regardless of what my clone does I’m still always better off defecting.
This is broken counterfactual reasoning.
Assume it’s not a perfect clone, it can defect with probability p even if you cooperate. Then apply CDT. You get “defect” for any p>0. So it is reasonable to implicitly assume continuity and declare that CDT forces you to defect when p=0. However, if you apply CDT for the case p=0 directly, you get “cooperate” instead.
In other words, the conterfactual reasoning gets broken when the map CDT(p, PD) is not continuous at the point p=0.
I disagree. If the agent has a 95% probability of doing the same thing as me and a 5% chance of defecting, I still cooperate. (With 95% probability, most likely, because you gotta punish defectors.)
Indeed, consider the following game: You give me a program that must either output “give” or “keep”. I roll a 20 sided die. On a 20, I play your program against a program that always keeps its token. Otherwise, I play your program against itself. I give you the money that (the first instance of) your program wins. Are you willing to pay me $110 to play? I’d be happy to pay you $110 for this opportunity.
I don’t cooperate with myself because P(TheirChoice=Defect)=0, I cooperate with myself because I don’t reason as if p is independent from my action.
Suppose you have to submit the source code of a program X, and I will play Y = “run X, then do what X did with probability 0.99 and the reverse with probability 0.01” against Y’ which is the same as Y but with a different seed for the RNG, and pay you according to how Y does.
Then “you” (i.e. Y) are not a perfect clone of your opponent (i.e. Y’).
Assume it’s not a perfect clone, it can defect with probability p even if you cooperate. Then apply CDT. You get “defect” for any p>0. So it is reasonable to implicitly assume continuity and declare that CDT forces you to defect when p=0. However, if you apply CDT for the case p=0 directly, you get “cooperate” instead.
In other words, the conterfactual reasoning gets broken when the map CDT(p, PD) is not continuous at the point p=0.
I disagree. If the agent has a 95% probability of doing the same thing as me and a 5% chance of defecting, I still cooperate. (With 95% probability, most likely, because you gotta punish defectors.)
Indeed, consider the following game: You give me a program that must either output “give” or “keep”. I roll a 20 sided die. On a 20, I play your program against a program that always keeps its token. Otherwise, I play your program against itself. I give you the money that (the first instance of) your program wins. Are you willing to pay me $110 to play? I’d be happy to pay you $110 for this opportunity.
I don’t cooperate with myself because P(TheirChoice=Defect)=0, I cooperate with myself because I don’t reason as if p is independent from my action.
It’s not entirely clear what you’re saying, but I’ll try to take the simplest interpretation. I’m guessing that:
If you’re going to defect, your clone always defects.
If you’re going to cooperate, your clone cooperates with probability 1-p and defects with probability p
In that case, I don’t see how it is that you get “defect” for p>0; the above formulation gives “cooperate” for 0<=p<0.5.
Suppose you have to submit the source code of a program X, and I will play Y = “run X, then do what X did with probability 0.99 and the reverse with probability 0.01” against Y’ which is the same as Y but with a different seed for the RNG, and pay you according to how Y does.
Then “you” (i.e. Y) are not a perfect clone of your opponent (i.e. Y’).
What do you do?