In general, I suspect, but I haven’t proved it, that if there are n otherwise iddle applicants, the Shapley value assigned to the organization is (n-1)/n. This suggests that a lot of the impact of the position goes to whomever created the position.
I think you meant the fraction of the total Shapley value assigned to the organisation is S_o/S = n/(n + 1), which equals 1⁄2 for 1 applicant (n = 1). I have confirmed this is the case:
The number of players is n + 1.
The only coalition for which the marginal contribution of each applicant is not null is one having solely the organisation. In this case, the marginal contribution of each applicant is 100, and the coalition size is 1.
There are n coalitions with size 1, n − 1 coalitions containing only one applicant, and 1 containing only the organisation.
So the Shapley value of each applicant is S_a = 100/n/(n + 1).
The total Shapley value of S = 100 should be equal to n*S_a + S_o, therefore S_o = 100 - n*S_a = 100*(1 − 1/(n + 1)) = 100*n/(n + 1).
In other words, the fraction of the total Shapley value assigned to the organisation is S_o/S = S_o/100 = n/(n + 1).
It is interesting to note that the relative contribution of each applicant S_a/S = 1/n/(n+1) is roughly proportional to n^-2. This suggests the contribution of each applicant would become 0.25 times as large if the number of applicants doubled.
However, all of this assumes applicants would be equally good at their job (or that the selected applicant would be chosen randomly). If the factual impact of the selected applicant (i.e. the total Shapley value) equals S = n^alpha, where alpha >= 0, the relative contribution of each applicant would be:
Sa=1n+1∑ni=1iα−(i−1)α(ni).
I have demonstrated this is equal to n^(alpha − 1)/(n + 1), but the proof is too long to fit in this comment… Just kidding, the formula is probably not that simple.
I really like this post!
I think you meant the fraction of the total Shapley value assigned to the organisation is S_o/S = n/(n + 1), which equals 1⁄2 for 1 applicant (n = 1). I have confirmed this is the case:
The number of players is n + 1.
The only coalition for which the marginal contribution of each applicant is not null is one having solely the organisation. In this case, the marginal contribution of each applicant is 100, and the coalition size is 1.
There are n coalitions with size 1, n − 1 coalitions containing only one applicant, and 1 containing only the organisation.
So the Shapley value of each applicant is S_a = 100/n/(n + 1).
The total Shapley value of S = 100 should be equal to n*S_a + S_o, therefore S_o = 100 - n*S_a = 100*(1 − 1/(n + 1)) = 100*n/(n + 1).
In other words, the fraction of the total Shapley value assigned to the organisation is S_o/S = S_o/100 = n/(n + 1).
It is interesting to note that the relative contribution of each applicant S_a/S = 1/n/(n+1) is roughly proportional to n^-2. This suggests the contribution of each applicant would become 0.25 times as large if the number of applicants doubled.
However, all of this assumes applicants would be equally good at their job (or that the selected applicant would be chosen randomly). If the factual impact of the selected applicant (i.e. the total Shapley value) equals S = n^alpha, where alpha >= 0, the relative contribution of each applicant would be:
Sa=1n+1∑ni=1iα−(i−1)α(ni).
I have demonstrated this is equal to n^(alpha − 1)/(n + 1), but the proof is too long to fit in this comment… Just kidding, the formula is probably not that simple.
> I have confirmed this is the case
Oh, nice!