<p>Clearly, there are three pure strategy equilibria of the given game: (B,A), (B,B) and (C,C).</p>
<p>Sam's strategy A is strictly dominated by any of the Sam's mixed strategy <span class="math-container">$\sigma^S=(\sigma_A^S,\sigma_B^S,\sigma_C^S)=(0,p,1-p)$</span>, where <span class="math-container">$0&lt;p&lt;\frac{1}{3}$</span>. So, Sam will never play strategy A in equilibrium.</p>
<p>Different types of mixed-strategy Nash Equilibrium in this game are as follows:</p>
<ul>
<li>Sam plays <span class="math-container">$\sigma^S=(\sigma_A^S,\sigma_B^S,\sigma_C^S)=(0,1,0)$</span> and Jane plays <span class="math-container">$\sigma^J=(\sigma_A^J,\sigma_B^J,\sigma_C^J)=(q,1-q,0)$</span>, where <span class="math-container">$q\in (0,1)$</span>.</li>
<li>Sam plays <span class="math-container">$\sigma^S=(\sigma_A^S,\sigma_B^S,\sigma_C^S)=(0,p,1-p)$</span>, where <span class="math-container">$\frac{3}{5}\leq p&lt;1$</span> and Jane plays <span class="math-container">$\sigma^J=(\sigma_A^J,\sigma_B^J,\sigma_C^J)=(0,1,0)$</span>.</li>
<li>Sam plays <span class="math-container">$\sigma^S=(\sigma_A^S,\sigma_B^S,\sigma_C^S)=(0,\frac{3}{5},\frac{2}{5})$</span> and Jane plays <span class="math-container">$\sigma^J=(\sigma_A^J,\sigma_B^J,\sigma_C^J)=(\frac{3\theta}{4},1-\theta,\frac{\theta}{4})$</span> where <span class="math-container">$\theta\in [0,1]$</span></li>
</ul>

CC BY-SA 4.0
2024-11-20T12:32:17.660Z
2024-11-20T12:32:17.660Z
Amit
<p>Just because there exist pure strategy NEs does not mean there is no mixed strategy NE. The game you show very likely has a MSNE.</p>
<p>You also made inconsistent assertions: If a game is dominance solvable, then it cannot have multiple NEs (in fact, there must be a unique NE in pure strategies).</p>

