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Wei_Dai comments on Freaky unfairness - Less Wrong Discussion

17 Post author: benelliott 12 January 2011 10:08AM

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Comment author: Wei_Dai 12 January 2011 06:30:03PM 0 points [-]

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...

Comment author: benelliott 12 January 2011 06:35:28PM *  0 points [-]

"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.

Comment author: Wei_Dai 12 January 2011 07:02:31PM *  0 points [-]

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.

Comment author: benelliott 12 January 2011 07:36:08PM 0 points [-]

Changed it, and I think you are correct about it being an ordinary Nash Equilibrium (one of infinitely many).

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