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Here's a payoff matrix for a Prisoners Dilemma:

enter image description here

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?")

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  • $\begingroup$ 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... youtube.com/watch?v=nM3rTU927io $\endgroup$ Aug 7, 2023 at 15:45

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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 $$3>1.$$

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  • $\begingroup$ Oh lort, I misspoke. Let's see... if I play 𝛼 and my pair plays 𝛼, I get 0. But if I play 𝛽 and my pair plays 𝛽, I get 1. That makes me think that playing 𝛽 can be better than 𝛼, so 𝛼 can't be a strictly dominating strategy. $\endgroup$ Aug 7, 2023 at 19:45
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    $\begingroup$ 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. $\endgroup$
    – Giskard
    Aug 7, 2023 at 19:55
  • $\begingroup$ Ahh I get it, I wasn't looking at pair's strategy as having to be fixed. Thank you! $\endgroup$ Aug 8, 2023 at 6:06
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Going by the definition of dominant strategy from here:

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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