What model are you using to conclude the workers capture less value?
I’m reasonably sure Shapley values are preserved under agent decomposition. That is, if you split agent A into multiple agents B+C who collectively make the same contribution as A, their combined share should be the same as A’s original share.
Though I don’t know whether/how you can extend this model to handle the fact that players can join and leave the “game”—workers can quit, other workers can be hired, the owner could decide not to build the factory but a different tycoon could build one instead, etc. This means you aren’t dealing with a fixed pool of “players”. (One theoretical option is to treat the entire world as one giant game, but then the problem becomes intractable.)
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.
First, I note that this comment thread was previously about Shapley values, and you don’t seem to have done any Shapley calculations. If this is meant to be the same rule, but explained from a different angle, then I don’t see how to establish equivalence. If this is meant to be a new system, then I don’t see how it generalizes to more complex examples, such as where the factory output scales with the number of workers (rather than being all-or-nothing). (I also don’t see why you’d choose this particular comment to start promoting your alternative system.)
Second, you’re analyzing a situation where a required input can be provided by any of multiple parties; that is, if there are 2 owners, you only need 1 owner to agree in order to make the factory run. But the story problem above was about a situation where you need all of multiple parties; i.e. replace the 1 owner with 1 capitalist + 1 technologist and you need both of them to make the factory run.
If I came to you and said, here’s a game with 3 people (1 worker + 1 capitalist + 1 technologist), you need all 3 people working together to produce anything, how do they split the profits? I suspect you’d say an even 3-way split. But that implies that the owner from the 2-person can divide himself into 2 dummy agents (1 capitalist + 1 technologist) and then demand 2⁄3 of the profit (up from 1⁄2) because he’s now (nominally) doing 2 out of 3 jobs.
How do you prevent this exploit?
Third, I don’t buy your claim in your advanced examples that “any lesser split has a better alternative that one party can force”. For instance, in the 1 owner + 2 workers example, if the owner offers worker A a 70-30 split, that’s better for both the owner and worker A than your proposed split, and I don’t see what worker B can do about it.
You seem to be arguing that worker A should reject this split on some sort of timeless logic (?) where A reasons that there was an equal chance the offer would have been made to B and so if A+B are the sort of people who accept this offer then they each get 15 in expectation across all counterfactuals. Even if you buy the timeless logic, this only works if A and B use correlated strategies such that A is effectively choosing for both of them; otherwise, after A rejects this split, the owner proposes it to B and A gets nothing in all counterfactuals. So that seems to me like a coordinated solution, not a solution that a single party can unilaterally force.
In fact, it looks to me like you’ve said something pretty close to “my system rewards monopolies, so A and B are incentivized to form a cartel and act like a single agent, and therefore I assume they do so.”
I did a really simply and naive thing in an online calculator, and I’d like to state explicitly that I’ll be quite intrigued if you show me I did it wrong as I feel I must have made a mistake looking back over this comment.
Here I did a trivial thing where you need all owners and 2 workers here, and included only those people. You can see that they trivially are going to telescope to each getting 1/n of the value out.
Apologies for the bad example, but I really don’t want to enter all those cells again and am going to only do the better computations if I write some code for it later.
Sounds like you set up a model where the factory produces nothing unless it has 100% participation from all players, which means that all players are in equivalent positions (any one of them can veto the entire operation and leave the other players with nothing). Under this model, “owners” and “workers” are actually equivalent—you’ve assumed that if 1 worker goes on strike, the entire factory shuts down.
Obviously, if all players are in equivalent positions, then by symmetry they’ll have equal payouts.
In the original story-example, the factory owner can claim a higher Shapley value on the basis that if you removed just 1 worker, the factory would still produce quite a bit, but if you removed the owner, you wouldn’t have a factory anymore.
A very simple model for this might be something like: owner + any N workers produces N utility, while any subset not including the owner produces 0 utility. Then, with 10 workers + 1 owner, if you add people in a random order, each worker produces an average of 0.5 marginal utils (= 0 if added before the owner, 1 if added after) while the owner produces an average of 5 marginal utils (= the average number of workers added before the owner).
If you double the number of workers to 20 (and thus double the max output to 20 as well), then each worker still adds an average of 0.5 marginal utils, but the owner now adds an average of 10 marginal utils. So the worker wages stay the same, but the owner earns more from the larger operation.
Now suppose we go back to 10 workers, but now we have 2 owners, with the rule that the factory produces 0 unless both owners participate. (Thus, these 2 owners collectively do the same job that the 1 owner previously did.)
