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

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

Aug 21, 2022 834 likes •1.34k views

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

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PHPE 400 Individual and Group Decision Making

Eric Pacuit University of Maryland

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The Majority Relation

Given a profile P for voters V and candidates X, ◮ For x, y ∈ X, let P(x, y) = {i ∈ V | xPiy}. ◮ We write NP(x, y) for the number of voters in P ranking x above y, i.e., NP(x, y) = |P(x, y)|. ◮ For x, y ∈ X, let MarginP(x, y) = NP(x, y) − NP(y, x)

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Majority Relation

We say that a majority prefers x to y in P, denoted x >M

P y, when

NP(x, y) > NP(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 x

α

−→P y if α = MarginP(x, y) > 0.

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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

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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

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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

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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

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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

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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

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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.

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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.

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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?

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

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

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

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

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

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

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

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

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Condorcet

The Condorcet winner in a profile P is a candidate x ∈ C that is the maximum

  • f the majority ordering, i.e., for all y ∈ C, if x y, then x >M

P y. The Condorcet

voting method is: Condorcet(P) =        {x} if x is the Condorcet winner in P C if there is no Condorcet winner.

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Copeland

For each P and x ∈ C, let wlP(x) = |{z | NetP(x, z) > 0}| − |{z | NetP(z, x) > 0}|. Copeland(P) = argmaxx∈c(wlP(x)).

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

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Not all cycles are created equal...

c b a 3 3 1 c b a 3 1 1 b a c d 1 1 1 3 1 1

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Minimax

For each P and x ∈ X, let supp(x, P) = max({NP(y, x) | y ∈ X, y x}). Minimax(P) = argminx∈X(supp(x, P)).

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Beat Path

For candidates a, b ∈ X, the strength of a path a = x1, x2, . . . , xn = b in M(P) is min{MarginP(xk, xk+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.

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b a c d 3 3 1 1 3 3

◮ 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.

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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.

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b a c d 3 3 1 1 3 3 b c d 3 1 3 b a c 3 3 3 b a c d 3 1 3 3

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b a c d 3 3 1 1 3 3 b c d 3 1 3 b a c 3 3 3 b a c d 3 1 3 3

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b a c d 3 3 1 3 3 b c d 3 1 3 b a c 3 3 3 b a c d 3 1 3 3

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b a c d 1 3 b c d 3 1 3 b a c 3 3 3 b a c d 3 1 3 3

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b a c d 3

The Split Cycle winners are a, b and d

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

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

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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 b >Bc a >Bc c a >M b >M c

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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 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

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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 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 a >M b >M c

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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 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 a >M b >M c

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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 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 a >M b >M c

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Condorcet’s Other Paradox

# voters 30 1 29 10 10 1 s2 a a b b c c s1 b c a c a b s0 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 b >Bc a >Bc c a >M b >M c

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Condorcet’s Other Paradox

# voters 30 1 29 10 10 1 s2 a a b b c c s1 b c a c a b s0 c b c a b a Condorcet’s Other Paradox: No scoring rule will work... Score(a) = s2 × 31 + s1 × 39 + s0 × 11 Score(b) = s2 × 39 + s1 × 31 + s0 × 11 b >Bc a >Bc c a >M b >M c

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Condorcet’s Other Paradox

# voters 30 1 29 10 10 1 s2 a a b b c c s1 b c a c a b s0 c b c a b a Condorcet’s Other Paradox: No scoring rule will work... Score(a) = s2 × 31 + s1 × 39 + s0 × 11 Score(b) = s2 × 39 + s1 × 31 + s0 × 11 Score(a) > Score(b) ⇒ 31s2 + 39s1 > 39s2 + 31s1 ⇒ s1 > s2 b >Bc a >Bc c a >M b >M c

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Condorcet’s Other Paradox

# voters 30 1 29 10 10 1 s2 a a b b c c s1 b c a c a b s0 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.

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Condorcet Triples

n n n a b c b c a c a b n n n a c b c b a b a c In both profiles, any reasonable voting method should select all candidates as winners

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2 2 2 a b c b c a c a b 1 2 a b b a c c − no Condorcet winner in the left profile − b is the Condorcet winner in the right profile − a is the Condorcet winner in the combined profiles

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Is the Condorcet winner the “best” choice?

# voters 47 47 3 3 a b c c c c a b b a b a c is the Condorcet winner Is there “positive support” for c?

  • or-

Are the supporters of the main rivals a and b using c to “separate” the rankings of a and b?

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Is the Condorcet winner the “best” choice?

# voters 47 47 3 3 a b c c c c a b b a b a c is the Condorcet winner Is there “positive support” for c?

  • or-

Are the supporters of the main rivals a and b using c to “separate” the rankings of a and b?

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