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dankane comments on A Thought on Pascal's Mugging - Less Wrong Discussion

12 Post author: komponisto 10 December 2010 06:08AM

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Comment author: dankane 12 May 2012 12:03:22AM 0 points [-]

I stumbled across this fix and unfortunately discovered what I consider to be a massive problem with it - it would imply that your utility function is non-computable.

OK. So in order for this to work, it needs to be the case that your prior has the property that: P(3^^^3 disutility | I fail to give him $5) << 1/3^^^3.

Unfortunately, if we have an honest Kolmogorov prior and utility is computable via a complexity << 3^^^3 Turing machine, this cannot possibly be the case. In particular, it is a Theorem that for any computable function C (whose Turing machine has complexity K(C)), so that there are x with C(x) > N, then under the Kolmogorov prior for x we have that: P( C(x) > N ) >> 2^{ - K(C) - K(N) } Now, since K(3^^^3) is small, as long as utility is computed by a small Turing machine, and it is possible to have 3^^^3 disutility, such a circumstance will not be too unlikely under a Kolmogorov prior.

For those interested, here's how the theorem is proved. I will produce a Turing machine of size K(C) + K(N) +O(1) that outputs an x (in fact, the smallest x) so that C(x) > N. By definition, I can encode C and N in size K(C) + K(N) +O(1). I then have a Turing machine enumerate all x until it finds one so that C(x) > N and output's that x. This provides a lower bound.

I guess the problem is that if you just have a Kolmogorov prior, there is a relatively simple universe that is actually out to get you. In fact, being the shortest computation causing 3^^^3 disutility is actually a pretty simple condition.