Equilibria and Matrices [closed]

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How do I find the equilibria of a matrix game with payoff matrix :

beginpmatrix
0 & -2 & 3& 0 \
2 & 0 &0&-3 \
-3 & 0 &0&4 \
0 & 3 &-4&0\
endpmatrix

I can't seem to find any solutions...

game-theory

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edited Aug 2 at 12:34

asked Aug 1 at 11:41

MathematicianP

3614

closed as off-topic by amWhy, Adrian Keister, John Ma, Isaac Browne, José Carlos Santos Aug 1 at 22:41

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Adrian Keister, John Ma, Isaac Browne, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

up vote
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How do I find the equilibria of a matrix game with payoff matrix :

beginpmatrix
0 & -2 & 3& 0 \
2 & 0 &0&-3 \
-3 & 0 &0&4 \
0 & 3 &-4&0\
endpmatrix

I can't seem to find any solutions...

game-theory

share|cite|improve this question

edited Aug 2 at 12:34

asked Aug 1 at 11:41

MathematicianP

3614

closed as off-topic by amWhy, Adrian Keister, John Ma, Isaac Browne, José Carlos Santos Aug 1 at 22:41

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Adrian Keister, John Ma, Isaac Browne, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

How do I find the equilibria of a matrix game with payoff matrix :

beginpmatrix
0 & -2 & 3& 0 \
2 & 0 &0&-3 \
-3 & 0 &0&4 \
0 & 3 &-4&0\
endpmatrix

I can't seem to find any solutions...

game-theory

share|cite|improve this question

edited Aug 2 at 12:34

asked Aug 1 at 11:41

MathematicianP

3614

How do I find the equilibria of a matrix game with payoff matrix :

beginpmatrix
0 & -2 & 3& 0 \
2 & 0 &0&-3 \
-3 & 0 &0&4 \
0 & 3 &-4&0\
endpmatrix

I can't seem to find any solutions...

game-theory

share|cite|improve this question

edited Aug 2 at 12:34

asked Aug 1 at 11:41

MathematicianP

3614

share|cite|improve this question

share|cite|improve this question

edited Aug 2 at 12:34

edited Aug 2 at 12:34

edited Aug 2 at 12:34

asked Aug 1 at 11:41

MathematicianP

3614

asked Aug 1 at 11:41

MathematicianP

3614

asked Aug 1 at 11:41

MathematicianP

3614

3614

closed as off-topic by amWhy, Adrian Keister, John Ma, Isaac Browne, José Carlos Santos Aug 1 at 22:41

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Adrian Keister, John Ma, Isaac Browne, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by amWhy, Adrian Keister, John Ma, Isaac Browne, José Carlos Santos Aug 1 at 22:41

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Adrian Keister, John Ma, Isaac Browne, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

add a comment | 

2 Answers
2

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oldest

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Your matrix, I call it $A$, is invertible, hence the equation $Ax=0$ has the unique solution $x=0$.

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answered Aug 1 at 11:54

Fred

37k1237

add a comment | 

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This game has no pure Nash equilibrium so you need to use Indifference Principle for finding the mixed ones. In fact Indifference Principle says that if a player namely player I has two pure strategies $A$ and $B$ where both appear with positive probability in a NE namely $(x*,y*)$ then $$U_I(A,y*)=U_I(B,y*)$$The reason is simple. If one is more that the other for example $U_I(A,y*)>U_I(B,y*)$ the player can easily deviate to $A$ for more benefit by reducing the weight of strategy $B$ in his set of strategies which contradicts the definition of NE. This principle is known as a necessary condition for NE existence albeit is sufficient either. Here I assumed that a NE may exist where both players use all of their own strategies of positive probability.

To demonstrate how such an equilibrium was derived, let we name the columns of the matrix with $$a_1,b_1,c_1,d_1$$ from top-down as the strategies of player I and the columns of the matrix with $$a_2,b_2,c_2,d_2$$ from left to right as the strategies of player II, for example $$U(b_1,d_2)=-3\ U(b_1,a_2)=2$$now in an equilibrium if we denote the optimal strategies of players I and II by $x^*=[x_1a_1,x_2b_1,x_3c_1,x_4d_1]$ and $y^*=[y_1a_2,y_2b_2,y_3c_2,y_4d_2]$ respectively where $x_1+x_2+x_3+x_4=y_1+y_2+y_3+y_4=1$then according to indifference principle we would have $$U(a_1,y^*)=U(b_1,y^*)=U(c_1,y^*)=U(d_1,y^*)$$one equation (for example) is as following$$-2y_2+3y_3=2y_1-3y_4$$solving this system of linear equations leads to the same optimal strategies found before.

