GROUP DECISION MAKING UNDER MULTIPLE CRITERIA Content

Arrow's conditions for social walfare functions

Criteria to evaluate the methods

Single transferable vote The Coombs method Arrow's conditions for social walfare functions

Axiom 1: For all x and y, either x R y or y R x. [connected ] Axiom 2: For all x, y, and z, x R y and y R z imply x R z. [transitive]

­ x R y means «x is preferred or indifferent to y»

Conditions: 1. domain properties: ­ The number of alternatives in A is greater than or equal to three. ­ The social welfare function f is defined for all possible profiles of individual orderings. ­ There are at least two individuals. 2. Positive Association of Social and Individual Values 3. Independence of Irrelevant Alternatives 4. Citizen's Sovereignty 5. Non-dictatorship Arrow's conditions

POSITIVE ASSOCIATION OF SOCIAL AND INDIVIDUAL VALUES

If the welfare function asserts that x is preferred to y for a given profile of individual preferences, it shall assert the same when the profile is modified as follows: ­ (a) The individual paired comparisons between alternatives other than x are not changed, ­ (b) Each individual paired comparison between x and any other alternative either remains unchanged or it is modified in x's favor.

Monotonocity failure!* ­ Upward monotonicity failure: Given the use of method and a profile in which candidate X is the winner, X may nevertheless lose if some voters change their preference in the favor of X. ­ Downward monotonicity failure: Given the use of voting method and a ballot profile B in which candidate X is a loser, X may nevertheless win if some voters change their preference against the favor of X.

*Felsenthal and Tideman, (2014) 64 D.S. Felsenthal, N. Tideman / Mathematical Social Sciences 67 (2014) 57–66 64 D.S. Felsenthal, N. Tideman / Mathematical Social Sciences 67 (2014) 57–66

