I don’t think the counterexample is really a counterexample to cousin_it’s claim. Quoting from Wikipedia:
The Nash equilibrium defines stability only in terms of unilateral deviations. In cooperative games such concept is not convincing enough. Strong Nash equilibrium allows for deviations by every conceivable coalition [4]. Formally, a Strong Nash equilibrium is a Nash equilibrium in which no coalition, taking the actions of its complements as given, can cooperatively deviate in a way that benefits all of its members [5]. However, the Strong Nash concept is some times perceived too “strong” in that the environment allows for unlimited private communication. In fact, Strong Nash equilibrium has to be Pareto efficient. As a result of these requirements, Strong Nash almost never exists.
The counterexample shows that ‘all players run Freaky Fairness’ is not a Strong Nash equilibrium but that wasn’t the claim...
Ok, after re-reading his post, I agree he probably did mean “strong Nash Equilibrium”. Can you make a note in your post that both of you are really talking about the Strong Nash equilibrium? As far as I can tell, ‘all players run Freaky Fairness’ is actually a (plain) Nash equilibrium so the current wording is rather confusing.
I don’t think the counterexample is really a counterexample to cousin_it’s claim. Quoting from Wikipedia:
The counterexample shows that ‘all players run Freaky Fairness’ is not a Strong Nash equilibrium but that wasn’t the claim...
“any coalition of players that decides to deviate (collectively or individually) cannot win total payoff greater than their group security value”
This phrase was what caused me to assume that a strong Nash Equilibrium was meant.
Ok, after re-reading his post, I agree he probably did mean “strong Nash Equilibrium”. Can you make a note in your post that both of you are really talking about the Strong Nash equilibrium? As far as I can tell, ‘all players run Freaky Fairness’ is actually a (plain) Nash equilibrium so the current wording is rather confusing.
Changed it, and I think you are correct about it being an ordinary Nash Equilibrium (one of infinitely many).