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cousin_it comments on Freaky unfairness - Less Wrong Discussion

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

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Comment author: cousin_it 12 January 2011 12:05:46PM *  8 points [-]

You're right and I was unbelievably stupid. (And I even was aware of the setup you mentioned, but failed to connect it to FF!) Added an update to the post. Thanks! Please do more of this.

Comment author: benelliott 12 January 2011 04:12:50PM *  3 points [-]

If its any consolation, I've found that the nasty games actually don't have Nash Equilibria if you view algorithms as strategies, so FF is as good as it gets.

Comment author: cousin_it 12 January 2011 09:11:29PM 1 point [-]

Could you give a proof of that?

Comment author: benelliott 12 January 2011 10:17:39PM *  1 point [-]

By the definition of a nasty game, in any possible mixed (or pure) strategy equilibrium at least one subset S of players must be receiving less than their security value. Therefore, if they were to each pick the following algorithm:

  1. Play my part in the strategy that gives S our security value.
  2. Share any pay-offs necessary to ensure everybody in S gets more than they would have done before.

They will all get more than they got before.

I hope I didn't confuse anyone by just saying Nash Equilibrium without specifying that it had to be strong (which are the kind of Nash Equilibria that are referred to in the original post). The theorem that Nash Equilibria always exist only applies to Weak Nash Equilibria.

Incidentally, at the time I made the above post my proof was hideously long and complicated, I am slightly ashamed of how long it took me to spot this one.

Comment author: DanielLC 14 January 2011 12:45:24AM 0 points [-]

Do you mean Strong Nash Equilibria? I've read that it's been proven that every possible game has at least one Nash Equilibrium.

Comment author: benelliott 14 January 2011 07:18:47AM 0 points [-]

Yes, I do mean that, although I'm not sure the proof extends to games with infinite numbers of strategies.