<p>Mixed strategy Nash equilibrium cannot involve strictly dominated strategies. In particular, Cooperate is strictly dominated for player 1 (<span class="math-container">$6&lt;8$</span> and <span class="math-container">$3&lt;4$</span>). Therefore, no <span class="math-container">$b\in[0,1]$</span> can make player 1 indifferent between Cooperate and Non-cooperate. You made a mistake by trying to solve for <span class="math-container">$b$</span> by equating player 1's expected payoffs from his two pure strategies. </p>

CC BY-SA 4.0
2020-01-04T04:39:01.490Z
2020-01-04T04:39:01.490Z
Herr K.
<p>I have a game represented by following table:</p>

<p><a href="https://i.sstatic.net/iGCP1.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/iGCP1.png" alt="enter image description here"></a></p>

<p>It is clear that there is a pure Nash equilibrium at 4,2 (both players do not cooperate, player 1 awarded 4 points and player 2 awarded 2 points). </p>

<p>Now if I used mixed strategy Nash equlibrium algorithm I get a = 1 and b = -1 as the probability of player 1 plays Cooperate and player 2 plays Cooperate respectively. To my understanding, a and b should be equal to 0 so that both players should play the above pure equilibrium strategy.</p>

<p>Why is there such a difference in pure and mixed strategy? </p>

<p>Edit 1: mixed strategy Nash equlibrium algorithm:</p>

<ul>
<li>The expected ultility for player 2 playing cooperate is: a*2 + (1-a)*1 (This means that some of the time player 1 plays Cooperate, so the probability is a).</li>
<li>The expected ultility for player 2 playing non-cooperate is: a*2 + (1-a)*2.</li>
<li>By equating the two formuala I have (1-a) = 2*(1-a) so a = 1.</li>
<li>Simmilarly, the expected ultility for player 1 playing cooperate is: 6*b + 3*(1 - b).</li>
<li>The expected ultility for player 1 playing non-cooperate is: 8*b + 4*(1 - b).</li>
<li>By equating the two formuala I have 0 = 2*b + (1-b) so b = -1.</li>
</ul>

CC BY-SA 4.0
2020-01-03T20:35:02.033Z
You should probably go into detail about your "mixed strategy Nash equlibrium algorithm". E.g. what do the starting equations mean, etc.
Giskard
CC BY-SA 4.0
2020-01-03T21:12:11.730Z
@Giskard added to the post. Apologise for not making the question clear.
Leo Long Vu
<game-theory><nash-equilibrium>
Pure and Mixed Nash Equilibrium algorithm gives different results
Leo Long Vu
# Pure and Mixed Nash Equilibrium algorithm gives different results

(Asked by: username_0 on 2020-01-03)

I have a game represented by following table:

[![enter image description here](https://i.sstatic.net/iGCP1.png)](https://i.sstatic.net/iGCP1.png)

It is clear that there is a pure Nash equilibrium at 4,2 (both players do not cooperate, player 1 awarded 4 points and player 2 awarded 2 points).

Now if I used mixed strategy Nash equlibrium algorithm I get a = 1 and b = -1 as the probability of player 1 plays Cooperate and player 2 plays Cooperate respectively. To my understanding, a and b should be equal to 0 so that both players should play the above pure equilibrium strategy.

Why is there such a difference in pure and mixed strategy?

Edit 1: mixed strategy Nash equlibrium algorithm:

* The expected ultility for player 2 playing cooperate is: a\*2 + (1-a)\*1 (This means that some of the time player 1 plays Cooperate, so the probability is a).
* The expected ultility for player 2 playing non-cooperate is: a\*2 + (1-a)\*2.
* By equating the two formuala I have (1-a) = 2\*(1-a) so a = 1.
* Simmilarly, the expected ultility for player 1 playing cooperate is: 6\*b + 3\*(1 - b).
* The expected ultility for player 1 playing non-cooperate is: 8\*b + 4\*(1 - b).
* By equating the two formuala I have 0 = 2\*b + (1-b) so b = -1.

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

> **Comments:**
> **username_1**: You should probably go into detail about your "mixed strategy Nash equlibrium algorithm". E.g. what do the starting equations mean, etc.  
> **username_0**: @username_1 added to the post. Apologise for not making the question clear.  

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## 1 answers

### Answer by username_2 on 2020-01-04. Score: 2 | Accepted answer

Mixed strategy Nash equilibrium cannot involve strictly dominated strategies. In particular, Cooperate is strictly dominated for player 1 ($6<8$ and $3<4$). Therefore, no $b\in[0,1]$ can make player 1 indifferent between Cooperate and Non-cooperate. You made a mistake by trying to solve for $b$ by equating player 1's expected payoffs from his two pure strategies.


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