These methods assume input of ranked ballots. The ballot A>B>C means the voter prefers candidate A to candidate B, B to C and A to C. The line

4:A>B>C |

means four of that same ballot. Most of the methods allow input of tied preferences, which are shown using an equals sign, so

3:A>B=C>D |

means three voters like A most and D least, but have no preference between B and C.

All of these methods rightly choose the majority winner when there are only two candidates, but they often differ when there are three or more.

**Borda** is a point-count system. Each candidate is given a score equal to the
total number of times that candidate is ranked over another minus the total number of times that candidate is ranked
under another. For example, if there are 32 voters who vote

14:Alan>Beth>Carl |

11:Beth>Carl>Alan |

7:Carl>Alan>Beth |

Alan is ranked over another candidate 35 times (21 over Beth and 14 over Carl) and under another candidate 29 times (11 under Beth and 18 under Carl), so his Borda score is 35 minus 29, or 6. Similarly, Beth receives a score of 8 (11+25-21-7) and Carl a score of -14 (18+7-14-25). (Equivalently, the following could be used: a candidate gets no points for a last-place vote, one point for a second-to-last-place vote, on up to first place.) So Borda picks Beth in this election. Borda is sometimes considered the best ranked-ballot method when voters are sincere, but it’s probably the most easily manipulated. It also fails to be independent of clones: Consider an election with three candidates. Eric is right-wing and Fran and Gary are left-wing. 63% of the voters are right-wing, 37% are left-wing and all of the voters prefer Fran to Gary because of a scandal involving Gary, so the sincere votes are

63:Eric>Fran>Gary |

37:Fran>Gary>Eric |

Borda gives the win to Fran even though Eric received a strong majority of first-place votes and would have won handily if Gary hadn’t run, showing that parties have incentive to saturate an election with their candidates under Borda. Fran and Gary are called clones in this election because they’re next to each other on every ballot. Clone-independent methods give the same results if a group of clones is replaced by one candidate; any reasonable clone-independent method would choose Eric in this election, removing the candidate-saturation incentive.

**Nanson**, **Baldwin** and
**Rouse** are recursive elimination methods that use Borda. At each step, Nanson
eliminates all candidates that have a negative Borda score, then calculates new Borda scores based on the uneliminated
candidates and repeats. Baldwin eliminates only the candidate with the single lowest Borda score before
recalculating. Rouse is like Baldwin with an extra step of recursion: it successively eliminates the candidate with
the *highest* Borda score until one is left, then it eliminates that candidate from the original list; this step
is repeated until one is left. For example, given the votes

14:Jana>Kurt>Lisa |

7:Kurt>Lisa>Jana |

11:Lisa>Jana>Kurt |

Nanson would eliminate Kurt and Lisa, who have Borda scores of -8 and -6, making Jana the winner. Baldwin would eliminate only Kurt at first; recalculating the Borda scores without Kurt gives Jana -4 and Lisa 4, so Lisa would win. Rouse would excuse Jana from being eliminated first since she has the best Borda score; without Jana, Kurt would beat Lisa 10 to -10, so Lisa would be eliminated first. With Lisa gone, Jana and Kurt would have Borda scores of 18 and -18, so Rouse would pick Jana as the winner.

**Hare**, **Carey** and
**Coombs** are also elimination methods. Hare is the original name of the
infamous Instant Runoff Voting, which seems to many the most intuitive ranked-ballot extension of our current plurality
system. At first only the first-rank votes are counted. The candidate with the smallest first-rank total is
eliminated from the ballots, possibly adding to other candidates’ first-rank totals. This step is repeated
until a winner is left. Carey (a generalization of Craig Carey’s three-candidate IFPP method) has a similar
approach, but instead of eliminating only one candidate per step, it eliminates all candidates with below-average
first-rank totals. Coombs is much like Hare in reverse: The candidate with the largest last-rank total is
eliminated. The last-rank totals are recalculated and the step repeated until only one remains. For example,
in the election

9:Katy>Luke>Mary |

4:Luke>Mary>Katy |

6:Mary>Luke>Katy |

Hare eliminates Luke (with a first-rank total of 4) and chooses Mary over Katy (10 to 9), while Carey eliminates both Luke and Mary in the first step (since the average first-rank total is 6.333) and picks Katy. Coombs eliminates Katy and chooses Luke over Mary. Notice that under Hare, the Katy-first voters are punished for voting sincerely; they could have prevented their last choice from winning by insincerely voting Luke>Katy>Mary instead. Also, under Carey, the Luke- and Mary-first voters are punished for voting sincerely; they would be better served by deciding collectively which candidate to support, thus keeping Katy from winning, much as in our current plurality system. Of all the methods, Hare, Carey and Coombs consider the smallest amount of ballot information at any one time.

