twanvl comments on Outlawing Anthropics: An Updateless Dilemma - Less Wrong

26 Post author: Eliezer_Yudkowsky 08 September 2009 06:31PM

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Comment author: twanvl 08 September 2009 10:28:24PM 1 point [-]

Before the experiment, you calculate the general utility of the conditional strategy "Reply 'Yes' to the question if you wake up in a green room" as (50% * ((18 * +$1) + (2 * -$3))) + (50% * ((18 * -$3) + (2 * +$1))) = -$20

This assumes that the question is asked only once, but then, to which of the 20 copies will it be asked?

If all 20 copies get asked the same question (or equivalently if a single copy chosen at random is) then the utility is (50% * 18/20 * ((18 * +$1) + (2 * -$3))) + (50% * 2/20 * ((18 * -$3) + (2 * +$1))) = 2.8$ = 50% * 5.6$.

Consider the similar thought experiments:

  • I flip a fair coin to determine whether to switch to my headdy coin or my tailly coin, which have a 90% and 10% probability of heads respectively.
  • Now I flip this biased coin. If it comes up heads then I paint the room green, if it comes up tails I paint it red.
  • You then find yourself in a green room.
  • Then I flip the biased coin again, and repaint the room.
  • Before this second flip, I offer you the bet of +1$ if the room stays green and -3$ if it becomes red.

The prior expected utility before the experiment is:

E(util|headdy) = 90% * 1$ + 10% * -3$ = 0.6$
E(util|tailly) = 10% * 1$ + 90% * -3$ = -2.6$
E(util) = 50% * E(util|headdy) + 50% * E(util|tailly) = -1$

Given that you find yourself in a green room after the first flip, you can determine the probability that the headdy coin is used:

P(green) = 0.5
P(green|headdy) = 0.9
P(headdy|green) = 0.9

Which gives a posterior utility:

E(util|green) = 0.9 * E(util|headdy) + 0.1 * E(util|tailly) = 0.28$
Comment author: DanArmak 08 September 2009 10:41:50PM 3 points [-]

This assumes that the question is asked only once, but then, to which of the 20 copies will it be asked?

Every copy that is in a green room is asked the question (so either 2 or 18 copies total are asked). If all answer Play, we play. If all answer Don't Play, we don't. In any other case we fine all 20 copies some huge amount; this is intended to make them agree beforehand on what answer to give. (This is reworded from the OP.)

For your other thought experiment - if there aren't actual N copies being asked the question, then there's no dilemma; you (the only copy) simply update on the evidence available (that the room is green). So yes, the original problem requires copies being asked in parallel to introduce the possibility that you're hurting other copies of yourself by giving a self-serving answer. Whereas if you're the only copy, you always give a self-serving answer, i.e. play only if the room is green.

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