Case 1: Players I,II use strategies $a_1,b_1$ and $a_2,b_2$ respectively in a
NE

At this case, we must have have $$-2(1-y)=2y$$which has no answers so this case is impossible.

Case 2: Players I,II use strategies $a_1,b_1$ and $b_2,c_2$ respectively in a NE

At this case, we must have have $$-2y+3(1-y)=0\-2x=3(1-x)$$which doesn't make sense since $x=0$

Case 3: Players I,II use strategies $a_1,b_1$ and $c_2,d_2$ respectively in a NE

there is no NE with the same argument as of case 1.

Case 4: Players I,II use strategies $a_1,b_1$ and $a_2,c_2$ respectively in a NE

At this case we have $$2y=3(1-y)\3x=2(1-x)$$which leads to valid NE$$x^*=beginbmatrixdfrac25& dfrac35& 0& 0endbmatrix\y^*=beginbmatrixdfrac35 &0 &dfrac25 &0endbmatrix$$

All other cases can be calculated similarly.

share|cite|improve this answer

edited Aug 2 at 17:06

answered Aug 1 at 18:46

Mostafa Ayaz

8,5183630

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am not familiar with this principle... how did you get the above answers?
  – MathematicianP
  Aug 2 at 10:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well! That took long to find this equilibria. In fact Indifference Principle says that if a player namely player I has two pure strategies $A$ and $B$ where both appear with positive probability in a NE namely $(x*,y*)$ then $$U_I(A,y*)=U_I(B,y*)$$. The reason is simple. If one is more that the other for example $U_I(A,y*)>U_I(B,y*)$ the player can easily deviate to $A$ for more benefit by reducing the weight of strategy $B$ in his set of strategies which contradicts the definition of NE. This principle is known as a necessary condition for NE existence.
  – Mostafa Ayaz
  Aug 2 at 10:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I also included more explanations up in my answer. Hope it works! Ask if still some ambiguity persists.....
  – Mostafa Ayaz
  Aug 2 at 10:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for the explanation! However I still don't understand how you got the answer above since the expected payoffs for each row are different...
  – MathematicianP
  Aug 2 at 10:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Another equation would be -3y1 + 4y4 = 3y2 - 4y3 but this is not satisfied by your solution
  – MathematicianP
  Aug 2 at 12:32
  

  
  
  
  

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote

Your matrix, I call it $A$, is invertible, hence the equation $Ax=0$ has the unique solution $x=0$.

share|cite|improve this answer

answered Aug 1 at 11:54

Fred

37k1237

add a comment | 

up vote
0
down vote

Your matrix, I call it $A$, is invertible, hence the equation $Ax=0$ has the unique solution $x=0$.

share|cite|improve this answer

answered Aug 1 at 11:54

Fred

37k1237

add a comment | 

up vote
0
down vote

up vote
0
down vote

Your matrix, I call it $A$, is invertible, hence the equation $Ax=0$ has the unique solution $x=0$.

share|cite|improve this answer

answered Aug 1 at 11:54

Fred

37k1237

Your matrix, I call it $A$, is invertible, hence the equation $Ax=0$ has the unique solution $x=0$.

share|cite|improve this answer

answered Aug 1 at 11:54

Fred

37k1237

share|cite|improve this answer

share|cite|improve this answer

answered Aug 1 at 11:54

Fred

37k1237

answered Aug 1 at 11:54

Fred

37k1237

answered Aug 1 at 11:54

Fred

37k1237

37k1237

add a comment | 

add a comment | 

up vote
0
down vote

This game has no pure Nash equilibrium so you need to use Indifference Principle for finding the mixed ones. In fact Indifference Principle says that if a player namely player I has two pure strategies $A$ and $B$ where both appear with positive probability in a NE namely $(x*,y*)$ then $$U_I(A,y*)=U_I(B,y*)$$The reason is simple. If one is more that the other for example $U_I(A,y*)>U_I(B,y*)$ the player can easily deviate to $A$ for more benefit by reducing the weight of strategy $B$ in his set of strategies which contradicts the definition of NE. This principle is known as a necessary condition for NE existence albeit is sufficient either. Here I assumed that a NE may exist where both players use all of their own strategies of positive probability.

