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Week 4: Weighted Voting Systems Part 1

by: Amy Brogan

Week 4: Weighted Voting Systems Part 1 MATH 1014

Marketplace > University of Cincinnati > Mathematics (M) > MATH 1014 > Week 4 Weighted Voting Systems Part 1
Amy Brogan
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Calculating the power of each voter in a permutation voting system.
Mathematics of Social Choice
Mary Koshar
Class Notes
permutation voting system power quota
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This 4 page Class Notes was uploaded by Amy Brogan on Friday February 5, 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 21 views. For similar materials see Mathematics of Social Choice in Mathematics (M) at University of Cincinnati.


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Date Created: 02/05/16
Koshar Amy Brogan February 1 & 3, 2016 Weighted Voting Systems In the 1960’s, Kings, Queens, Nasser, and Suffolk made up seven municipals. Reprisentatives from each voted yes or no on motions about various things. Eventually they began to use the system that different groupd would have mutiple votes (population, income tax, whatever.) Thirty votes were spread out among them. Hempstead 1 9 A budding young mathematician named John Hempstead 2 9 Banzhaf III, who graduated college at the age of North Hempstead 7 15, claimed and proved that this was a bad Oyster Bay 3 system. Glen Corner 1 Long Beach 1 Majority Vote: 16 Possible Voting outcomes are shown with {_,_} such as: {9, 7, 1} or {9,3} or {9,9,7,1,1}. If the majority is 16, do the motions pass for each sets of votes? (Written: [16: 9, 9, 7, 3, 1, 1] {9, 7, 1} does with 17 votes, {9,3} does not with 12 votes out of the needed 16, and {9,9,7,1,1} passes easily. When you were counting up the votes, did you notice anything about how much power each voter has? It’s not Hempstead 1 and 2; they do have a lot of power, but motions can be passed without their vote. But Oyster Bay, Glen Cove, and Long Island don’t have to vote at all to change the outcome of any motion. Ever. Weighted Voting Systems: voters do not necessarily have the same weight (voter situations where voters vote yes or no to a motion. 1. [15: 8, 6, 5, 5, 3, 1] 15 is the majority/quota/minimum number of votes needed to pass. Possible outcomes: {8, 5, 5, 1} (passes 19/15), {8, 3} (does not pass 11/15), or {8, 6, 5, 3, 1} which can also be written {6, 5, 8, 3, 1} since order does not matter. (Passes easily. Once you reach the majority vote, you don’t have to keep counting unless you want to check your work.) 2. [6: 4, 4, 2, 1]  {4} (does not pass), {4, 2, 1} (passes), {4, 4, 2, 1} (passes) The third outcome is a combination, while in the second we have to note which weight-4 voter voted for the motion. {2, 4a} and {4a, 2} are the same outcome, but {4b, 2} is a different outcome. Mark as needed with the alphabet through the [set] if there are multiple voters with the same weight, or if you wish to identify the first, second, etc. voter by more than their weight. Generalize: [q: w1, w2, w3, …, wn] where q is the quota, and n is the number of voters, and w is the weight of each vote. Kinds of Power: Dictator: has enough to pass the vote by themselves, and can’t pass the vote without them Ex: [14: 15, 7, 5] – 15 votes is more than the quota, and 7+5 < 14 so the first voter has all the power Dummy: an unnecessary voter who has no power, the vote can pass without them in any situation Ex 1: [14: 10, 5, 5, 2] – even if A or a combination of B and C vote yes, D voting for the motion will not get it passed Ex 2: [14: 8, 6, 5, 5, 3, 1] – The F voter can sway the vote to pass or deny the motion, so this example does not have a dummy vote Veto Power: Does not have the power to pass a vote on its own, but motion need their vote to pass every