# Divide and conquer The Shapley Shubik Power Index

With Lloyd Shapley having just won the Nobel prize  it seems like a good time to look at the Shapley-Shubik power index.

This is a way to measure how much power different voters have in a vote where they control a different number of votes. It is not necessarily the case that someone controlling 20% of the votes has twice as much power as someone controlling 10% of the votes. They may have more or they may have less depending on the overall situation.

How it works

The way that the Shapley Shubik power index works is by taking every possible combination of voters and working out how many times a particular voter is pivotal. With this index the order matters so if there are three voters A, B, and C then there are six possible combinations to consider (ABC, ACB, BAC, BCA, CAB, CBA).

To work out the pivotal voter we look at how many votes the first voter has. If this is enough to reach a majority (or whatever level is needed) then they are the pivotal voter. If they are not then we move on to the second voter in the combination. If they voted the same was as the first voter and this took the total number of votes passed a majority then the second voter would be the pivotal one. It is whichever voter is the one that tips the combination passes the number of votes needed to win.

For example, assume that there are 100 votes to be cast and voter A controls 15 of them, voter B controls 40, and voter C controls 45. In the ABC combination B is the pivotal voter. This is because A has 15 votes so they do not have enough votes to get to 50 on their own. Once we add in B’s 40 votes then we get up to 55 votes so A and B together control more than 50% of the votes. This means that with this order of voters, B is the pivotal voter that takes the total over 50%.

The pivotal voter for the other five combinations is as follows:

ACB – Pivotal voter is C

BAC – Pivotal voter is A

BCA – Pivotal voter is C

CAB – Pivotal voter is A

CBA – Pivotal voter is B

So across all six combinations, A, B and C are all pivotal twice, so they are considered to have the same amount of power, even though they have a different amount of votes. This is because they each need one other player to vote with them to gain a majority. It doesn’t matter whether it is a large majority, like when B and C combine or a small majority when A combines with either B or C.

Strategy – Divide and conquer

A player with an understanding of where power really lies in a vote can use a strategy of ‘divide and conquer’ to take a disproportionate amount of power from their rivals.

If we start with the game that is described above and then imagine that B is able to change the game by introducing a new player D who takes 11 of C’s votes. This leaves A still with 15 votes; B with 40; C is reduced to 34; and D with 11.

There are now 24 possible combinations shown below with the pivotal voter in each marked in bold.

DABC; DACB; DBAC; DBCA; DCAB; DCBA

A is pivotal 4 times out of the 24 combinations and so has 16.7% of the power. B is pivotal 12 times out of 24, or 50%. C and D are both pivotal four times, the same as A.

Remember in the first game each voter had an equal amount of power, so they had 33.3% each. By managing to introduce a new player to take some of C’s votes B has greatly increased their own power from 33% to 50%. With three players they all had an equal amount of power despite their different number of votes. Now B has reduced A and C’s power while increasing their own.

In the first game any two pairs of voters could form a coalition and win the vote. By giving some of C’s votes to a new player, D, it changed the game so that B could pair up with anyone else to get a majority but none of the other players could pair up to reach a majority. This put B in a much stronger position than the first game.

Divide and conquer in action!

Image courtesy of taoty / FreeDigitalPhotos.net

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### 1 Response to Divide and conquer

1. Matt says:

I’ve only just discovered your blog, so pardon me if this is an obvious question– it seems like B’s power depends on voting sequentially, which is a circumstance that comes up rarely, if ever. Aren’t most votes cast simultaneously? Or, at least secretly, which amounts to the same thing.

In that case, you can divide the vote on any two-sided issue as follows:
A-BCD, B-ACD, C-ABD, D-ABC, AB-CD, AC-BD, AD-BC.

B is on the winning side of any issue, as long as she doesn’t stand alone. In other words, B can stand with any other single player and have a majority. The other three players must team up to outweigh her 40%. That seems like B has more than 50% of the power, but more like 86%.

Or am I missing something?