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Test 1 Review: Voting Systems

by: Amy Brogan

Test 1 Review: Voting Systems MATH 1014

Marketplace > University of Cincinnati > Mathematics (M) > MATH 1014 > Test 1 Review Voting Systems
Amy Brogan
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A review of voting systems, and weighted voting. Includes a terms list, practice with answers, and important points to remember.
Mathematics of Social Choice
Mary Koshar
Study Guide
math social change power index permutation n! voting weighted systems
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This 9 page Study Guide was uploaded by Amy Brogan on Saturday February 6, 2016. The Study Guide belongs to MATH 1014 at University of Cincinnati taught by Mary Koshar in Spring 2016. Since its upload, it has received 68 views. For similar materials see Mathematics of Social Choice in Mathematics (M) at University of Cincinnati.


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Date Created: 02/06/16
Koshar Amy Brogan Test 1: Voting Systems Terms:  Borda Count: point system for ranking vote o Last place gets zero points, increasing by one up from there  Hare “Elimination” Method st o Eliminate the choice with the least amount of 1 place votes and choose from there o Number of rounds = number of candidates  Plurality o Most votes in first place o Can have a tie because it’s not for majority  Majority = more than 50%  Plurality Run-Off o Only two rounds  Top two 1 place candidates (eliminate all the rest)  Choose from the new most first place  Condercet’s “Head-to-head” Method: o Compare two choices at a time, looking for one who is the best overall o When comparing two, eliminate all others (A vs. B only, then B vs. C only, etc)  Rank Method o Spread out possible points  Approval Voting: each voter can give 1 vote to as many/few candidates as they choose, not a rank o Approval voting Drawbacks:  No distinction between top choice and other choices  Approving everybody advances no one  Equal to not voting  Compare with plurality: approval is not the majority, but possibly acceptable  Disapproval Voting: o each voter can disapprove (anti-vote) as many/few candidates as they want  Coombs: o elimination method, but deleting the candidate with the most last place votes  Permutation: o an arrangement of objects (such as candidates) where order is important  Combination: o an arrangement where order is not important  Weighted Voting Systems: o voters do not necessarily have the same weight (voter situations where voters vote yes or no to a motion.)  Dictator: o has enough to pass the vote by themselves, and can’t pass the vote without them  Dummy: o an unnecessary voter who has no power, the vote can pass without them in any situation  Veto Power: o Does not have the power to pass a vote on its own, but motion need their vote to pass every time  Shapely-Shubik Power Index (1954) o Power: a voter has power when he/she is pivotal in some voting permutation o Pivotal: the first person who’s input puts the vote up to or over the margin  Preference list ballot: o A ballot that ranks the candidates from most preferred to least preferred, with no ties  May’s Theorem: o For two alternatives and an odd number of voters, majority rule is the only voting system that satisfies the three natural properties  All votes are treated equally  If B beats A, and more people switch to B in a recast, then B should still win  A tie cannot occur unless there is an even number of voters Desirable Traits of Voting - Should be a winner - Every Ballot is Equal- Equal candidates (no secret states) - Advantage for actual want vs. manipulation (voting only for political party, or just to beat one candidate)  Condorcet winner: candidate beats all others in head to head matches  Monotonicity: if B wins the election, and a recast makes B’s ranking higher, B should still win  Independent Irrelevant Alternatives (IIA): change in rankings of irrelevant alternates should not change the outcome o If B wins election 1, and in the recast C beats A, B should still win  Pareto Condition: everyone prefers A to B, then B should not win Drawbacks:  Condorcet: Possible to have