A worker produces 1 marginal value if added after both owners, 0 otherwise. Since there are 2 owners, on average 1⁄3 will be added before both, 1⁄3 between, and 1⁄3 after both. So the worker average is 1⁄3 util.
The first owner added produces 0, the second produces as many as there are workers. On average 2⁄3 of workers are added before the second owner, so the second owner produces 6.67 marginal utils. Each owner is the second owner half the time, so each gets 3.33.
...um. That is not the result I was expecting.
Can anyone either point out an error in my math, or explain to me philosophically why I ought to be ok with this result? It seems concerning to me if the factory gets a different share depending on whether it’s owned by 1 person or by 2 people.
Yup, I wrote the program and still get owner-value-portion-grabbing. If it’s any help, I can now generate any simple pattern you want very quickly and easily.
Only the second owner to join adds any value, so they get the same slice as the sole owner in the first example, ie, half. Since each owner has a 50% chance of being the second owner, they each get 25%. With 3 owners it’s 1/3*.5=1/6. Etc. the math works out very neatly this way.
Only the second owner to join adds any value, so they get the same slice as the sole owner in the first example, ie, half.
Show your work and/or explain what’s wrong with my reasoning above? I already took that into account, but got a different answer, because there’s more permutations where the second owner is added late in the sequence than where the second owner is added early in the sequence.
(Note that if my reasoning from the previous comment is wrong, it has to be wrong about both the owners and the workers, since they need to sum to the correct total.)
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.
If I understand you, you’re saying that this simplified calculation should give the same answer as the full calculation, and that this should be obvious to me after I try it both ways.
But you’re also saying that the simplified calculation is equivalent to the earlier problem with only 1 owner, which means I already worked it both ways (one way in the 1-owner problem and the other way in the 2-owner problem) and I got different answers. So, no, the equivalence is not obvious after running it both ways.
My suspicion is that this half-remembered theorem you are trying to apply actually applies to a different situation and you are using it wrong. But if you don’t think that’s the case, then yes, I’d like to see a more precise explanation. (And ideally, an explanation of what you think I did wrong in my own calcs.)
What model are you using to conclude the workers capture less value?
I’m reasonably sure Shapley values are preserved under agent decomposition. That is, if you split agent A into multiple agents B+C who collectively make the same contribution as A, their combined share should be the same as A’s original share.
Though I don’t know whether/how you can extend this model to handle the fact that players can join and leave the “game”—workers can quit, other workers can be hired, the owner could decide not to build the factory but a different tycoon could build one instead, etc. This means you aren’t dealing with a fixed pool of “players”. (One theoretical option is to treat the entire world as one giant game, but then the problem becomes intractable.)
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.
First, I note that this comment thread was previously about Shapley values, and you don’t seem to have done any Shapley calculations. If this is meant to be the same rule, but explained from a different angle, then I don’t see how to establish equivalence. If this is meant to be a new system, then I don’t see how it generalizes to more complex examples, such as where the factory output scales with the number of workers (rather than being all-or-nothing). (I also don’t see why you’d choose this particular comment to start promoting your alternative system.)
Second, you’re analyzing a situation where a required input can be provided by any of multiple parties; that is, if there are 2 owners, you only need 1 owner to agree in order to make the factory run. But the story problem above was about a situation where you need all of multiple parties; i.e. replace the 1 owner with 1 capitalist + 1 technologist and you need both of them to make the factory run.
If I came to you and said, here’s a game with 3 people (1 worker + 1 capitalist + 1 technologist), you need all 3 people working together to produce anything, how do they split the profits? I suspect you’d say an even 3-way split. But that implies that the owner from the 2-person can divide himself into 2 dummy agents (1 capitalist + 1 technologist) and then demand 2⁄3 of the profit (up from 1⁄2) because he’s now (nominally) doing 2 out of 3 jobs.
How do you prevent this exploit?
Third, I don’t buy your claim in your advanced examples that “any lesser split has a better alternative that one party can force”. For instance, in the 1 owner + 2 workers example, if the owner offers worker A a 70-30 split, that’s better for both the owner and worker A than your proposed split, and I don’t see what worker B can do about it.
You seem to be arguing that worker A should reject this split on some sort of timeless logic (?) where A reasons that there was an equal chance the offer would have been made to B and so if A+B are the sort of people who accept this offer then they each get 15 in expectation across all counterfactuals. Even if you buy the timeless logic, this only works if A and B use correlated strategies such that A is effectively choosing for both of them; otherwise, after A rejects this split, the owner proposes it to B and A gets nothing in all counterfactuals. So that seems to me like a coordinated solution, not a solution that a single party can unilaterally force.
