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Anja comments on A utility-maximizing varient of AIXI - Less Wrong
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I think then you would count that twice, wouldn't you? Because my original formula already contains the Solomonoff probability...
Oh right. But you still want the probability weighting to be inside the sum, so you would actually need=\frac{1}{\xi\left(\dot{y}\dot{x}_{%3Ck}y\underline{x}_{k:m_{k}}\right)}\sum_{q:q(y_{1:m_k})=x_{1:m_k}}%20U(q,y_{1:m_k})2%5E{-\ell\left(q\right)}%0A)
True. :)