<p>If you are talking pure strategies Nash equilibria, then yes. Otherwise, you could have infinitely many mixed strategies Nash equilibria. Take a matrix in which all cells have the payoff vector $(1,1)$. So any convex combination of pure strategies is a Nash equilibrium.</p>

CC BY-SA 3.0
2017-12-25T18:03:06.003Z
2017-12-25T18:03:06.003Z
ml0105
<p>Let's say I'm given a matrix ixj with n games, is there a finite amount of Nash equilibria for those games?</p>

CC BY-SA 3.0
2017-12-25T20:09:04.740Z
With n games? What does that mean?
Michael Greinecker
<game-theory>
For any given non-null game, is there always a finite amount of Nash equilibria?
Nathan Furnal
# For any given non-null game, is there always a finite amount of Nash equilibria?

(Asked by: username_0 on 2017-12-25)

Let's say I'm given a matrix ixj with n games, is there a finite amount of Nash equilibria for those games?

**Tags:** game-theory
> **Comments:**
> **username_1**: With n games? What does that mean?  

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## 1 answers

### Answer by username_2 on 2017-12-25. Score: 3 | Accepted answer

If you are talking pure strategies Nash equilibria, then yes. Otherwise, you could have infinitely many mixed strategies Nash equilibria. Take a matrix in which all cells have the payoff vector $(1,1)$. So any convex combination of pure strategies is a Nash equilibrium.


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