Potential Game: Solving Prisoner's Dilemma

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I am currently trying to get a hang of potential games. However I am struggeling with the Potential Functions. Especially I cant figure out how the Prisoners Dilemma Potentials are computed as it is given this question:
HERE

Could anyone please explain me how @Rahul Savani arrived at this Potential Matrix ?

game-theory algorithmic-game-theory

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asked Aug 6 at 7:37

Astraeus

32

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up vote
0
down vote

favorite

I am currently trying to get a hang of potential games. However I am struggeling with the Potential Functions. Especially I cant figure out how the Prisoners Dilemma Potentials are computed as it is given this question:
HERE

Could anyone please explain me how @Rahul Savani arrived at this Potential Matrix ?

game-theory algorithmic-game-theory

share|cite|improve this question

asked Aug 6 at 7:37

Astraeus

32

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

I am currently trying to get a hang of potential games. However I am struggeling with the Potential Functions. Especially I cant figure out how the Prisoners Dilemma Potentials are computed as it is given this question:
HERE

Could anyone please explain me how @Rahul Savani arrived at this Potential Matrix ?

game-theory algorithmic-game-theory

share|cite|improve this question

asked Aug 6 at 7:37

Astraeus

32

I am currently trying to get a hang of potential games. However I am struggeling with the Potential Functions. Especially I cant figure out how the Prisoners Dilemma Potentials are computed as it is given this question:
HERE

Could anyone please explain me how @Rahul Savani arrived at this Potential Matrix ?

game-theory algorithmic-game-theory

share|cite|improve this question

asked Aug 6 at 7:37

Astraeus

32

share|cite|improve this question

share|cite|improve this question

asked Aug 6 at 7:37

Astraeus

32

asked Aug 6 at 7:37

Astraeus

32

asked Aug 6 at 7:37

Astraeus

32

32

add a comment | 

add a comment | 

1 Answer
1

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accepted

We want to define a function $P$ from the strategy space $(c, c), (q, c), (c, q), (q, q)$ to $mathbb R$ such that when either player deviates unilaterally i.e., changes from $q$ to $c$, or from $c$ to $q$ while the other player keeps the same strategy, the change in the potential function is the same as the change in that players utility.

Say the row player plays $q$ and column player plays $c$ which means we are in $(q, c)$. The row player has payoff 10 and the column player has payoff 0. Now, if the column player changes her strategy from $c$ to $q$ that means we end up in $(q, q)$ and hence her payoff increases from 0 to 2. Then the potential should also increase by 2 i.e., $P((q, q)) - P((q, c)) = 2$. If instead the row player switched from $q$ to $c$, thus we go from $(q, c)$ to $(c,c)$, her payoff changes by -4 from 10 to 6 and since the potential has to reflect this it must be that $P((c,c)) - P((q, c)) = -4$.

Doing this also for the situation $(c, q)$ to $(q,q)$ we obtain the equations
beginalign*
P((q, q)) - P((q, c)) &= 2 \
P((c,c)) - P((q, c)) &= -4 \
P((q, q)) - P((c, q)) &= 2 \
endalign*

Observe that if one of the terms is fixed, all others are determined by the equations. Also, we only care about the change in potential and not about the value of the potential.

Suppose we set $P((c, c)) = C$. Then, by the second equality, $P((q, c)) = C+4$. By the first equality we obtain $P((q,q)) = C+6$ and finally by the last equality $P((c, q)) = C+4$. In the matrix notation from Rahul Savani's post
$$
beginpmatrix
C & C+4 \
C+4 & C+6
endpmatrix
$$

by taking $C=0$ we get the same potential matrix as Rahul Savani, but we could have taken any other real value.

share|cite|improve this answer

answered Aug 13 at 14:21

Ruben

532

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much thats the explanation I was looking for and very well structured thank you.
  – Astraeus
  Aug 14 at 23:46
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

We want to define a function $P$ from the strategy space $(c, c), (q, c), (c, q), (q, q)$ to $mathbb R$ such that when either player deviates unilaterally i.e., changes from $q$ to $c$, or from $c$ to $q$ while the other player keeps the same strategy, the change in the potential function is the same as the change in that players utility.

