<p>I'm given a 4x4 payoff matrix</p>
<p><span class="math-container">\begin{bmatrix}(0,7)&amp;(2,5)&amp;(7,0)&amp;(0,1)\\(5,2)&amp;(3,3)&amp;(5,2)&amp;(0,1)\\(7,0)&amp;(2,5)&amp;(0,7)&amp;(0,1)\\(0,0)&amp;(0,-2)&amp;(0,0)&amp;(10,-1)\end{bmatrix}</span></p>
<p>There isn't any strategy that is dominated for both players.
As a combination of best responses, I see that (3,3) is a pure strategy Nash equilibrium. I tried to compute mixed strategy Nash equilibrium by setting probabilities of row player's actions as p,q,r,s and equating the payoffs for the column player, which yields p= -1/3</p>
<p>Does this mean there is no mixed strategy Nash equilibrium? Thus only one Nash equilibrium exists?</p>

CC BY-SA 4.0
2020-10-07T09:57:48.263Z
It only means that there is no equilibrium in which the row player plays each strategy with a positive probability.
Michael Greinecker
CC BY-SA 4.0
2020-10-07T10:01:46.227Z
So if I limit the strategies, there can be mixed equilibriums? But then how do I know which strategy to discard?
ph8ndstne
CC BY-SA 4.0
2020-10-07T10:07:21.223Z
In principle, you have to do work through every subset of strategies for each player. But here the problem is a bit simpler. The fourth strategy of the column player, for example, is strictly dominated by a mixed strategy.
Michael Greinecker
CC BY-SA 4.0
2020-10-07T10:09:54.423Z
Oh I understand now. So to reduce a strategy, I need to find distributions of other strategies that dominate it right?
ph8ndstne
CC BY-SA 4.0
2020-10-07T10:11:11.260Z
If that is possible, the strategy will certainly never be played with positive probability. So you can discard the strategy when looking for Nash equilibria.
Michael Greinecker
CC BY-SA 4.0
2020-10-07T10:20:19.247Z
So I'm left with (1,2,3) rows x (1,2,3) columns. But if I assign again p, q, r to the rows, q is given a negative number. I don't think the matrix can be reduced anymore since no combination of 2 strategies dominates the other. Is it safe then to assume there is no mixed equilibrium?
ph8ndstne
<game-theory>
One pure Nash equilibrium, no mixed equilibrium?
ph8ndstne
# One pure Nash equilibrium, no mixed equilibrium?

(Asked by: username_0 on 2020-10-07)

I'm given a 4x4 payoff matrix

\begin{bmatrix}(0,7)&(2,5)&(7,0)&(0,1)\\(5,2)&(3,3)&(5,2)&(0,1)\\(7,0)&(2,5)&(0,7)&(0,1)\\(0,0)&(0,-2)&(0,0)&(10,-1)\end{bmatrix}

There isn't any strategy that is dominated for both players.
As a combination of best responses, I see that (3,3) is a pure strategy Nash equilibrium. I tried to compute mixed strategy Nash equilibrium by setting probabilities of row player's actions as p,q,r,s and equating the payoffs for the column player, which yields p= -1/3

Does this mean there is no mixed strategy Nash equilibrium? Thus only one Nash equilibrium exists?

**Tags:** game-theory
> **Comments:**
> **username_1**: It only means that there is no equilibrium in which the row player plays each strategy with a positive probability.  
> **username_0**: So if I limit the strategies, there can be mixed equilibriums? But then how do I know which strategy to discard?  
> **username_1**: In principle, you have to do work through every subset of strategies for each player. But here the problem is a bit simpler. The fourth strategy of the column player, for example, is strictly dominated by a mixed strategy.  
> **username_0**: Oh I understand now. So to reduce a strategy, I need to find distributions of other strategies that dominate it right?  
> **username_1**: If that is possible, the strategy will certainly never be played with positive probability. So you can discard the strategy when looking for Nash equilibria.  
> **username_0**: So I'm left with (1,2,3) rows x (1,2,3) columns. But if I assign again p, q, r to the rows, q is given a negative number. I don't think the matrix can be reduced anymore since no combination of 2 strategies dominates the other. Is it safe then to assume there is no mixed equilibrium?  

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## 0 answers

