Oh, come on. According to Janes, the marginal probability P(Omega is correct | Omega predicts something) is supposed to be additionally conditioned on everything you know about the situation. If you know that Omega always predicts “two-box”, then P(Omega is correct | Omega predicts something) is equal to the relative frequency of two-boxers in the population. If you know that Omega first always predicts correctly and then modifies its answer in 10% cases, then it’s something completely different. If you have no knowledge about whether the first or the second is true, then what can you do? Presumably, try Solomonoff induction, too bad it’s incomputable.
Oh, come on. According to Janes, the marginal probability P(Omega is correct | Omega predicts something) is supposed to be additionally conditioned on everything you know about the situation. If you know that Omega always predicts “two-box”, then P(Omega is correct | Omega predicts something) is equal to the relative frequency of two-boxers in the population. If you know that Omega first always predicts correctly and then modifies its answer in 10% cases, then it’s something completely different. If you have no knowledge about whether the first or the second is true, then what can you do? Presumably, try Solomonoff induction, too bad it’s incomputable.