Voting and social choice Looking at a problem from the designers point of view.

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Presentation transcript:

Voting and social choice Looking at a problem from the designers point of view

Voting over alternatives > > > > voting rule (mechanism) determines winner based on votes Can vote over other things too –Where to go for dinner tonight, other joint plans, …

Voting (rank aggregation) Set of m candidates (aka. alternatives, outcomes) n voters; each voter ranks all the candidates –E.g., for set of candidates {a, b, c, d}, one possible vote is b > a > d > c –Submitted ranking is called a vote A voting rule takes as input a vector of votes (submitted by the voters), and as output produces either: –the winning candidate, or –an aggregate ranking of all candidates Can vote over just about anything –political representatives, award nominees, where to go for dinner tonight, joint plans, allocations of tasks/resources, … –Also can consider other applications: e.g., aggregating search engine’s rankings into a single ranking

Example voting rules Each voter gives a vector of ranked choices. Scoring rules are defined by a vector (a 1, a 2, …, a m ); being ranked i th in a vote gives the candidate a i points –Plurality is defined by (1, 0, 0, …, 0) (winner is candidate that is ranked first most often, only first choice votes are counted) –Veto (or anti-plurality) is defined by (1, 1, …, 1, 0) (winner is candidate that is ranked last the least often) –Borda is defined by (m-1, m-2, …, 0)

Nanson (Borda variant) Candidate with the lowest Borda score is eliminated, then we re-compute Borda counts and continue.

Runoff voting rules proceeds in stages Plurality with (2-candidate) runoff: top two candidates in terms of plurality score proceed to runoff; whichever is ranked higher than the other by more voters, wins How would you describe the idea behind a runoff? Single Transferable Vote (STV, aka. Instant Runoff): candidate with lowest plurality score drops out; if you voted for that candidate (as your first choice), your vote transfers to the next (live) candidate on your list; repeat until one candidate remains Similar runoffs can be defined for rules other than plurality

Pairwise elections select pairwise comparison from complete list > > > > > > > two votes prefer Obama to McCain > two votes prefer Obama to Nader > two votes prefer Nader to McCain > >

Pairwise elections select pairwise comparison from complete list > two votes prefer Obama to McCain > two votes prefer Obama to Nader > two votes prefer Nader to McCain > > 2 22 May care about numbers of voters or only winner

Pairwise elimination Candidates given a schedule of pairwise competitions Loser is eliminated at each stage. Winner goes on to compete at next round Like a single elimination athletic event (but no parallel competitions) Not every pair is considered

Sensitivity to agenda setter: order of elimination matters 35 agents a > c > b 33 agents b > a > c 32 agents c > b > a Who is the winner in the following pairings? (a,b) c (a,c) b (b,c) a

Condorcet Condorcet winner of an election: –A candidate who wins every pairwise election Condorcet methods are named for the eighteenth-century French mathematician and philosopher Marie Jean Antoine Nicolas Caritat, the Marquis de Condorcet (took their title from the town of Condorcet in Dauphine) Wikipedia Oct 2010 Clearly, there may not be a Condorcet winner Condorcet condition: if there is a Condorcet winner, he must be the winner.

Condorcet cycles > > > > > > > two votes prefer McCain to Obama > two votes prefer Obama to Nader > two votes prefer Nader to McCain ?

Condorcet cycles > two votes prefer McCain to Obama > two votes prefer Obama to Nader > two votes prefer Nader to McCain ? 2 2 2

Condorcet cycles 500 voters > 251 votes prefer McCain to Obama > 400 votes prefer Obama to Nader > 330 votes prefer Nader to McCain ? Number of voters who are unhappy with ranking or number of winning arcs who are unhappy

Preference profile: a tuple giving a preference ordering for each agent #(o>o’) is the number of agents who prefer o to o’ Smith set: is the smallest set S  O having the property that  o’  S, #(o> o’)  #(o’>o) In other words, every outcome in the Smith set is preferred (by at least half of the agents) to every outcome outside the set.

