This result should be easy to extend to games with more than two players
I suspect that there are difficulties when more than two players are involved in the decision making, and a coalition between a majority can leave a minority powerless.
For example, suppose that three players iteratively play a game in which they vote as to how to divide up a jackpot. Four choices:
Split evenly between 1 and 2
Split evenly between 2 and 3
Split evenly between 3 and 1
Split evenly among all three players.
Majority rule. If no proposed split receives a majority, then nobody gets anything.
There are three stable bargaining solutions—each of which is pareto optimal but forces one of your θ factors to be zero. There needs to be some added structure to the problem to force the three-way split as the unique solution. As I described it, the three-way split is not even a solution.
ETA: In the 3-way split option, assume that Omega increases the jackpot by 5%, so that total payoff is maximized by the split. But it is still better to split two ways, if you can find a partner.
Yes coallitions and things like that can happen. But the result still applies—every Pareto optimal solution is a weighing of utilites and a maximising of the sum. They don’t have to be fair or nice, though.
If the players all have utilities that scale linearly with money for the jackpot, then all four of your choices correspond to valuing everyone’s utilities equally and maximising the sum; only because the solution space is a flat plane, we need an extra “tie-break” method for deciding which to go for.
every Pareto optimal solution is a weighing of utilities and a maximizing of the sum.
all four of your choices correspond to valuing everyone’s utilities equally and maximizing the sum;
Not true after my edit. The three-way split is the unique maximum when all players are valued equally; but the two way splits are also Pareto optimal, are each favored by a majority of players, and are bargaining solutions (for some bargaining protocols).
But, now that I look at it, these two-way splits are also scaled utility maxima—for example, using a (49, 49, 2) scaling.
You have me convinced that every point on the Pareto frontier is a maximum of some weighted sum of utilities. But you don’t yet have me convinced that, in every case, all of the weights can be non-zero.
ETA: I am a bit more convinced after re-reading the last paragraph of your posting.
No, you are right. For certain setups (say when the set of outcomes is a perfect circle/sphere), then if one player gets everything they want, this only happens if the weight of the other player’s utility is zero.
I didn’t write up that case, because there I would have to let the thetas be zero and infinity, a subtlety that would distract from the main point.
I suspect that there are difficulties when more than two players are involved in the decision making, and a coalition between a majority can leave a minority powerless.
For example, suppose that three players iteratively play a game in which they vote as to how to divide up a jackpot. Four choices:
Split evenly between 1 and 2
Split evenly between 2 and 3
Split evenly between 3 and 1
Split evenly among all three players.
Majority rule. If no proposed split receives a majority, then nobody gets anything.
There are three stable bargaining solutions—each of which is pareto optimal but forces one of your θ factors to be zero. There needs to be some added structure to the problem to force the three-way split as the unique solution. As I described it, the three-way split is not even a solution.
ETA: In the 3-way split option, assume that Omega increases the jackpot by 5%, so that total payoff is maximized by the split. But it is still better to split two ways, if you can find a partner.
Yes coallitions and things like that can happen. But the result still applies—every Pareto optimal solution is a weighing of utilites and a maximising of the sum. They don’t have to be fair or nice, though.
If the players all have utilities that scale linearly with money for the jackpot, then all four of your choices correspond to valuing everyone’s utilities equally and maximising the sum; only because the solution space is a flat plane, we need an extra “tie-break” method for deciding which to go for.
Not true after my edit. The three-way split is the unique maximum when all players are valued equally; but the two way splits are also Pareto optimal, are each favored by a majority of players, and are bargaining solutions (for some bargaining protocols).
But, now that I look at it, these two-way splits are also scaled utility maxima—for example, using a (49, 49, 2) scaling.
You have me convinced that every point on the Pareto frontier is a maximum of some weighted sum of utilities. But you don’t yet have me convinced that, in every case, all of the weights can be non-zero.
ETA: I am a bit more convinced after re-reading the last paragraph of your posting.
No, you are right. For certain setups (say when the set of outcomes is a perfect circle/sphere), then if one player gets everything they want, this only happens if the weight of the other player’s utility is zero.
I didn’t write up that case, because there I would have to let the thetas be zero and infinity, a subtlety that would distract from the main point.