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by Ryan Carey 73 days ago | link | parent

[Note: This comment is three years later than the post]

The “obvious idea” here unfortunately seems not to work, because it is vulnerable to so-called “infinite improbability drives”. Suppose \(B\) is a shutdown button, and \(P(b|e)\) gives some weight to \(B=pressed\) and \(B=unpressed\). Then, the AI will benefit from selecting a Q such that it always chooses an action \(a\), in which it enters a lottery, and if it does not win, then it the button B is pushed. In this circumstance, \(P(b|e)\) is unchanged, while both \(P(c|b=pressed,a,e)\) and \(P(c|b=unpressed,a,e)\) allocate almost all of the probability to great \(C\) outcomes. So the approach will create an AI that wants to exploit its ability to determine \(B\).



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