CC BY-SA 4.0
2024-10-21T00:25:02.613Z
Hi Herr K., thank you for your answers. Indeed, I have solved and found a MSNE. However, the payoff for both Sam and Jane is not better off that if they had played pure strategies. Playing pure strategies would have gotten them the outcome C,C. This payoff is better than their payoff for MSNE. So can I say they should just play pure strategies and that a MSNE is inexistent?
HelpPls
CC BY-SA 4.0
2024-10-21T02:14:27.297Z
@HelpPls: An equilibrium exists regardless of whether you think it "should" be played. You're probably confusing the Pareto dominance of equilibrium outcomes with equilibrium itself. For example, $(B,A)$ is a pure strategy NE despite being Pareto dominated by $(C,C)$.  An equilibrium is a pair of *mutually best responding strategies*, while a Pareto dominant equilibrium outcome is one that cannot be Pareto improved upon by another equilibrium outcome.
Herr K.
CC BY-SA 4.0
2024-10-21T08:49:48.297Z
Hi Herr K., I've added another image in my original post. When I mean dominance solvable, I mean that I can eliminate dominated strategies to arrive at an unique NE. Is this the Pareto dominance you mentioned?
HelpPls
CC BY-SA 4.0
2024-10-21T13:17:19.567Z
@HelpPls: Pareto dominance is defined [here](https://www.gametheory.net/dictionary/ParetoDominated.html#:~:text=An%20outcome%20of%20a%20game,by%20at%20least%20one%20player.). It's also called [payoff dominance](https://en.wikipedia.org/wiki/Risk_dominance#:~:text=Strategy%20pair%20(H%2C%C2%A0H)%20payoff%20dominates%20(G%2C%C2%A0G)%20if%20A%C2%A0%E2%89%A5%C2%A0D%2C%20a%C2%A0%E2%89%A5%C2%A0d%2C%20and%20at%20least%20one%20of%20the%20two%20is%20a%20strict%20inequality%3A%20A%C2%A0%3E%C2%A0D%20or%20a%C2%A0%3E%C2%A0d.). However, neither is related to dominance solvability.
Herr K.
CC BY-SA 4.0
2024-10-21T13:19:41.870Z
@HelpPls: Can you explain the steps that led you to $(C,C)$? I don't see how you could eliminate $B$ for Sam and either $A$ or $B$ for Jane.
Herr K.
CC BY-SA 4.0
2024-10-23T05:57:40.670Z
sorry i made a mistake, i now understand that weakly dominated strategies cannot be completely eliminated. Hence, only strategy A for Sam can be eliminated. May I ask how you would go about solving this problem?
HelpPls
CC BY-SA 4.0
2024-10-23T14:27:12.773Z
@HelpPls: You've already found 3 pure NEs. There should be a continuum of MSNEs, varying by the probabilities assigned to each of Jane's pure strategies.
Herr K.
CC BY-SA 4.0
2024-10-20T12:51:49.867Z
2024-10-20T12:51:49.867Z
Herr K.
<p><a href="https://i.sstatic.net/IYR283QW.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/IYR283QW.png" alt="enter image description here" /></a></p>
<p><a href="https://i.sstatic.net/VCUPlq5t.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/VCUPlq5t.jpg" alt="enter image description here" /></a></p>
<p>Analyse and find all the Nash Equilibria (including pure and mixed strategy NE) for the following game table. Explain why if there is none. (Note: You need to present in a clear and easy-to-understand manner.)</p>
<p>I understand that with best response analysis, you get 3 Nash Equilibrium (B,A), (B,B) and (C,C).</p>
<p>However, I also understand that the game is dominance solvable through the iterated elimination of strictly dominated strategies. Resulting pure Nash Equilibrium is (C,C)</p>
<p>Hence, I conclude that there is no mixed strategy nash equilibrium because a pure nash equilibrium exists.</p>
<p>But how should I prove this? How do I explain where there is no mixed strategy nash equilibrium?</p>
<p>How would you do this?</p>

<game-theory><nash-equilibrium><probability>
How to prove that a mixed strategy nash equlibrium does not exist, when a game is dominance solvable?
HelpPls
# How to prove that a mixed strategy nash equlibrium does not exist, when a game is dominance solvable?

(Asked by: username_0 on 2024-10-20)

[![enter image description here](https://i.sstatic.net/IYR283QW.png)](https://i.sstatic.net/IYR283QW.png)

[![enter image description here](https://i.sstatic.net/VCUPlq5t.jpg)](https://i.sstatic.net/VCUPlq5t.jpg)

Analyse and find all the Nash Equilibria (including pure and mixed strategy NE) for the following game table. Explain why if there is none. (Note: You need to present in a clear and easy-to-understand manner.)

I understand that with best response analysis, you get 3 Nash Equilibrium (B,A), (B,B) and (C,C).

However, I also understand that the game is dominance solvable through the iterated elimination of strictly dominated strategies. Resulting pure Nash Equilibrium is (C,C)

Hence, I conclude that there is no mixed strategy nash equilibrium because a pure nash equilibrium exists.

But how should I prove this? How do I explain where there is no mixed strategy nash equilibrium?

How would you do this?

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


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## 2 answers

### Answer by username_1 on 2024-11-20. Score: 0

Clearly, there are three pure strategy equilibria of the given game: (B,A), (B,B) and (C,C).

Sam's strategy A is strictly dominated by any of the Sam's mixed strategy $\sigma^S=(\sigma\_A^S,\sigma\_B^S,\sigma\_C^S)=(0,p,1-p)$, where $0<p<\frac{1}{3}$. So, Sam will never play strategy A in equilibrium.

