Let’s say I build my Omega by using a perfect predictor plus a source of noise that’s uncorrelated with the prediction. It seems weird that you’d deterministically two-box against such an Omega, even though you deterministically one-box against a perfect predictor. Are you sure you did the math right?
Let’s say I build my Omega by using a perfect predictor plus a source of noise that’s uncorrelated with the prediction. It seems weird that you’d deterministically two-box against such an Omega, even though you deterministically one-box against a perfect predictor. Are you sure you did the math right?
Even in the case when the random noise dominates and the signal is imperceptibly small?
I think the more relevant case is when the random noise is imperceptibly small. Of course you two-box if it’s basically random.