Mixing
What is the format of a mixed strategy nash equilibrium?
$begingroup$
Take this game where player 1 has choices T and B, while player 2 has choices L and R.
We need to find the Mixed Strategy Nash Equilibria. As a side note, it seems like (B,L), and (T,R) are Pure Strategy Nash Equilibria (correct me if I'm wrong).
For player 1, I find the expected payout if he chooses T or B, assuming P2 (player 2) chooses L and R with probability q and 1-q, respectively. The payout then becomes 0. E(B) subsequently becomes 2q. So a MSNE occurs if E(B) = E(T), which implies q = 0.
Doing the same process, E(L)= E(R) if p = 1/2.
So I get as MSNE ((.5, .5), (0,1)). My concern is that player 2 chooses R always here, which means it's a Pure Strategy(?). What then is the MSNE? I am not sure what is the format response when a question asks for a Mixed Strategy Nash Equilibria. A final question is why is the question phrased in plurality (equilibria)? It seems like there is only one mixed strategy.
Apologies for the many questions, but I think I understand the process for solving, but need some intuition on the terminology and the formatting of MSNE.
nash-equilibrium
asked Jan 27 at 5:17
MathGuyForLifeMathGuyForLife
729
$endgroup$
add a comment |
$begingroup$
Take this game where player 1 has choices T and B, while player 2 has choices L and R.
We need to find the Mixed Strategy Nash Equilibria. As a side note, it seems like (B,L), and (T,R) are Pure Strategy Nash Equilibria (correct me if I'm wrong).
For player 1, I find the expected payout if he chooses T or B, assuming P2 (player 2) chooses L and R with probability q and 1-q, respectively. The payout then becomes 0. E(B) subsequently becomes 2q. So a MSNE occurs if E(B) = E(T), which implies q = 0.
Doing the same process, E(L)= E(R) if p = 1/2.
So I get as MSNE ((.5, .5), (0,1)). My concern is that player 2 chooses R always here, which means it's a Pure Strategy(?). What then is the MSNE? I am not sure what is the format response when a question asks for a Mixed Strategy Nash Equilibria. A final question is why is the question phrased in plurality (equilibria)? It seems like there is only one mixed strategy.
Apologies for the many questions, but I think I understand the process for solving, but need some intuition on the terminology and the formatting of MSNE.
nash-equilibrium
asked Jan 27 at 5:17
MathGuyForLifeMathGuyForLife
729
$endgroup$
add a comment |
$begingroup$
Take this game where player 1 has choices T and B, while player 2 has choices L and R.
We need to find the Mixed Strategy Nash Equilibria. As a side note, it seems like (B,L), and (T,R) are Pure Strategy Nash Equilibria (correct me if I'm wrong).
For player 1, I find the expected payout if he chooses T or B, assuming P2 (player 2) chooses L and R with probability q and 1-q, respectively. The payout then becomes 0. E(B) subsequently becomes 2q. So a MSNE occurs if E(B) = E(T), which implies q = 0.
Doing the same process, E(L)= E(R) if p = 1/2.
So I get as MSNE ((.5, .5), (0,1)). My concern is that player 2 chooses R always here, which means it's a Pure Strategy(?). What then is the MSNE? I am not sure what is the format response when a question asks for a Mixed Strategy Nash Equilibria. A final question is why is the question phrased in plurality (equilibria)? It seems like there is only one mixed strategy.
Apologies for the many questions, but I think I understand the process for solving, but need some intuition on the terminology and the formatting of MSNE.
nash-equilibrium
asked Jan 27 at 5:17
MathGuyForLifeMathGuyForLife
729
$endgroup$
Take this game where player 1 has choices T and B, while player 2 has choices L and R.
We need to find the Mixed Strategy Nash Equilibria. As a side note, it seems like (B,L), and (T,R) are Pure Strategy Nash Equilibria (correct me if I'm wrong).
For player 1, I find the expected payout if he chooses T or B, assuming P2 (player 2) chooses L and R with probability q and 1-q, respectively. The payout then becomes 0. E(B) subsequently becomes 2q. So a MSNE occurs if E(B) = E(T), which implies q = 0.
Doing the same process, E(L)= E(R) if p = 1/2.
So I get as MSNE ((.5, .5), (0,1)). My concern is that player 2 chooses R always here, which means it's a Pure Strategy(?). What then is the MSNE? I am not sure what is the format response when a question asks for a Mixed Strategy Nash Equilibria. A final question is why is the question phrased in plurality (equilibria)? It seems like there is only one mixed strategy.
Apologies for the many questions, but I think I understand the process for solving, but need some intuition on the terminology and the formatting of MSNE.
nash-equilibrium
nash-equilibrium
asked Jan 27 at 5:17
MathGuyForLifeMathGuyForLife
729
asked Jan 27 at 5:17
MathGuyForLifeMathGuyForLife
729
asked Jan 27 at 5:17
MathGuyForLifeMathGuyForLife
729
asked Jan 27 at 5:17
MathGuyForLifeMathGuyForLife
729
asked Jan 27 at 5:17
MathGuyForLifeMathGuyForLife
729
729
add a comment |
add a comment |
1 Answer
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$begingroup$
First off, you are indeed correct that $(B,L)$ and $(T,R)$ are pure equilibria. Next, note that according to the definition in Wikipedia:
https://en.wikipedia.org/wiki/Strategy_(game_theory)#Mixed_strategy
a mixed strategy equilibrium is one in which at least one player is using a mixed strategy (so it's OK if the other is playing a pure strategy).
