<p>No, this is merely an artifact of a method of calculating equilibria in mixed strategies.</p>
<p>Formally, a Nash equilibrium is defined in terms of inequalities. These inequalities state that the expected payoff of the (possibly pure, degenerate) equilibrium mixed strategy is at least as large as that of any other mixed strategy given, the mixed strategies of the other players.</p>
<p>This gives us infinitely many inequalities, but one can show that it is enough to check that the expected payoff of the equilibrium strategy is at least as high as any alternative pure strategy. The linearity of expected payoffs in probabilities implies that the set of (possibly pure) mixed best replies to a profile of strategies of the others is the set of mixed strategies that play pure best replies with positive probability.</p>
<p>If it is optimal to mix between two pure strategies, playing any of the two pure strategies is optimal too. Proper mixed best replies are never strict. This means that when we look at equilibria in which a player mixes two pure strategies, we have to specify the strategies of the other player so that both pure strategies give the same payoff. Here, we work with equalities instead of inequalities. But that is only so because we are looking for a player who actually mixes.</p>
<p>There are many settings in which mixed strategy equilibria are less reasonable than pure strategy equilibria. Consider the game of choosing on which side of the street to drive, left or right- modeled as a pure coordination game.</p>
<p><span class="math-container">$$\begin{array} {|r|r|}\hline  &amp; L &amp; R \\ \hline L &amp; 1,1 &amp; 0,0 \\ \hline R &amp; 0,0 &amp; 1,1 \\ \hline  \end{array}$$</span></p>
<p>If everyone drives on the left, driving on the left is optimal. If everyone drives on the right, driving on the right is optimal. Frontal crashes into other cars are bad. There is also a mixed strategy equilibrium in which everyone randomizes with probability <span class="math-container">$1/2$</span> for both pure strategies. We would not expect that equilibrium to be a good predictor of what happens in reality (though the Harsanyi-Selten theory of equilibrium selection would actually choose it). Indeed, if we think of equilibria as the outcome of some learning process, we would expect only the equilibria in pure strategies to persist. If people are slightly more likely to drive on the left, driving on the left is optimal, and vice versa. In the long run, the behavior should converge to everyone driving on the same side.</p>

CC BY-SA 4.0
2022-11-29T09:15:20.933Z
As a student, I had a professor joking "Some say that the Greek drivers play the mixed equilibrium, though." I used this joke myself but stopped doing so a few years ago after having been told by students that this joke is "slightly racist".
VARulle
CC BY-SA 4.0
2022-11-29T10:11:49.447Z
@VARulle replace Greek by blonds and the joke will still work :)
1muflon1
CC BY-SA 4.0
2022-11-29T10:35:13.923Z
@1muflon1, well, a blonde already plays a major role in the "bar scene" clip from "A Beautiful Mind" I used as part of a homework project. I have been advised to stop using this clip b/c it is sexist...
VARulle
CC BY-SA 4.0
2022-11-29T21:08:00.063Z
@VARulle what country are you from? I hear that people complain about this in US but I did not though it is so bad...
csilvia
CC BY-SA 4.0
2022-11-29T22:17:34.530Z
"If it is optimal to mix between two pure strategies, playing any of the two pure strategies is optimal too. Proper mixed best replies are never strict."  How is this statement consistent with Rock, Paper, Scissors?  Isn't mixing strictly better than playing a pure strategy there?  Or if you are assuming the game is to be played only once with no possibility for learning, then any mixed or pure strategy in the driving game seems equally good?
user10478
CC BY-SA 4.0
2022-11-29T23:52:45.783Z
@user10478 That game has a unique equilibrium in which both players randomize with equal probability. And if the other player randomizes this way, it doesn't matter what you do. Your payoff is always the same.
Michael Greinecker
CC BY-SA 4.0
2022-11-30T00:57:55.357Z
@csilvia, I'm from Austria. These attitudes have increasingly been imported from the US during the last few years I think.
VARulle
CC BY-SA 4.0
2022-11-29T09:01:37.853Z
2022-11-29T09:01:37.853Z
Michael Greinecker
<p>I am a bit confused as to why pure strategy Nash Equilibria are often used in analyzing a game alongside mixed strategy Nash Equilibria.  While I understand that pure strategy Nash Equilibria can be thought of as special cases of mixed strategy Nash Equilibria, the pure strategy Nash Equilibria of a game typically do not pop out of the computation used to find mixed strategy Nash Equilibria.  This leads me to believe that pure strategy Nash Equilibira are generally valid only when mixing is not allowed.</p>
<p>Is this correct?  For example, in the Battle of the Sexes game, do the pure strategy Nash Equilibria go away when mixing is allowed, or should players continue to consider pure strategy options (assuming no alterations to the game such as randomness or burning money)?  Should the player always prefer to used the Mixed Strategy Nash Equilibirum over the pure strategy Nash Equilibria as long as mixing is possible?  Does this result generalize to all games?</p>

<game-theory><nash-equilibrium>
Does introducing mixed strategy Nash Equilibria override the pure strategy Nash Equilibria?
user10478
# Does introducing mixed strategy Nash Equilibria override the pure strategy Nash Equilibria?

