Game theory: Is there always only one dominant strategy?

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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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  • Are you talking about two person zero sum games with finitely many strategies for each player or about something more general?
    – Hans Engler
    Sep 10 at 20:19














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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?










share|cite|improve this question





















  • Are you talking about two person zero sum games with finitely many strategies for each player or about something more general?
    – Hans Engler
    Sep 10 at 20:19












up vote
0
down vote

favorite









up vote
0
down vote

favorite











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?










share|cite|improve this question













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






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asked Sep 10 at 20:13









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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?
    – 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?
    – 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?
– 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?
– Hans Engler
Sep 10 at 20:19










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
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    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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      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.






      share|cite|improve this answer






















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        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.






        share|cite|improve this answer












        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 10 at 20:18









        Pete Caradonna

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