**Bucklin** tries to find a majority for some candidate by counting only the
first-place votes. If no candidate has a majority of votes, all second-place votes are included in the count, then
third-place, etc., until some candidate has more than half the number of voters. Consider the following election:

7:Mark>Nell>Owen |

2:Nell>Mark>Owen |

3:Owen>Mark>Nell |

5:Owen>Nell>Mark |

No candidate has a majority of first-place votes, so second-place votes are added to the counts and Nell wins with 14 votes. Note that Mark’s 12 votes also constitute a majority at that point, but Nell is still the sole Bucklin winner (even though Mark would have beaten Nell in a one-on-one race).

Probably the most desirable property of a method is resistance to manipulation; as often as possible, the best strategic vote should be a sincere vote. (Gibbard and Satterthwaite proved that no ranked-ballot method can be strategy-free in all situations, but insincere voting can be made difficult.) The above methods are often easy for voters to manipulate when there are more than two candidates, but with only two candidates the best voting strategy is always sincere. This insight suggests a new approach: comparing candidates one-on-one, that is, treating the multicandidate election as a series of two-candidate elections. The rest of the methods use a pairwise matrix, calculated in one pass through the ranked ballots, that tells how often each candidate was ranked over each other candidate (tied ranks are counted as half-votes for each side). For example, consider the following five-candidate election:

98:Abby>Cora>Erin>Dave>Brad |

64:Brad>Abby>Erin>Cora>Dave |

12:Brad>Abby>Erin>Dave>Cora |

98:Brad>Erin>Abby>Cora>Dave |

13:Brad>Erin>Abby>Dave>Cora |

125:Brad>Erin>Dave>Abby>Cora |

124:Cora>Abby>Erin>Dave>Brad |

76:Cora>Erin>Abby>Dave>Brad |

21:Dave>Abby>Brad>Erin>Cora |

30:Dave>Brad>Abby>Erin>Cora |

98:Dave>Brad>Erin>Cora>Abby |

139:Dave>Cora>Abby>Brad>Erin |

23:Dave>Cora>Brad>Abby>Erin |

The corresponding pairwise matrix is

against | ||||||
---|---|---|---|---|---|---|

Abby | Brad | Cora | Dave | Erin | ||

for | Abby | 458 | 461 | 485 | 511 | |

Brad | 463 | 461 | 312 | 623 | ||

Cora | 460 | 460 | 460 | 460 | ||

Dave | 436 | 609 | 461 | 311 | ||

Erin | 410 | 298 | 461 | 610 |

So 609 voters preferred Dave to Brad and 312 preferred Brad to Dave. (The winning total of each one-on-one matchup is emphasized.)

Very often, there is a candidate who wins all of his one-on-one matchups, called the Condorcet winner. All of the following pairwise methods (plus Nanson, Baldwin and Rouse from above) choose the Condorcet winner when one exists. The example above has no Condorcet winner, and the following methods pick different winners.

**Black** chooses the Condorcet winner when one exists and the Borda winner
otherwise. Abby would win the above election using Black since she is the Borda winner
and no Condorcet winner exists.

**Copeland** counts pairwise victories for each candidate, counting a pairwise tie as
half a victory, and picks the candidate with the largest result. For the above example, Copeland would pick
Abby and Brad as winners, each with a score of 3. Copeland
ignores the relative strengths of the pairwise comparisons and is indecisive relatively often.

**Small** calculates scores for the candidates exactly like Copeland. If more than
one candidate ties with the best Copeland score, the other candidates are eliminated and the scores are recalculated;
This step is repeated until no more candidates can be eliminated. In the above example,
Abby and Brad tie with a Copeland score of 3, so
Cora, Dave and Erin are eliminated, then
Brad wins over Abby 1 to 0.

