Pure Nash equilibrium in Zero sum [closed]
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Do zero sum games have a pure Nash equilibrium and if so how do I find the pure Nash equilibrium
game-theory nash-equilibrium algorithmic-game-theory
asked Dec 20 '18 at 20:18
Tommy jones Tommy jones
12
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closed as off-topic by mrtaurho, Frpzzd, Henrik, amWhy, user10354138 Dec 20 '18 at 23:12
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$begingroup$
Do zero sum games have a pure Nash equilibrium and if so how do I find the pure Nash equilibrium
game-theory nash-equilibrium algorithmic-game-theory
asked Dec 20 '18 at 20:18
Tommy jones Tommy jones
12
$endgroup$
closed as off-topic by mrtaurho, Frpzzd, Henrik, amWhy, user10354138 Dec 20 '18 at 23:12
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, Frpzzd, Henrik, amWhy, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Do zero sum games have a pure Nash equilibrium and if so how do I find the pure Nash equilibrium
game-theory nash-equilibrium algorithmic-game-theory
asked Dec 20 '18 at 20:18
Tommy jones Tommy jones
12
$endgroup$
Do zero sum games have a pure Nash equilibrium and if so how do I find the pure Nash equilibrium
game-theory nash-equilibrium algorithmic-game-theory
game-theory nash-equilibrium algorithmic-game-theory
asked Dec 20 '18 at 20:18
Tommy jones Tommy jones
12
asked Dec 20 '18 at 20:18
Tommy jones Tommy jones
12
asked Dec 20 '18 at 20:18
Tommy jones Tommy jones
12
asked Dec 20 '18 at 20:18
Tommy jones Tommy jones
12
asked Dec 20 '18 at 20:18
Tommy jones Tommy jones
12
12
closed as off-topic by mrtaurho, Frpzzd, Henrik, amWhy, user10354138 Dec 20 '18 at 23:12
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, Frpzzd, Henrik, amWhy, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by mrtaurho, Frpzzd, Henrik, amWhy, user10354138 Dec 20 '18 at 23:12
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, Frpzzd, Henrik, amWhy, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
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1 Answer
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Unfortunately not every zero sum game needs to have a pure Nash equilibrium; you can see this quite easily from the example of matching pennies.
Computing Nash equilibria is a hard problem in general, but for pure equilibria it turns out to be quite easy. From the definition, a pure Nash equilibrium is a strategy profile in which no player can increase their utility by unilaterally changing their strategy (i.e. no one can deviate by themselves in order to get a better outcome). If we only consider pure strategies (and two player games in bimatrix form with finite strategy spaces), then a particular strategy profile is a pure equilibrium if the row players payout is the largest in that column while simultaneously the column players payout is the largest in that row. You can generalize that idea to more than two players readily enough.
Finding all pure equilibria then becomes polynomial in the size of the game, which is about as much as you can hope for.
answered Dec 20 '18 at 20:33
DeficientMathDudeDeficientMathDude
1212
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Unfortunately not every zero sum game needs to have a pure Nash equilibrium; you can see this quite easily from the example of matching pennies.
Computing Nash equilibria is a hard problem in general, but for pure equilibria it turns out to be quite easy. From the definition, a pure Nash equilibrium is a strategy profile in which no player can increase their utility by unilaterally changing their strategy (i.e. no one can deviate by themselves in order to get a better outcome). If we only consider pure strategies (and two player games in bimatrix form with finite strategy spaces), then a particular strategy profile is a pure equilibrium if the row players payout is the largest in that column while simultaneously the column players payout is the largest in that row. You can generalize that idea to more than two players readily enough.
Finding all pure equilibria then becomes polynomial in the size of the game, which is about as much as you can hope for.
answered Dec 20 '18 at 20:33
DeficientMathDudeDeficientMathDude
1212
$endgroup$
add a comment |
$begingroup$
Unfortunately not every zero sum game needs to have a pure Nash equilibrium; you can see this quite easily from the example of matching pennies.
Computing Nash equilibria is a hard problem in general, but for pure equilibria it turns out to be quite easy. From the definition, a pure Nash equilibrium is a strategy profile in which no player can increase their utility by unilaterally changing their strategy (i.e. no one can deviate by themselves in order to get a better outcome). If we only consider pure strategies (and two player games in bimatrix form with finite strategy spaces), then a particular strategy profile is a pure equilibrium if the row players payout is the largest in that column while simultaneously the column players payout is the largest in that row. You can generalize that idea to more than two players readily enough.
Finding all pure equilibria then becomes polynomial in the size of the game, which is about as much as you can hope for.
answered Dec 20 '18 at 20:33
DeficientMathDudeDeficientMathDude
1212
$endgroup$
add a comment |
$begingroup$
Unfortunately not every zero sum game needs to have a pure Nash equilibrium; you can see this quite easily from the example of matching pennies.
Computing Nash equilibria is a hard problem in general, but for pure equilibria it turns out to be quite easy. From the definition, a pure Nash equilibrium is a strategy profile in which no player can increase their utility by unilaterally changing their strategy (i.e. no one can deviate by themselves in order to get a better outcome). If we only consider pure strategies (and two player games in bimatrix form with finite strategy spaces), then a particular strategy profile is a pure equilibrium if the row players payout is the largest in that column while simultaneously the column players payout is the largest in that row. You can generalize that idea to more than two players readily enough.
Finding all pure equilibria then becomes polynomial in the size of the game, which is about as much as you can hope for.
answered Dec 20 '18 at 20:33
DeficientMathDudeDeficientMathDude
1212
$endgroup$
Unfortunately not every zero sum game needs to have a pure Nash equilibrium; you can see this quite easily from the example of matching pennies.
Computing Nash equilibria is a hard problem in general, but for pure equilibria it turns out to be quite easy. From the definition, a pure Nash equilibrium is a strategy profile in which no player can increase their utility by unilaterally changing their strategy (i.e. no one can deviate by themselves in order to get a better outcome). If we only consider pure strategies (and two player games in bimatrix form with finite strategy spaces), then a particular strategy profile is a pure equilibrium if the row players payout is the largest in that column while simultaneously the column players payout is the largest in that row. You can generalize that idea to more than two players readily enough.
Finding all pure equilibria then becomes polynomial in the size of the game, which is about as much as you can hope for.
answered Dec 20 '18 at 20:33
DeficientMathDudeDeficientMathDude
1212
answered Dec 20 '18 at 20:33
DeficientMathDudeDeficientMathDude
1212
answered Dec 20 '18 at 20:33
DeficientMathDudeDeficientMathDude
1212
answered Dec 20 '18 at 20:33
DeficientMathDudeDeficientMathDude
1212
1212
add a comment |
add a comment |