To demonstrate how such an equilibrium was derived, let we name the columns of the matrix with $$a_1,b_1,c_1,d_1$$ from top-down as the strategies of player I and the columns of the matrix with $$a_2,b_2,c_2,d_2$$ from left to right as the strategies of player II, for example $$U(b_1,d_2)=-3\ U(b_1,a_2)=2$$now in an equilibrium if we denote the optimal strategies of players I and II by $x^*=[x_1a_1,x_2b_1,x_3c_1,x_4d_1]$ and $y^*=[y_1a_2,y_2b_2,y_3c_2,y_4d_2]$ respectively where $x_1+x_2+x_3+x_4=y_1+y_2+y_3+y_4=1$then according to indifference principle we would have $$U(a_1,y^*)=U(b_1,y^*)=U(c_1,y^*)=U(d_1,y^*)$$one equation (for example) is as following$$-2y_2+3y_3=2y_1-3y_4$$solving this system of linear equations leads to the same optimal strategies found before.

Case 1: Players I,II use strategies $a_1,b_1$ and $a_2,b_2$ respectively in a
NE

At this case, we must have have $$-2(1-y)=2y$$which has no answers so this case is impossible.

Case 2: Players I,II use strategies $a_1,b_1$ and $b_2,c_2$ respectively in a NE

At this case, we must have have $$-2y+3(1-y)=0\-2x=3(1-x)$$which doesn't make sense since $x=0$

Case 3: Players I,II use strategies $a_1,b_1$ and $c_2,d_2$ respectively in a NE

there is no NE with the same argument as of case 1.

Case 4: Players I,II use strategies $a_1,b_1$ and $a_2,c_2$ respectively in a NE

At this case we have $$2y=3(1-y)\3x=2(1-x)$$which leads to valid NE$$x^*=beginbmatrixdfrac25& dfrac35& 0& 0endbmatrix\y^*=beginbmatrixdfrac35 &0 &dfrac25 &0endbmatrix$$

All other cases can be calculated similarly.

share|cite|improve this answer

edited Aug 2 at 17:06

answered Aug 1 at 18:46

Mostafa Ayaz

8,5183630

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am not familiar with this principle... how did you get the above answers?
  – MathematicianP
  Aug 2 at 10:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well! That took long to find this equilibria. In fact Indifference Principle says that if a player namely player I has two pure strategies $A$ and $B$ where both appear with positive probability in a NE namely $(x*,y*)$ then $$U_I(A,y*)=U_I(B,y*)$$. The reason is simple. If one is more that the other for example $U_I(A,y*)>U_I(B,y*)$ the player can easily deviate to $A$ for more benefit by reducing the weight of strategy $B$ in his set of strategies which contradicts the definition of NE. This principle is known as a necessary condition for NE existence.
  – Mostafa Ayaz
  Aug 2 at 10:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I also included more explanations up in my answer. Hope it works! Ask if still some ambiguity persists.....
  – Mostafa Ayaz
  Aug 2 at 10:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for the explanation! However I still don't understand how you got the answer above since the expected payoffs for each row are different...
  – MathematicianP
  Aug 2 at 10:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Another equation would be -3y1 + 4y4 = 3y2 - 4y3 but this is not satisfied by your solution
  – MathematicianP
  Aug 2 at 12:32
  

  
  
  
  

add a comment | 

up vote
0
down vote

This game has no pure Nash equilibrium so you need to use Indifference Principle for finding the mixed ones. In fact Indifference Principle says that if a player namely player I has two pure strategies $A$ and $B$ where both appear with positive probability in a NE namely $(x*,y*)$ then $$U_I(A,y*)=U_I(B,y*)$$The reason is simple. If one is more that the other for example $U_I(A,y*)>U_I(B,y*)$ the player can easily deviate to $A$ for more benefit by reducing the weight of strategy $B$ in his set of strategies which contradicts the definition of NE. This principle is known as a necessary condition for NE existence albeit is sufficient either. Here I assumed that a NE may exist where both players use all of their own strategies of positive probability.

To demonstrate how such an equilibrium was derived, let we name the columns of the matrix with $$a_1,b_1,c_1,d_1$$ from top-down as the strategies of player I and the columns of the matrix with $$a_2,b_2,c_2,d_2$$ from left to right as the strategies of player II, for example $$U(b_1,d_2)=-3\ U(b_1,a_2)=2$$now in an equilibrium if we denote the optimal strategies of players I and II by $x^*=[x_1a_1,x_2b_1,x_3c_1,x_4d_1]$ and $y^*=[y_1a_2,y_2b_2,y_3c_2,y_4d_2]$ respectively where $x_1+x_2+x_3+x_4=y_1+y_2+y_3+y_4=1$then according to indifference principle we would have $$U(a_1,y^*)=U(b_1,y^*)=U(c_1,y^*)=U(d_1,y^*)$$one equation (for example) is as following$$-2y_2+3y_3=2y_1-3y_4$$solving this system of linear equations leads to the same optimal strategies found before.

