<p>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.</p>

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

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2019-05-18T22:12:07.130Z
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?
Eden Hazard
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2019-05-18T22:29:07.267Z
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?
Eden Hazard
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2019-05-19T08:22:09.367Z
Most people are you not your bros, neither am I.
Giskard
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2019-05-19T08:24:08.133Z
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](https://economics.stackexchange.com/questions/1864/are-symmetric-equilibria-continuous-with-respect-to-the-payoff-matrix/3034#3034). That would mean pure strategy equilibria also exist.
Giskard
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2019-05-19T12:10:23.817Z
Chill, no need to make a big deal out of calling you ‘bro’ . But anyway I really appreciate your help Thanks!!
Eden Hazard
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2019-05-18T21:51:10.670Z
2019-05-18T21:51:10.670Z
Giskard
<p>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?</p>

<p>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? </p>

<game-theory>
Construction of a 3x3 simultaneous game
Eden Hazard
# Construction of a 3x3 simultaneous game

(Asked by: username_0 on 2019-05-18)

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?

**Tags:** game-theory

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## 1 answers

### Answer by username_1 on 2019-05-18. Score: 2

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.

> **Comments:**
> **username_0**: 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?  
> **username_0**: 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?  
> **username_1**: Most people are you not your bros, neither am I.  
> **username_1**: 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](https://economics.stackexchange.com/questions/1864/are-symmetric-equilibria-continuous-with-respect-to-the-payoff-matrix/3034#3034). That would mean pure strategy equilibria also exist.  
> **username_0**: Chill, no need to make a big deal out of calling you ‘bro’ . But anyway I really appreciate your help Thanks!!  

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