endoself comments on Observed Pascal's Mugging - Less Wrong

25 [deleted] 28 June 2011 03:53PM

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Comment author: endoself 28 June 2011 09:30:39PM *  1 point [-]

Take a bunch of events of the form "N people get tortured". if you use Solomonoff induction, your prior probability for these events will roughly decrease in log(N).

That isn't correct; the prior will decrease more slowly than any computable function that monotonically decreases to zero.

Comment author: Bongo 29 June 2011 06:25:26PM 1 point [-]

Doesn't it not decrease at all? After all, 3^^^^3 people get tortured is more likely than m people get tortured where m is some complicated integer between 0 and 3^^^^3 admitting no short description.

Comment author: endoself 30 June 2011 02:20:46AM 1 point [-]

It doesn't monotonically decrease but, for any probability p, there exists a number of people m such that the probability of n people get tortured, for any n > m, is less than p.

Comment author: MixedNuts 29 June 2011 06:42:13AM 0 points [-]

Huh, I don't understand Solomonoff induction then. Explain?

Comment author: benelliott 29 June 2011 10:53:19AM *  2 points [-]

Very roughly, the idea is that the prior probability that the universe is a (n+1) state Turing machine is half the prior probability that the universe is a n state Turing machine. Whereas the most anyone can offer you in a n state machine is BB(n) but the most they can offer you in a (n+1) state machine is BB(n+1)

So, again very roughly, the probability I can offer you BB(n) is roughly k2^(-n) where k is a very small constant. So the probability I can offer you m utility is roughly k2^(-inverseBB(n)).

InverseBB(n) is a monotonically increasing function that increases more slowly than any montonically increasing computable function, so k2^(-inverseBB(n)) is a monotonically decreasing function that decreases more slowly than any computable monotonically decreasing function.

Comment author: MixedNuts 29 June 2011 01:17:55PM 0 points [-]

Ooo, I get it! I was thinking of writing out the utility (log_base), but this is more general. Thanks!

Comment author: endoself 30 June 2011 02:27:01AM 1 point [-]

To add to that explanation, you can prove that the number of people who can be simulated on a halting n-state Turing machine has no computable upper bound by considering a Turing machine that alternates between some computation and simulating humans every fixed number of steps. If we could compute an upper bound on the number of humans simulates we could compute whether the TM would halt by waiting for that many people to be simulated, similarly to how we could use BB(n) to determine whether any n-state TM will halt if we knew its value.