We can then follow the process you suggest to find the other mixed equilibria. We can first observe that if player 2 plays $L$ with positive probability, then player 1's unique best response is to play $B$. Continuing this line of thought, if player 1 plays $B$ with probability $1$, then player 2's unique best response is to play $L$, leading to the pure equilibrium $(B,L)$ that you already found.
So remaining mixed equilibria must have player 2 playing $R$ with probability $1$. In this case, player 1 is indifferent between his options, and so any mixed strategy profile of the form $(p,1-p),(0,1)$ with $p in (0,1)$ is a mixed equilibrium.
To summarize: The mixed equilibria are ${((p,1-p),(0,1)) : p in (0,1) }$. In particular, there are infinitely many MSNEs.
answered Feb 5 at 18:50
MicappsMicapps
1,15739
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$begingroup$
First off, you are indeed correct that $(B,L)$ and $(T,R)$ are pure equilibria. Next, note that according to the definition in Wikipedia:
https://en.wikipedia.org/wiki/Strategy_(game_theory)#Mixed_strategy
a mixed strategy equilibrium is one in which at least one player is using a mixed strategy (so it's OK if the other is playing a pure strategy).
We can then follow the process you suggest to find the other mixed equilibria. We can first observe that if player 2 plays $L$ with positive probability, then player 1's unique best response is to play $B$. Continuing this line of thought, if player 1 plays $B$ with probability $1$, then player 2's unique best response is to play $L$, leading to the pure equilibrium $(B,L)$ that you already found.
So remaining mixed equilibria must have player 2 playing $R$ with probability $1$. In this case, player 1 is indifferent between his options, and so any mixed strategy profile of the form $(p,1-p),(0,1)$ with $p in (0,1)$ is a mixed equilibrium.
To summarize: The mixed equilibria are ${((p,1-p),(0,1)) : p in (0,1) }$. In particular, there are infinitely many MSNEs.
answered Feb 5 at 18:50
MicappsMicapps
1,15739
$endgroup$
add a comment |
$begingroup$
First off, you are indeed correct that $(B,L)$ and $(T,R)$ are pure equilibria. Next, note that according to the definition in Wikipedia:
https://en.wikipedia.org/wiki/Strategy_(game_theory)#Mixed_strategy
a mixed strategy equilibrium is one in which at least one player is using a mixed strategy (so it's OK if the other is playing a pure strategy).
We can then follow the process you suggest to find the other mixed equilibria. We can first observe that if player 2 plays $L$ with positive probability, then player 1's unique best response is to play $B$. Continuing this line of thought, if player 1 plays $B$ with probability $1$, then player 2's unique best response is to play $L$, leading to the pure equilibrium $(B,L)$ that you already found.
So remaining mixed equilibria must have player 2 playing $R$ with probability $1$. In this case, player 1 is indifferent between his options, and so any mixed strategy profile of the form $(p,1-p),(0,1)$ with $p in (0,1)$ is a mixed equilibrium.
To summarize: The mixed equilibria are ${((p,1-p),(0,1)) : p in (0,1) }$. In particular, there are infinitely many MSNEs.
answered Feb 5 at 18:50
MicappsMicapps
1,15739
$endgroup$
add a comment |
$begingroup$
First off, you are indeed correct that $(B,L)$ and $(T,R)$ are pure equilibria. Next, note that according to the definition in Wikipedia:
https://en.wikipedia.org/wiki/Strategy_(game_theory)#Mixed_strategy
a mixed strategy equilibrium is one in which at least one player is using a mixed strategy (so it's OK if the other is playing a pure strategy).
We can then follow the process you suggest to find the other mixed equilibria. We can first observe that if player 2 plays $L$ with positive probability, then player 1's unique best response is to play $B$. Continuing this line of thought, if player 1 plays $B$ with probability $1$, then player 2's unique best response is to play $L$, leading to the pure equilibrium $(B,L)$ that you already found.
So remaining mixed equilibria must have player 2 playing $R$ with probability $1$. In this case, player 1 is indifferent between his options, and so any mixed strategy profile of the form $(p,1-p),(0,1)$ with $p in (0,1)$ is a mixed equilibrium.
To summarize: The mixed equilibria are ${((p,1-p),(0,1)) : p in (0,1) }$. In particular, there are infinitely many MSNEs.
answered Feb 5 at 18:50
MicappsMicapps
1,15739
$endgroup$
First off, you are indeed correct that $(B,L)$ and $(T,R)$ are pure equilibria. Next, note that according to the definition in Wikipedia:
https://en.wikipedia.org/wiki/Strategy_(game_theory)#Mixed_strategy
a mixed strategy equilibrium is one in which at least one player is using a mixed strategy (so it's OK if the other is playing a pure strategy).
We can then follow the process you suggest to find the other mixed equilibria. We can first observe that if player 2 plays $L$ with positive probability, then player 1's unique best response is to play $B$. Continuing this line of thought, if player 1 plays $B$ with probability $1$, then player 2's unique best response is to play $L$, leading to the pure equilibrium $(B,L)$ that you already found.
So remaining mixed equilibria must have player 2 playing $R$ with probability $1$. In this case, player 1 is indifferent between his options, and so any mixed strategy profile of the form $(p,1-p),(0,1)$ with $p in (0,1)$ is a mixed equilibrium.
To summarize: The mixed equilibria are ${((p,1-p),(0,1)) : p in (0,1) }$. In particular, there are infinitely many MSNEs.
answered Feb 5 at 18:50
MicappsMicapps
1,15739
edited Feb 5 at 18:58
answered Feb 5 at 18:50
MicappsMicapps
1,15739
answered Feb 5 at 18:50
MicappsMicapps
1,15739
answered Feb 5 at 18:50
MicappsMicapps
1,15739
1,15739
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