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Week 3: Voting Systems Concluded

by: Amy Brogan

Week 3: Voting Systems Concluded MATH 1014

Marketplace > University of Cincinnati > Mathematics (M) > MATH 1014 > Week 3 Voting Systems Concluded
Amy Brogan
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About this Document

The introduction of Approval and Disproval voting, Coomb's elimination method, a review of all methods up till this point, and the difference between Permutation and Combination arrangements.
Mathematics of Social Choice
Mary Koshar
Class Notes
voting systems Coomb's method elimination permutation combination Subway
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This 7 page Class Notes was uploaded by Amy Brogan on Saturday January 30, 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 14 views. For similar materials see Mathematics of Social Choice in Mathematics (M) at University of Cincinnati.


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Date Created: 01/30/16
Koshar Amy Brogan January 25, 27, & 29, 2016 Monday Approval Voting Looking at this map, what do the different colored areas tell you? What do they represent? If you said the House of Representatives districts, you were right. Each of these colored places shows the location of the groups of people that each representative for Ohio is in Congress for. How else does this come into play in politics? When it comes time to vote for the next president, Ohio has 18 electoral college votes. But there are only 16 distinctions? The senators, each state gets the same mount, counts for the other two. The population decides how many representatives Ohio gets in Congress. This number continued to grow for all the states until they established a ceiling. Now there can only be 435 representatives for the entire United States, and they have to be equally spread out. But if you are trying to do this by conventional methods, most of the time you’ll get something like 18.2. But you can’t give a state .2 representatives. We will get more into how they spread out whole representatives later in the semester. Now back to voting methods:  Approval Voting: each voter can give 1 vote to as many/few candidates as they choose, not a rank o The Baseball Hall of Fame uses this, as does the UN, and he National Academy of Science.  Disapproval Voting: each voter can disapprove (anti-vote) as many/few candidates as they want 4 Mathmeticions: 4 Candidates for the Head of the Academy of Science (in our example) Newton, Liebniz, Euler, Fermat In approval voting, the ballot can look like this: Candidate Voter 1 Voter 2 Newton X X Liebniz Euler X Fermat X In another: (5) (6) (10) (11) A X X B X X X C X X A: 5 + 11 = 16 B: 5 + 6 + 10 = 21 * most approval votes C: 5 + 10 = 15 But 5 people pretty much threw their votes by voting for everyone. It would have the same effect as not voting at all.  Approval voting Drawbacks: o No distinction between top choice and other choices o Approving everybody advances no one  Equal to not voting o Compare with plurality: approval is not the majority, but possibly acceptable 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 Okay, 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 st st C st 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. Bonus: Coombs Method Clyde Coombs: physicist: 1900’s – elimination method, but deleting the candidate with the most last place votes This method repeatedly eliminates candidates who are least preferred. But what definition do you prefer? “Least disliked”, or Hare’s “Most Liked?” 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 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. Wednesday (based on the worksheet we did in groups in class. Not the exact same charts!) # of voters (4) (3) (2) st 1nd A C B 2 B B C 3rd C A A 1. Using the Hare method, find the winner for this voting table. 2. Since the Hare method does not satisfy the independence of irrelevant methods, what would switching two of the candidates in one of these columns do to the outcome? Do this in a way that alters the outcome of the first question. 3. Use the Coombs method to find a winner. Then find out if it satisfies the Condorcet method. 4. There are 65 voters in a committee and they want to make a two-member sub-committee using approval voting. With the chart below, who would be on this sub-committee? (8) (9) (10) (12) (12) (14) Bryan X X X Stacie X X X X Dan X X Adrianna X X X Nicko X X X Answers: 1. Hare method A B C 4 2 3 4 / 5* C wins by Hare’s elimination 2. Possible alternative chart: 4 3 2 1st A C A 2nd B B C rd 3 C A B New Hare outcome: A B C 6 0 2 A wins by Hare’s elimination in this alternate chart 3. Coombs and Condorcet Coombs: Condorcet: A B C A (2+2=4) vs. B (2+3=5)  B wins 5 0 4 B (2+2+2=6) vs. C (3)  B wins / 3 6 B wins by Coombs’ elimination B wins by Condorcet In this instance (because this may not be the same outcome every time), the Coombs method of elimination satisfies the Condorcet Head-to-Head comparison. 4. Sub-committees Bryan: 8+10+12 = 30 Stacie: 9+10+12+14 = 45* Dan: 12+14 = 26 Adrianna: 9+12+14 = 35* Nicko: 8+9+14 = 31 Stacie and Adrianna should be the two members on the sub-committee according to approval voting. Friday Borda Count Points Review If there are 21 voters for 5 candidates, how many Borda points are there? nd Remember the last place doesn’t get points, so first place has a total possible 4 points, 2 has 3, and so on, and then all that multiplied by the number of voters. (4+3+2+1+0) x 21  10x21 = 210 Borda points Now for 76 voters for 7 candidates? (6+5+4+3+2+1+0) x 7 = _______* 1 How would we generalize this “equation” for v amount of voters and c amount of candidates?  V x (C-1)! o Why won’t this work? o Because “!” means that we would be multiplying each (C-1) level, and we need to add them. But the multiplying Vat the front is correct.  V x ([c-1]+[c-2]+…2+1) o There isn’t really an easier way to do it, we have to write it out Permutation vs. Combination Permutation: an arrangement of objects (such as candidates) where order is important Combination: an arrangement where order is not important Consider two comparisons for remembering the difference. Permutation is like baking a cake. First you mix your dry ingredients, then your wet, and then together, and then you bake it. Once you’ve let it cool out of the oven, then you can decorate. If you tried to bake the wets and dry’s separately before mixing, it wouldn’t make much sense. Combination on the other hand, is when the order does not matter, like when mixing a salad. It doesn’t really make a difference if you put in carrots or the tomatos or lettuce in the bowl first, once it’s mixed, it’s a salad. Examples: 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 1147 Answers: 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. But what about finding how many ways to rank the top 3 candidates out of 7? To just rank 3, we would use 3! for 3x2x1 and get 6 ways to rank them. A_ _ B_ _ C_ _ BC AC AB CB CA BA With that in mind, we would use 7x6x5 =210 because for 1 place we have 7 possible candidates, for 2 nd rd place we have 6 candidates left, and for 3 we have 5 candidates. How would we write this out? The number of ways to choose n candidates among n(order matters) = nPn = n! where the first n is total number of candidates, the P stands for permutation, and the second n is the number of choices. But what if there are k items?  nPk = n!/(n-k)! So in our ranking 3 top candidates for among 7: 7!/(7-3)!  7!/4!  and we still get 120 Which is more? Permutation or combination? Permutation is because for 3 candidates, there are six ways to rank them versus combination, where order does not matter so three candidates are the same three, no matter what order. The fourth question above brought up an interesting question: how many ways can you build a subway sandwich? Subway has 11 types of bread, 8 types of cheese, 14 types of toppings, 19 types of sauces, and 6 types of meat. (had to look it up) If we leave everything other than bread as a “yes” or “no” (2) option for putting it on our sandwich (because we don’t care about what order they go on) then we get something like this: 11 x (2^8) x (2^14) x (2^19) x (2^6) - then we can add all our exponents together: 11 x 2^47 – and solve. 11x2^47 = 1.548112e15 or 1,548,112,000,000,000 possible subway sandwiches from Subway. How cool is that?


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