NO winner because no one beats everyone head-to-head  Hare: winner doesn’t beat every other candidate  Plurality Run-off: not the same outcome as Condorcet  Plurality: If B&C switch ranking, the original winner (A) may not stay the winner  Sequential Pairwise: does not satisfy Pareto condition Kenneth Arrow: Arrow’s Impossibility Theorem: 1951: Proved that finding an absolute fair and decisive voting system is impossible With 3+ candidates and any number of voters, there does not exist there never will exist a voting system that always produces a winner, satisfies Pareto condition, is independent of irrelevant alternatives, and is not a dictatorship Notation for quota and voting weights: [q: w1, w2, …, wn] Permutations; n factorial: where order matters for n voters = n! outcomes (7! = 7x6x5x4x3x2x1) Practice Part 1: Do Borda Count, Hare’s Method, Condorcet, Plurality, and Plurality Run-Off for chart 1 and 2. Chart 1 (5) (4) (3) (3) (2) st 1 E A C D B 2nd B B B B C 3rd C C D C D 4th D D A A A th 5 A E E E E Chart 2 (17) (12) (9) (4) (4) (2) (1) First Q T E W Q E Q Second W R Y R T W E Third E Q R Y R Q T Fourth R W W Q Y R Y Fifth T Y T T E T W Sixth Y E Q E W Y R Example 3:How would you chart this?  26% want A as their first choice, but would also approve of B  25% want A as their first choice and approve neither B or C  15% want B as their first choice and approve neither A or C  18% want C as their first choice, but would also approve of B  16% want C as their first choice and approve neither A or B Part 2: Chart 3: Do both Coombs and Hare’s Method with this chart and compare. No. Voters 1 1 1 1 1 1 1 1st C D C B E D C 2nd A A E D D E A rd 3 E E D A A A E 4th B C A E C B B 5th D B B C B C D Examples 2: Permutation or Combination? 1. Rob and Mary are planning a trip around the world. They are deciding on which 9 cities to visit. 2. Batting order for 7 players on a twelve-person team? 3. A committee of 10 members wants to choose a chair and assistant chair. 4. All possible ways to build a Subway sandwich. 5. All possible outcomes when ranking 3 candidates 6. Choosing 3 applicants out of 445 people who applied for a programming position Sets and Power: Identify any dictators, dummies, or veto power holders in these sets. [15: 11, 5, 5, 2] [15: 16, 7, 5] [14: 9, 6, 5, 5, 3, 1] [6: 5, 3, 1] For these sets calculate the power of each voter: [6: 3, 1, 1, 1, 1, 1, 1, 1, 1] [7: 5, 4, 3, 1] Answers Part 1: Chart 1: Hare Method: eliminate those with least amount of 1 place votes A B C D E st 4 2 (least 1 place) 3 3 5 4 / 5 3 (least new 1 ) 5 4 (least 1 ) / 8 / 5 / / 12 *winner / 5 Condorcet: Chart 3: 17 votes = 9 votes for majority  E (5) vs. B (4+3+3)  B  B (5+4) vs. C  B has majority already, C eliminated  B (5+4) vs D  D Eliminated  B (5+3+3+2) vs. A (4)  B is winner overall Plurality Run-off: A B C D E 4* top 2 2 3 3 5* top 2 4+3+3+2 = 9* win 5 Plurality Problems: 6 4 3 1st A C B nd 2 B A C 3rd C B A A wins with 6 votes, but  6 4 3 1st A C C 2nd B A B rd 3 C B A If in the recast, 3 people vote C over B, then C wins with 7 votes total instead of A(6) Chart 2: Condorcet: Head to Head Comparisons Q W Q(24) does not beat R(25) T Q 17 9 12 17 12 4 So let’s try it with R: 9 4 4 2 R E 4 12 17 1 2 Q wins against W’s 15 1 4 9 1 votes with 34 4 3 T (21) was beaten by Q(28), R(20) does not beat E(29), Q E who was already beaten by and since Q was beaten 17 9 earlier, there is not a 12 2 W, so R and E can’t be the Condorcet winner. Condorcet winner. 4 1 T Y Q (38) beats E (11) 17 9 Q R 12 4 17 12 4 4 9 2 2 4 1 1 T (36) beats Y(13) Hare’s Elimination Method: Top votes Q W E R T Y 22 4 11 0 (eliminate) 12 0 (eliminate) 22 4 (eliminate) 11 12 26 11 (eliminate) 12 28 21 Q wins with Hare’s method. Remember: We are only looking at who is the top vote per round, per column of votes. As candidates are eliminated their votes have to go to a new candidate because there will be a new candidate in the top ranking. Borda Count: st nd rd th th th Borda Count Points: 1 place = 5; 2 = 2; 3 = 3; 4 = 2; 5 = 1; 6 = 0 (Votes in Column x Borda count for rank) + (votes in next column x Borda count for rank) + etc. (17x_) + (12x_) + (9x_) + (4x_) + (4x_) + (2x_) + (1x_) = ___ Q: (17x5) + (12x3) + (9x0) + (4x2) + (4x5) + (2x3) + (1x5) = 160 W: (17x4) + (12x2) + (9x2) + (4x5) + (4x0) + (2x4) + (1x1) = 139 E: (17x3) + (12x0) + (9x5) + (4x0) + (4x1) + (2x5) + (1x4) = 114 R: (17x2) + (12x4) + (9x3) + (4x4) + (4x3) + (2x2) + (1x0) = 141 T: (17x1) + (12x5) + (9x1) + (4x1) + (4x4) + (2x1) + (1x3) = 111 Y: (17x0) + (12x1) + (9x4) + (4x3) + (4x2) + (2x0) + (1x2) = 70 To check your math, you can add up all the totals you got here^ and compare it with the total Borda points. st nd Borda points for 1 place + Borda Points for 2 + … all multiplied by total number of voters 5+4+3+2+1 = 15  Total Voters: 17+12+9+4+4+2+1= 49  15 x 49 = 735 Borda points 160+139+114+141+111+70 = 735 Plurality: You only need to look at the first place ranking: (17) (12) (9) (4) (4) (2) (1) First Q T E W Q E Q Q has 22 first place votes, and the most, so Q wins by Plurality. Plurality Run-Off As there wasn’t a tie with Plurality, there is no need for a run-off. We would only need to do a run-off if there was a tie of two or more candidates. Example 3: since a percentage is out of 100, we can make each 1% the same as 1 vote: 26 25 15 18 16 A X X B X X X C X X A: 26 + 25 = 51 B: 26 + 15 + 18 = 59 *wins by approval voting C: 18 + 16 = 34 But when we remember that Approval Voting doesn’t take into account which is the top vote, problems arise. Check with plurality: 26 25 15 18 16 A X (1 ) X (1 ) B X X X C X (1 ) X (1 ) A: 51 * most 1 place votes B: 15 – fewest first place votes C: 39 This is why Approval Voting is not more widely used. Answers Part 2: Chart 3: th Hare’s Method: Coomb’s Method: Starting in 5 and moving up A B C D E A B C D E 0 1 3 2 1 0 3 2 2 0 / 1 3 2 1 1 / 4 2 0 / / 3 4* / 3 / / 2 2 D is the most liked / / / 4 3* E is the least disliked? Hare over Coombs: it’s widely accepted that it’s better to eliminate from the top than to eliminate candidates placed on the bottom possibly for emotional dislike. Examples 2: 1. Since they are just choosing their countries to visit, it’s combination. Mary can write down China, and then Rob can add Brazil. The only time when this example could be permutation is when they are planning their route. (England to France to Italy to Israel to India to China etc. If you mixed these up, the flying back and forth would be time consuming and expensive.) 2. This one is permutation because it asking for the order of the batters in a team. 3. Choosing a chair and assistant chair is permutation because the order matters. If they were simply making a sub-committee, then it would be a combination. 4. In class we came to the decision that this is combination because a BLT is still a BLT no matter if the bacon goes above or below the tomato. 5. Because we are ranking candidates, it is permutation. 6. Choosing 3 candidates would still be the same three, no matter what order their names were called, so this is a combination. Sets: [15: 11, 5, 5, 2] – dummy (2) – The last voter will never be needed to pass a motion. [15: 16, 7, 5] – dictator (16) – The first voter can pass the motion by him/herself and the others can’t vote together to pass it [14: 9, 6, 5, 5, 3, 1] - none [6: 5, 3, 1] – veto (5) – The first voter is needed in every possibility to pass the motion, the other two can’t do it without them, but they can’t pass the motion by themselves. Power: {1, 1, 1, 3, 1, 1, 1, 1, 1} and {1, 1, 1, 1, 3, 1, 1, 1, 1} and {1, 1, 1, 1, 1, 3, 1, 1, 1} W3 = 3/9 = 1/3, and all the rest have a total of 2/3 power combined. (2/3) x (1/8) = 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) Ex: [7: 5, 4, 3, 1]  [Q: A, B, C, D] Permutations Pivotal Voter A(B)CD – 5+4=9 > 7 B ABDC B ACBD C ACDB C AD(B)C – 5+1 = 6+3=9 > 7 C ADCB B BCDA A BCAD A BDCA C BDAC C BACD C BADC A CDAB A CDBA A CADB B CABD B CBAD B CBDA A DABC B DACB C DCAB C DCBA A DBCA B DBAC A I have filled in a couple of the Permutation blanks so you can see how to get the pivotal voter. Once the total reaches 7 or more, the last voter added in is the pivotal one. A: 8; B: 8; C: 8; D: 0  out of 24: (8/24, 8/24, 8/24, 0). This can also be written with simplified fractions or in decimal form.


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