Is the game of rock,paper scissors considered solved according to game theory?Game Theory - Centipede GameNash Equilibria for zero-sum games (Rock Paper Scissors)Mixed Strategy Nash Equilibrium of Rock Paper Scissors with 3 players?Is there no Pareto-optimal Shotgun game?rock, paper, scissors, wellOptimal strategy for Jackpot Rock Paper ScissorsIs there a name for the ratio between the optimal social-welfare equilibrium and the worst social-welfare equilibrium of a strategic game?“Unbalanced” variation of rock-paper-scissorsOptimal strategy for meta paper-scissor-rockSimple continuous game or not?
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Is the game of rock,paper scissors considered solved according to game theory?
Game Theory - Centipede GameNash Equilibria for zero-sum games (Rock Paper Scissors)Mixed Strategy Nash Equilibrium of Rock Paper Scissors with 3 players?Is there no Pareto-optimal Shotgun game?rock, paper, scissors, wellOptimal strategy for Jackpot Rock Paper ScissorsIs there a name for the ratio between the optimal social-welfare equilibrium and the worst social-welfare equilibrium of a strategic game?“Unbalanced” variation of rock-paper-scissorsOptimal strategy for meta paper-scissor-rockSimple continuous game or not?
$begingroup$
Rock, paper, scissors definitely has an optimal strategy of just choose randomly for each toss. And two players using that strategy are in Nash equilibrium.
What I'm wondering is if that's the same as the game being "solved", or is there a distinction between the two concepts? After all you still can't predict the outcome of the game
game-theory nash-equilibrium
asked Mar 30 at 2:35
user1984974user1984974
61
$endgroup$
add a comment |
$begingroup$
Rock, paper, scissors definitely has an optimal strategy of just choose randomly for each toss. And two players using that strategy are in Nash equilibrium.
What I'm wondering is if that's the same as the game being "solved", or is there a distinction between the two concepts? After all you still can't predict the outcome of the game
game-theory nash-equilibrium
asked Mar 30 at 2:35
user1984974user1984974
61
$endgroup$
$begingroup$
Nash equilibria are not optimal! If you know your opponent always plays rock, you should always play paper, not the Nash equilibrium.
$endgroup$
– Qiaochu Yuan
Mar 30 at 8:36
add a comment |
$begingroup$
Rock, paper, scissors definitely has an optimal strategy of just choose randomly for each toss. And two players using that strategy are in Nash equilibrium.
What I'm wondering is if that's the same as the game being "solved", or is there a distinction between the two concepts? After all you still can't predict the outcome of the game
game-theory nash-equilibrium
asked Mar 30 at 2:35
user1984974user1984974
61
$endgroup$
Rock, paper, scissors definitely has an optimal strategy of just choose randomly for each toss. And two players using that strategy are in Nash equilibrium.
What I'm wondering is if that's the same as the game being "solved", or is there a distinction between the two concepts? After all you still can't predict the outcome of the game
game-theory nash-equilibrium
game-theory nash-equilibrium
asked Mar 30 at 2:35
user1984974user1984974
61
asked Mar 30 at 2:35
user1984974user1984974
61
asked Mar 30 at 2:35
user1984974user1984974
61
asked Mar 30 at 2:35
user1984974user1984974
61
asked Mar 30 at 2:35
user1984974user1984974
61
61
$begingroup$
Nash equilibria are not optimal! If you know your opponent always plays rock, you should always play paper, not the Nash equilibrium.
$endgroup$
– Qiaochu Yuan
Mar 30 at 8:36
add a comment |
$begingroup$
Nash equilibria are not optimal! If you know your opponent always plays rock, you should always play paper, not the Nash equilibrium.
$endgroup$
– Qiaochu Yuan
Mar 30 at 8:36
$begingroup$
Nash equilibria are not optimal! If you know your opponent always plays rock, you should always play paper, not the Nash equilibrium.
$endgroup$
– Qiaochu Yuan
Mar 30 at 8:36
$begingroup$
Nash equilibria are not optimal! If you know your opponent always plays rock, you should always play paper, not the Nash equilibrium.
$endgroup$
– Qiaochu Yuan
Mar 30 at 8:36
add a comment |
1 Answer
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Yes, it is solved because it has been proven that no other strategy is superior to random choice on the assumption that you do not rely on the play of your opponent. Say your opponent plays rock the first $20$ times. There is a temptation to assume that your opponent will always play rock, so you should play paper. The game theory I learned considers that out of bounds and passes that question to the psychologists or somebody else. Game theory is predicated on the best strategy for you, either ignoring what your opponent does or assuming your opponent is seeking his/her best advantage. It does not ask that you predict the outcome of one game, but does ask that you predict the average outcome of a long series of games.
answered Mar 30 at 2:44
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
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Yes, it is solved because it has been proven that no other strategy is superior to random choice on the assumption that you do not rely on the play of your opponent. Say your opponent plays rock the first $20$ times. There is a temptation to assume that your opponent will always play rock, so you should play paper. The game theory I learned considers that out of bounds and passes that question to the psychologists or somebody else. Game theory is predicated on the best strategy for you, either ignoring what your opponent does or assuming your opponent is seeking his/her best advantage. It does not ask that you predict the outcome of one game, but does ask that you predict the average outcome of a long series of games.
answered Mar 30 at 2:44
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
$begingroup$
Yes, it is solved because it has been proven that no other strategy is superior to random choice on the assumption that you do not rely on the play of your opponent. Say your opponent plays rock the first $20$ times. There is a temptation to assume that your opponent will always play rock, so you should play paper. The game theory I learned considers that out of bounds and passes that question to the psychologists or somebody else. Game theory is predicated on the best strategy for you, either ignoring what your opponent does or assuming your opponent is seeking his/her best advantage. It does not ask that you predict the outcome of one game, but does ask that you predict the average outcome of a long series of games.
answered Mar 30 at 2:44
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
$begingroup$
Yes, it is solved because it has been proven that no other strategy is superior to random choice on the assumption that you do not rely on the play of your opponent. Say your opponent plays rock the first $20$ times. There is a temptation to assume that your opponent will always play rock, so you should play paper. The game theory I learned considers that out of bounds and passes that question to the psychologists or somebody else. Game theory is predicated on the best strategy for you, either ignoring what your opponent does or assuming your opponent is seeking his/her best advantage. It does not ask that you predict the outcome of one game, but does ask that you predict the average outcome of a long series of games.
answered Mar 30 at 2:44
Ross MillikanRoss Millikan
301k24200375
$endgroup$
Yes, it is solved because it has been proven that no other strategy is superior to random choice on the assumption that you do not rely on the play of your opponent. Say your opponent plays rock the first $20$ times. There is a temptation to assume that your opponent will always play rock, so you should play paper. The game theory I learned considers that out of bounds and passes that question to the psychologists or somebody else. Game theory is predicated on the best strategy for you, either ignoring what your opponent does or assuming your opponent is seeking his/her best advantage. It does not ask that you predict the outcome of one game, but does ask that you predict the average outcome of a long series of games.
answered Mar 30 at 2:44
Ross MillikanRoss Millikan
301k24200375
answered Mar 30 at 2:44
Ross MillikanRoss Millikan
301k24200375
answered Mar 30 at 2:44
Ross MillikanRoss Millikan
301k24200375
answered Mar 30 at 2:44
Ross MillikanRoss Millikan
301k24200375
301k24200375
add a comment |
add a comment |
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Nash equilibria are not optimal! If you know your opponent always plays rock, you should always play paper, not the Nash equilibrium.
$endgroup$
– Qiaochu Yuan
Mar 30 at 8:36