VoterVoter No. No. Ranking Ranking abcdefabcdefSum Sum 1 1 a b ad cb ed f c e f a – 9 7 10a 6 10– 42 9 7 10 6 10 42 2 a d c f b e b 7 – 7 10 10 8 42 2 a d c f b e b 7 – 7 10 10 8 42 3 3 b d bc ed ac f e a f c 9 9 – 5c 9 99 41 9 – 5 9 9 41 4 b d e c a f d 6 6 11 – 8 9 40 4 b d e c a f d 6 6 11 – 8 9 40 5 b e a f d c e 10 6 7 8 – 9 40 6 5 c b be ae da f f d c f 6 8 7 7e 7 –10 35 6 7 8 – 9 40 7 6 c d ca fb be e a d f Totalf 6 240 8 7 7 7 – 35 8 d a c b e f 7 c d a f b e Total 240 9 d c b e a f As the average Borda score remains 40 (240/6), candidates d, e, 8 d a c b e f 10 e a b d c f and f are eliminated and a revised matrix of paired comparisons 11 9 e f db dc c b a e a f among the remaining threeAs the candidates average is Borda computed: score remains 40 (240/6), candidates d, e, 12 10 f a ec ea bb d d c f and f are eliminated and a revised matrix of paired comparisons 13 f c a b d e abcSum 11 e f b d c a among the remaining three candidates is computed: 14 f e a d c b 12 f a c e b d a – 9 7 16 15 f e b c a d b 7 – 7 14 16 13 f e fd c aa b b d e abcSum c 9 9 – 18 14 f e a d c b Total 48 a – 9 7 16 In view of these15 rankings, thef Bordae scoreb of eachc a of thed six b 7 – 7 14 Example candidates is equal16 to the sum of hisf respectivee d row inc thea followingb Here the average Borda score of the three voters a–c is 16 c 9 9 – 18 matrix of paired comparisons: (48/3), so all candidates except c are eliminated and c becomes the winner. Note that, on the basis of his or her originalTotal ranking, voter 48 In view of theseabcdef rankings, the BordaSum score of each#7 of benefited the six from the upward move of a up in his or her ranking candidatesa is equal– to 9 the 7 sum 9 of 6 his 10 respective 41 row in thebecause, followingceteris paribusHere, s/he obtainedthe average as a result Borda the election score of of c the three voters a–c is 16 which s/he ranked initially above a. 64 Apply Nanson’s functionD.S. Felsenthal,: N. Tidemanmatrix of / Mathematical pairedb 7 comparisons: – Social 7 10 Sciences 10 8 67 42 (2014) 57–66 (48/3), so all candidates except c are eliminated and c becomes the c 9 9 – 5 9 9 41 winner. Note that, on the basis of his or her original ranking, voter d 7 6 11 – 8 9 41 3.5.3. Upward monotonicity failure — dynamic voters worse off [UW] abcdef Sum #7 benefited from the upward move of a up in his or her ranking Voter No. Ranking e 10 6 7 8 – 9 40 abcdefSuppose instead that, ceterisSum paribus, voter #8 changes his or her f a 6 8– 7 9 7 7 7 – 9 35 6 10 41 ranking from d because,a c toceterisa d paribusc ,thereby s/he obtainedincreasing as a result the election of c Total 240 ··· ··· 1 a b d c e f b 7 – 7 10a 10 8– 42 9a’s 7 support. 10 So we 6which now 10 obtain s/he 42 the ranked same matrices initially and above final resultsa. as in Section 3.5.1, i.e., here too candidate c is elected. The only 2 a d c f b e c 9 9 – 5b 9 919 7 41 – 7 10 10 8 42 The sum of the Borda scores of all six candidates is 240, hence difference is that, on the basis of his or her original ranking, here 3 b d c e a f the average Bordad score is 407 (240 6/6). 11 According –c 8to the 9 Nanson9 41 9 –voter #8 5 is worse 93.5.3. off 9 as Upwarda result 41 of monotonicity the rise in the failure ranking — of dynamica voters worse off [UW] method, one eliminatese at the10 end 6 of every 7 counting 8 – round 9 those 40 because s/he obtains the election of c which s/he ranked originally 64 4 b d e c a D.S.f Felsenthal, N. Tideman / Mathematical Social Sciencesd 67 (2014)6 57–66 6 11 – 8Suppose 9 40 instead that, ceteris paribus, voter #8 changes his or her candidates whose Bordaf score6 does 8 not 7 exceed 7 the 7 average – Borda 35 lower than a. ranking from d a c to a d c thereby increasing 5 b e a f d c score of all the candidates participating in thate round.10 So here 6 7 8 – 9 40 ··· ··· Total 240 a’s support. So we now obtain the same matrices and final results 6 Voterc No.b Rankinge a d f candidates e and f are eliminated and a revisedf matrix of pairedabcdef6 83.5.4. 