It becomes a lot more obvious when you consider two factors; 1, that every agent wants to maximize its utility, and 2, that no split will be agreed to if it makes one agent worse off for participating in the trade than their bull action. Consider our simplest factory case: a factory that requires 1 owner to supply capital and one worker to provide labor. Since both are required to obtain the output, you split the gains from trade by taking the income from the factory, paying out the costs to bring both agents to net 0 from agreeing to the trade (ie, maintenance on the capital, taxes, insurance against work injuries, food to replace calories expended in labor, etc). Then whatever is left, the profit, gets split between both.
Now expand this. If there is one supplier of Capital and 2 workers, but it still only needs 1 worker to function, how does the split go? If the workers have good decision theory, they realize their best option is to coordinate. The Capital provider and the Laborer each get 50-50 split, and since each laborer has a 50% chance to be chosen that comes out to 50-25-25. This comes from a few factors. First, the Capital owner won’t accept less than half. If he is better off refusing to deal with B entirely and just trade with A, obviously he will do that, so our split has to be at least 50 to him. However, the workers also won’t accept less than 25 in expectation, or else they are both better off with one of them going to the other and saying “I won’t compete with you in exchange for half your profit.” A 50% chance at a 70-30 split (since there is a half chance each is chosen) is worse than that in expectation, even if the nominal payout is higher. Thus if the owner won’t accept less than half and our workers won’t accept less than a quarter, the only possible split is 50-25-25. Since our position is symmetric, 2 Capital providers and one laborer get a 25-25-50 split. Assuming there are 2 of each, that comes down to 2 50-50 splits again, since any lesser split has a better alternative that one party can force. Thus in any trade that requires one member of party A to contribute and one of Party B, each member of party A gets 1/2*#A and each member of B gets 1/2*#B. If the factory requires 2 workers to function, you can treat those two workers as a single agent that distributes half of the profit between them.
This has a few desirable outcomes.
For one, you are never better off for having more competition
You are never worse off for having more people to trade with
There is no coordination between agents that can reduce this to a simpler form that gets in expectation a higher payout for that group
it is agnostic of which of a set of interchangeable options is chosen.
There is no incentive to create dummy agents
And most importantly,
it is very simple to calculate. However many workers there are split between them half of the profit, irrespective of how large the labor pool is. This gets them in expectation less money since they have less chance of getting picked, but doesn’t change the amount changing hands in each actual timeline.
You’re right, this was less clear than I intended it to be; my apologies. The proposal I offer here is that since the first owner to join offers nothing, they aren’t included in the calculation; you just run them as though the second owner was the only owner, and get the same result for the value of the second owner as you did for the sole owner the first time. This has the advantage of being very computationally simple and gets you the same results as the full calculation for this case. It’s a specific case of a broader way you can simplify shapely calculations, where when the payout can be calculated by (A1+A2+A3…AN)*(B1+B2+B3…BM)+…+(X1+X2+…XO), you’ll get the same values for combining “Like Terms” (Not exactly the same concept but I don’t remember what the real word was, been a while) and running the shapely calculation for A, B,…X and then dividing the total of A by which fraction of A each A1, A2, etc terms makes up. For more complicated payout amounts it’s a bit more annoying to do this way but you can still do it if you’re careful to set it up properly. This applies even when one of the terms, eg A2, is 0, which then gets us our specific case here. I don’t remember the exact setup for the proof my professor used, but I don’t think it’s neccesary? It feels like one of those things where it’s obvious once you run it through both ways for an example, but if it isn’t lmk and I’ll see about digging through my old notes.