Gaming O-F iterated prisoner's dillemma tournament - rules finalised, START YOUR AGENTS

I will send you my "Agent" this weekend. What are the proposed bounds on the number of turns? I'd like to run some basic tests to familiarize myself with the dynamics of the problem...
 
I will send you my "Agent" this weekend. What are the proposed bounds on the number of turns? I'd like to run some basic tests to familiarize myself with the dynamics of the problem...

There will be between 5 and 1000 iterations. If you wish to run tests, the entire framework is in the OP.
 
I guess I could also use psuedocode, if that's easier for you? I've never worked with Python and I'd rather be able to participate now than in two weeks after I've found time to learn the basics. :)

If you have any programming experience, you'll pick up Python in an afternoon. Very simple.
 
Reading through the OP, I see that. My problem is that I really don't have too many free afternoons, me being an engineering grad student. Regardless, I think I've figured out enough to decode the snippet in the OP.

That said, I've run a bit of an analysis using the code in the OP and it seems that the only way to "win" is to always go with Defect. I set up an experiment in which Player 2 always defects and Player 1 selects an action randomly. In no case did Player 1 ever beat Player 2, or even lose by a small margin. I slowly adjusted the probability such that Player 1 would have higher and higher chances of selecting Defect, and my scores converged on Player 2's scores.

My only thought is to ask if the code in the OP is up-to-date. I know there were discussions (above) that were talking about Jackpots and whatnot, but I didn't quite catch how that factored into each round of the tournament and the like. Some clarification would be handy.

EDIT: Here's a plot I made that describes what I'm talking about.

ipd.png


Notice that as Player 1 begins to defect more and more, the score inequality gets lower and lower. Since I've not found a strategy that can beat the Always Defect tactic, I guess I'd have to go with that. :P
 
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The code is the final version.

Consider the following: Constantly defecting means the lowest possible score. Your ranking will be based on the sum of your scores from all matches. Agents that don't defect constantly will achieve far higher scores than those who do - two agents cooperating perfectly will achieve a score of 10 each, while two constantly defecting agents will only achieve a score of 5 each. As long as there are at least two agents capable of mutual cooperation while punishing defection, they will come out on top of defectors, because they'll only lose by a small margin to them.
 
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The code is the final version.

Consider the following: Constantly defecting means the lowest possible score.

The lowest possible score is 1 per turn, the result of choosing to cooperate, when the opponent defects - even when this happens every turn.

The trick is to play nice when it pays out, punish, when you need to punish (= minimize losses against cheating players), and cheat, when it offers you advantages. Make Machiavelli proud of you. :lol:
 
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