Case 1: Players I,II use strategies $a_1,b_1$ and $a_2,b_2$ respectively in a
NE

At this case, we must have have $$-2(1-y)=2y$$which has no answers so this case is impossible.

Case 2: Players I,II use strategies $a_1,b_1$ and $b_2,c_2$ respectively in a NE

At this case, we must have have $$-2y+3(1-y)=0\-2x=3(1-x)$$which doesn't make sense since $x=0$

Case 3: Players I,II use strategies $a_1,b_1$ and $c_2,d_2$ respectively in a NE

there is no NE with the same argument as of case 1.

Case 4: Players I,II use strategies $a_1,b_1$ and $a_2,c_2$ respectively in a NE

At this case we have $$2y=3(1-y)\3x=2(1-x)$$which leads to valid NE$$x^*=beginbmatrixdfrac25& dfrac35& 0& 0endbmatrix\y^*=beginbmatrixdfrac35 &0 &dfrac25 &0endbmatrix$$

All other cases can be calculated similarly.

share|cite|improve this answer

edited Aug 2 at 17:06

answered Aug 1 at 18:46

Mostafa Ayaz

8,5183630

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am not familiar with this principle... how did you get the above answers?
  – MathematicianP
  Aug 2 at 10:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well! That took long to find this equilibria. In fact Indifference Principle says that if a player namely player I has two pure strategies $A$ and $B$ where both appear with positive probability in a NE namely $(x*,y*)$ then $$U_I(A,y*)=U_I(B,y*)$$. The reason is simple. If one is more that the other for example $U_I(A,y*)>U_I(B,y*)$ the player can easily deviate to $A$ for more benefit by reducing the weight of strategy $B$ in his set of strategies which contradicts the definition of NE. This principle is known as a necessary condition for NE existence.
  – Mostafa Ayaz
  Aug 2 at 10:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I also included more explanations up in my answer. Hope it works! Ask if still some ambiguity persists.....
  – Mostafa Ayaz
  Aug 2 at 10:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for the explanation! However I still don't understand how you got the answer above since the expected payoffs for each row are different...
  – MathematicianP
  Aug 2 at 10:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Another equation would be -3y1 + 4y4 = 3y2 - 4y3 but this is not satisfied by your solution
  – MathematicianP
  Aug 2 at 12:32
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

This game has no pure Nash equilibrium so you need to use Indifference Principle for finding the mixed ones. In fact Indifference Principle says that if a player namely player I has two pure strategies $A$ and $B$ where both appear with positive probability in a NE namely $(x*,y*)$ then $$U_I(A,y*)=U_I(B,y*)$$The reason is simple. If one is more that the other for example $U_I(A,y*)>U_I(B,y*)$ the player can easily deviate to $A$ for more benefit by reducing the weight of strategy $B$ in his set of strategies which contradicts the definition of NE. This principle is known as a necessary condition for NE existence albeit is sufficient either. Here I assumed that a NE may exist where both players use all of their own strategies of positive probability.

To demonstrate how such an equilibrium was derived, let we name the columns of the matrix with $$a_1,b_1,c_1,d_1$$ from top-down as the strategies of player I and the columns of the matrix with $$a_2,b_2,c_2,d_2$$ from left to right as the strategies of player II, for example $$U(b_1,d_2)=-3\ U(b_1,a_2)=2$$now in an equilibrium if we denote the optimal strategies of players I and II by $x^*=[x_1a_1,x_2b_1,x_3c_1,x_4d_1]$ and $y^*=[y_1a_2,y_2b_2,y_3c_2,y_4d_2]$ respectively where $x_1+x_2+x_3+x_4=y_1+y_2+y_3+y_4=1$then according to indifference principle we would have $$U(a_1,y^*)=U(b_1,y^*)=U(c_1,y^*)=U(d_1,y^*)$$one equation (for example) is as following$$-2y_2+3y_3=2y_1-3y_4$$solving this system of linear equations leads to the same optimal strategies found before.