What is the Smith set? 35 agents a > c> b >d 33 agents b > a > d > c 32 agents c >d > b > a What is the relationship between Smith set and Condorcet? a d b c 65 (35) 67(33) 68(32) 67(33) 68(32)

Cumulative voting: Each voter is given k votes which can be cast arbitrarily (voting for any set of candidates he wants). The candidate with the most votes is selected. Approval voting: Each voter can cast a single vote for as many of the candidates as he wishes; the candidate with the most votes is selected.

Copeland: candidate gets two points for each pairwise election it wins, one point for each pairwise election it ties Second order Copeland: sum of Copeland scores of alternatives you defeat. (once used by NFL as tie-breaker) 35 agents c > a> b >d 33 agents b > a > d > c 32 agents c >d > b > a What is Copeland Score? What is second order Copeland Score? a d b c 65 (35) 67(33) 68(32) 67(33) 68 (32) Voting rule based on pairwise elections

Another voting rule based on pairwise elections Maximin (aka. Simpson): candidate whose worst pairwise result is the best among candidates – wins. So if there are four candidates and 10 voters and between pairs (me,opponent): (9,1), (10,0), (8,2), and (5,5). If others had a worse pairwise vote than (5,5), I would be the winner.

Slater: create an overall ranking of the candidates that is inconsistent with as few pairwise elections as possible –NP-hard! Consider all orders and count inconsistencies. An instance of the Slater problem can be represented by a “pairwise election” graph whose vertices are the candidates, and which has a directed edge from a to b if and only if a defeats b in their pairwise election. The goal, then, is to minimize the number of edges that must be flipped in order to make the graph acyclic. Cup/pairwise elimination: pair candidates, losers of pairwise elections drop out, repeat

Slater on pairwise election graphs Final ranking = acyclic tournament graph Slater ranking seeks to minimize the number of inverted edges a b d c a b d c pairwise election graph Slater ranking (a > b > d > c) What about b >d>c>a?

Even more voting rules… Kemeny: create an overall ranking of the candidates that has as few disagreements as possible (where a disagreement is with a vote on a pair of candidates). For each pair of voters (X,Y) count how many times X is preferred to Y. Test all possible order-of-preference sequences, calculate a sequence score for each sequence, and compare the scores. Each sequence score equals the sum of the pairwise counts that are “honored by” the sequence (a is preferred to b and a precedes b in the sequence). The sequence with the highest score is identified as the overall ranking –NP-hard! –Similar to Slater – but looks at actual numbers of votes not just result of pairing. Bucklin: start with k=1 and increase k gradually until some candidate is among the top k candidates in more than half the votes; that candidate wins

Kemeny on pairwise election graphs Final ranking = acyclic tournament graph –Edge (a, b) means a ranked above b –Acyclic = no cycles, tournament = edge between every pair Kemeny ranking seeks to minimize the total weight of the inverted edges a b d c pairwise election graph Kemeny ranking a b d c 2 2 (b > d > c > a)

So… SO… how do we choose a rule from all of these rules? How do we know that there does not exist another, “perfect” rule? Let us look at some criteria that we would like our voting rule to satisfy

Condorcet criterion A candidate is the Condorcet winner if it wins all of its pairwise elections Does not always exist… … but the Condorcet criterion says that if it does exist, it should win Many rules do not satisfy this simple criterion Consider plurality voting: –b > a > c > d –c > a > b > d –d > a > b > c a is the Condorcet winner, but it does not win under plurality. Explain

Majority criterion If a candidate is ranked first by majority of votes that candidate should win –Relationship to Condorcet criterion? a > b > c > d > e e > a > b > c > d c > b > d > a > e Some rules do not even satisfy this E.g. Borda: –a > b > c > d > e –c > b > d > e > a a is the majority winner, but it does not win under Borda (b wins under Borda, right?)