In fact, it looks to me like you’ve said something pretty close to “my system rewards monopolies, so A and B are incentivized to form a cartel and act like a single agent, and therefore I assume they do so.”
I did a really simply and naive thing in an online calculator, and I’d like to state explicitly that I’ll be quite intrigued if you show me I did it wrong as I feel I must have made a mistake looking back over this comment.
I used http://shapleyvalue.com/ with several (ANNOYING TO INPUT) sets of values (10 utility in each).
Here I did a trivial thing where you need all owners and 2 workers here, and included only those people. You can see that they trivially are going to telescope to each getting 1/n of the value out.
1 owner / 2 worker = ( 3.33333 / owner, 3.33333 / worker) 2 owner / 2 worker = ( 2.5 / owner, 2.5 / worker) 3 owner / 2 worker = ( 2 / owner, 2 / worker)
Apologies for the bad example, but I really don’t want to enter all those cells again and am going to only do the better computations if I write some code for it later.
Sounds like you set up a model where the factory produces nothing unless it has 100% participation from all players, which means that all players are in equivalent positions (any one of them can veto the entire operation and leave the other players with nothing). Under this model, “owners” and “workers” are actually equivalent—you’ve assumed that if 1 worker goes on strike, the entire factory shuts down.
Obviously, if all players are in equivalent positions, then by symmetry they’ll have equal payouts.
In the original story-example, the factory owner can claim a higher Shapley value on the basis that if you removed just 1 worker, the factory would still produce quite a bit, but if you removed the owner, you wouldn’t have a factory anymore.
A very simple model for this might be something like: owner + any N workers produces N utility, while any subset not including the owner produces 0 utility. Then, with 10 workers + 1 owner, if you add people in a random order, each worker produces an average of 0.5 marginal utils (= 0 if added before the owner, 1 if added after) while the owner produces an average of 5 marginal utils (= the average number of workers added before the owner).
If you double the number of workers to 20 (and thus double the max output to 20 as well), then each worker still adds an average of 0.5 marginal utils, but the owner now adds an average of 10 marginal utils. So the worker wages stay the same, but the owner earns more from the larger operation.
Now suppose we go back to 10 workers, but now we have 2 owners, with the rule that the factory produces 0 unless both owners participate. (Thus, these 2 owners collectively do the same job that the 1 owner previously did.)
A worker produces 1 marginal value if added after both owners, 0 otherwise. Since there are 2 owners, on average 1⁄3 will be added before both, 1⁄3 between, and 1⁄3 after both. So the worker average is 1⁄3 util.
The first owner added produces 0, the second produces as many as there are workers. On average 2⁄3 of workers are added before the second owner, so the second owner produces 6.67 marginal utils. Each owner is the second owner half the time, so each gets 3.33.
...um. That is not the result I was expecting.
Can anyone either point out an error in my math, or explain to me philosophically why I ought to be ok with this result? It seems concerning to me if the factory gets a different share depending on whether it’s owned by 1 person or by 2 people.
Yup, I wrote the program and still get owner-value-portion-grabbing. If it’s any help, I can now generate any simple pattern you want very quickly and easily.
Only the second owner to join adds any value, so they get the same slice as the sole owner in the first example, ie, half. Since each owner has a 50% chance of being the second owner, they each get 25%. With 3 owners it’s 1/3*.5=1/6. Etc. the math works out very neatly this way.
Show your work and/or explain what’s wrong with my reasoning above? I already took that into account, but got a different answer, because there’s more permutations where the second owner is added late in the sequence than where the second owner is added early in the sequence.
(Note that if my reasoning from the previous comment is wrong, it has to be wrong about both the owners and the workers, since they need to sum to the correct total.)
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.
If I understand you, you’re saying that this simplified calculation should give the same answer as the full calculation, and that this should be obvious to me after I try it both ways.
But you’re also saying that the simplified calculation is equivalent to the earlier problem with only 1 owner, which means I already worked it both ways (one way in the 1-owner problem and the other way in the 2-owner problem) and I got different answers. So, no, the equivalence is not obvious after running it both ways.
My suspicion is that this half-remembered theorem you are trying to apply actually applies to a different situation and you are using it wrong. But if you don’t think that’s the case, then yes, I’d like to see a more precise explanation. (And ideally, an explanation of what you think I did wrong in my own calcs.)