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Most references seem to state that Nash showed every symmetric game has a symmetric equilibrium point, but as far as I can tell from Nash's paper, he actually showed the much more general statement that every finite game has a symmetric equilibrium point.

I realise the references wouldn't be wrong, but why don't they state the more general result?

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  • $\begingroup$ As far as I remember, Nash's paper only talks about equilibrium, there is no reference to any symmetry whatsoever. Sorry if I am wrong. $\endgroup$
    – GH from MO
    Apr 14, 2014 at 16:13
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    $\begingroup$ To be clear, I'm referring to his PhD thesis, a copy of which is located at princeton.edu/mudd/news/faq/topics/… $\endgroup$
    – user17474
    Apr 14, 2014 at 16:15
  • $\begingroup$ It might also be worth reminding people that Nash's definition of a symmetric game is more general than most people use, it includes any fair games like matching pennies. $\endgroup$
    – user17474
    Apr 14, 2014 at 16:18
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    $\begingroup$ Hi Nick! Asymmetric finite games need not have symmetric equilibria -- the players need not even have e.g. the same number of strategies. Did you ask what you meant to ask? $\endgroup$
    – Noah Stein
    Apr 14, 2014 at 16:43
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    $\begingroup$ @GHfromMO There are two papers. Equilibrium Points in n-Person Games from 1950 only contained a proof that an equilibrium exists. Non-Cooperative Games from 1951 (essentially the published thesis) contains a proof that every game has a symmetric equilibrium. $\endgroup$
    – Michael Greinecker
    Apr 14, 2014 at 17:01

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Nash defined symmetries of finite games and proved existence of an equilibrium point that is invariant under all symmetries. He called such an equilibrium a symmetric equilibrium. For this to make a difference, a game needs to have nontrivial symmetries. Generic games don't.

A clearer source for his result would be the published version in the Annals of Mathematics. Theorem 2 there says: Any finite game has a symmetric equilibrium point.

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  • $\begingroup$ > For this to make a difference, a game needs to have nontrivial symmetries. Generic games don't. >Theorem 2 there says: Any finite game has a symmetric equilibrium point. $\endgroup$
    – user17474
    Apr 14, 2014 at 20:52
  • $\begingroup$ Don't those two statements contradict each other? $\endgroup$
    – user17474
    Apr 14, 2014 at 20:54
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    $\begingroup$ @NickHam No. A symmetric equilibrium is an equilibrium invariant under all symmetries of the game. In a generic game, the only such symmetry is the identity (there is a open dense, full measure set of games with no nontrivial symmetry). Nash did not define symmetric games. $\endgroup$
    – Michael Greinecker
    Apr 14, 2014 at 20:58

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