My Math Forum > Math I've worked out how to beat Nash's Equilibrium strategies without co-operation...

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 December 20th, 2017, 07:43 AM #1 Newbie   Joined: Apr 2015 From: England Posts: 16 Thanks: 0 I've worked out how to beat Nash's Equilibrium strategies without co-operation... ... It is very simple logic, but I think it is new. (It only works in certain games). I cant find anybody who understands it. Can you guys?? I'm a poker player, and in Poker we use either equilibrium strategies or something called exploitative strategies. It's well known that exploitative strategies will take chips from imperfect players quicker than N.E. And so, if we are in a tournament with a mixed field of players, an exploitative player will take chips from the imperfect players faster than the N.E player will. I will exploit the weaker players, take their chips, and then when I face the N.E. player I will have more chips and so will hold the advantage. Instead of co-operating with the other players to beat N.E. you can USE the other players to beat it.
 December 20th, 2017, 07:23 PM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 642 Thanks: 406 Math Focus: Dynamical systems, analytic function theory, numerics I think there isn't much for me to say here that people didn't already tell you on reddit. https://www.reddit.com/r/badmathematics/ In short: At present, you clearly do not understand what a Nash equilibrium is whatsoever. If you want to learn math, then spend time reading math.
 December 20th, 2017, 07:40 PM #3 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 This guy is a nut case...see his previous posts...
 December 21st, 2017, 07:00 AM #4 Newbie   Joined: Apr 2015 From: England Posts: 16 Thanks: 0 Lol I just looked at some of my older posts on here. All my curiosities find their way to threads like these... But anyway... On Reddit they seem to say that this is clearly co-operative. Perhaps this falls under implicit co-operation in repeated games??? To confirm we are on the same page... There is only one N.E. solution to Holdem Poker? If someone were to be using this solution then I would not be able to deviate from an N.E. strategy with any benefit to myself? Even if a weaker player were involved. I still would not be able to deviate without any co-operation from anyone else for any kind of gain to myself? Now... If we are in a repeated game, and the N.E player was not involved in this particular 'hand', we ourselves can divert from N.E. to take advantage of the weaker player. When we then come to face the N.E. player. We will hold the advantage. If my terminology is somehow wrong do tell me where? It seems to me like mathematicians dont have a clue how to use their formulas in real life situations.
 December 21st, 2017, 08:40 AM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,681 Thanks: 2659 Math Focus: Mainly analysis and algebra I would be very surprised if poker had a (known) equilibrium strategy that wasn't in part a random choice. But I don't think you actually understand what an equilibrium strategy is. Nor do I think you know what you are actually doing: namely you take more risks earlier on with a view to having a stronger position from which to play conservatively. I'm sure you don't have enough solid data on which to base any claim that your approach is better in the long run. I'm also sure that any equilibrium strategy will assume perfect players (no "tells" and no errors in probabilistic calculations) who simply don't exist. It's certainly very clear that you don't understand the process of mathematical modelling that you are attacking.

 Tags beat, cooperation, equilibrium, game theory, nash, strategies, worked

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