7 Downward 7 7 monotonicity –Sum 35 failure — dynamic voters better off comparisons among the remaining four candidates is computed: [DB] as in Section 3.5.1, i.e., here too candidate c is elected. The only 7 1 c d aa bf db c e e f Totala – 9 719 10 6 10 42 240 The sum of the Borda scores of all six candidates is 240, Ashence candidate bdifferencewas not elected is that, initially, on the suppose basis now of his that, or her original ranking, here 8 2 d a ac db c e f f b thee average Borda scoreabc is 40 d (240Sum/6).b According7 to – theceteris 7 Nanson 10 paribus, 10 votervoter 8 #5 downgrades #8 42 is worse candidate off asb by a result changing of his the rise in the ranking of a rank order from b e a to e b a . As a result, 9 3 d c bb de ca e f a method,f one eliminatesa – 9As at 7 the the 9 end average 25 of everyc Borda counting9 score 9 round – remains those 5 409 (240because 9/ 416), s/he··· candidates obtains the d election,···e, of c which s/he ranked originally ceteris paribus, the initial Borda score of b decreases from 42 to 41 4 b d e c a f b 7 – 7 10 24 d 6 6 11 – 8lower 9 than40 a. 10 e a b d c f candidates whose Bordaand scoref are does eliminated not exceed and the averagea revisedand the Borda initial matrix Borda of score paired of e increases comparisons from 40 to 41, while the 5 b e a f d c c 9 9 – 5 23 e 10 6 7 8 – 9 40 11 e f b d c a score of all thed candidates7among 6 11 participating –the 24 remaining in that three round. candidatesBorda So scores here of is all computed: the other candidates and the average Borda score 6 c b e a d candidatesf e andTotalf are eliminated 96 and af revised6 matrix 8(40) of 7 pairedremain 7 the 7 same.3.5.4. – So according Downward 35 to Nanson’s monotonicity method candidate failure — dynamic voters better off 12 f a c e b d f is eliminated after the first count, and we obtain the following 7 c d a f b comparisonse among the remaining four candidatesTotal is computed: [DB] 240 (revised) matrix of paired comparisons and Borda scores of the 13 8 f c da ab cd b e e Heref the average Borda score of the four voters a d is 24 abcSum remaining five candidates:As candidate b was not elected initially, suppose now that, (96 /4), so all candidates except a are eliminated and a becomes 14 9 f e ad cd bc e b a f abcAs the average d Sum Borda score remains 40 (240ceteris/6), paribus, candidates voterd, #5e, downgrades candidate b by changing his the winner. a – 9 7 16 15 10 f e be ac ba d d c f and f are eliminated and a revised matrixrank of pairedabcde order comparisons from b Sume a to e b a . As a result, a – 9 7 9 25 b 7 – 7 14 ··· ··· 16 11 f e de fc ba d b c a among the remaining three candidates is computed:aceteris– paribus 9 7, the 9 6initial 31 Borda score of b decreases from 42 to 41 3.5.2. Upward monotonicityb failure7 — dynamic – 7 voters 10 better 24 offc [UB] 9 9 – 18b 7 – 7 10 9 33 12 f a c e b d and the initial Borda score of e increases from 40 to 41, while the Suppose now that, ceterisc paribus9, voter 9 – #7 (who 5 would 23 rather c 9 9 – 5 9 32 13 f c a b d e TotalabcSum 48Borda scores of all the other candidates and the average Borda score In view of these rankings, the Borda score of eachhave ofc or thed than sixa, whod will be7 elected 6 11 if all – voters 24 rank the d 7 6 11 – 8 32 14 f e a d c candidatesb sincerely) changes his or her ranking from c d e(40)10 remain 7 7the 8same. – So 32 according to Nanson’s method candidate Total 96 a – 9 7 16 candidates is equal to the15 sum of hisf respectivee b rowc ina a thed to followingc a d therebyHereincreasing the averagea’s support. SoBorda we score of theTotalf threeis eliminated voters aftera–c the 160is first 16 count, and we obtain the following Suppose 7th voter changes from c >··· d > a > f··· > b >e b 7 – 7 14 matrix of paired comparisons:16 f e d c a nowb obtain the following matrix/ of paired comparisons and Borda (revised) matrix of paired comparisons and Borda scores of the scores: Here the average(48 Borda3), score so all of candidates the four votersc excepta 9cThed 9areis (revised) – 24 eliminated 18 average Borda and c scorebecomes is now 32 the (160/5), so all candidates except bremainingare eliminated five and candidates:b becomes the winner. Note to c >(96 a/ 4),> sod all> candidates f >winner. b > except e Notea are that, eliminated on theTotal