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Week 5: Weighted Voting Systems Part 2

by: Amy Brogan

Week 5: Weighted Voting Systems Part 2 MATH 1014

Marketplace > University of Cincinnati > Mathematics (M) > MATH 1014 > Week 5 Weighted Voting Systems Part 2
Amy Brogan
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About this Document

The Banzhaf Power Index and how to quantify the number of sets depending on the number of voters.
Mathematics of Social Choice
Mary Koshar
Class Notes
Banzhaf power index combinations permutations sets voting weighted nCk
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This 2 page Class Notes was uploaded by Amy Brogan on Saturday February 13, 2016. The Class Notes belongs to MATH 1014 at University of Cincinnati taught by Mary Koshar in Spring 2016. Since its upload, it has received 11 views. For similar materials see Mathematics of Social Choice in Mathematics (M) at University of Cincinnati.


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Date Created: 02/13/16
Koshar Amy Brogan February 12, 2016 Week 5: Weighted Voting Systems Part 2 We have learned that a voting group with 5 voters has 120 permutation sets. With the Shapely-Shubik index, power is the number of times a voter is pivotal in a permutation. But what if the voters are only voting yes/no, and the order doesn’t matter. How do we quantify a combination voting set? This is called a coalition, and it’s used in the Banzhaf Power Index.  Coalition: o (general) a group of people with a common cause o (math) a combination of votes (vs. a permutation) Ex 1: [16: 10, 8, 6, 5, 3] Some Outcomes: {10, 6, 3} {3} {8, 6, 5, 3} Are these winning combinations (do they add up or over the quota)? Yes, no, yes. Which voters are critical to pass the motion? In the first set, if w10 changed their mind and voted no, then it would not pass, so w10 is critical. If w6 changed their mind, the same thing, so w6 is critical. But if w3 changed their mind, the motion would still carry, so they are not critical. Ex 2: [13: 11, 8, 5] What are all winning sets for this group? (They’re combination, so order doesn’t matter) {11, 8} {11, 5} {8, 5} and {11, 8, 5} What about listing all possible coalitions? This is easiest to do it by number of voters: 1 Voter Combination: {11} {8} {5} 2 Voters Combination: {11, 8} {11, 5} {8, 5} 3 Voters Combination: {11, 8, 5} That’s 7 combinations, but we also have to remember there is an empty set { } where nobody votes yes, so 8 combinations. With Banzhaf the number of times each voter is critical is 11:2, 8:2, and 5:2.  Critical: when the voter’s weight in a winning coalition is required to make it winning  Power: a voter has power when s/he is critical in some winning coalition  Goal: find the number of times a voter is critical Since there are only two possible answers (yes and no) we can calculate the possible number of coalitions with 2n 2^n n is the number of voters Ex 3: [6: 5, 4, 2] All coalitions: 1: {5} {4} {2} 2: {5, 4} {5, 2} {4, 2} 3: {5, 4, 2} +{ } for 8; the winning coalitions are highlighted, and the critical voters are underlined 5 is critical 2 times, as are 4 and 2. The total number critical voters is 6 (2+2+2) so the power index for each is 2/6, or simplified 1/3. The end “fractions” have to add up to one. If they don’t, check your work. The power index is written as before: (1/3, 1/3, 1/3) Ex 4: [10: 10, 6, 5] 1: {10} {6} {5} 2: {10, 6} {10, 5} {6, 5} 3: {10, 6, 5} 10: 3, 6:1, 5:1  5 critical positions  (3/5, 1/5, 1/5) Ex 5: [18: 12, 8, 8, 6] 2 4= 16 coalitions 1 voter has n sets, and n voters has 1 set when we are writing out all possible coalitions, but how do we calculate the middle sets? nC k = # of permutations of k from n voters = n! (multiplied by) 1 = n! __ ways to order k (n-k)! k! k!(n-k!) Ex 6: C 3 = There are 20 coalitions for sets of 3 voters. Ex 7: [12: 8, 6, 3, 3, 1, 1] – a lot of coalitions, what about just finding the number of sets of 4? 7C 4 = There are 35 coalitions for sets of 4 voters.


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