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?")

  • $\begingroup$ 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

2 Answers 2


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 $$3>1.$$

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.