Case 1: Players I,II use strategies $a_1,b_1$ and $a_2,b_2$ respectively in a
NE

At this case, we must have have $$-2(1-y)=2y$$which has no answers so this case is impossible.

Case 2: Players I,II use strategies $a_1,b_1$ and $b_2,c_2$ respectively in a NE

At this case, we must have have $$-2y+3(1-y)=0\-2x=3(1-x)$$which doesn't make sense since $x=0$

Case 3: Players I,II use strategies $a_1,b_1$ and $c_2,d_2$ respectively in a NE

there is no NE with the same argument as of case 1.

Case 4: Players I,II use strategies $a_1,b_1$ and $a_2,c_2$ respectively in a NE

At this case we have $$2y=3(1-y)\3x=2(1-x)$$which leads to valid NE$$x^*=beginbmatrixdfrac25& dfrac35& 0& 0endbmatrix\y^*=beginbmatrixdfrac35 &0 &dfrac25 &0endbmatrix$$

All other cases can be calculated similarly.

share|cite|improve this answer

edited Aug 2 at 17:06

answered Aug 1 at 18:46

Mostafa Ayaz

8,5183630

This game has no pure Nash equilibrium so you need to use Indifference Principle for finding the mixed ones. In fact Indifference Principle says that if a player namely player I has two pure strategies $A$ and $B$ where both appear with positive probability in a NE namely $(x*,y*)$ then $$U_I(A,y*)=U_I(B,y*)$$The reason is simple. If one is more that the other for example $U_I(A,y*)>U_I(B,y*)$ the player can easily deviate to $A$ for more benefit by reducing the weight of strategy $B$ in his set of strategies which contradicts the definition of NE. This principle is known as a necessary condition for NE existence albeit is sufficient either. Here I assumed that a NE may exist where both players use all of their own strategies of positive probability.

To demonstrate how such an equilibrium was derived, let we name the columns of the matrix with $$a_1,b_1,c_1,d_1$$ from top-down as the strategies of player I and the columns of the matrix with $$a_2,b_2,c_2,d_2$$ from left to right as the strategies of player II, for example $$U(b_1,d_2)=-3\ U(b_1,a_2)=2$$now in an equilibrium if we denote the optimal strategies of players I and II by $x^*=[x_1a_1,x_2b_1,x_3c_1,x_4d_1]$ and $y^*=[y_1a_2,y_2b_2,y_3c_2,y_4d_2]$ respectively where $x_1+x_2+x_3+x_4=y_1+y_2+y_3+y_4=1$then according to indifference principle we would have $$U(a_1,y^*)=U(b_1,y^*)=U(c_1,y^*)=U(d_1,y^*)$$one equation (for example) is as following$$-2y_2+3y_3=2y_1-3y_4$$solving this system of linear equations leads to the same optimal strategies found before.

Case 1: Players I,II use strategies $a_1,b_1$ and $a_2,b_2$ respectively in a
NE

At this case, we must have have $$-2(1-y)=2y$$which has no answers so this case is impossible.

Case 2: Players I,II use strategies $a_1,b_1$ and $b_2,c_2$ respectively in a NE

At this case, we must have have $$-2y+3(1-y)=0\-2x=3(1-x)$$which doesn't make sense since $x=0$

Case 3: Players I,II use strategies $a_1,b_1$ and $c_2,d_2$ respectively in a NE

there is no NE with the same argument as of case 1.

Case 4: Players I,II use strategies $a_1,b_1$ and $a_2,c_2$ respectively in a NE

At this case we have $$2y=3(1-y)\3x=2(1-x)$$which leads to valid NE$$x^*=beginbmatrixdfrac25& dfrac35& 0& 0endbmatrix\y^*=beginbmatrixdfrac35 &0 &dfrac25 &0endbmatrix$$

All other cases can be calculated similarly.

