<p>For the reason you mentioned, if a player randomizes in equilibrium over some pure strategies, then these strategies must give the same payoff against the strategy choices of the other players. The set of mixed best replies consists of all randomization over pure best replies.</p>
<p>The point is that expected payoffs are linear ina a player's own strategy:
<span class="math-container">$$\sum_{(s_1,\ldots,s_n)}\sigma_1(s_1)\cdots\sigma_n(s_n)~u_i(s_1,\ldots,s_n)=\sum_{s_i}\sigma_i(s_i)\sum_{(s_1,\ldots,s_{i-1},s_{i+1},\ldots,s_n)}u_i(s_1,\ldots,s_n).$$</span>
So the player's expected payoff will be a weighted average of the payoffs of the pure strategies, and it can only be a best response if it mixes between pure best responses.</p>
<p>The slightly odd consequence is that in a Nash equilibrium, a player does not randomize because it provides them with a higher payoff, they randomize so that the other players become indifferent between the pure strategies that they randomize over.</p>

CC BY-SA 4.0
2023-04-11T17:34:18.097Z
Thank you, if I understand correctly, then it is not possible that the payments are different, right? What happens in the matching pennies game then?
Haus
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2023-04-11T18:46:57.907Z
In matching pennies, both players are indifferent between both pure strategies in equilibrium. That is actually how one usually calculates mixed strategy equilibria: One finds the mixture that makes the other player indifferent.
Michael Greinecker
CC BY-SA 4.0
2023-04-11T17:24:11.943Z
2023-04-11T17:24:11.943Z
Michael Greinecker
<p>I have a question with mixed strategies.
The question is as follows, if we're in a strategy profile that is a Nash equilibrium and a player is playing a mixed strategy, can the pure strategies that are part of such a mixed strategy give different payoffs?</p>
<p>My first impression is no. If, for example, we consider a game such that the mixed strategy assigns <span class="math-container">$\mu$</span> to pure strategy <span class="math-container">$A$</span> and <span class="math-container">$1-\mu$</span> to pure strategy <span class="math-container">$B$</span> and if <span class="math-container">$A$</span> gives higher payouts than <span class="math-container">$B$</span>, then another mixed strategy that gives probability 1 to playing <span class="math-container">$A$</span> and 0 playing <span class="math-container">$B$</span> would give higher payoffs for said player, contrary to it being a Nash equilibrium.</p>
<p>I don't know if I'm right or if I have a mistake, I would really appreciate your comments or if you have a better way of arguing it or have a counterexample.</p>
<p>Update:
For example, in matching pennies, we are in an equilibrium when we play the strategy (0.5,0.5) for each player and yet the payoffs of the associated pennies are different. I don't know if this serves to exemplify that it is possible.</p>

CC BY-SA 4.0
2023-04-11T16:45:36.127Z
Yes, I think you should generally not delete answered questions. If they are poor questions they are supposed to be closed/deleted by the community.
Giskard
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2023-04-11T16:46:58.773Z
Done, an apology, on other stack exchange sites they have sometimes closed me, they ask like that.
Haus
<game-theory><nash-equilibrium><dominant-strategy>
Different payouts of pure strategies in mixed strategies
Haus
# Different payouts of pure strategies in mixed strategies

(Asked by: username_0 on 2023-04-11)

I have a question with mixed strategies.
The question is as follows, if we're in a strategy profile that is a Nash equilibrium and a player is playing a mixed strategy, can the pure strategies that are part of such a mixed strategy give different payoffs?

My first impression is no. If, for example, we consider a game such that the mixed strategy assigns $\mu$ to pure strategy $A$ and $1-\mu$ to pure strategy $B$ and if $A$ gives higher payouts than $B$, then another mixed strategy that gives probability 1 to playing $A$ and 0 playing $B$ would give higher payoffs for said player, contrary to it being a Nash equilibrium.

I don't know if I'm right or if I have a mistake, I would really appreciate your comments or if you have a better way of arguing it or have a counterexample.

Update:
For example, in matching pennies, we are in an equilibrium when we play the strategy (0.5,0.5) for each player and yet the payoffs of the associated pennies are different. I don't know if this serves to exemplify that it is possible.

**Tags:** game-theory, nash-equilibrium, dominant-strategy
> **Comments:**
> **username_1**: Yes, I think you should generally not delete answered questions. If they are poor questions they are supposed to be closed/deleted by the community.  
> **username_0**: Done, an apology, on other stack exchange sites they have sometimes closed me, they ask like that.  

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## 1 answers

### Answer by username_2 on 2023-04-11. Score: 3

For the reason you mentioned, if a player randomizes in equilibrium over some pure strategies, then these strategies must give the same payoff against the strategy choices of the other players. The set of mixed best replies consists of all randomization over pure best replies.

The point is that expected payoffs are linear ina a player's own strategy:
$$\sum\_{(s\_1,\ldots,s\_n)}\sigma\_1(s\_1)\cdots\sigma\_n(s\_n)~u\_i(s\_1,\ldots,s\_n)=\sum\_{s\_i}\sigma\_i(s\_i)\sum\_{(s\_1,\ldots,s\_{i-1},s\_{i+1},\ldots,s\_n)}u\_i(s\_1,\ldots,s\_n).$$
So the player's expected payoff will be a weighted average of the payoffs of the pure strategies, and it can only be a best response if it mixes between pure best responses.

The slightly odd consequence is that in a Nash equilibrium, a player does not randomize because it provides them with a higher payoff, they randomize so that the other players become indifferent between the pure strategies that they randomize over.

> **Comments:**
> **username_0**: Thank you, if I understand correctly, then it is not possible that the payments are different, right? What happens in the matching pennies game then?  
> **username_2**: In matching pennies, both players are indifferent between both pure strategies in equilibrium. That is actually how one usually calculates mixed strategy equilibria: One finds the mixture that makes the other player indifferent.  

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