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.
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.
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:
Play my part in the strategy that gives S our security value.
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.
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.
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.
Could you give a proof of that?
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:
Play my part in the strategy that gives S our security value.
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.
Do you mean Strong Nash Equilibria? I’ve read that it’s been proven that every possible game has at least one Nash Equilibrium.
Yes, I do mean that, although I’m not sure the proof extends to games with infinite numbers of strategies.