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MrHen comments on My Fundamental Question About Omega - Less Wrong
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Comments (151)
This is true but it doesn't change how frequently you would give Omega $5. It changes Omega's success rate, but only in the sense that it won't play the game if you aren't willing to give $5.
If A = You pay Omega $5 and O = Omega asks for $5:
p(A|O) = p(O|A) * p(A) / (p(O|A) * p(A) + p(O|~A) * p(~A))
Making Omega a perfect predictor sets p(Omega asks|You don't pay) to 0, so p(O|~A) = 0.
p(A|O) = p(O|A) * p(A) / (p(O|A) * p(A) + 0 * p(~A))
p(A|O) = p(O|A) * p(A) / p(O|A) * p(A)
p(A|O) = 1
Therefore, p(You pay Omega $5|Omega asks for $5) is 1. If Omega asks, you will pay. Big whoop. This is a restriction on Omega asking, not on you giving.
Yes, but consider what happens when you start conditioning on the statement B="I do not intend to give Omega $5". If Omega is perfect, this is irrelevant; you will hand over the cash.
If Omega is not perfect, then the situation changes. Use A and O as above; then a relvant question is: how many of Omega's errors have B (nearly all of them) versus how many of Omega's successes have B (nearly none of them). Basically, you're trying to estimate the relative sizes of (B&A)|O versus (B&~A)|O.
Now A|O is very large while ~A|O is very small, but (B&A)|O is tiny in A|O while (B&~A)|O makes up most of ~A|O. So I'd crudely estimate that those two sets are generally of pretty comparable size. If Omega is only wrong one in a million, I'd estimate I'd have even odds of handing him the $5 if I didn't want to.
Right, when Omega is perfect, this isn't really a useful distinction. The correlation between B and A is irrelevant for the odds of p(A|O). It does get more interesting when asking:
p(A|B)
p(~A|B)
p(O|B)
These are still interesting even when Omega is perfect. If, as you suggest, we look at the relationship between A, B, and O when Omega isn't perfect, your questions are dead on in terms of what matters.