I play  and my pair plays , I get 0. But if I play  and my pair plays , I get 1. That makes me think that playing  can be better than , so  can't be a strictly dominating strategy.
user5646514
CC BY-SA 4.0
2023-08-07T19:55:18.357Z
No matter what fixed strategy your partner plays, you are better off playing $\alpha$. This is all that the definition is. It does not say anything about what happens if both of you change strategies.
Giskard
CC BY-SA 4.0
2023-08-08T06:06:51.890Z
Ahh I get it, I wasn't looking at pair's strategy as having to be fixed. Thank you!
user5646514
CC BY-SA 4.0
2023-08-07T15:57:34.780Z
2023-08-07T15:57:34.780Z
Giskard
<p>Here's a payoff matrix for a Prisoners Dilemma:</p>
<p><a href="https://i.sstatic.net/HL8PB.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/HL8PB.png" alt="enter image description here" /></a></p>
<p>Alpha is supposed to be the dominant strategy, which means that &quot;no matter what my pair chooses, I will be better off playing Alpha than playing Beta.&quot;</p>
<p>I'm having trouble understanding this, because if my pair plays Beta, I'm better off playing Beta than Alpha.</p>
<p>My question is: &quot;Why is Alpha the dominant strategy?&quot;
(or &quot;Why is defecting the dominant strategy in the prisoners dilemma?&quot;)</p>

CC BY-SA 4.0
2023-08-07T15:45:14.570Z
I just noticed that this matrix doesn't follow the T>R>P>S criteria of the Prisoners Dilemma. Here it seems that R<P. 
But in this lecture, the teacher says that Alpha is the dominant strategy! So the question remains... https://www.youtube.com/watch?v=nM3rTU927io
user5646514
<game-theory>
Trouble Understanding Dominant Strategy in the Prisoners Dilemma
user5646514
# Trouble Understanding Dominant Strategy in the Prisoners Dilemma

(Asked by: username_0 on 2023-08-07)

Here's a payoff matrix for a Prisoners Dilemma:

[![enter image description here](https://i.sstatic.net/HL8PB.png)](https://i.sstatic.net/HL8PB.png)

Alpha is supposed to be the dominant strategy, which means that "no matter what my pair chooses, I will be better off playing Alpha than playing Beta."

I'm having trouble understanding this, because if my pair plays Beta, I'm better off playing Beta than Alpha.

My question is: "Why is Alpha the dominant strategy?"
(or "Why is defecting the dominant strategy in the prisoners dilemma?")

**Tags:** game-theory

> **Comments:**
> **username_0**: I just noticed that this matrix doesn't follow the T>R>P>S criteria of the Prisoners Dilemma. Here it seems that R<P.  But in this lecture, the teacher says that Alpha is the dominant strategy! So the question remains... https://www.youtube.com/watch?v=nM3rTU927io  

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## 2 answers

### Answer by username_2 on 2023-08-07. Score: 2 | Accepted answer

> if my pair plays Beta, I'm better off playing Beta than Alpha.

In case of $(\beta,\beta)$, you get 1, but in case of $(\alpha,\beta)$, meaning you play $\alpha$ against pair's $\beta$, you get 3, and $$3>1.$$

> **Comments:**
> **username_0**: Oh lort, I misspoke. Let's see... if I play  and my pair plays , I get 0. But if I play  and my pair plays , I get 1. That makes me think that playing  can be better than , so  can't be a strictly dominating strategy.  
> **username_2**: No matter what fixed strategy your partner plays, you are better off playing $\alpha$. This is all that the definition is. It does not say anything about what happens if both of you change strategies.  
> **username_0**: Ahh I get it, I wasn't looking at pair's strategy as having to be fixed. Thank you!  

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### Answer by username_1 on 2023-09-21. Score: 0

Going by the definition of dominant strategy from [here](https://web.mit.edu/14.12/www/02F_lecture3-602.pdf):

A strategy $s\_i^∗$ strictly dominates $s\_i$ iff $u\_i(s^∗\_i , s\_{−i}) > u\_i(s\_i, s\_{−i})$, $∀s\_{−i} \in S\_{−i}$.

Here we have $S\_{-me}=\{\alpha, \beta\}$ and $s\_{me} \in \{\alpha, \beta\} = S\_{me}$, with

$u\_{me}(\alpha, \alpha) = 0 > -1 = u\_{me}(\beta, \alpha)$,

$u\_{me}(\alpha, \beta) = 3 > 1 = u\_{me}(\beta, \beta)$

Combining the above two,

$u\_{me}(\alpha , s\_{−me}) > u\_{me}(s\_{me}, s\_{−me})$, $∀s\_{−me} \in S\_{−me}$ and $s\_{me} \in S\_{me} - \{\alpha\}$

$\implies s^∗\_{me} = \alpha$, i.e., $\alpha$ is the dominant strategy for me.


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