and basisa becomes of his or 48 her original ranking, voter that, on the basis of his or her original ranking, voter #5 is better In view of these rankings, the Borda score19 of each of the six abcdef Sum theNote winner. that the sum of all the#7 voters’ benefited Borda scores can from also be the obtained upward by off move from the of falla ofupb in inhis or his her or ranking, her becauseranking as a result s/he multiplying the number of voters (16 in this example) by the number of paired abcdeSum candidates is equal to the sum of his respective row in the following obtains the election of b whom s/he prefers over the election of the comparisons among all the candidatesbecause, (6 5/Here2 ceteris15the paired average comparisons paribus in Borda, thiss/he obtainedscore of the as three a result voters thea election–c is 16 of c Upwarda –monotonicity 9 7 9failure 6 10! 41See Felsenthal and Tideman· = (2014) ( ) a – 9 7 9 6 31 matrix of paired comparisons: example). (48/3), so all candidates exceptoriginalc are winner eliminateda . and c becomes the 3.5.2. Upward monotonicitywhich failure s/he ranked — dynamic initially voters better above offa [UB]. b 7 – 7 10 9 33 b 7 – 7 10 10 8 42 winner. Note that, on the basis of his or her original ranking, voter Suppose now that, ceteris paribus, voter #7 (who would rather c 9 9 – 5 9 32 c 9 9 –abcdef 5 9 9 41 Sum #7 benefited from the upward move of a up in his or her ranking have c or d than a, who will be elected if all voters rank the d 7 6 11 – 8 32 d 7 6 11 – 8 9 41 3.5.3.because, Upwardceteris monotonicity paribus, s/he failure obtained — dynamic as a result voters the election worse off of [UW]c a – 9 7 9 6 10candidates 41 sincerely) changes his or her ranking from c d e 10 7 7 8 – 32 e 10 6 7 8 – 9 40 which s/he ranked initially abovea. b 7 – 7 10 10 8a 42 to c a d Supposethereby insteadincreasing that, a’sceteris support. paribus So we , voter #8 changesTotal his or her 160 ··· ··· f 6c 8 79 9 7 – 7 5 – 9 35 9now 41 obtain the followingranking matrix from of pairedd comparisonsa c to anda Bordad c thereby increasing ··· ···The (revised) average Borda score is now 32 (160/5), so all Totald 7 6 11 – 8 240 9scores: 41 a’s support.3.5.3. Upward So we monotonicity now obtain failure the — same dynamic matrices voters worse and final off [UW] results candidates except b are eliminated and b becomes the winner. Note e 10 6 7 8 – 9 40 Suppose instead that, ceteris paribus, voter #8 changes his or her as in Section 3.5.1, i.e., here too candidatethat,c is on elected. the basis ofThe his only or her original ranking, voter #5 is better f 6 8 7 7 7 –19 3519 ranking from d a c to a d c thereby increasing The sum of the Borda scores of all six candidates is 240,Notehence that the sum ofdifference all the voters’ is Borda that, scores on can the also basis··· be obtained of his by or heroff··· original from the fallranking, of b in hishere or her ranking, because as a result s/he Totalmultiplying 240 the number of votersa’s support.(16 in thisexample) So we now by the obtain number the of paired same matrices and final results the average Borda score is 40 (240/6). According tocomparisons the Nanson among all thevoter candidates #8 (6 is worse5/2 15 off paired as comparisons a result in of this the riseobtains in the the election ranking of b ofwhoma s/he prefers over the election of the as in Section· =3.5.1, i.e., here too candidate c is elected. The only example). because s/he obtains the election of c whichoriginal s/he winnerranked(a originally). method, one eliminatesThe sum of atthe the Borda end scores of every of all sixcounting candidates round is 240, those19 hence difference is that, on the basis of his or her original ranking, here candidates whosethe average Borda Borda score score does is not 40 (240 exceed/6). According the average to the Borda Nanson lowervoter than #8a is. worse off as a result of the rise in the ranking of a score of allmethod, the candidates one eliminates participating at the end in of everythat round. counting So round here those because s/he obtains the election of c which s/he ranked originally candidates ecandidatesand f are whose eliminated Borda