**Dodgson** simply sums each candidate’s margins of defeat and chooses the
candidate with the smallest sum. In a way, it picks the candidate “closest” to being a Condorcet
winner. The Dodgson winner for the above example is Cora, with a defeat-margin sum of
4. Notice that although Cora is a Condorcet loser, that is, she loses all of her
one-on-one matchups, she loses them by the smallest of margins. Dodgson can also be thought of as a sort of
compromise between Copeland and Simpson (below).

**Simpson** picks the candidate with the smallest maximum pairwise defeat. The
Simpson winner for the above example is also Cora; Cora’s
largest defeat is by only one vote, and the other candidates have greater defeats. Simpson can be thought of as
successively ignoring the smallest defeat until one candidate is unbeaten.

**Raynaud** is an elimination method based on Simpson. The candidate with the
largest single pairwise defeat among remaining candidates is eliminated until only one remains. With her 623
defeat, Erin is the first to be eliminated in the example above; Brad
is next, then Dave and Cora, so Abby is the
winner.

**Schulze** uses the concept of beatpaths to resolve pairwise cycles. A beatpath
is a sequence of pairwise victories linking one candidate to another.
Abby>Erin>Dave>Brad is an example from the above election; it’s formed from the
victories Abby>Erin, Erin>Dave and
Dave>Brad. Each beatpath has a strength equal to the strength of the weakest victory
it includes, so Abby>Erin>Dave>Brad has a strength of 511, the strength of the
Abby>Erin victory. For each pair of candidates, the strengths of the strongest
beatpaths each way are calculated:

Abby>Erin>Dave>Brad | 511 | 463 | Brad>Abby |

Abby>Cora | 461 | 460 | Cora>Abby |

Abby>Erin>Dave | 511 | 463 | Dave>Brad>Abby |

Abby>Erin | 511 | 463 | Erin>Dave>Brad>Abby |

Brad>Cora | 461 | 460 | Cora>Brad |

Brad>Erin>Dave | 610 | 609 | Dave>Brad |

Brad>Erin | 623 | 609 | Erin>Dave>Brad |

Cora>Dave | 460 | 461 | Dave>Cora |

Cora>Erin | 460 | 461 | Erin>Cora |

Dave>Brad>Erin | 609 | 610 | Erin>Dave |

The candidate who always has a stronger beatpath to another than the other has to her, in this case Abby (511-463 over Brad, Dave and Erin, and 461-460 over Cora), is chosen as the Schulze winner. Schulze usually agrees with Simpson, but Schulze avoids Simpson’s paradoxical results as in this example.

**Tideman** orders the pairwise victories from strongest to weakest and locks them in
order, ignoring those that contradict stronger victories. In the above example, the
Brad>Erin victory is strongest and is locked. Erin>Dave
is the next to be locked. Dave>Brad is the next-strongest victory, but it’s
skipped since it would contradict Brad>Erin and Erin>Dave
(which imply Brad>Dave). The other victories can all be locked in order, generating
the ordering Brad>Abby>Erin>Dave>Cora; Tideman would pick
Brad as the winner since he has locked defeats against all other candidates. Note that
Copeland, Schulze and Tideman find Cora to be the worst candidate, but Dodgson and Simpson pick
Cora as the winner. (Dodgson and Simpson very rarely choose a Condorcet loser, as in this
example, but Copeland, Schulze and Tideman never do.)

Unavoidably, these methods will sometimes be unable to choose only one candidate. For example, given the election

25:Ryan>Sara>Todd |

25:Ryan>Todd>Sara |

25:Sara>Ryan>Todd |

25:Sara>Todd>Ryan |

no unbiased method would be able to choose one candidate decisively. A tiebreaking ranking must be used to decide between Ryan and Sara. A ballot can be chosen randomly to provide the tiebreaking ranking. The methods vary widely in their decisiveness and thus in their reliance on tiebreakers: Black is the most decisive, whereas Copeland is the least, having to rely on a tiebreaker much more frequently.

Three more procedures, **Smith**, **Schwartz** and
**Landau**, are so indecisive that they’re usually thought of as producing sets
of candidates rather than winners. . . .

Back to Ranked-ballot voting methods

Robert Hampton “Honky” LeGrand III ’98