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ocr-fork comments on Metaphilosophical Mysteries - Less Wrong
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Oh.
I feel stupid now.
EDIT: Wouldn't it also break even by predicting the next Busy Beaver number? "All 1's except for BB(1...2^n+1)" is also only slightly less likely. EDIT: I feel more stupid.
The next number in the sequence is BB(2^(n+1)), not BB(2^n+1).
ETA: In case more explanation is needed, it takes O(2^n) more bits to computably describe BB(2^(n+1)), even if you already have BB(2^n). (It might take O(2^n) more bits to describe BB(2^n+1) as well, but I wasn't sure so I used BB(2^(n+1)) in my example instead.)
Since K(BB(2^(n+1)) | BB(2^n)) > 100 for large n, AIXI actually will not bet on 0 when BB(2^(n+1) comes around, and all those 0s that it does bet on are simply "wasted".
You can find it by emulating the Busy Beaver.