score and does a revised not exceed matrix the of average paired Borda 3.5.4.lower Downward than a. monotonicity failure — dynamic voters better off comparisonsscore among of all the the remaining candidates four participating candidates in is that computed: round. So here [DB] candidates e and f are eliminated and a revised matrix of paired As3.5.4. candidate Downwardb monotonicitywas not elected failure initially, — dynamic suppose voters better now off that, comparisons among the remaining four candidates is computed: [DB] abc d Sum ceteris paribus, voter #5 downgrades candidate b by changing his As candidate b was not elected initially, suppose now that, rank order from b e a to e b a . As a result, a – 9 7abc 9 25 d Sum ceteris paribus, voter #5 downgrades··· candidate b by··· changing his b 7 – 7 10 24 ceterisrank paribus order from, theb initiale Bordaa scoreto e of bbdecreasesa . As from a result, 42 to 41 a – 9 7 9 25 and the initial Borda score of e···increases from 40··· to 41, while the c 9b 9 –7 – 5 723 10 24 ceteris paribus, the initial Borda score of b decreases from 42 to 41 Borda scores of all the other candidates and the average Borda score d 7c 6 119 – 9 – 24 5 23 and the initial Borda score of e increases from 40 to 41, while the Totald 7 6 11 96 – 24 (40)Borda remain scores the of same. all the So other according candidates to and Nanson’s the average method Borda candidate score Total 96 f is(40) eliminated remain the after same. the So first according count, to and Nanson’s we obtain method the candidate following (revised)f is eliminated matrixof after paired the first comparisons count, and we and obtain Borda the scores following of the Here the average Borda score of the four voters a d is 24 (revised) matrix of paired comparisons and Borda scores of the Here the average Borda score of the four voters a d is 24 remaining five candidates: (96/4), so all candidates except a are eliminated and a becomes remaining five candidates: (96/4), so all candidates except a are eliminated and a becomes the winner. the winner. abcdeabcdeSumSum a a –– 9 9 7 7 9 9 6 6 31 31 3.5.2. Upward3.5.2. monotonicity Upward monotonicity failure — failure dynamic — dynamic voters bettervoters better off [UB] off [UB] b b 77 – – 7 7 10 10 9 9 33 33 Suppose nowSuppose that, ceteris now that, paribusceteris, voterparibus #7, voter (who #7 would (who would rather rather c c 99 9 9 – 5 5 9 9 32 32 have c or dhavethanc ora,d whothan willa, who be elected will be elected if all voters if all voters rank rank the the d d 77 6 6 11 – – 8 8 32 32 candidates sincerely) changes his or her ranking from c d e 10 7 7 8 – 32 candidates sincerely) changes his or her ranking from c d e 10 7 7 8 – 32 a to c a d thereby increasing a’s support. So we Total 160 a to c ···a d thereby ··· increasing a’s support. So we Total 160 ··· now obtain the··· following matrix of paired comparisons and Borda now obtainscores: the following matrix of paired comparisons and Borda The (revised) average Borda score is now 32 (160/5), so all scores: Thecandidates (revised) except averageb are eliminated Borda score and b becomes is now the 32 winner. (160/5), Note so all candidatesthat, on theexcept basisb ofare his eliminated or her original and ranking,b becomes voter the #5 winner. is better Note 19 Note that the sum of all the voters’ Borda scores can also be obtained by that,off on from the the basis fall ofb hisin hisor her or her original ranking, ranking, because voter as a result #5 is s/he better 19 Note thatmultiplying the sum of the all number the voters’ of voters Borda (16 in scores this example) can also by be the obtained number of by paired comparisons among all the candidates (6 5/2 15 paired comparisons in this off fromobtains the the fall election of b in of hisb whom or her s/he ranking, prefers over because the election as a result of the s/he multiplying the number of voters (16 in this example)· by= the number of paired example). original winner (a). comparisons among all the candidates (6 5/2 15 paired comparisons in this obtains the election of b whom s/he prefers over the election of the · = example). original winner (a). Arrow's conditions

