But, how??

Explaining all possible positional, pairwise voting paradoxes & prop.

Don SaariInstitute for Math Behavioral Sciences

University of California, Irvine CA 92697-5100

3 A>B>C4 A>C>B6 C>B>A2 B>C>A2 B>A>C

Vote for one (1, 0, 0):A

Vote for two (1, 1, 0):BBorda (2, 1, 0):C

Paired comparisons?

B>A, A>C, C>B, a cycle

How would you explain, create, this example?

Approval voting? Cumulative voting?Any of the 13 possible rankings is a

sincere outcomeStandard approach since Borda, Condorcet,

Arrow:“Proof” by example

Put forth “desirable properties that a voting rule “must” satisfy

Example: “Will voting rule always elect Condorcet winner?”

What have we accomplished?Many impressive results (and difficult), but …

Answer above challenges? Why paradoxes? Arrow? Consensus -- even among those in voting theory?

My approach was influenced by:Fishburn -- paradoxes, which are properties, n=7, 1050

Nurmi -- Nothing goes rightFor a price …..

Need new approach

Rather than “select rule and find supporting properties”Science approach is to find all properties and identify

appropriate rules.

New approachWell, not so new (1999, 2000) Symmetry

groups, new way to achieve objective for any number

of candidatesFind appropriate coordinate system

that will identify all properties

Polar coordinates

y = A xEigenvectors

My goal: find “profile coordinate system” so that

we can explain all differences in paired comparisons,

in positional outcomes, in everything

Coordinate system for morning coffeeWaterCoffeeCreamSugar

A B

C

Tallying ballots

3 A>B>C4 A>C>B6 C>B>A2 B>C>A2 B>A>C

34

0 62

2

7 10

7

108

9

Positional voting(w1, w2, 0)

Normalize; i.e., divide each weight by w1

So, weights are (1, s, 0)s=0, plurality, s=½,

Borda, s=1, vote-for-two7+2s 4 + 9s

6+6s

Coordinates:Water, nothing happens

A B

C

x

xx

xx

x

More alternatives, more complicated

and more dimensions

A B

C

Paired comparisonsA>B

-1-1

1

11

-1

A B

C

Paired comparisonsB>C

11

-1

-11

-1

A B

C

Paired comparisonsA>C

1-1

-1

11

-1

Basis for all pairwise votesIs there a better one?

n=6, Kernel has dimension 720- 130= 590

Better (by giving more information) pairwise basis

A B

C

Paired comparisonsA>B

-1-1

1

11

-1

A B

C

Paired comparisonsA>C

1-1

-1

11

-1+ = (well, after taking ½)

A B

C

-1-11

1

2 -2

-2

2

0

0

ABasic

2 + 0s

-1+0s

-1+0s

A B

C

11

-1-1

BBasic

Complete coordinate system for pairs;

source of all properties of any rule depending on paired

comparisons such as Dodgsen, Borda, Condorcet, Kemeny,

Arrow, Sen, etc.Nothing else is needed

A B

C

-11

1

1 -1-1

Condorcet

1

1 -1

1

-1

-1

A>B>C, B>C>A, C>A>BCriticism: Borda need not elect Condorcet winner

Nothing goes wrong; complete agreement

Source of all problemsAnalysis depends on how rule treats

this feature

More accurate criticism: Condorcet winner need not be Borda

winnerKelly: n=3, Ostrogorski paradox implies no Condorcet winner: Now trivial. Conjecture true for

odd n>3

(Add A>B, B>C, C>A)

A B

C

-1-11

1

ABasicA B

C

11

-1-1

BBasic

A B

C

-11

1

1 -1-1

Condorcet

1

1 -1

1

-1

-1

For n candidatesSame idea--start with “n choose 2” X>Y vectors

XBasic; n-1 everywhere X is top

ranked, n-3 everywhere X is

second ranked, ….

n-(2j-1) everywhere X is jth ranked

Basis for n-1 dimensional space

Condorcet, (n-1)!/2 of themA>B>...>Y>Z,B>...>Y>Z>A,

…

-Z>Y>...>B>A-A>Z>Y>...>B

Source of all possible

properties or rules that are based on

paired comparisons;

this is all that is needed

Arrow, Sen, likelihood of cycles, agendas, Kemeny, Borda, Condorcet,

etc.

Example of how answer is easily found for typical

problem (problem, from earlier

analysis, is flawed): What voting rules will always elect a Condorcet winner?

Kelly Conjecture: Ostrogorski paradox implies no Condorcet winner for odd n>3

True for all n >2.

n=6, dimension is 60

Back to n=3 and finding coordinate system

A B

C

-1-11

1

ABasicA B

C

11

-1-1

BBasicA B

C

-11

1

1 -1-1

Condorcet

1

1 -1

1

-1

-1

Positional (1, s, 0)What does not affect pairs and is not included?A>B>C, C>B>A

A B

C

1

111

-2

-2

AReversal

2-4s

-1+2s

-1+2sA B

C

11

11

-2

-2

BReversal

That is all there is;all three alternative properties follow from these coordinates

Includes runoffs, etc.

All positional outcome problems and differences are caused by how rule

(Plurality, AV, Cumulative, etc.) react to reversal

(Only Borda is not affected by Reversal)

Example: Which positional rules have rankings that coincide (in any way) with

paired comparisions?Borda

Source of paradoxes

A B

C

Creating initial example

3 A>B>C4 A>C>B6 C>B>A2 B>C>A2 B>A>C

1

This C>B>A preference will be Borda Outcome

Next, to get A to be Plurality winner and B vote-for-two winner

add reversal components

xxy

y

So, solvex+y>1+y>x to get plurality A>C>B;

e.g., x=2, y=3

A B

C

223

4

To obtain cyclic effect, add Condorcet term

z

zzSolve:

5+z < 6+2z5+z<6+2z6+z< 5+2ze.g., z=2

6

224

3

Cambridge University press

Coordinates for n=4 (same idea holds for all n>3)

A B

C

11

11

-2

-2

BReversal

-2

-2A B

C

1

111

AReversal

Need this structure for triplets, but also need to ensure that it

does not affect paired comparisons (already done) and

set of all 4A>B>C>D, D>C>B>AB>A>D>C, C>D>A>B

Same approach: Sum all of the A-reversals to find coordinate over all triplets (as with the coordinate over all pairs)

then find “cyclic part” of “A>B>C, B>C>D, C>D>A, D>A>C” kind.

Finally, behavior for set of all four candidates

that cannot for triplets, and pairs

This coordinate system, then, can be used to identify all possible properties of four candidate election rules

where positional and paired comparisons are involvedSame approach continues for n>4