3 ms·

I’m pretty sure that’s incorrect. I haven’t had time to look through the text yet but it sounds like you’re confusing the one-shot equilibrium (1/3 probability

by qubex 9y ago

I’m pretty sure that’s incorrect. I haven’t had time to look through the text yet but it sounds like you’re confusing the one-shot equilibrium (1/3 probability of playing any one of the three available strategies), each deploying simultaneously, and the iterated game scenario (in which one player could be random, and you argue the other could continually play the same strategy repeatedly, but then the other player would latch on and play the matching countermove, pushing them both back towards randomness). EDIT: It’s not nice to vote people down just because they politely point out you’re demonstrably wrong.

deleted 9y ago
[deleted]
bluetwo 9y ago
I haven't down voted anyone. Setup your own experiment and let me know what you find.
- nostromo95 9y ago
  You’re mixing up equilibrium payoff and Nash Equilibrium (which is a strategy profile). Yes, the payoffs of player A randomizing and player B playing scissors is the same as both players randomizing, but observe that this profile of strategies fails the ex-post facto definition of Nash Equlibrium, i.e. in retrospect player A would want to change their strategy choice to “play rock every time.”
- qubex 9y ago
  Then I apologise for thinking it had been you. I’m going to sound very arrogant but: I don’t need to set up the scenario you describe because optimal strategies in identical games have been proven since the 1940s when Oskar Morgenstern and John Von Neumann published their seminal Theory of Games and Economic Behaviour (1944). If you are getting results that are in conflict with that which is mathematically proven (provided certain circumstantial criteria are met), then you can be sure that you have not exactly matched the circumstantial criteria assumed as a perquisite for the proof. So basically you’re in another situation where the theorem makes no claim to hold and, unsurprisingly, doesn’t.
- Yeikoff 9y ago
  Lets do then. Player 1 randomizes (1/3, 1/3, 1/3). Player 2 randomizes (1, 0, 0). Does player 1 want to deviate? Yes! Randomizing is not the best strategy if player 2 is going to be playing the same pure strategy all the time, player 1 would be better off if she were to play the best response to whatever (1, 0, 0) is. -> Not an equilibrium.
privong 9y ago
> EDIT: It’s not nice to vote people down just because they politely point out you’re demonstrably wrong. HN doesn't let you downvote the replies to your own comments or top-level comments for your own article submissions.

by qubex 9y ago

deleted 9y ago

[deleted]

bluetwo 9y ago

I haven't down voted anyone. Setup your own experiment and let me know what you find.

nostromo95 9y ago
You’re mixing up equilibrium payoff and Nash Equilibrium (which is a strategy profile). Yes, the payoffs of player A randomizing and player B playing scissors is the same as both players randomizing, but observe that this profile of strategies fails the ex-post facto definition of Nash Equlibrium, i.e. in retrospect player A would want to change their strategy choice to “play rock every time.”
qubex 9y ago
Then I apologise for thinking it had been you. I’m going to sound very arrogant but: I don’t need to set up the scenario you describe because optimal strategies in identical games have been proven since the 1940s when Oskar Morgenstern and John Von Neumann published their seminal Theory of Games and Economic Behaviour (1944). If you are getting results that are in conflict with that which is mathematically proven (provided certain circumstantial criteria are met), then you can be sure that you have not exactly matched the circumstantial criteria assumed as a perquisite for the proof. So basically you’re in another situation where the theorem makes no claim to hold and, unsurprisingly, doesn’t.
Yeikoff 9y ago
Lets do then. Player 1 randomizes (1/3, 1/3, 1/3). Player 2 randomizes (1, 0, 0). Does player 1 want to deviate? Yes! Randomizing is not the best strategy if player 2 is going to be playing the same pure strategy all the time, player 1 would be better off if she were to play the best response to whatever (1, 0, 0) is. -> Not an equilibrium.

privong 9y ago

> EDIT: It’s not nice to vote people down just because they politely point out you’re demonstrably wrong. HN doesn't let you downvote the replies to your own comments or top-level comments for your own article submissions.