Oh, I had a typo in my last comment. I wrote: “non-zero, infinitesimal chance” instead of “non-zero, non-infinitesimal chance”.
I wasn’t claiming that it goes wrong for a bounded problem (although it might). This is hard for me to answer as I don’t know what you mean by “bounded problem”. I’ll admit that this isn’t an issue for typical example problems as you are just told a scenario.
But when you are just given a bunch of observations as per real life and have to derive the situation yourself, you will have small amounts of probability on all kinds on weird and wacky theories. And the transitive closure means that an absolutely tiny probability of one scenario can dramatically change the reference class.
Not sure I understand. When simplifying a real life situation into UDT terms, why would you go from “Alice and Bob have a tiny chance of having a common ancestor” to “Alice and Bob are definitely part of the same reference class”?
Because UDT cares about possibilities that don’t occur, no matter how small the odds. Ie. Counterfactual Mugging with a 1 in a million chance of the coin coming up heads is still included in the possibility tree. Similarly, A and B are part of the same reference class if C is included in the possibility tree, no matter how small the odds. Not sure if that is clear though.
When I said “If two agents have a common ancestor, they are in the same reference class” I meant that they must have a common ancestor for certain, not just with nonzero probability. That’s also discontinuous (what if two agents have 99% probability of having a common ancestor?) but it works fine on many problems that are simplified from the real world, because simplification often involves certainty.
Well, I suspect that the discontinuity will lead to strange results in practice, such as when you are uncertain of the past. For example, in Counterfactual Mugging, if there is any chance that you are a Boltzmann Brain who was created knowing/believing the coin was tails then you won’t have a common ancestor with brains that actually experienced the problem and saw tails, so you shouldn’t pay.But perhaps you don’t actually have to have an experience, but only believe that you had such an experience in the past?
All theories have limits of applicability. For example, Von Neumann-Morgenstern expected utility maximization requires the axiom of independence, which means you can’t be absent-minded (forgetting something and ending up in a previous mental state, like in the Absent-Minded Driver problem). If there’s even a tiny chance that you’re absent-minded, the problem can no longer be cast in VNM terms. That’s where UDT comes in, it can deal with absent-mindedness and many other things. But if there’s even a tiny chance of having more than one reference class, the problem can no longer be cast in UDT terms either. With multiple reference classes you need game theory, not decision theory.
I suppose the difference is that VNM states the limits within it operates, while I haven’t seen the limits of UDT described anywhere apart from this conversation.
I don’t know, maybe you’re right. Can you describe a bounded problem where the idea of common ancestors goes wrong?
Oh, I had a typo in my last comment. I wrote: “non-zero, infinitesimal chance” instead of “non-zero, non-infinitesimal chance”.
I wasn’t claiming that it goes wrong for a bounded problem (although it might). This is hard for me to answer as I don’t know what you mean by “bounded problem”. I’ll admit that this isn’t an issue for typical example problems as you are just told a scenario.
But when you are just given a bunch of observations as per real life and have to derive the situation yourself, you will have small amounts of probability on all kinds on weird and wacky theories. And the transitive closure means that an absolutely tiny probability of one scenario can dramatically change the reference class.
Not sure I understand. When simplifying a real life situation into UDT terms, why would you go from “Alice and Bob have a tiny chance of having a common ancestor” to “Alice and Bob are definitely part of the same reference class”?
Because UDT cares about possibilities that don’t occur, no matter how small the odds. Ie. Counterfactual Mugging with a 1 in a million chance of the coin coming up heads is still included in the possibility tree. Similarly, A and B are part of the same reference class if C is included in the possibility tree, no matter how small the odds. Not sure if that is clear though.
When I said “If two agents have a common ancestor, they are in the same reference class” I meant that they must have a common ancestor for certain, not just with nonzero probability. That’s also discontinuous (what if two agents have 99% probability of having a common ancestor?) but it works fine on many problems that are simplified from the real world, because simplification often involves certainty.
Do you mean that they must have a common ancestor if they exist or that they must have a common ancestor full stop?
Common ancestor full stop sounds more right.
Well, I suspect that the discontinuity will lead to strange results in practice, such as when you are uncertain of the past. For example, in Counterfactual Mugging, if there is any chance that you are a Boltzmann Brain who was created knowing/believing the coin was tails then you won’t have a common ancestor with brains that actually experienced the problem and saw tails, so you shouldn’t pay.But perhaps you don’t actually have to have an experience, but only believe that you had such an experience in the past?
All theories have limits of applicability. For example, Von Neumann-Morgenstern expected utility maximization requires the axiom of independence, which means you can’t be absent-minded (forgetting something and ending up in a previous mental state, like in the Absent-Minded Driver problem). If there’s even a tiny chance that you’re absent-minded, the problem can no longer be cast in VNM terms. That’s where UDT comes in, it can deal with absent-mindedness and many other things. But if there’s even a tiny chance of having more than one reference class, the problem can no longer be cast in UDT terms either. With multiple reference classes you need game theory, not decision theory.
I suppose the difference is that VNM states the limits within it operates, while I haven’t seen the limits of UDT described anywhere apart from this conversation.