INDEPENDENCE OF IRRELEVANT ALTERNATIVES

Let A1 be a subset of alternatives in A. If a profile of orderings is modified in such a way that each individual's paired comparisons among the alternatives of A1 are unchanged, the social preference orderings resulting from the original and modified profiles of individual orderings should be identical for the alternatives in A1 . Example: Borda Score of the following preferences:

­ 3 votes: w P x P y P z; 2 votes: y P z P w P x. w = 9+2=11 w = 6+0=6 ­ Delete x and calculate again. x = 6+0=6 y = 3+6=9 y = 3+4=7 z = 0 + 4 = 4 z = 0 + 2 = 2 Goodman and Markowitz (1952) states: «irrelevant alternative is not necessarily irrelevant.» ­ Tea and Coffee example Arrow's conditions

CITIZEN'S SOVEREIGNTY For each pair of alternatives x and y, there is some profile of individual orderings in R(n) such that society prefers x to y. Consider a social welfare function which asserts that x P y regardless of the preferences of any of the individuals in the society. Such an undesirable social function is said to be imposed.

The Pareto criterion says that the election method must respect unanimous opinions. ­ The first version: when every voter strictly prefers alternative a to alternative b, then alternative a must perform better than alternative b. ­ The second version: when no voter strictly prefers alternative b to alternative a, then alternative b must not perform better than alternative a.

NON-DICTATORSHIP There is no individual with the property that whenever he prefers x to y, for any x and y, society does likewise, regardless of the preferences of other individuals. Arrow's possibility theorem for two alternatives

THEOREM: Arrow's Possibility Theorem for Social Welfare Function for Two Alternatives If the total number of alternatives is two, the method of decision is a social welfare function which satisfies Conditions 2 through 5 and yields a social preference ordering of the two alternatives for every set of individual orderings. Arrow's impossibility theorem

(General Possibility Theorem, or Arrow’s paradox)

UnfortunateIy, for three or more alternatives, Kenneth J. Arrow in his book Social Choice and Individual Values shows that there does not exist a social welfare function which satisfies conditions 1 through 5 by yielding a social ordering relation consistent with Axioms I and II.

General Possibility Theorem: ­ If there are at least three alternatives which the members of the society are free to order in any way, then every social welfare function satisfying Conditions 2 and 3 and yielding a social ordering satisfying Axioms I and II must be either imposed or dictatorial. Criteria to evalaute election methods*

Transitivity (Arrow’s 2nd axiom) ­ If A P B and B P C then A P C. Resolvability ­ Resolvability basically says that usually there is a unique winner ­ Formulation 1: An election method satisfies the first version of the resolvability criterion if (for every given number of alternatives) the proportion of profiles without a unique winner tends to zero as the number of voters in the profile tends to infinity. ­ Formulation 2: When there is more than one winner, then, for every alternative, it is sufficient to add a single ballot so that alternative a becomes the unique winner. ­ When preference order is reversed for all voters, then also the result of the must be reversed. Pareto (Arrow’s Condition: Citizen’s Sovereignty) Monotonicity (Arrow’s Condition: Positive Association of Social and individual values) TheSmith criterion and Smith-IIA (Arrow’s condition: independence of irrelevant alternatives)

*depends on Schulze (2011) Criteria to evalaute election methods (cont.)

