PHPE 400 Individual and Group Decision Making Eric Pacuit - PowerPoint PPT Presentation

PHPE 400 Individual and Group Decision Making Eric Pacuit University of Maryland 1 / 22

The Majority Relation Given a profile P for voters V and candidates X , ◮ For x , y ∈ X , let P ( x , y ) = { i ∈ V | x P i y } . ◮ We write N P ( x , y ) for the number of voters in P ranking x above y , i.e., N P ( x , y ) = | P ( x , y ) | . ◮ For x , y ∈ X , let Margin P ( x , y ) = N P ( x , y ) − N P ( y , x ) 2 / 22

Majority Relation We say that a majority prefers x to y in P , denoted x > M P y , when N P ( x , y ) > N P ( y , x ) . The margin graph of P , M ( P ) , is the weighted directed graph whose set of vertices is C with an edge from a to b weighted by Margin ( x , y ) when Margin ( x , y ) > 0. We write α x −→ P y if α = Margin P ( x , y ) > 0 . 3 / 22

The Problem Voter 1 Voter 2 Voter 3 a c b b a c c b a Does the group prefer a over b ? Yes Does the group prefer b over c ? Yes Does the group prefer a over c ? No 4 / 22

The Problem Voter 1 Voter 2 Voter 3 a c b b a c c b a ◮ Does the group prefer a over b ? Yes Does the group prefer b over c ? Yes Does the group prefer a over c ? No 4 / 22

The Problem Voter 1 Voter 2 Voter 3 a c b b a c c b a ◮ Does the group prefer a over b ? Yes Does the group prefer b over c ? Yes Does the group prefer a over c ? No 4 / 22

The Problem Voter 1 Voter 2 Voter 3 a c b b a c c b a ◮ Does the group prefer a over b ? Yes ◮ Does the group prefer b over c ? Yes Does the group prefer a over c ? No 4 / 22

The Problem Voter 1 Voter 2 Voter 3 a c b b a c c b a ◮ Does the group prefer a over b ? Yes ◮ Does the group prefer b over c ? Yes ◮ Does the group prefer a over c ? No 4 / 22

The Problem Voter 1 Voter 2 Voter 3 a c b b a c c b a The majority relation > M is not transitive! There is a Condorcet cycle : a > M b > M c > M a 4 / 22

How bad is this? ◮ Final decisions are extremely sensitive to institutional features such as who can set the agenda, arbitrary time limits place on deliberation, who is permitted to make motions, etc. 5 / 22

How bad is this? ◮ Final decisions are extremely sensitive to institutional features such as who can set the agenda, arbitrary time limits place on deliberation, who is permitted to make motions, etc. ◮ Is there empirical evidence that Condorcet cycles have shown up in real elections? W. Riker. Liberalism against Populism . Waveland Press, 1982. G. Mackie. Democracy Defended . Cambridge University Press, 2003. 5 / 22

How bad is this? ◮ Final decisions are extremely sensitive to institutional features such as who can set the agenda, arbitrary time limits place on deliberation, who is permitted to make motions, etc. ◮ Is there empirical evidence that Condorcet cycles have shown up in real elections? W. Riker. Liberalism against Populism . Waveland Press, 1982. G. Mackie. Democracy Defended . Cambridge University Press, 2003. ◮ How likely is a Condorcet cycle? 5 / 22

A voting method is Condorcet consistent if it selects the Condorcet winner if it exists. 6 / 22

7 5 4 3 a b a b d c b c b d c d c a d a a b c d 7 / 22

7 5 4 3 a b a b d c b c b d c d c a d a a b c d 7 / 22

7 5 4 3 a b a b d c b c b d c d c a d a a b c d 7 / 22

7 5 4 3 a b a b d c b c b d c d c a d a a b c d 7 / 22

7 5 4 3 a b a b d c b c b d c d c a d a a b c d 7 / 22

7 5 4 3 a 1 b a b d c b c b d 5 5 c d c a d a a b c d 7 / 22

7 5 4 3 a 1 b a b d c b c b d 5 5 5 c d c a 13 d a a b c 11 d 7 / 22

Condorcet The Condorcet winner in a profile P is a candidate x ∈ C that is the maximum of the majority ordering, i.e., for all y ∈ C , if x � y , then x > M P y . The Condorcet voting method is:  { x } if x is the Condorcet winner in P   Condorcet ( P ) =   C if there is no Condorcet winner .   8 / 22

Copeland For each P and x ∈ C , let wl P ( x ) = |{ z | Net P ( x , z ) > 0 }| − |{ z | Net P ( z , x ) > 0 }| . Copeland ( P ) = argmax x ∈ c ( wl P ( x )) . 9 / 22

7 5 4 3 a b a b d c b c b d c d c a d a a b c d 10 / 22

Not all cycles are created equal... a 1 b a a 1 1 1 3 1 3 1 1 c c 3 1 b b c 3 d 11 / 22

Minimax For each P and x ∈ X , let supp ( x , P ) = max ( { N P ( y , x ) | y ∈ X , y � x } ) . Minimax ( P ) = argmin x ∈ X ( supp ( x , P )) . 12 / 22

Beat Path For candidates a , b ∈ X , the strength of a path a = x 1 , x 2 , . . . , x n = b in M ( P ) is min { Margin P ( x k , x k + 1 ) | 1 ≤ k ≤ n − 1 } Then, a defeats b in P according to Beat Path if the strength of the strongest path from a to b is greater than the strength of the strongest path from b to a . 13 / 22

c 3 3 3 a b 3 1 1 d ◮ d defeats a according to Beat Path ◮ d defeats b according to Beat Path ◮ d defeats c according to Beat Path ◮ Since no candidate defeats d according to Beat Path, d is a Beat Path winner. 14 / 22

