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by Stuart Armstrong 613 days ago | Jessica Taylor and Patrick LaVictoire like this | link | parent

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?



by Jessica Taylor 612 days ago | link

This looks correct; thanks for doing this analysis!

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