Game theory: Is there always only one dominant strategy?
Clash Royale CLAN TAG#URR8PPP
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I get the impression that there are two classes of games, ones that have been solved and for whom a dominant strategy exists, and ones that have not been solved and for whom a myriad of strong but not necessarily dominant strategies exist.
Is it ever the case that a game has more than one non-equivalent dominant strategy? Or do all cases of games with multiple strong strategies belong to the set of games that have not been solved?
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I get the impression that there are two classes of games, ones that have been solved and for whom a dominant strategy exists, and ones that have not been solved and for whom a myriad of strong but not necessarily dominant strategies exist.
Is it ever the case that a game has more than one non-equivalent dominant strategy? Or do all cases of games with multiple strong strategies belong to the set of games that have not been solved?
game-theory
Are you talking about two person zero sum games with finitely many strategies for each player or about something more general?
â Hans Engler
Sep 10 at 20:19
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up vote
0
down vote
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I get the impression that there are two classes of games, ones that have been solved and for whom a dominant strategy exists, and ones that have not been solved and for whom a myriad of strong but not necessarily dominant strategies exist.
Is it ever the case that a game has more than one non-equivalent dominant strategy? Or do all cases of games with multiple strong strategies belong to the set of games that have not been solved?
game-theory
I get the impression that there are two classes of games, ones that have been solved and for whom a dominant strategy exists, and ones that have not been solved and for whom a myriad of strong but not necessarily dominant strategies exist.
Is it ever the case that a game has more than one non-equivalent dominant strategy? Or do all cases of games with multiple strong strategies belong to the set of games that have not been solved?
game-theory
game-theory
asked Sep 10 at 20:13
Ingolifs
1215
1215
Are you talking about two person zero sum games with finitely many strategies for each player or about something more general?
â Hans Engler
Sep 10 at 20:19
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Are you talking about two person zero sum games with finitely many strategies for each player or about something more general?
â Hans Engler
Sep 10 at 20:19
Are you talking about two person zero sum games with finitely many strategies for each player or about something more general?
â Hans Engler
Sep 10 at 20:19
Are you talking about two person zero sum games with finitely many strategies for each player or about something more general?
â Hans Engler
Sep 10 at 20:19
add a comment |Â
1 Answer
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One cannot have two strictly dominant strategies $sigma_i$ and $sigma_i'$ because as $sigma_i$ is dominant, for all opposing strategies tuples $sigma_-i$:
$$
u_i(sigma_i, sigma_-i) > u_i(sigma_i', sigma_-i)
$$
but as $sigma_i'$ is dominant, for all $sigma_-i$:
$$
u_i(sigma_i', sigma_-i) > u_i(sigma_i, sigma_-i)
$$
which poses a contradiction. If you mean 'weakly dominant,' so long as your definition of weak dominance is standard (i.e. requires at least one strict comparison), the same argument holds.
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
One cannot have two strictly dominant strategies $sigma_i$ and $sigma_i'$ because as $sigma_i$ is dominant, for all opposing strategies tuples $sigma_-i$:
$$
u_i(sigma_i, sigma_-i) > u_i(sigma_i', sigma_-i)
$$
but as $sigma_i'$ is dominant, for all $sigma_-i$:
$$
u_i(sigma_i', sigma_-i) > u_i(sigma_i, sigma_-i)
$$
which poses a contradiction. If you mean 'weakly dominant,' so long as your definition of weak dominance is standard (i.e. requires at least one strict comparison), the same argument holds.
add a comment |Â
up vote
1
down vote
One cannot have two strictly dominant strategies $sigma_i$ and $sigma_i'$ because as $sigma_i$ is dominant, for all opposing strategies tuples $sigma_-i$:
$$
u_i(sigma_i, sigma_-i) > u_i(sigma_i', sigma_-i)
$$
but as $sigma_i'$ is dominant, for all $sigma_-i$:
$$
u_i(sigma_i', sigma_-i) > u_i(sigma_i, sigma_-i)
$$
which poses a contradiction. If you mean 'weakly dominant,' so long as your definition of weak dominance is standard (i.e. requires at least one strict comparison), the same argument holds.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
One cannot have two strictly dominant strategies $sigma_i$ and $sigma_i'$ because as $sigma_i$ is dominant, for all opposing strategies tuples $sigma_-i$:
$$
u_i(sigma_i, sigma_-i) > u_i(sigma_i', sigma_-i)
$$
but as $sigma_i'$ is dominant, for all $sigma_-i$:
$$
u_i(sigma_i', sigma_-i) > u_i(sigma_i, sigma_-i)
$$
which poses a contradiction. If you mean 'weakly dominant,' so long as your definition of weak dominance is standard (i.e. requires at least one strict comparison), the same argument holds.
One cannot have two strictly dominant strategies $sigma_i$ and $sigma_i'$ because as $sigma_i$ is dominant, for all opposing strategies tuples $sigma_-i$:
$$
u_i(sigma_i, sigma_-i) > u_i(sigma_i', sigma_-i)
$$
but as $sigma_i'$ is dominant, for all $sigma_-i$:
$$
u_i(sigma_i', sigma_-i) > u_i(sigma_i, sigma_-i)
$$
which poses a contradiction. If you mean 'weakly dominant,' so long as your definition of weak dominance is standard (i.e. requires at least one strict comparison), the same argument holds.
answered Sep 10 at 20:18
Pete Caradonna
1,3791720
1,3791720
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Are you talking about two person zero sum games with finitely many strategies for each player or about something more general?
â Hans Engler
Sep 10 at 20:19