Independence of clones ­ Running a large number of similar alternatives, so-called clones, must not have any impact on the result of the elections. ­ Example: In 1969, when the Canadian city that is now known as Thunder Bay was amalgamating, there was some controversy over what the name should be. In opinion polls, a majority of the voters preferred the name The Lakehead to the name Thunder Bay. But when the polls opened, there were three names on the referendum ballot: Thunder Bay, Lakehead, and The Lakehead. As the were counted using voting, it was not a surprise when Thunder Bay won. The votes were as follows: Thunder Bay 15870, Lakehead 15302, The Lakehead 8377. Condorcet Criterion ­ A voting system satisfies the Condorcet criterion if it always chooses the Condorcet winner when one exists Please also learn: MinMax Set Prudence Consistency Mojority Homework 3

For each of the following criteria, please find at least one method that does not satisfy the criterion and provide an example to show it. Please find different methods for the criteria. ­ Monotonocity ­ Condorcet ­ Independence of irrelevant alternatives ­ Prudence ­ Consistency ­ Mojority ­ Independence of clones ­ Reversal symmetry

­ Due date: Next Week Recent Preference Aggregation Techniques with Respect to Citizens

Single transferable vote The Coombs method

Respect to citizens means (satisfy ): –Select a candidate if it has a majority vote –Do not select a candidate which is not wanted by the majority Contradictory example

Our present municipality election is based on the system where the candidate having the highest vote wins

Ex. (A,B,C) %33 of votes APBPC %33 of votes BPAPC %34 of votes CPAPB

According to our current system-C wins But notice that %66 of voters accept it as the worst candidate Single Transferable Vote

Single transferable vote is found by Hare, as a reaction to the high level of vote losses in the PR system based on list system (ex. D’Honts, Greatest Remainder etc)

In STV, each voter ranks the list of candidates in order of preference

When used in a single winner election, it is called alternate vote /instant runoff /Hare /majority in a preference Instant runoff voting

Each ballot is initially assigned to candidates who are listed as the first choice on that ballot. If any candidate already has a majority of the votes at this point, then they automatically win the election If no one has a majority yet, then the candidate with the fewest top choice votes is eliminated, and the votes for them are transferred to the next choice on each ballot The process continues until one candidate achieves a majority, or until only one candidate remains Instant runoff voting - Example

23 votes : A B C D 23 Votes : A C B D 12 Votes : B C D A 22 Votes : D B C A 20 Votes: C B D A Instant runoff voting

Problems with IRV: May not always find the Condorcet winner when it exists

It can also violate monotonicity (positive association) and reversal symmetry. The Coombs method

It is similar to Instant runoff, Except; the candidate who is eliminated in a given round under the Coombs method is the candidate who is ranked last by the largest number of voters (instead of the candidate who is ranked first by the smallest number of voters as under the AV method) The Coombs method - Example

23 votes : A B C D 23 Votes : A C B D 12 Votes : B C D A 22 Votes : D B C A 20 Votes: C B D A

A is eliminated in the first round, D is eliminated in the second round, B is the winner! Single Transferable Vote

Hare records that when he was teaching in his father’s school, his students were asked to elect a committee by standing beside the boy they liked best. –This first produced a number of unequal groups but soon, ­ the boys in the largest groups came to the conclusion that not all of them were necessary for the election of their favorite and some moved on to help another candidate while, on the other hand, ­ the few supporters of an unpopular boy gave him up as a hopeless and transferred themselves to the candidate they considered the next best Single Transferable Vote

STEPS: Any candidate that receives more than a certain number (quota) of first place votes is elected. If the elected candidate receives more votes than necessary for election, their excess votes are distributed to the other candidates in accordance with the second choice preference of voters. If the redistribution procedures reaches a point where there are no more votes to be redistributed and there are still more elected positions to be filled, the candidate with the least votes is eliminated and the votes for that candidate are redistributed This process repeats until either a winner is found for every seat or there are as many seats as remaining candidates. Single Transferable Vote

SETTING THE QUOTA In its standard form, the is used to determine the quota needed for election (New Zealand) D(n,k) =[n/(k+1)]+ =(votes/(seat +1))+1

Other possible quotas that can be used : (votes/seat). ­ Originally used by STV but it is now generally considered to be technically inferior ­ Newland Britton quota or rational Droop quota : (votes/(seat+1)) Single Transferable Vote - Example

23 votes : A B C D 23 Votes : A C B D 12 Votes : B C D A 22 Votes : D B C A 20 Votes: C B D A

