Roko comments on A problem with Timeless Decision Theory (TDT) - Less Wrong

36 Post author: Gary_Drescher 04 February 2010 06:47PM

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Comment deleted 04 February 2010 09:55:43PM *  [-]
Comment author: Vladimir_Nesov 05 February 2010 12:30:18AM *  2 points [-]

The logical/physical distinction itself can be seen as ad-hoc: you can consider the whole set-up Q as a program that is known to you (R), because the rules of the game were explained, and also consider yourself (R) as a program known to Q. Then, Q can reason about R in interaction with various situations (that is, run, given R, but R is given as part of Q, so "given R" doesn't say anything about Q), and R can do the same with Q (and with the R within that Q, etc.). Prisoner's dilemma can also be represented in this way, even though nobody is pulling Omega in that case.

When R is considering "the past", it in fact considers facts about Q, which is known to R, and so facts about the past can be treated as "logical facts". Similarly, when these facts within Q reach R at present and interact with it, they are no more "physical facts" than anything else in this setting (these interactions with R "directed from the past" can be seen as what R predicts Q-within-R-within-Q-... to do with R-within-Q-within-R-...).

Comment author: Jonathan_Lee 04 February 2010 10:37:54PM 2 points [-]

This ad-hoc fix breaks as soon as Omega makes a slightly messier game, wherein you receive a physical clue as to a computation output, and this computation and your decision determine your reward.

Suppose that for any output of the computation there is a a unique best decision, and that furthermore this set of (computation output, predicted decision) pairs are mapped to distinct physical clues. Then given the clue you can infer what decision to make and the logical computation, but this requires that you infer from a logical fact (the predictor of you) to the physical state to the clue to the logical fact of the computation.

Comment deleted 04 February 2010 11:16:19PM *  [-]
Comment author: Jonathan_Lee 04 February 2010 11:55:31PM *  1 point [-]

The game is to pick a box numbered from 0 to 2; there is a hidden logical computation E yielding another value 0 to 2. Omega has a perfect predictor D of you. You choose C.

The payout is 10^((E+C)mod 3), and there is a display showing the value of F = (E-D)mod 3.

If F = 0, then:

  • D = 0 implies E = 0 implies optimal play is C = 2; contradiction
  • D = 1 implies E = 1 implies optimal play is C = 1; no contradiction
  • D = 2 implies E = 2 implies optimal play is C = 0; contradiction

And similarly for F = 1, F = 2 play C = F+1 as the only stable solution (which nets you 100 per play)

If you're not allowed to infer anything about E from F, then you're faced with a random pick from winning 1, 10 or 100, and can't do any better...

Comment author: Wei_Dai 05 February 2010 11:14:16AM 0 points [-]

I'm not sure this game is well defined. What value of F does the predictor D see? (That is, it's predicting your choice after seeing what value of F?)

Comment author: Jonathan_Lee 05 February 2010 01:53:46PM 0 points [-]

The same one that you're currently seeing; for all values of E there is a value of F such that this is consistent, ie that D has actually predicted you in the scenario you currently find yourself in.