# Voting power

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After the 2020 American Presidential election, with the usual post-election analyses and (in this case) vast numbers of lawsuits, I started looking at the Electoral College, and trying to work out how it worked in terms of power. Although power is often conflated simply with the number of votes, that's not necessarily the case. We consider power as the ability of any state to affect the outcome of an election. Clearly a state with more votes: such as California with 55, will be more powerful than a state with fewer, for example Wyoming with 3. But often power is not directly correlated with size.

For example, imagine a version of America with just 3 states, Alpha, Beta, and Gamma, with electoral votes 49, 49, 2 respectively, and 51 votes needed to win.

The following table shows the ways that the states can join to reach (or exceed) that majority, and in each case which state is "necessary" for the win:

Winning Coalitions Votes won Necessary States
Alpha, Beta 98 Alpha, Beta
Alpha, Gamma 51 Alpha, Gamma
Beta, Gamma 51 Beta, Gamma
Alpha, Beta, Gamma 100 No single state

By "necessary states" we mean a state whose votes are necessary for the win. And in looking at that table, we see that in terms of influencing the vote, Gamma, with only 2 electors, is equally as powerful as the other two states.

To give another example, the Treaty Of Rome in the 1950's established the first version of the European Common Market, with six member states, each allocated a number of votes for decision making:

1 France 4
2 West Germany 4
3 Italy 4
4 The Netherlands 2
5 Belgium 2
6 Luxembourg 1

The treaty determined that a quota of 12 votes was needed to pass any resolution. At first this table might seem manifestly unfair: West Germany with a population of over 55 million compared with Luxembourg's roughly 1/3 of a million, thus with something like 160 times the population, West Germany got only 4 times the number of votes of Luxembourg.

But in fact it's even worse: since 12 votes are required to win, and all the other numbers of votes are even, there is no way that Luxembourg can influence any vote at all: its voting power was zero. If another state joined, also with a vote of 1, then it and Luxembourg together can influence a vote, and so Luxembourg's voting power would increase.

A power index is some numerical value attached to a weighted vote which describes its power in this sense. Although there are many such indices, there are two which are most widely used. The first was developed by Lloyd Shapley (who would win the Nobel Prize for Economics in 2012) and Martin Shubik in 1954; the second by John Banzhaf in 1965.

## Basic definitions

First, some notation. In general we will have $$n$$ voters each with a weight $$w_i$$, and a quota $$q$$ to be reached. For the American Electoral College, the voters are the states, the weights are the numbers of Electoral votes, and $$q$$ is the number of votes required: 238. This is denoted as

$[q; w_1, w_2,\ldots,w_n].$

The three state example above is thus denoted

$[51; 49, 49, 2]$

$[12; 4,4,4,2,2,1].$

### The Shapley-Shubik index

Suppose we have $$n$$ votes with weights $$w_1$$, $$w_2$$ up to $$w_n$$, and a quote $$q$$ required. Consider all permutations of $$1,2, \ldots,n$$. For each permutation, add up the weights starting at the left, and designate as the pivot voter the first voter who causes the cumulative sum to equal or exceed the quota. For each voter $$i$$, let $$s_i$$ be the number of times that voter has been chosen as a pivot. Then its power index is $$s_i/n!$$. This means that the sum of all power indices is unity.

Consider the three state example above, where $$w_1=w_2=49$$ and $$w_3=2$$, and where we compute cumulative sums only up to reaching or exceeding the quota:

Permutation Cumulative sum of weights Pivot Voter
1 2 3 49, 98 2
1 3 2 49, 51 3
2 1 3 49, 98 1
2 3 1 49, 51 3
3 1 2 2, 51 1
3 2 1 2, 51 2

We see that $$s_1=s_2=s_3=2$$ and so the Shapley-Shubik power indices are all $$1/3$$.

### The Banzhaf index

For the Banzhaf index, we consider the winning coalitions: these are any subset $$S$$ of voters for which the sum of weights is not less than $$q$$. It's convenient to define a function for this:

$v(S) = \begin{cases} 1 & \text{if } \sum_{i\in S}w_i\ge q \cr 0 & \text{otherwise} \end{cases}$

A voter $$i$$ is necessary for a winning coalition $$S$$ if $$S-\{i\}$$ is not a winning coalition; that is, if $$v(S)-v(S-\{i\})=1$$. If we define

$p(i) =\sum_S v(S)-v(s-\{i\})$

then $$b(i)$$ is a measure of power, and the (normalized) Banzhaf power indices are defined as

$b(i) = \frac{p(i)}{\sum_i p(i)}$

so that the sum of all indices (as for the Shapley-Shubik index) is again unity.

Considering the first table above, we see that $$p(1)=p(2)=p(3)=2$$ and the Banzhf power indices are all $$1/3$$. For this example the Banzhaf and Shapley-Shubik values agree. This is not always the case.

For the EEC example, the winning coalitions are, with necessary voters:

1,2,3 12 1,2,3
1,2,4,5 12 1,2,4,5
1,3,4,5 12 1,3,4,5
2,3,4,5 12 2,3,4,5
1,2,3,6 13 1,2,3
1,2,4,5,6 13 1,2,4,5
1,3,4,5,6 13 1,3,4,5
2,3,4,5,6 13 2,3,4,5
1,2,3,4 14 1,2,3
1,2,3,5 14 1,2,3
1,2,3,4,6 15 1,2,3
1,2,3,5,6 15 1,2,3
1,2,3,4,5 16 No single voter
1,2,3,4,5,6 17 No single voter

Counting up the number of times each voter appears in the rightmost column, we see that

$p(1) = p(2) = p(3) = 10,\quad p(4) = p(5) = 6,\quad p(6) = 0$

and so

$b(1) = b(2) = b(3) = \frac{5}{21},\quad b(4) = b(5) = \frac{1}{7}.$

Note that the power of the three biggest states is in fact only 5/3 times that of the smaller states, in spite of having twice as many votes. This is a striking example of how power is not proportional to voting weight.

Note that computing the Shapley-Shubik index could be unwieldy; there are

$\frac{6!}{3!2!} = 60$

different permutations of the weights, and clearly as the number of weights increases, possibly with very few repetitions, the number of permutations will be excessive. For the Electoral College, with 51 members, and a few states with the same numbers of voters, the total number of permutations will be

$\frac{51!}{(2!)^4(3!)^3(4!)(5!)(6!)(8!)} = 5368164393879631593058456306349344975896576000000000$

which is clearly far too large for enumeration. But as we shall see, there are other methods.