EXAMPLE 8 Zoë’s Power After the Revolt

Zoë will be the pivotal voter of a permutation if and only if the voters coming before her in the permutation have a combined weight of exactly 5. There are two kinds of voting permutations that meet this condition ( is Zoë):

Counting these permutations will involve the fundamental principle of counting. Let’s count the voting permutations of the first type. There are three places that Alice could occupy before Zoë, and two places that Bill could occupy after Zoë. By the fundamental principle of counting, there are ways we could position Alice and Bill in a voting permutation with Alice before Zoë and Bill after Zoë. We can count the number of permutations in each of the six groups, where Alice, Zoë, and Bill have already been positioned. The three other committee members can be ordered in ways, and put accordingly into the three open spaces. Using the fundamental principle of counting (again), we see that the number of voting permutations of the first type is . The number of permutations of the second type, where Bill comes before Zoë and Alice after, is the same. Therefore, there are voting permutations in which Zoë is pivotal. The Shapley–Shubik index of Zoë is therefore . Before the revolt, Zoë had of the power according to the Shapley–Shubik model, so her power has increased a bit.