Split Cycle Candidate a is a Split Cycle winner if a is not defeated after deleting the edges with the smallest margin in every cycle in the margin graph. 15 / 22

c 3 3 3 a b 3 1 1 d c c 3 3 3 3 c a b b 3 1 3 3 1 a d b 3 d 16 / 22

c 3 3 3 a b 3 1 1 d c c 3 3 3 3 c a b b 3 1 3 3 1 a d b 3 d 16 / 22

c 3 3 3 a b 3 1 d c c 3 3 3 3 c a b b 3 1 3 3 1 a d b 3 d 16 / 22

c 3 a b 1 d c c 3 3 3 3 c a b b 3 1 3 3 1 a d b 3 d 16 / 22

c 3 a b d The Split Cycle winners are a , b and d 16 / 22

7 5 4 3 a 1 b a b d c b c b d 5 5 5 c d c a 13 d a a b c 11 d Condorcet winners a , b , c , d Copeland winners b , c Minimax winners b Beat Path winners b Split Cycle winners b 17 / 22

Should we select a Condorcet winner (when one exists)? 18 / 22

Condorcet’s Other Paradox # voters 30 1 29 10 10 1 a a b b c c b c a c a b c b c a b a BS ( a ) = 2 × 31 + 1 × 39 + 0 × 11 = 101 BS ( b ) = 2 × 39 + 1 × 31 + 0 × 11 = 109 BS ( c ) = 2 × 11 + 1 × 11 + 0 × 59 = 33 a > M b > M c b > Bc a > Bc c 19 / 22

Condorcet’s Other Paradox # voters 30 1 29 10 10 1 2 a a b b c c 1 b c a c a b 0 c b c a b a BS ( a ) = 2 × 31 + 1 × 39 + 0 × 11 = 101 BS ( b ) = 2 × 39 + 1 × 31 + 0 × 11 = 109 BS ( c ) = 2 × 11 + 1 × 11 + 0 × 59 = 33 b > Bc a > Bc c 19 / 22

Condorcet’s Other Paradox # voters 30 1 29 10 10 1 2 a a b b c c 1 b c a c a b 0 c b c a b a BS ( a ) = 2 × 31 + 1 × 39 + 0 × 11 = 101 BS ( b ) = 2 × 39 + 1 × 31 + 0 × 11 = 109 BS ( c ) = 2 × 11 + 1 × 11 + 0 × 59 = 33 a > M b > M c b > Bc a > Bc c 19 / 22

Condorcet’s Other Paradox # voters 30 1 29 10 10 1 2 a a b b c c 1 b c a c a b 0 c b c a b a BS ( a ) = 2 × 31 + 1 × 39 + 0 × 11 = 101 BS ( b ) = 2 × 39 + 1 × 31 + 0 × 11 = 109 BS ( c ) = 2 × 11 + 1 × 11 + 0 × 59 = 33 a > M b > M c b > Bc a > Bc c 19 / 22

Condorcet’s Other Paradox # voters 30 1 29 10 10 1 2 a a b b c c 1 b c a c a b 0 c b c a b a BS ( a ) = 2 × 31 + 1 × 39 + 0 × 11 = 101 BS ( b ) = 2 × 39 + 1 × 31 + 0 × 11 = 109 BS ( c ) = 2 × 11 + 1 × 11 + 0 × 59 = 33 a > M b > M c b > Bc a > Bc c 19 / 22

Condorcet’s Other Paradox # voters 30 1 29 10 10 1 s 2 a a b b c c s 1 b c a c a b s 0 c b c a b a Condorcet’s Other Paradox : No scoring rule will work... BS ( b ) = 2 × 39 + 1 × 31 + 0 × 11 = 109 BS ( c ) = 2 × 11 + 1 × 11 + 0 × 59 = 33 a > M b > M c b > Bc a > Bc c 19 / 22

Condorcet’s Other Paradox # voters 30 1 29 10 10 1 s 2 a a b b c c s 1 b c a c a b s 0 c b c a b a Condorcet’s Other Paradox : No scoring rule will work... Score ( a ) = s 2 × 31 + s 1 × 39 + s 0 × 11 Score ( b ) = s 2 × 39 + s 1 × 31 + s 0 × 11 a > M b > M c b > Bc a > Bc c 19 / 22

Condorcet’s Other Paradox # voters 30 1 29 10 10 1 s 2 a a b b c c s 1 b c a c a b s 0 c b c a b a Condorcet’s Other Paradox : No scoring rule will work... Score ( a ) = s 2 × 31 + s 1 × 39 + s 0 × 11 Score ( b ) = s 2 × 39 + s 1 × 31 + s 0 × 11 Score ( a ) > Score ( b ) ⇒ 31 s 2 + 39 s 1 > 39 s 2 + 31 s 1 ⇒ s 1 > s 2 a > M b > M c b > Bc a > Bc c 19 / 22

Condorcet’s Other Paradox # voters 30 1 29 10 10 1 s 2 a a b b c c s 1 b c a c a b s 0 c b c a b a Theorem (Fishburn 1974) . For all m ≥ 3, there is some voting situation with a Condorcet winner such that every scoring rule will have at least m − 2 candidates with a greater score than the Condorcet winner. P. Fishburn. Paradoxes of Voting . The American Political Science Review, 68:2, pgs. 537 - 546, 1974. 19 / 22

Condorcet Triples n n n n n n a b c a c b b c a c b a c a b b a c In both profiles, any reasonable voting method should select all candidates as winners 20 / 22