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by Stuart Armstrong 833 days ago | Jessica Taylor likes this | link | parent

Let \(p\) be the probability the shutdown button is pressed, given \(\pi\). If the AI produces paperclips for \(n\) turns then shuts down, it gets utility \(p(10-n)+(1-p)n\); pressing the shutdown button itself or destroying it just wastes a turn and reduces the utility, so we’ll ignore those two options.

This utility is \(10p+(1-2p)n\). For fixed \(p>0.5\), this is maximised for \(n=0\), for \(p<0.5\) this is maximised for \(n=10\). However, \(n=10\) implies \(p=1\) and \(n=0\) implies \(p=0\), so there are no compatible solutions there.

Thus \(p=0.5\), and the utility is \(10p=5\), independent of \(n\). All that is needed is to ensure that \(p=0.5\) (without the AI pressing the button itself), which means \(n<8\) with \(0.5\) probability and \(n\geq 8\) with \(0.5\) probability. This thus extends your solution 5.



by Jessica Taylor 833 days ago | link

I think this almost works. Suppose the AI constructs 7 paperclips 50% of the time, and 8 paperclips 50% of the time (shutting down after producing the last paperclip). This means the button is pushed 50% of the time after step 8, and never pushed 50% of the time. Given this distribution of button pushes, what’s the best-response policy?

I think the best-response policy is to make 8 paperclips, then see if the shutdown button has been pressed; if it has been pressed, then shut down, else make 2 more paperclips. When the button is not pressed, this makes 10 paperclips; when the button is pressed, this shuts down for 2 steps. Thus the expected utility is 6, which is better than the original 5. So I think it’ll be more difficult to find an equilibrium policy; the uncertainty about when the shutdown button is pressed must be such that the agent is indifferent between making a paperclip and shutting down on step 8.

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by Stuart Armstrong 830 days ago | Jessica Taylor and Patrick LaVictoire like this | link

You’re right, let’s be more careful.

First, let’s list the optimal policies. Pressing or destroying the button remain suboptimal policies. Once the button is seen to be pressed, then anything but shutdown is suboptmial. Therefore there are \(11\) potentially optimal policies, labelled by \(n=0\) to \(n=10\). These involve producing \(n\) paperclips then shutting down for \(10-n\) turns, unless it sees the shutdown button has been pressed, in which case it shuts down (this is only relevant for \(n=9,10\)).

Let \(p\) be the total shutdown probability. Note that for all the optimal policies, either the shutdown button is never pressed, or it is pressed on turn \(8\), so we can ignore the possibilities of it being pressed later.

The utility gain for \(n\leq 8\) is \(n(1-p)+(10-n)p=10p+(1-2p)n\). For \(n=9\) it is that quantity, plus \(p\) (it shuts down one turn earlier if the shutdown button is pressed on turn \(8\), and gains one extra utility for it). For \(n=10\) it is that quantity, plus \(2p\)).

For \(p\leq 0.5\), the optimal policy is \(n=10\), which ensures \(p=1\), hence contradiction.

For \(p>0.5\), the only possible optimal policies are \(n=0\) (utility \(10p\)), \(n=9\) (utility \(9-7p\)), and \(n=10\) (utility \(10-8p\)). Subtracting the last two gives \(1-p >0\), hence \(n=9\) is never optimal.

The agent is indifferent between \(n=0\) and \(n=10\) for \(10p=10-8p\) ie \(p=5/9\approx 0.5555\ldots\).

Therefore the result should be that the agent pursues the mixed policy \(5/9(n=10) + 4/9(n=0)\), with probability of shutdown being equal to \(5/9\), and pseudo expected utility \(50/9\) (real expected utility \(10/9\)).

Have I got this right?

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by Jessica Taylor 829 days ago | link

This looks correct; thanks for doing this analysis!

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