This is my attempt at a pedagogical exposition of “the solution”. It’s overly long, and I've lost perspective completely about what is understood by the group here and what isn't. But since I've written up this solution for myself, I'll go ahead and share it.
The cases I'm describing below are altered from the OP so that they completely non-metaphysical, in the sense that you could implement them in real life with real people. Thus there is an objective reality regarding whether money is collectively lost or won, so there is finally no ambiguity about what the correct calculation actually is.
Suppose that there are twenty different graduate students {Amy, Betty, Cindy, ..., Tony} and two hotels connected by a breezeway. Hotel Green has 18 green rooms and 2 red rooms. Hotel Red has 18 red rooms and 2 green rooms. Every night for many years, students will be assigned a room in either Hotel Green or Hotel Red depending on a coin flip (heads --> Hotel Green for the night, tails --> Hotel Red for the night). Students won’t know what hotel they are in but can see their own room color only. If a student sees a green room, that student correctly deduces they are in Hotel Green with 90% probability.
Case 1: Suppose that every morning, Tony is allowed to bet that he is in a green room. If he bets ‘yes’ and is correct, he pockets $12. If he bets ‘yes’ and is wrong, he has to pay $52. (In other words, his payoff for a correct vote is $12, the payoff for a wrong vote is -$52.) What is the expected value of his betting if he always says ‘yes’ if he is in a green room?
For every 20 times that Tony says ‘yes’, he wins 18 times (wins $12x18) and he loses twice (loses $52x2), consistent with his posterior. One average he wins $5.60 per bet , or $2.80 per night. (He says “yes” to the bet 1 out of every 2 nights, because that is the frequency with which he finds himself in a green room.) This is a steady money pump in the student’s favor.
The correct calculation for Case 1 is:
average payoff per bet = (probability of being right)x(payoff if right)+ (probability of being wrong)x(payoff if wrong) = .9x18+.1x-52 =5.6.
Case 2: Suppose that Tony doesn’t pocket the money, but instead the money is placed in a tip jar in the breezeway. Tony’s betting contributes $2.80 per night on average to the tip jar.
Case 3: Suppose there is nothing special about Tony, and all the students get to make bets. They will all make bets when they wake in green rooms, and add $2.80 per night to the tip jar on average. Collectively, the students add $56 per night to the tip jar on average. (If you think about it a minute, you will see that they add $216 to the tip jar on nights that they are assigned to hotel Green and lose $104 on nights that they are assigned to hotel Red.) If the money is distributed back to the students, they each are making $2.80 per night, the same steady money pump in their favor that Tony took advantage of in Case 1.
Case 4: Now consider the case described in the OP. We already understand that the students will vote “yes” if they wake in a green room and that they expect to make money doing so. Now the rules are going to change, however, so that when all the green roomers unanimously vote “yes”, $12 are added to the tip jar if they are correct and $52 are subtracted if they are wrong. Since the students are assigned to Hotel Green half the time and to Hotel Red half the time, on average the tip jar loses $20 every night. Suddenly, the students are losing $1 a night!
Each time a student votes correctly, it is because they are all in Hotel Green, as per the initial set up of the problem in the OP. So all 18 green roomer votes are correct and collectively earn $12 for that night. The payoff is $12/18 per correct vote. Likewise, the payoff per wrong vote is -$52/2.
So the correct calculation for case 4 is as follows:
average payoff per bet = (probability of being right)x(payoff if right)+ (probability of being wrong)x(payoff if wrong) = .9x(18/12)+.1x(-52/2) = -2.
So in conclusion, in the OP problem, the green roomer must recognize that he is dealing with case #4 and not Case #1, in which the payoff is different (but not the posterior).
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"Suppose you have ten ideal game-theoretic selfish agents and a pie to be divided by majority vote. "
Well then, the statistical expected (average) share any agent is going to get long-term is 1/10th of the pie. The simplest solution that ensures this is the equal division; anticipating this from the start cuts down on negotiation costs, and if a majority agrees to follow this strategy (i.e agrees to not realize more than their "share"), it is also stable - anyone who ponders upsetting it risks to be the "odd man out" who eats the loss of an unsymmetric strategy.
In practice (i.e. in real life) there are other situations that are relatively stable, i.e. after a few rounds of "outsiders" bidding low to get in, there might be two powerful "insiders" who get large shares in liaison with four smaller insiders who agree to a very small share because it is better than nothing; the best the insiders can do then is to offer the four outsiders small shares also, so that each small-share individual wil be faced with the choice of cooperating and receiving a small share, or not cooperating and receiving nothing. Whether the two insiders can pull this off will depend on how they frame the problem, and how they present themselves ("we are the stabilizers that ensure that "social justice" is done and nobody has to starve").
How you can get an AI to understand setups like this (and if it wants to move past the singularity, it probably will have to) seems to be quite a problem; to recognize that statistically, it can realize no more than 1/10th, and to push for the simplest solution that ensures this seems far easier (and yet some commentators seem to think that this solution of "cutting everyone in" is somehow "inferior" as a strategy - puny humans ;-).