But three people should do already. Im fairly convinced that this game is unstable in the sense it would not make sense for any of them to agree to get ^{1}⁄_{3} as they can always guarantee themselves more by defecting with someone (even by offeing them ^{1}⁄_{6} - epsilon which is REALLY hard to turn down). It seems that a given majority getting ^{1}⁄_{2} each would be a more probable solution but you would really need to formalize the rules before this can be proven. Im a cryptologist so this is sadly not really my area...

I almost posted on the three-person situation earlier, but what I wrote wasn’t cogent enough. It does seem like it should work as an archetype for any N > 2.

The problem is how the game is iterated. Call the players A, B, and C. If A says, “B, let’s go 50-50,” and you assume C doesn’t get to make a counter-offer and they vote immediately, 50-50-0 is clearly the outcome. This is probably also the case for the 10-person if there’s no protracted bargaining.

If there is protracted bargaining, it turns into an infinite regression as long as there is an out-group, and possibly even without an outgroup. Take this series of proposals, each of which will be preferred to the one prior (format is Proposer:A gets-B gets-C gets):

A:50-50-0

C:0-55-45

A:50-0-50

B: 55-45-0

C:0-55-45

A:50-0-50 …

There’s clearly no stable equilibrium. It seems (though I’m not sure how to prove this) that an equal split is the appropriate efficient outcome. Any action by any individual will create an outgroup that will spin them into an infinite imbalance. Moreover, if we are to arbitrarily stop somewhere along that infinite chain, the expected value for each player is going to be ^{100}⁄_{3} (they get part of a two-way split twice which should average to 50 each time overall, and they get zero once per three exchanges). Thus, at 33-33-33, one can’t profitably defect. At 40-40-20, C could defect and have a positive expected outcome.

If the players have no bargaining costs whatsoever, and always have the opportunity to bargain before a deal is voted on, and have an infinite amount of time and do not care how long it takes to reach agreement (or if agreement is reached), then it does seem like you get an infinite loop, because there’s always going to be an outgroup that can outbid one of the ingroup. This same principle should also apply to the 10-person model; with infinite free time and infinite free bargaining, no equilibrium can be reached. If there is some cost to defecting, or a limitation on bargaining, there should be an even N/2+1-way split (depending admittedly on how those costs and limits are defined). If there is no limitation on bargaining and no cost to defecting, but time has a cost or time will be arbitrarily “called,” an even N-way split seems like the most likely/efficient outcome. The doubly-infinite situations is so far divorced from reality that it does not seem worth losing sleep over.

Also, the problem may stem from our limitation of thinking of this as a linear series of propositions, because that’s how people would have to actually bargain. In the no-repeated bargaining game, whether it’s 50-50-0 or 0-50-50 all depends on who asks first, which seems like an improper and unrealistic determining factor. This linear, proposer-centered view may not be how such beings would actually bargain.

The example of the Rubinstein bargaining model suggests that you could make players alternate offers and introduce exponential temporal discounting. An equal split isn’t logically necessary in this case: a player’s payoff will likely depend on their personal rate of utility discounting, also known as “impatience”, and others’ perceptions of it. The search keyword is “n-person bargaining”; there seems to be a lot of literature that I’m too lazy and stupid to quickly summarize.

But three people should do already. Im fairly convinced that this game is unstable in the sense it would not make sense for any of them to agree to get

^{1}⁄_{3}as they can always guarantee themselves more by defecting with someone (even by offeing them^{1}⁄_{6}- epsilon which is REALLY hard to turn down). It seems that a given majority getting^{1}⁄_{2}each would be a more probable solution but you would really need to formalize the rules before this can be proven. Im a cryptologist so this is sadly not really my area...I almost posted on the three-person situation earlier, but what I wrote wasn’t cogent enough. It does seem like it should work as an archetype for any N > 2.

The problem is how the game is iterated. Call the players A, B, and C. If A says, “B, let’s go 50-50,” and you assume C doesn’t get to make a counter-offer and they vote immediately, 50-50-0 is clearly the outcome. This is probably also the case for the 10-person if there’s no protracted bargaining.

If there is protracted bargaining, it turns into an infinite regression as long as there is an out-group, and possibly even without an outgroup. Take this series of proposals, each of which will be preferred to the one prior (format is Proposer:A gets-B gets-C gets):

A:50-50-0

C:0-55-45

A:50-0-50

B: 55-45-0

C:0-55-45

A:50-0-50 …

There’s clearly no stable equilibrium. It seems (though I’m not sure how to prove this) that an equal split is the appropriate efficient outcome. Any action by any individual will create an outgroup that will spin them into an infinite imbalance. Moreover, if we are to arbitrarily stop somewhere along that infinite chain, the expected value for each player is going to be

^{100}⁄_{3}(they get part of a two-way split twice which should average to 50 each time overall, and they get zero once per three exchanges). Thus, at 33-33-33, one can’t profitably defect. At 40-40-20, C could defect and have a positive expected outcome.If the players have no bargaining costs whatsoever, and always have the opportunity to bargain before a deal is voted on, and have an infinite amount of time and do not care how long it takes to reach agreement (or if agreement is reached), then it does seem like you get an infinite loop, because there’s always going to be an outgroup that can outbid one of the ingroup. This same principle should also apply to the 10-person model; with infinite free time and infinite free bargaining, no equilibrium can be reached. If there is some cost to defecting, or a limitation on bargaining, there should be an even N/2+1-way split (depending admittedly on how those costs and limits are defined). If there is no limitation on bargaining and no cost to defecting, but time has a cost or time will be arbitrarily “called,” an even N-way split seems like the most likely/efficient outcome. The doubly-infinite situations is so far divorced from reality that it does not seem worth losing sleep over.

Also, the problem may stem from our limitation of thinking of this as a linear series of propositions, because that’s how people would have to actually bargain. In the no-repeated bargaining game, whether it’s 50-50-0 or 0-50-50 all depends on who asks first, which seems like an improper and unrealistic determining factor. This linear, proposer-centered view may not be how such beings would actually bargain.

The example of the Rubinstein bargaining model suggests that you could make players alternate offers and introduce exponential temporal discounting. An equal split isn’t logically necessary in this case: a player’s payoff will likely depend on their personal rate of utility discounting, also known as “impatience”, and others’ perceptions of it. The search keyword is “n-person bargaining”; there seems to be a lot of literature that I’m too lazy and stupid to quickly summarize.