One candidate : C Wins Two Candidates : AC wins Single Transferable Vote

SURPLUS TRANSFER RULES Random transfers ­ When a candidate reaches a quota and generates a surplus, it is decided randomly which ballots will remain with that candidate and which will be transferred to subsequent choices

Fractional transfers ­ Instead of reducing only some of the ballots at a whole value, fractional transfer STV transfer the same portion of all the ballots which went into creating the quota and surplus ­ Gregory method

­ EXAMPLES on Word file! Single Transferable Vote

SURPLUS TRANSFER RULES Meek Transfer (Weighted Inclusive Gregory Method plus Transfers to Candidates Already Elected) ­ Although the basic fractional transfer rule is good, there remains a drawback that can allow some people to have more voting power than others ­ Suppose a ballot is to be transferred and its next preference is for a winner in a prior round. Hare ignore such preferences and transfer the ballot to the next preference. ­ Alternatively the vote could be transferred to that winner and the process continued. ­ Brian Meek proposed to correct this problem by continually recalculating the retention fractions of each elected candidate when new votes are added to the total, so that a part of the new votes are absorbed and yet the candidate still retains exactly one quota ­ Since the change in one candidate’s retention fraction might alter the retention fraction needed by another candidate, the Meek method needs to holistically compute the set retention fractions that will result in each elected candidate having exactly a quota ­ Currently Used in New Zealand Single Transferable Vote Application to the List system In the former situations none of the elements gets more than one seat while, in this case, this restriction will not be applied. The necessary modifications that should be made ­ First we specify how many voters support each party as their first place candidate and we allocate the seats among the parties according to the number of time they fill the Droop quota ­ If there is still any seat(s) not filled, we transfer the votes from the party having the smallest remaining vote to the others ­ The basic difference is that now the parties having already one seat may be eligible for a second seat Example: STV List system

23 votes : A B C D 23 Votes : A C B D 12 Votes : B C D A 22 Votes : D B C A 20 Votes: C B D A

3 seat will be distributed. Advantages of STV

STV, based on Meek or fractional transfers is an enormous improvement over virtually every other method of proportional representation It is a very effective system which gives true proportional representation without reliance on political parties Disadvantages of STV

The criticism of IRV is also applicable to STV in multi-winner election STV can eliminate a candidate who might have gone to win given another elimination order, that is , if another non-winning candidate had been eliminated earlier instead of later (Condorcet winner problem) In general this problem seems to be less severe the more seats there are to be filled by STV. That is it should not really effect the overall proportionality of a multi-seat STV election Although STV has respect to citizens it is not Pareto optimal Droop quota does not satisfy respect to majority so rational Droop quota is used SRV is sometimes criticized on the grounds that preference voting is unfamiliar in many societies STV count is quite complex. ­ This has been cited as one of the reasons why Estonia decided to abandon the system after its first election Disadvantages of STV

SRV can lead to a party with a plurality of votes nonetheless winning fewer seats than its rivals ­ – amended its system in the mid-1980s by providing for some extra compensatory seats to be awarded to a party in the event of this happening

It violates the monotonicity –i.e. Positive association: Butterfly effect of STV Presentation of Groups - 1

Two weeks later. I assign you one paper:

­ Schulze, M. (2011) “A new monotonic, clone-independent, reversal symmetric, and Condorcet- consistent single-winner election method”, Social Choice and Welfare, 36, pp 267–303.

Please find 3 more papers related to the topics: -Social Choice Functions (2 papers) -Process oriented approaches (1 paper) ­ See the list of papers at ninova. ­ Search on databases ­ Send me the papers for my approval. ­ Recent papers are preferred (>2018) Final list of the papers will be announced next week.

Notice that although the group members will make the presentation, other students are also responsible for the papers in order to facilitate class discussions. Please read the papers before the class. Next Week

Process oriented approches