(Asked by: username_0 on 2022-11-29)

I am a bit confused as to why pure strategy Nash Equilibria are often used in analyzing a game alongside mixed strategy Nash Equilibria. While I understand that pure strategy Nash Equilibria can be thought of as special cases of mixed strategy Nash Equilibria, the pure strategy Nash Equilibria of a game typically do not pop out of the computation used to find mixed strategy Nash Equilibria. This leads me to believe that pure strategy Nash Equilibira are generally valid only when mixing is not allowed.

Is this correct? For example, in the Battle of the Sexes game, do the pure strategy Nash Equilibria go away when mixing is allowed, or should players continue to consider pure strategy options (assuming no alterations to the game such as randomness or burning money)? Should the player always prefer to used the Mixed Strategy Nash Equilibirum over the pure strategy Nash Equilibria as long as mixing is possible? Does this result generalize to all games?

**Tags:** game-theory, nash-equilibrium


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## 1 answers

### Answer by username_1 on 2022-11-29. Score: 4 | Accepted answer

No, this is merely an artifact of a method of calculating equilibria in mixed strategies.

Formally, a Nash equilibrium is defined in terms of inequalities. These inequalities state that the expected payoff of the (possibly pure, degenerate) equilibrium mixed strategy is at least as large as that of any other mixed strategy given, the mixed strategies of the other players.

This gives us infinitely many inequalities, but one can show that it is enough to check that the expected payoff of the equilibrium strategy is at least as high as any alternative pure strategy. The linearity of expected payoffs in probabilities implies that the set of (possibly pure) mixed best replies to a profile of strategies of the others is the set of mixed strategies that play pure best replies with positive probability.

If it is optimal to mix between two pure strategies, playing any of the two pure strategies is optimal too. Proper mixed best replies are never strict. This means that when we look at equilibria in which a player mixes two pure strategies, we have to specify the strategies of the other player so that both pure strategies give the same payoff. Here, we work with equalities instead of inequalities. But that is only so because we are looking for a player who actually mixes.

There are many settings in which mixed strategy equilibria are less reasonable than pure strategy equilibria. Consider the game of choosing on which side of the street to drive, left or right- modeled as a pure coordination game.

$$\begin{array} {|r|r|}\hline & L & R \\ \hline L & 1,1 & 0,0 \\ \hline R & 0,0 & 1,1 \\ \hline \end{array}$$

If everyone drives on the left, driving on the left is optimal. If everyone drives on the right, driving on the right is optimal. Frontal crashes into other cars are bad. There is also a mixed strategy equilibrium in which everyone randomizes with probability $1/2$ for both pure strategies. We would not expect that equilibrium to be a good predictor of what happens in reality (though the Harsanyi-Selten theory of equilibrium selection would actually choose it). Indeed, if we think of equilibria as the outcome of some learning process, we would expect only the equilibria in pure strategies to persist. If people are slightly more likely to drive on the left, driving on the left is optimal, and vice versa. In the long run, the behavior should converge to everyone driving on the same side.

> **Comments:**
> **username_2**: As a student, I had a professor joking "Some say that the Greek drivers play the mixed equilibrium, though." I used this joke myself but stopped doing so a few years ago after having been told by students that this joke is "slightly racist".  
> **username_3**: @username_2 replace Greek by blonds and the joke will still work :)  
> **username_2**: @username_3, well, a blonde already plays a major role in the "bar scene" clip from "A Beautiful Mind" I used as part of a homework project. I have been advised to stop using this clip b/c it is sexist...  
> **username_4**: @username_2 what country are you from? I hear that people complain about this in US but I did not though it is so bad...  
> **username_0**: "If it is optimal to mix between two pure strategies, playing any of the two pure strategies is optimal too. Proper mixed best replies are never strict."  How is this statement consistent with Rock, Paper, Scissors?  Isn't mixing strictly better than playing a pure strategy there?  Or if you are assuming the game is to be played only once with no possibility for learning, then any mixed or pure strategy in the driving game seems equally good?  
> **username_1**: @username_0 That game has a unique equilibrium in which both players randomize with equal probability. And if the other player randomizes this way, it doesn't matter what you do. Your payoff is always the same.  
> **username_2**: @username_4, I'm from Austria. These attitudes have increasingly been imported from the US during the last few years I think.  

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