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Anatoly_Vorobey comments on Logical Uncertainty as Probability - Less Wrong Discussion

3 Post author: gRR 29 April 2012 10:26PM

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Comment author: Anatoly_Vorobey 29 April 2012 11:07:43PM 7 points [-]

You're sidestepping the whole point of cousin_it, which is that your mathematical knowledge is certainly enough to determine whether the millionth digit of pi is odd or even. One of these two statements is a trivial consequence of the Peano axioms and some well-known representation of pi as an infinite series. It's just that its being a trivial consequence is witnessed by a very long (and also trivial) proof which you're not aware of, and you don't know which of the two statements is backed by such a long proof.

Comment author: gRR 29 April 2012 11:48:30PM *  0 points [-]

I don't think I'm sidestepping the issue. The point of cousin_it's comment, as I understood it, was that assigning probabilities to "logically uncertain" statements results in inconsistencies. What I tried to show is that for probabilistic assignments to be consistent, it is only necessary to be logically omniscient at propositional calculus, not at full-power PA. And this is an important difference, because propositional calculus is decidable.

Comment author: Anatoly_Vorobey 30 April 2012 10:39:43AM 4 points [-]

Ah, well, if you're only closing under propositional tautologies, then you're not doing anything interesting. OLC(X) is for practical purposes the same as X (not just because it's decidable, as you say, but more importantly because it's so weak). So your suggestions boils down to assigning P=1 to axioms, P=0 to their negations, and trying to figure out non-trivial probabilities for everything else by constraining on propositional consistency. But propositional consistency is merely a very thin veneer over X.

Because propositional inference isn't going to be able to break down quantifiers and "peer" inside their clauses, any quantified statement that's not already related to X [in the sense of participating in X either as a member or as a clause of a Boolean member] is opaque to X. So if I write A = (exists Y)(Y=2) and B = (exists Z)(Z=2 and Z=3), you'll be forced to deduce P(A)=P(B) under your axioms, or pre-commit to include in your axioms A, not(B) and everything else in a smoothly growing (complexity-wise) list of arithmetical truths I can come up with. That doesn't seem very useful.

Comment author: gRR 30 April 2012 10:58:08AM *  0 points [-]

But propositional consistency is merely a very thin veneer over X.

That was my goal - to come up with a minimum necessary for consistency, but still sufficient to prove the 1/2 probability for digits of PI :) If you wish to make OLC stronger, you're free to do so, as long as it remains decidable. For example, you can define OLC(X) to be {everything provable from X by at most 10 steps of PA-power reasoning, followed by propositional calculus closure}.

Comment author: Anatoly_Vorobey 30 April 2012 11:07:03AM 1 point [-]

In your scheme you have P=1/2 for anything nontrivial and its negation that's not already in X. It just so happens that this looks reasonable in case of the oddity of a digit of pi, but that's merely a coincidence (e.g. take A="a millionth digit of pi is 3" rather than "...odd").

Comment author: gRR 30 April 2012 11:13:34AM *  0 points [-]

No, a statement and its negation are distinguishable, unless indeed you maliciously hide them under quantifiers and throw away the intermediate proof steps.