How could I construct a 3x3 simultaneous game with 2 players with no equilibrium in pure strategies but 1 is mixed strategies . I am able to write down one with no pure strategy Nash equilibrium but how would I then know if only one mixed strategy would show not more?

I was thinking if I could eliminate one row and one column ,I.e they are strictly dominated in mixed strategies then I would only get 1 MSNE otherwise I get a continuum of mixed strategy Nash equilibria’s is that right?


Eliminating a column and a row strategy is a nice idea. If the remaining 2x2 game has no pure strategy Nash equilibria (PSNE) than Nash's theorem guarantues you exactly one MSNE.

However there are also very trivial 3x3 games with zero PSNE and exactly one MSNE, e.g. rock-paper-scissors.

  • $\begingroup$ Thanks for your answer bro, just one thing is it always true that if you don't have any PSNE you end up with only 1 MSNE in a 2x2 game? $\endgroup$ – Eden Hazard May 18 '19 at 22:12
  • $\begingroup$ Is it because if you don’t have any PSNE in a 2x2 game then all strategies would be included in the best responses so you could eliminate rows and columns? $\endgroup$ – Eden Hazard May 18 '19 at 22:29
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    $\begingroup$ Most people are you not your bros, neither am I. $\endgroup$ – Giskard May 19 '19 at 8:22
  • $\begingroup$ In a 2x2 game all MSNE are fully mixed. You can only have multiple fully mixed equilibria if the payoff matrices are non-singular, see the second half of this answer. That would mean pure strategy equilibria also exist. $\endgroup$ – Giskard May 19 '19 at 8:24
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    $\begingroup$ Chill, no need to make a big deal out of calling you ‘bro’ . But anyway I really appreciate your help Thanks!! $\endgroup$ – Eden Hazard May 19 '19 at 12:10

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