It’s actually tough to predict what EDT would do, since it depends on picking the right reference class for yourself, and we have no idea how to formalize those. But the explanations of why EDT would one-box on Newcomb’s Problem appear isomorphic to explanations of why EDT would forgo smoking in the Smoking Lesion problem, so it appears that a basic implementation would fail one or the other.
That shouldn’t happen. If TDT knows that the boxes are being filled by simulating a CDT algorithm (even if that algorithm was its ancestor), then it will two-box.
It’s actually tough to predict what EDT would do, since it depends on picking the right reference class for yourself, and we have no idea how to formalize those. But the explanations of why EDT would one-box on Newcomb’s Problem appear isomorphic to explanations of why EDT would forgo smoking in the Smoking Lesion problem, so it appears that a basic implementation would fail one or the other.
That shouldn’t happen. If TDT knows that the boxes are being filled by simulating a CDT algorithm (even if that algorithm was its ancestor), then it will two-box.