The problem is no free lunch. Any decision theory is going to fail somewhere. The case for privileging Newcomb as a success goal over all other considerations has not, in fact, been made.
So I raised this problem too, and I got a convincing answer to it. The way I raised it was to say that it isn’t fair to fault CDT for failing to maximise expected returns in Newcomb’s problem, because Newcomb’s problem was designed to defeat CDT and we can design a problem to defeat any decision theory. So that can’t be a standard.
The response I got (at least, my interpretation of it) was this: It is of course possible to construct a problem in which any decision theory is defeated, but not all such problems are equal. We can distinguish in principle between problems that can defeat any decision procedure (such as ‘omega gives you an extra million for not using X’, where X is the decision procedure you wish to defeat) and problems which defeat certain decision procedures but cannot be constructed so as to defeat others. Call the former type 1 problems, and the latter type 2 problems. Newcomb’s problem is a type 2 problem, as is the prisoner’s dilemma against a known psychological twin. Both defeat CDT, but not TDT, and cannot be constructed so as to defeat TDT without becoming type 1. TDT is aimed (though I think not yet successful) at being able to solve all type 2 problems.
So if we have two decision theories, both of which fail type 1 problems, but only one of which fails type 2 problems, we should prefer the one that never fails type 2 problems. This would privilege Newcomb’s problem (as a type 2 problem) over any type 1 problems. It would cease to be an argument for the privileging of type 2 problems over type 1 problems if it turned out that every decision theory will always fail some set of type 2 problems. But that would, I think, be a hard case to make.
Can you construct a problem that defeats TDT that cannot be constructed to defeat CDT? (I think I can- The Pirates’ problem against psychological twins).
The CDT result is pretty well known: the first pirate gets almost everything.
The TDT result is hard for me, but if the first pirate gets anything then more than half of the other pirates had a strategy that could be trivially improved.
The problem is no free lunch. Any decision theory is going to fail somewhere. The case for privileging Newcomb as a success goal over all other considerations has not, in fact, been made.
So I raised this problem too, and I got a convincing answer to it. The way I raised it was to say that it isn’t fair to fault CDT for failing to maximise expected returns in Newcomb’s problem, because Newcomb’s problem was designed to defeat CDT and we can design a problem to defeat any decision theory. So that can’t be a standard.
The response I got (at least, my interpretation of it) was this: It is of course possible to construct a problem in which any decision theory is defeated, but not all such problems are equal. We can distinguish in principle between problems that can defeat any decision procedure (such as ‘omega gives you an extra million for not using X’, where X is the decision procedure you wish to defeat) and problems which defeat certain decision procedures but cannot be constructed so as to defeat others. Call the former type 1 problems, and the latter type 2 problems. Newcomb’s problem is a type 2 problem, as is the prisoner’s dilemma against a known psychological twin. Both defeat CDT, but not TDT, and cannot be constructed so as to defeat TDT without becoming type 1. TDT is aimed (though I think not yet successful) at being able to solve all type 2 problems.
So if we have two decision theories, both of which fail type 1 problems, but only one of which fails type 2 problems, we should prefer the one that never fails type 2 problems. This would privilege Newcomb’s problem (as a type 2 problem) over any type 1 problems. It would cease to be an argument for the privileging of type 2 problems over type 1 problems if it turned out that every decision theory will always fail some set of type 2 problems. But that would, I think, be a hard case to make.
Can you construct a problem that defeats TDT that cannot be constructed to defeat CDT? (I think I can- The Pirates’ problem against psychological twins).
No, I don’t have any such thing in mind. Could you explain how TDT and CDT get different results?
The CDT result is pretty well known: the first pirate gets almost everything.
The TDT result is hard for me, but if the first pirate gets anything then more than half of the other pirates had a strategy that could be trivially improved.