Hence if there are 2 instances of your decision algorithm in Green rooms, there are 2 runs of your decision algorithm, and if they vote to steal there is a loss of 3 from each red and gain 1 for each green, for a total gain of 12-318 = − 52.
If there are 18 instances in Green rooms, there are 18 runs of your decision algorithm, and if they vote to steal there is a loss of 3 from each red and a gain of 1for each green, for a total gain of 118-23 = 12
The “committal of a group of” is noting that there are 2 or 18 runs of your decision algorithm that are logically forced by the decision made this specific instance of the decision algorithm in a green room.
I’ve been watching for a while, but have never commented, so this may be horribly flawed, opaque or otherwise unhelpful.
I think the problem is entirely caused by the use of the wrong sets of belief, and that anything holding to Eliezer’s 1-line summary of TDT or alternatively UDT should get this right.
Suppose that you’re a rational agent. Since you are instantiated in multiple identical circumstances (green rooms) and asked identical questions, your answers should be identical. Hence if you wake up in a green room and you’re asked to steal from the red rooms and give to the green rooms, you either commit a group of 2 of you to a loss of 52 or commit a group of 18 of you to a gain of 12.
This committal is what you wish to optimise over from TDT/UDT, and clearly this requires knowledge about the likelyhood of different decision making groups. The distribution of sizes of random groups is not the same as the distribution of sizes of groups that a random individual is in. The probabilities of being in a group are upweighted by the size of the group and normalised. This is why Bostrom’s suggested 1/n split of responsibility works; it reverses the belief about where a random individual is in a set of decision making groups to a belief about the size of a random decision making group.
By the construction of the problem the probability that a random (group of all the people in green rooms) has size 18 is 0.5, and similarly for 2 the probability is 0.5. Hence the expected utility is (0.512)+(0.5-52)=-20.
If you’re asked to accept a bet on there being 18 people in green rooms, and you’re told that only you’re being offered it, then the decision commits exactly one instance of you to a specific loss or gain, regardless of the group you’re in. Hence you can’t do better than the 0.9 and 0.1 beliefs.
If you’re told that the bet is being offered to everyone in a green room, then you are committing to n times the outcome in any group of n people. In this case gains are conditional on group size, and so you have to use the 0.5-0.5 belief about the distribution of groups. It doesn’t matter because the larger groups have the larger multiplier and thus shutting up and multiplying yields the same answers as a single-shot bet.
ETA: At some level this is just choosing an optimal output for your calculation of what to do, given that the result is used variably widely.