Different types of mixed-strategy Nash Equilibrium in this game are as follows:

* Sam plays $\sigma^S=(\sigma\_A^S,\sigma\_B^S,\sigma\_C^S)=(0,1,0)$ and Jane plays $\sigma^J=(\sigma\_A^J,\sigma\_B^J,\sigma\_C^J)=(q,1-q,0)$, where $q\in (0,1)$.
* Sam plays $\sigma^S=(\sigma\_A^S,\sigma\_B^S,\sigma\_C^S)=(0,p,1-p)$, where $\frac{3}{5}\leq p<1$ and Jane plays $\sigma^J=(\sigma\_A^J,\sigma\_B^J,\sigma\_C^J)=(0,1,0)$.
* Sam plays $\sigma^S=(\sigma\_A^S,\sigma\_B^S,\sigma\_C^S)=(0,\frac{3}{5},\frac{2}{5})$ and Jane plays $\sigma^J=(\sigma\_A^J,\sigma\_B^J,\sigma\_C^J)=(\frac{3\theta}{4},1-\theta,\frac{\theta}{4})$ where $\theta\in [0,1]$


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### Answer by username_2 on 2024-10-20. Score: 0

Just because there exist pure strategy NEs does not mean there is no mixed strategy NE. The game you show very likely has a MSNE.

You also made inconsistent assertions: If a game is dominance solvable, then it cannot have multiple NEs (in fact, there must be a unique NE in pure strategies).

> **Comments:**
> **username_0**: Hi username_2., thank you for your answers. Indeed, I have solved and found a MSNE. However, the payoff for both Sam and Jane is not better off that if they had played pure strategies. Playing pure strategies would have gotten them the outcome C,C. This payoff is better than their payoff for MSNE. So can I say they should just play pure strategies and that a MSNE is inexistent?  
> **username_2**: @username_0: An equilibrium exists regardless of whether you think it "should" be played. You're probably confusing the Pareto dominance of equilibrium outcomes with equilibrium itself. For example, $(B,A)$ is a pure strategy NE despite being Pareto dominated by $(C,C)$.  An equilibrium is a pair of *mutually best responding strategies*, while a Pareto dominant equilibrium outcome is one that cannot be Pareto improved upon by another equilibrium outcome.  
> **username_0**: Hi username_2., I've added another image in my original post. When I mean dominance solvable, I mean that I can eliminate dominated strategies to arrive at an unique NE. Is this the Pareto dominance you mentioned?  
> **username_2**: @username_0: Pareto dominance is defined [here](https://www.gametheory.net/dictionary/ParetoDominated.html#:~:text=An%20outcome%20of%20a%20game,by%20at%20least%20one%20player.). It's also called [payoff dominance](https://en.wikipedia.org/wiki/Risk_dominance#:~:text=Strategy%20pair%20(H%2C%C2%A0H)%20payoff%20dominates%20(G%2C%C2%A0G)%20if%20A%C2%A0%E2%89%A5%C2%A0D%2C%20a%C2%A0%E2%89%A5%C2%A0d%2C%20and%20at%20least%20one%20of%20the%20two%20is%20a%20strict%20inequality%3A%20A%C2%A0%3E%C2%A0D%20or%20a%C2%A0%3E%C2%A0d.). However, neither is related to dominance solvability.  
> **username_2**: @username_0: Can you explain the steps that led you to $(C,C)$? I don't see how you could eliminate $B$ for Sam and either $A$ or $B$ for Jane.  
> **username_0**: sorry i made a mistake, i now understand that weakly dominated strategies cannot be completely eliminated. Hence, only strategy A for Sam can be eliminated. May I ask how you would go about solving this problem?  
> **username_2**: @username_0: You've already found 3 pure NEs. There should be a continuum of MSNEs, varying by the probabilities assigned to each of Jane's pure strategies.  

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