What about the agent using Solomonoff’s distribution? After seeing
BB(1),...,BB(2^n), the algorithmic complexity of BB(1),...,BB(2^n) is sunk,
so to speak. It will predict a higher expected payoff for playing 0 in any
round i where the conditional complexity K(i | BB(1),...,BB(2^n)) < 100.
This includes for example 2BB(2^n), 2BB(2^n)+1, BB(2^n)^2 * 3 + 4,
BB(2^n)^^^3, etc. It will bet on 0 in these rounds (erroneously, since
K(BB(2^(n+1)) | BB(2^n)) > 100 for large n), and therefore lose relative to
a human.
I don’t understand how the bolded part follows. The best explanation by round BB(2^n) would be “All 1′s except for the Busy Beaver numbers up to 2^n”, right?
Yes, that’s the most probable explanation according to the Solomonoff prior, but AIXI doesn’t just use the most probable explanation to make decisions, it uses all computable explanations that haven’t been contradicted by its input yet. For example, “All 1′s except for the Busy Beaver numbers up to 2^n and 2BB(2^n)” is only slightly less likely than “All 1′s except for the Busy Beaver numbers up to 2^n” and is compatible with its input so far. The conditional probability of that explanation, given what it has seen, is high enough that it would bet on 0 at round 2BB(2^n), whereas the human wouldn’t.
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”.
I don’t understand how the bolded part follows. The best explanation by round BB(2^n) would be “All 1′s except for the Busy Beaver numbers up to 2^n”, right?
Yes, that’s the most probable explanation according to the Solomonoff prior, but AIXI doesn’t just use the most probable explanation to make decisions, it uses all computable explanations that haven’t been contradicted by its input yet. For example, “All 1′s except for the Busy Beaver numbers up to 2^n and 2BB(2^n)” is only slightly less likely than “All 1′s except for the Busy Beaver numbers up to 2^n” and is compatible with its input so far. The conditional probability of that explanation, given what it has seen, is high enough that it would bet on 0 at round 2BB(2^n), whereas the human wouldn’t.
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.