time Ex 2: [14: 10, 5, 5, 2] The first voter can’t vin on their own, but the other three cannot band together to get the motion passed without the first voter Differences Weights does not equal the same power Dictator: [14: 15, 7, 5] – 15 has 100% of the power Dummy: [14: 10, 5, 5, 2] – 2 has 0% of the power Otherwise, how much power does each voter have if clearly defined? [15: 8, 6, 5, 5, 3, 1] In this chart, does voter E(3) have 3 times more power than F(1), or does A(8) have half the power of the group? Three mathematicians wanted to quantify the voting power and clearly define it. Banzhaf Power Index (1946, 1965, 1971) Shapely-Shubik Power Index (1954) Power: a voter has power when he/she is pivotal in some voting permutation Pivotal: the first person who’s input puts the vote up to or over the margin How many times is the voter pivotal? Ex: [6: 5, 3, 1] A(B)C (5+3=8>6) B(A)C (3+5=8>6) C(A)B (1+5=6 = 6) A(C)B (5+1=6 = 6) BC(A) (3+1+5=9>6) CB(A) (1+3+5=9>6) A is pivotal 4 out of 6 times: 4/6; B is pivotal 1/6, and C is pivotal 1/6. The answer for [6: 5, 3, 1] can be written I4/6, 1/6, 1/6) or (2/3, 1/3, 1/3) or (66.7%, 16%, 16%) Wednesday Alternate Pivotal Value Definition: Quota is reached when that candidate votes yes (in a permutation) If there are 3 candidates (ABC) there are 3! (6) possible order-matters/permutation sets. For 4 voters, there are 4! (24) sets, and with 5 voters there are 5! (120) sets. For each of these, or more, you would have to write out each set to find out which voter is pivotal. Ex: [7: 5, 4, 3, 1]  [Q: A, B, C, D] Just looking at this set, is there anything wrong with it? By looking at the count, the D voter with the weight of 1 is a dummy voter. Their count will never be important for passing a motion, but we will still have to write out their possibilities to see which of the others will be pivotal. Permutations Pivotal Voter CADB B A(B)CD – 5+4=9 > 7 B CABD B ABDC B CBAD B ACBD C CBDA A ACDB C DABC B AD(B)C – 5+1 = 6+3=9 > 7 C DACB C ADCB B DCAB C BCDA A DCBA A BCAD A DBCA B BDCA C DBAC A BDAC C I have filled in a couple of the Permutation BACD C blanks so you can see how to get the pivotal BADC A voter. Once the total reaches 7 or more, the last CDAB A voter added in is the pivotal one. CDBA A Now add up how many of each candidate to get the power degrees for each: A: 8; B: 8; C: 8; D: 0. Since there were 24 possible outcomes, each of these power counts is out of 24: (8/24, 8/24, 8/24, 0). This can also be written with simplified fractions or in decimal form. What if all of the votes but one have the same weight? Can this process be simpler? [6: 3, 1, 1, 1, 1, 1, 1, 1, 1] there are 9! Permutations. That’s 362,880 different ways to write out permutations of this since each voter who’s vote has a weight of one is separate voter from all the other w1’s. We’re not going to write out each. The only weight that is different is w3. It can have 9 different positions in the set such as {1, 1, 1, 1, 1, 1, 1, 1, 3} or {1, 1, 1, 1, 1, 1, 1, 3, 1}, and each of these can be written 8! different ways since the w1’s can switch places. What positions in the set would make 3 pivotal? The fourth place (1+1+1=3, need 3 more = 6, quota met), the fifth (1+1+1+1=4  +3=7, more than quota, but still needed) and the sixth position (where only w1 is needed, but w3 is also possible and helpful.) Anything past this makes the quota with different varieties of w1’s, and the w3 is no longer pivotal. Since w3 is pivotal 3 times out of the 9 different positions: 3/9 = 1/3, and all the rest have a total of 2/3 power combined. This is divided by each of the 8 w1’s, or multiplied by the reciprocal 1/8 to give each one a power of 2/3x1/8 = 2/24 = 1/12. This makes [6: 3, 1, 1, 1, 1, 1, 1, 1, 1] powerful by (1/3, 1/12, 1/12, …, 1/12)


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