But, how?? Explaining all possible positional, pairwise voting paradoxes & prop. Don Saari Institute for Math Behavioral Sciences University of California, Irvine CA A>B>C 4 A>C>B 6 C>B>A 2 B>C>A 2 B>A>C Vote for one (1, 0, 0):A Vote for two (1, 1, 0):B Borda (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 outcome Standard 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, Nurmi -- Nothing goes right For 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 approach Well, not so new (1999, 2000) Symmetry groups, new way to achieve objective for any number of candidates Find appropriate coordinate system that will identify all properties Polar coordinates y = A x Eigenvectors 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 coffee Water Coffee Cream Sugar

AB C Tallying ballots 3 A>B>C 4 A>C>B 6 C>B>A 2 B>C>A 2 B>A>C Positional voting (w 1, w 2, 0) Normalize; i.e., divide each weight by w 1 So, weights are (1, s, 0) s=0, plurality, s=½, Borda, s=1, vote-for-two 7+2s4 + 9s 6+6s Coordinates: Water, nothing happens AB C x x x x x x More alternatives, more complicated and more dimensions AB C Paired comparisons A>B AB C Paired comparisons B>C AB C Paired comparisons A>C Basis for all pairwise votes Is there a better one? n=6, Kernel has dimension = 590

Better (by giving more information) pairwise basis AB C Paired comparisons A>B AB C Paired comparisons A>C = (well, after taking ½) AB C A Basic 2 + 0s -1+0s AB C 1 1 B Basic 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 AB C Condorcet A>B>C, B>C>A, C>A>B Criticism: Borda need not elect Condorcet winner Nothing goes wrong; complete agreement Source of all problems Analysis depends on how rule treats this feature More accurate criticism: Condorcet winner need not be Borda winner Kelly: n=3, Ostrogorski paradox implies no Condorcet winner: Now trivial. Conjecture true for odd n>3 (Add A>B, B>C, C>A)

AB C 1 1 A Basic AB C 1 1 B Basic AB C Condorcet For n candidates Same idea--start with “n choose 2” X>Y vectors X Basic ; n-1 everywhere X is top ranked, n-3 everywhere X is second ranked, …. n-(2j-1) everywhere X is j th ranked Basis for n-1 dimensional space Condorcet, (n-1)!/2 of them A>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 AB C 1 1 A Basic AB C 1 1 B Basic AB C Condorcet Positional (1, s, 0) What does not affect pairs and is not included? A>B>C, C>B>A AB C A Reversal 2-4s -1+2s AB C B Reversal 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

AB C Creating initial example 3 A>B>C 4 A>C>B 6 C>B>A 2 B>C>A 2 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 xx y y So, solve x+y>1+y>x to get plurality A>C>B; e.g., x=2, y=3 AB C To obtain cyclic effect, add Condorcet term z z z Solve: 5+z < 6+2z 6+z< 5+2z e.g., z= Cambridge University press

Coordinates for n=4 (same idea holds for all n>3) AB C B Reversal -2 AB C A Reversal Need this structure for triplets, but also need to ensure that it does not affect paired comparisons (already done) and set of all 4 A>B>C>D, D>C>B>A B>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 involved Same approach continues for n>4