Monotonicity criteria Informally, monotonicity means that “ranking a candidate higher should help that candidate,” but there are multiple nonequivalent definitions

Monotonicity criteria A weak monotonicity requirement: if –candidate w wins given the current votes, –we then improve the position of w in some of the votes and leave everything else the same, then w should still win. E.g., Single Transferable Voting does not satisfy this: –7 votes b > c > a –7 votes a > b > c –6 votes c > a > b c drops out first (lowest plurality), its votes transfer to a (next candidate), a wins **But if 2 votes b > c > a change to a > b > c (we improve a’s ranking), b drops out first, its 5 votes transfer to c, and c wins –5 votes b > c > a –9 votes a > b > c –6 votes c > a > b

Monotonicity criteria… A strong monotonicity requirement: if –candidate w wins for the current votes, –we then change the votes in such a way that for each vote, if candidate c was ranked below w originally, c is still ranked below w in the new vote then w should still win. Note the other candidates can jump around in the vote, as long as they don’t jump ahead of w None of our rules satisfy this

Independence of irrelevant alternatives Independence of irrelevant alternatives criterion: if –the rule ranks a above b for the current votes, –we then change the votes but do not change which is ahead between a and b in each vote then a should still be ranked ahead of b. (The other votes are irrelevant to the relationship between a and b.) None of our rules satisfy this

Arrow’s impossibility theorem [1951] Suppose there are at least 3 candidates Then there exists no rule that is simultaneously: –Pareto efficient (if all votes rank a above b, then the rule ranks a above b), Explain use of term –nondictatorial (there does not exist a voter such that the rule simply always copies that voter’s ranking), and –independent of irrelevant alternatives

Weak Pareto efficient if there exist a pair of outcomes o1 and o2 such that  i o1 > i o2 then C([>])  o2 In other words, we cannot select any outcome that is dominated by another alternative for all agents Why is it called weak?

Muller-Satterthwaite impossibility theorem [1977] Suppose there are at least 3 candidates Then there exists no rule that simultaneously: –satisfies unanimity (if all votes rank a first, then a should win), –is nondictatorial (there does not exist a voter such that the rule simply always selects that voter’s first candidate as the winner), and –is monotone (in the strong sense).

The Shoham Leyton-Brown text has proofs of the impossibility theorems. We won’t go over them – but interested readers should take a look.

Manipulability Sometimes, a voter is better off revealing her preferences insincerely, aka. manipulating E.g. plurality –Suppose a voter prefers a > b > c –Also suppose she knows that the other votes are 2 voters b > c > a 2 voters c > a > b –Voting truthfully will lead to a tie between b and c –She would be better off voting e.g. b > a > c, guaranteeing b wins All our rules are (sometimes) manipulatable

Single-peaked preferences Suppose candidates are ordered on a line a1a1 a2a2 a3a3 a4a4 a5a5 Every voter prefers candidates that are closer to her most preferred candidate Let every voter report only her most preferred candidate (“peak”) v1v1 v2v2 v3v3 v4v4 v5v5 Choose the median voter’s peak as the winner –This will also be the Condorcet winner Nonmanipulable! Impossibility results do not necessarily hold when the space of preferences is restricted. Why would you guess this is true?

Some computational issues in social choice Sometimes computing the winner/aggregate ranking is hard –E.g. for Kemeny and Slater rules this is NP-hard For some rules (e.g., STV), computing a successful manipulation is NP-hard –Manipulation being hard is a good thing … But would like something stronger than NP-hardness –Also: work on the complexity of controlling the outcome of an election by influencing the list of candidates/schedule of the Cup rule/etc. Preference elicitation: –We may not want to force each voter to rank all candidates; –Rather, we can selectively query voters for parts of their ranking, according to some algorithm, to obtain a good aggregate outcome Combinatorial alternative spaces: –Suppose there are multiple interrelated issues that each need a decision –Exponentially sized alternative spaces Different models such as ranking webpages (pages “vote” on each other by linking)