<p>It depends on what you mean by generically. The Prisoner's Dilemma has exactly one equilibrium. If you generate the payoffs in your $2 \times 2$ game by drawing payoffs from continuous i.i.d.'s then there is a non-zero probability (seems to me that the exact number is 25%) that you will get a game similar to the Prisoner's Dilemma, i.e. a game with strictly dominant strategies for both players.</p>

<p>There is also a 'smaller' class of games with even number of equilibria, you can read about them here:</p>

<p><a href="http://mindyourdecisions.com/blog/2014/10/07/game-theory-tuesdays-why-most-games-have-an-odd-number-of-nash-equilibria/" rel="nofollow">http://mindyourdecisions.com/blog/2014/10/07/game-theory-tuesdays-why-most-games-have-an-odd-number-of-nash-equilibria/</a></p>

<blockquote>
  <p>The Oddness Theorem (Wilson 1971) states that nearly all finite games have an odd number of Nash equilibria. In this post, we’ll explain what the theorem means in practical terms.</p>
</blockquote>

<p>You can also have a game with an infinite number of equilibria. (E.g. a $2 \times 2$ game with all zeros.)</p>

<p>However if the number of equilibria is finite you cannot have more than three in a $2 \times 2$ game.</p>

CC BY-SA 3.0
2016-07-03T18:08:40.727Z
2016-07-03T18:08:40.727Z
Giskard
<p>I have seen that generically any finite games have odd number of equilibria. So for $2\times 2$ games can we say that generically they must have $3$ equilibria? If not can you give an example of games which have $5$ or $7$ equilibria? Many thanks!</p>

<game-theory>
Oddness of equilibrium points
user64066
# Oddness of equilibrium points

(Asked by: username_0 on 2016-07-03)

I have seen that generically any finite games have odd number of equilibria. So for $2\times 2$ games can we say that generically they must have $3$ equilibria? If not can you give an example of games which have $5$ or $7$ equilibria? Many thanks!

**Tags:** game-theory

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## 1 answers

### Answer by username_1 on 2016-07-03. Score: 4 | Accepted answer

It depends on what you mean by generically. The Prisoner's Dilemma has exactly one equilibrium. If you generate the payoffs in your $2 \times 2$ game by drawing payoffs from continuous i.i.d.'s then there is a non-zero probability (seems to me that the exact number is 25%) that you will get a game similar to the Prisoner's Dilemma, i.e. a game with strictly dominant strategies for both players.

There is also a 'smaller' class of games with even number of equilibria, you can read about them here:

<http://mindyourdecisions.com/blog/2014/10/07/game-theory-tuesdays-why-most-games-have-an-odd-number-of-nash-equilibria/>

> The Oddness Theorem (Wilson 1971) states that nearly all finite games have an odd number of Nash equilibria. In this post, we’ll explain what the theorem means in practical terms.

You can also have a game with an infinite number of equilibria. (E.g. a $2 \times 2$ game with all zeros.)

However if the number of equilibria is finite you cannot have more than three in a $2 \times 2$ game.


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