Say the row player plays $q$ and column player plays $c$ which means we are in $(q, c)$. The row player has payoff 10 and the column player has payoff 0. Now, if the column player changes her strategy from $c$ to $q$ that means we end up in $(q, q)$ and hence her payoff increases from 0 to 2. Then the potential should also increase by 2 i.e., $P((q, q)) - P((q, c)) = 2$. If instead the row player switched from $q$ to $c$, thus we go from $(q, c)$ to $(c,c)$, her payoff changes by -4 from 10 to 6 and since the potential has to reflect this it must be that $P((c,c)) - P((q, c)) = -4$.

Doing this also for the situation $(c, q)$ to $(q,q)$ we obtain the equations
beginalign*
P((q, q)) - P((q, c)) &= 2 \
P((c,c)) - P((q, c)) &= -4 \
P((q, q)) - P((c, q)) &= 2 \
endalign*

Observe that if one of the terms is fixed, all others are determined by the equations. Also, we only care about the change in potential and not about the value of the potential.

Suppose we set $P((c, c)) = C$. Then, by the second equality, $P((q, c)) = C+4$. By the first equality we obtain $P((q,q)) = C+6$ and finally by the last equality $P((c, q)) = C+4$. In the matrix notation from Rahul Savani's post
$$
beginpmatrix
C & C+4 \
C+4 & C+6
endpmatrix
$$

by taking $C=0$ we get the same potential matrix as Rahul Savani, but we could have taken any other real value.

share|cite|improve this answer

answered Aug 13 at 14:21

Ruben

532

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much thats the explanation I was looking for and very well structured thank you.
  – Astraeus
  Aug 14 at 23:46
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

We want to define a function $P$ from the strategy space $(c, c), (q, c), (c, q), (q, q)$ to $mathbb R$ such that when either player deviates unilaterally i.e., changes from $q$ to $c$, or from $c$ to $q$ while the other player keeps the same strategy, the change in the potential function is the same as the change in that players utility.

Say the row player plays $q$ and column player plays $c$ which means we are in $(q, c)$. The row player has payoff 10 and the column player has payoff 0. Now, if the column player changes her strategy from $c$ to $q$ that means we end up in $(q, q)$ and hence her payoff increases from 0 to 2. Then the potential should also increase by 2 i.e., $P((q, q)) - P((q, c)) = 2$. If instead the row player switched from $q$ to $c$, thus we go from $(q, c)$ to $(c,c)$, her payoff changes by -4 from 10 to 6 and since the potential has to reflect this it must be that $P((c,c)) - P((q, c)) = -4$.

Doing this also for the situation $(c, q)$ to $(q,q)$ we obtain the equations
beginalign*
P((q, q)) - P((q, c)) &= 2 \
P((c,c)) - P((q, c)) &= -4 \
P((q, q)) - P((c, q)) &= 2 \
endalign*

Observe that if one of the terms is fixed, all others are determined by the equations. Also, we only care about the change in potential and not about the value of the potential.

Suppose we set $P((c, c)) = C$. Then, by the second equality, $P((q, c)) = C+4$. By the first equality we obtain $P((q,q)) = C+6$ and finally by the last equality $P((c, q)) = C+4$. In the matrix notation from Rahul Savani's post
$$
beginpmatrix
C & C+4 \
C+4 & C+6
endpmatrix
$$

by taking $C=0$ we get the same potential matrix as Rahul Savani, but we could have taken any other real value.

share|cite|improve this answer

answered Aug 13 at 14:21

Ruben

532

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much thats the explanation I was looking for and very well structured thank you.
  – Astraeus
  Aug 14 at 23:46
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

We want to define a function $P$ from the strategy space $(c, c), (q, c), (c, q), (q, q)$ to $mathbb R$ such that when either player deviates unilaterally i.e., changes from $q$ to $c$, or from $c$ to $q$ while the other player keeps the same strategy, the change in the potential function is the same as the change in that players utility.