share|cite|improve this answer

edited Aug 2 at 17:06

answered Aug 1 at 18:46

Mostafa Ayaz

8,5183630

share|cite|improve this answer

share|cite|improve this answer

edited Aug 2 at 17:06

edited Aug 2 at 17:06

edited Aug 2 at 17:06

answered Aug 1 at 18:46

Mostafa Ayaz

8,5183630

answered Aug 1 at 18:46

Mostafa Ayaz

8,5183630

answered Aug 1 at 18:46

Mostafa Ayaz

8,5183630

8,5183630

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am not familiar with this principle... how did you get the above answers?
  – MathematicianP
  Aug 2 at 10:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well! That took long to find this equilibria. In fact Indifference Principle says that if a player namely player I has two pure strategies $A$ and $B$ where both appear with positive probability in a NE namely $(x*,y*)$ then $$U_I(A,y*)=U_I(B,y*)$$. The reason is simple. If one is more that the other for example $U_I(A,y*)>U_I(B,y*)$ the player can easily deviate to $A$ for more benefit by reducing the weight of strategy $B$ in his set of strategies which contradicts the definition of NE. This principle is known as a necessary condition for NE existence.
  – Mostafa Ayaz
  Aug 2 at 10:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I also included more explanations up in my answer. Hope it works! Ask if still some ambiguity persists.....
  – Mostafa Ayaz
  Aug 2 at 10:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for the explanation! However I still don't understand how you got the answer above since the expected payoffs for each row are different...
  – MathematicianP
  Aug 2 at 10:45
  

  
  
  
  


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  Another equation would be -3y1 + 4y4 = 3y2 - 4y3 but this is not satisfied by your solution
  – MathematicianP
  Aug 2 at 12:32
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am not familiar with this principle... how did you get the above answers?
  – MathematicianP
  Aug 2 at 10:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well! That took long to find this equilibria. In fact Indifference Principle says that if a player namely player I has two pure strategies $A$ and $B$ where both appear with positive probability in a NE namely $(x*,y*)$ then $$U_I(A,y*)=U_I(B,y*)$$. The reason is simple. If one is more that the other for example $U_I(A,y*)>U_I(B,y*)$ the player can easily deviate to $A$ for more benefit by reducing the weight of strategy $B$ in his set of strategies which contradicts the definition of NE. This principle is known as a necessary condition for NE existence.
  – Mostafa Ayaz
  Aug 2 at 10:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I also included more explanations up in my answer. Hope it works! Ask if still some ambiguity persists.....
  – Mostafa Ayaz
  Aug 2 at 10:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for the explanation! However I still don't understand how you got the answer above since the expected payoffs for each row are different...
  – MathematicianP
  Aug 2 at 10:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Another equation would be -3y1 + 4y4 = 3y2 - 4y3 but this is not satisfied by your solution
  – MathematicianP
  Aug 2 at 12:32
  

  
  
  
  

I am not familiar with this principle... how did you get the above answers?
– MathematicianP
Aug 2 at 10:16

I am not familiar with this principle... how did you get the above answers?
– MathematicianP
Aug 2 at 10:16

Well! That took long to find this equilibria. In fact Indifference Principle says that if a player namely player I has two pure strategies $A$ and $B$ where both appear with positive probability in a NE namely $(x*,y*)$ then $$U_I(A,y*)=U_I(B,y*)$$. The reason is simple. If one is more that the other for example $U_I(A,y*)>U_I(B,y*)$ the player can easily deviate to $A$ for more benefit by reducing the weight of strategy $B$ in his set of strategies which contradicts the definition of NE. This principle is known as a necessary condition for NE existence.
– Mostafa Ayaz
Aug 2 at 10:24

Well! That took long to find this equilibria. In fact Indifference Principle says that if a player namely player I has two pure strategies $A$ and $B$ where both appear with positive probability in a NE namely $(x*,y*)$ then $$U_I(A,y*)=U_I(B,y*)$$. The reason is simple. If one is more that the other for example $U_I(A,y*)>U_I(B,y*)$ the player can easily deviate to $A$ for more benefit by reducing the weight of strategy $B$ in his set of strategies which contradicts the definition of NE. This principle is known as a necessary condition for NE existence.
– Mostafa Ayaz
Aug 2 at 10:24

I also included more explanations up in my answer. Hope it works! Ask if still some ambiguity persists.....
– Mostafa Ayaz
Aug 2 at 10:27

I also included more explanations up in my answer. Hope it works! Ask if still some ambiguity persists.....
– Mostafa Ayaz
Aug 2 at 10:27

Thank you for the explanation! However I still don't understand how you got the answer above since the expected payoffs for each row are different...
– MathematicianP
Aug 2 at 10:45

Thank you for the explanation! However I still don't understand how you got the answer above since the expected payoffs for each row are different...
– MathematicianP
Aug 2 at 10:45

Another equation would be -3y1 + 4y4 = 3y2 - 4y3 but this is not satisfied by your solution
– MathematicianP
Aug 2 at 12:32

Another equation would be -3y1 + 4y4 = 3y2 - 4y3 but this is not satisfied by your solution
– MathematicianP
Aug 2 at 12:32

add a comment |