# Shapley Value for Voters in Electoral College

Consider (a simplified version of) the USA electoral college. Each state has millions of voters and tens of electors. The candidate with the most electors wins and each state allocates electors according the popular vote in that state (some are winner take all and some are proportional). There is lots of work on algorithms for computing the Shapley-Shubik value (voting power) for each state, but how do you compute the Shapley-Shubik value by voters? ie how much power does a voter in one state have compared to another? Note that this is not simply a weighted voting game.

• It is less than entirely true that in every state the winner of the majority gets all of the electors. In Maine and Nebraska that candidate gets two of the electors and then another elector is chosen in each congressional district, so that there may not be unanimity. In 1800, North Carolina was divided into presidential election districts, so that one elector was chosen in each district, and again there didn't have to be unanimity. (I don't know if there was.) Of course there had to be two more such districts than the number of congressional districts. In some states.... Dec 23, 2020 at 19:36
• In some states in the past all presidential electors were chosen by the legislature. What the Constitution of the United States says about it is that in each state, the legislature decided how the electors are chosen. Dec 23, 2020 at 19:37
• @MichaelHardy: yes, I simplified the explanation, by just mentioning proportional and winner take all. In the end I would like a computational method where I can take all the precise rules into account, but for now would be happy to understand a simplified version. Dec 23, 2020 at 19:39

## 1 Answer

I think the Shapley-Shubik power of any given voter is the Shapley-Shubik power of the district the voter is in, divided by the amount of voters in the district.

## Definition

First, let's define what we're talking about. We will define the Shapley-Shubik power of a voter $$i$$ as the proportion of permutations of voters such that a coalition of voters up to and including $$i$$ is sufficient to win the election. In other words, if every voter up to and including $$i$$ votes for candidate A and all others vote for B, A should win.

Note that this is a slight deviation of the "real" Shapley-Shubik power, or alternatively of the United States electoral system. The problem is that there is no set point at which one wins an election, as one's candidate might still win without a majority of the electoral votes, either because there is a tie, because a third party has won some states, or because some electors don't properly do their job. A faithful representation of the American system would have to account for this by giving a value to any given coalition based on the probability of winning with that coalition, after which we could calculate the Shapley value for each voter, instead of the Shapley-Shubik power. Clearly however this is not feasible, so we instead assume that there are only two candidates, ties don't occur, and everything works "nicely".

## Which voter is the pivot voter?

Now then, we need to find the pivot voter for a given permutation. One way to do so is to find the pivot voter for each separate district, throw out all the other ones, and then consider the remaining voters as being districts. The pivot "district" in this new scenario will be the pivot voter of the old scenario.

For example, consider a system with three districts ($$a$$, $$b$$, $$c$$), all with three voters and with the same amount of votes/power. One possible permutation would be:

abbcaccab

First, we only consider the middle voter of each district:

abbcaccab $$\rightarrow$$ bac

Now, because each district has the same amount of power, district $$a$$ is the pivot district, and the second voter from this district is the pivot voter in the original permutation.

## What is the probability of a district being pivotal?

This is the most important part of this answer, as I argue that the probability of a district being pivotal is the same, regardless of whether we look at districts or at voters as the entities to which we ascribe Shapley-Shubik power.

This is equivalent to saying that for a permutation of voters from a number of districts, the permutations of districts derived by only considering the pivotal voters for each district will yield all possible permutations of districts equally often. I believe the truth of this statement is fairly clear for 2 districts, and I'm pretty sure you can prove it through induction for more districts, but I have not done so. However, if you disagree with this, or if you cannot find a proof, you might want to ask this subquestion on this very site again.

As for this answer, we will state without proof that the probability of a voter from district $$a$$ being pivotal is equal to district $$a$$ being pivotal if we consider the district as voters themselves.

## Conclusion

Because the probability of a voter from a district being pivotal is equal to the district itself being pivotal, the sum of the Shapley-Shubik powers of the voters must equal the power of the district. Furthermore, as the voters are all identical, they must through symmetry all have the same power. Therefore, the power of a voter is the power of a district divided by the number of voters.