Say the row player plays $q$ and column player plays $c$ which means we are in $(q, c)$. The row player has payoff 10 and the column player has payoff 0. Now, if the column player changes her strategy from $c$ to $q$ that means we end up in $(q, q)$ and hence her payoff increases from 0 to 2. Then the potential should also increase by 2 i.e., $P((q, q)) - P((q, c)) = 2$. If instead the row player switched from $q$ to $c$, thus we go from $(q, c)$ to $(c,c)$, her payoff changes by -4 from 10 to 6 and since the potential has to reflect this it must be that $P((c,c)) - P((q, c)) = -4$.

Doing this also for the situation $(c, q)$ to $(q,q)$ we obtain the equations
beginalign*
P((q, q)) - P((q, c)) &= 2 \
P((c,c)) - P((q, c)) &= -4 \
P((q, q)) - P((c, q)) &= 2 \
endalign*

Observe that if one of the terms is fixed, all others are determined by the equations. Also, we only care about the change in potential and not about the value of the potential.

Suppose we set $P((c, c)) = C$. Then, by the second equality, $P((q, c)) = C+4$. By the first equality we obtain $P((q,q)) = C+6$ and finally by the last equality $P((c, q)) = C+4$. In the matrix notation from Rahul Savani's post
$$
beginpmatrix
C & C+4 \
C+4 & C+6
endpmatrix
$$

by taking $C=0$ we get the same potential matrix as Rahul Savani, but we could have taken any other real value.

share|cite|improve this answer

answered Aug 13 at 14:21

Ruben

532

We want to define a function $P$ from the strategy space $(c, c), (q, c), (c, q), (q, q)$ to $mathbb R$ such that when either player deviates unilaterally i.e., changes from $q$ to $c$, or from $c$ to $q$ while the other player keeps the same strategy, the change in the potential function is the same as the change in that players utility.

Say the row player plays $q$ and column player plays $c$ which means we are in $(q, c)$. The row player has payoff 10 and the column player has payoff 0. Now, if the column player changes her strategy from $c$ to $q$ that means we end up in $(q, q)$ and hence her payoff increases from 0 to 2. Then the potential should also increase by 2 i.e., $P((q, q)) - P((q, c)) = 2$. If instead the row player switched from $q$ to $c$, thus we go from $(q, c)$ to $(c,c)$, her payoff changes by -4 from 10 to 6 and since the potential has to reflect this it must be that $P((c,c)) - P((q, c)) = -4$.

Doing this also for the situation $(c, q)$ to $(q,q)$ we obtain the equations
beginalign*
P((q, q)) - P((q, c)) &= 2 \
P((c,c)) - P((q, c)) &= -4 \
P((q, q)) - P((c, q)) &= 2 \
endalign*

Observe that if one of the terms is fixed, all others are determined by the equations. Also, we only care about the change in potential and not about the value of the potential.

Suppose we set $P((c, c)) = C$. Then, by the second equality, $P((q, c)) = C+4$. By the first equality we obtain $P((q,q)) = C+6$ and finally by the last equality $P((c, q)) = C+4$. In the matrix notation from Rahul Savani's post
$$
beginpmatrix
C & C+4 \
C+4 & C+6
endpmatrix
$$

by taking $C=0$ we get the same potential matrix as Rahul Savani, but we could have taken any other real value.

share|cite|improve this answer

answered Aug 13 at 14:21

Ruben

532

share|cite|improve this answer

share|cite|improve this answer

answered Aug 13 at 14:21

Ruben

532

answered Aug 13 at 14:21

Ruben

532

answered Aug 13 at 14:21

Ruben

532

532

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much thats the explanation I was looking for and very well structured thank you.
  – Astraeus
  Aug 14 at 23:46
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much thats the explanation I was looking for and very well structured thank you.
  – Astraeus
  Aug 14 at 23:46
  

  
  
  
  

Thank you very much thats the explanation I was looking for and very well structured thank you.
– Astraeus
Aug 14 at 23:46

Thank you very much thats the explanation I was looking for and very well structured thank you.
– Astraeus
Aug 14 at 23:46

add a comment | 

 

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