Weighted Voting Systems - W.H. Freeman

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$Microscopy Math - W.H. Freeman$

CHAPTER 11 Weighted Voting Systems 399and D is the pivotal voter in 2 permutations. Therefore, the Shapley–Shubik powerindex for this weighted voting system is 1 0 6 6 2, , , 242 4 2 4 2 4 11 00 EXAMPLE 7How to Compute the Shapley–Shubik Power IndexFor voting systems with no more than four voters, the Shapley–Shubik power indexmay be calculated by making a list of all the voting permutations and identifyingthe pivotal voter in each.Listing all of the permutations is the brute force way of calculating the Shapley–Shubik power index. As we saw in connection with the traveling salesperson problemin Chapter 2, brute force methods can be impossible to carry out, due to thecombinatorial explosion. If we were to study a 10-voter system by the brute forcemethod, we would have to list 10! permutations—that’s more than 3 1 million. If2 we made the list using the same size page and same typeface as this text, we couldprobably get about 350 permutations on each page. It would take 10,000 of thesepages to list all of the permutations. If we used a computer, each permutationwould occupy 10 bytes, and 3 1 million of them would take 35 megabytes, a man-2 ageable size. But even a computer would not be able to handle a 100-voter systembecause the number of permutations is a 158-digit number. Even by usingall the memory devices—chips, disk drives, magnetic tape, flash memory, and soon—in the world, you would not be able to save a file containing all of the permutations.If all the voters have the same voting weight, you don’t need to make a listof all the permutations, because each has the same share of power. If there are100 voters, each with 1 vote, the Shapley–Shubik index of each is . If all but oneor two of the voters have equal power, we can still calculate the Shapley–Shubikpower index of each without making a list of permutations. We use two principlesto do this:Voters with the same voting weight have the same Shapley–Shubik powerindex. The sum of the Shapley–Shubik power indices of all the voters is 1.A Seven-Person CommitteeThe chairperson of a committee has 3 votes, and there are six other memberswith 1 vote. The quota for passing a measure is a simple majority, 5 of the 9votes. In our notation, this voting system is [5 : 3, 1, 1, 1, 1, 1, 1].Each ordinary member has the same power index. Our strategy is to computethe index of the chair, and then divide the share of power that the chairdoes not have equally among the ordinary members.There are 7! 5040 permutations to consider. We will group them by theposition occupied by the chairperson. Thus, CMMMMMM, in which the chair-
400 PART III Voting and Social Choiceperson is first, is the first group. Counting from the left, we see that the votes areaccumulated in the sequence 3, 4, 5, 6, 7, 8, 9. In these permutations (there are6! of them), the chairperson is not the pivot; an ordinary member is. In the secondgroup, MCMMMMM, the votes accumulate in the sequence 1, 4, 5, 6, 7,8, 9, and again, the chairperson is not the pivot.The chairperson is the pivot in the next three groups of permutations,MMCMMMM, MMMCMMM, and MMMMCMM, with vote accumulations1, 2, 5, 6, 7, 8, 9; 1, 2, 3, 6, 7, 8, 9; and 1, 2, 3, 4, 7, 8, 9, respectively. In the finaltwo groups, MMMMMCM and MMMMMMC, with vote accumulations1, 2, 3, 4, 5, 8, 9 and 1, 2, 3, 4, 5, 6, 9, an ordinary member, not the chairperson,is the pivot again.Each of the 7 groups of permutations is of the same size, 6!, because the 6ordinary members can appear in any order in each. The chairperson is the pivot inthree groups, for a total of 3 7 of the total number of permutations. His Shapley–Shubik power index is therefore 3 . The remaining 4 7 7 of the voting power is sharedequally by the 6 ordinary members. Therefore each hasof the power. 4 7 6 2 2 1The Shapley–Shubik power index of this weighted voting system is therefore 3 7 , 2 2 2 2 2 2, , , , , 212 1 2 1 2 1 2 1 2 1Because 3 , the Shapley–Shubik model indicates that the chairperson7 2 4 1 2 2 is 4 1 times as powerful as an ordinary member, although his voting power is only2 3 times as much. EXAMPLE 8A Committee with Two Co-ChairsA committee has 7 members: two co-chairs who each have 3 votes, and five othermembers with 1 vote each. The quota is 7, and thus the weighted voting systemis [7 : 3, 3, 1, 1, 1, 1, 1]. Our objective is to determine the Shapley–Shubik powerindex of each member of the committee. Our strategy this time is to determinethe voting power of A, a weight-1 voter. A is pivotal when the voters coming beforehim in the permutation have a combined weight of exactly 6. There are twoways to meet this condition: C 1 C 2 AX 1 X 2 X 3 X 4 , where C 1 and C 2 represent co-chairs, and X 1 , . . . , X 4represent members who are not co-chairs and are not A.Y 1 Y 2 Y 3 Y 4 AY 5 Y 6 , where one of Y 1 , . . . Y 4 is a co-chair, one of Y 5 , Y 6 is aco-chair, and the remaining Y ’s are weight-1 voters.For the first type of permutation, there are 2 ways to order the co-chairs (ifthe co-chairs are P and Q , then C 1 could be P or Q and C 2 would be the remainingco-chair), and there are 4! ways of ordering the other 4 members. There arethus 2 4! 48 permutations of the first type.
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Weighted Voting Systems - W.H. Freeman

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