Ranked Choice Including Pairwise Elimination

Ranked Choice Including Pairwise Elimination (abbreviated as RCIPE which is pronounced "recipe") is an election vote-counting method that uses ranked ballots and eliminates pairwise losing candidates (Condorcet losers) when they occur, and otherwise eliminates the candidate who currently has the smallest top-choice count.

This method modifies instant runoff voting (IRV) by adding the elimination of Condorcet losers. This addition would have prevented the failure of instant-runoff voting to elect the most popular candidate in the 2009 mayoral election in Burlington, Vermont.

Description

Voters rank the candidates using as many ranking levels as there are candidates, or at least five ranking levels. Using fewer ranking levels than candidates is useful on paper ballots where space is limited, and in elections where there are numerous candidates who are unlikely to win.

This method eliminates one candidate at a time. The final remaining (not-yet-eliminated) candidate is declared to be the winner.

If an elimination round has a Condorcet loser (a pairwise losing candidate), this candidate is eliminated as the least-popular candidate. The Condorcet loser is the candidate who would lose every one-on-one contest against each and every other candidate.

If an elimination round does not have a Condorcet loser, the candidate who has the smallest top-choice count is eliminated. A candidate's top-choice count is the count of how many ballots rank that candidate highest compared to the other remaining candidates.

Unlike instant-runoff voting, which ends when a candidate reaches majority support, the eliminations continue until only a single candidate remains.

The last candidate to be eliminated is the runner-up candidate. If this counting method is used in the primary election of a major political party, and if the runoff or "general" election is counted in a way that is not vulnerable to vote splitting, then ideally the runner-up candidate would move to the runoff or general election along with the primary-election winner. Small political parties would not qualify to move their runner-up candidate to the runoff or general election.

Importantly, the runner-up candidate does not deserve to win any kind of elected seat. This means this method is not suitable for filling multiple seats, such as on a city council or in a multi-member district.

To avoid spoiled ballots in elections where the voter uses a pen or marker to mark their paper ballot, more than one candidate can be marked at the same ranking level. When an elimination round involves a ballot that has two or more remaining highest-ranked candidates, that ballot's single vote is split equally among these candidates. This splitting of a single vote can be done using fractions or decimal numbers that do not exceed a total of one vote per ballot. If a law does not permit the use of fractions or decimal numbers, the ballots that have the same shared ranking can be distributed uniformly among the same-ranked candidates, such as alternating which candidate gets each successive ballot on which the same two candidates are highest-ranked at the same level. Regardless of which method is used, each elimination round re-calculates which ballots support which candidates.

Also to avoid spoiled ballots, if a voter marks more than one ranking level for the same candidate, only the highest-marked ranking level is used during counting.

If the voter does not mark any ovals for a candidate, that candidate is ranked at the lowest ranking level, as if the voter marked the oval for the lowest ranking level.

The ranking level below the lowest ranking level is reserved for write-in candidates whose names do not appear on the ballot being counted.

Tie breaker

If two or more candidates have the same smallest top-choice count, this tie is resolved by eliminating the candidate with the largest pairwise opposition count, which is determined by counting on each ballot the number of not-yet-eliminated tied candidates who are ranked above that candidate, and adding these numbers across all the ballots.

If there is a tie for the largest pairwise opposition count, this tie is resolved by eliminating the candidate with the smallest pairwise support count, which is determined by counting on each ballot the number of not-yet-eliminated tied candidates who are ranked above that candidate, and adding these numbers across all the ballots.

Note that the pairwise opposition count and pairwise support count are calculated using only the candidates who are currently tied. This means that ballot information about eliminated candidates and not-tied candidates is ignored when resolving ties.

If there is also a tie for the smallest pairwise support count, then another tie-breaking method is needed to identify which of the still-tied candidates to eliminate.

Example

Tennessee's four cities are spread throughout the state

Imagine that Tennessee is having an election on the location of its capital. The population of Tennessee is concentrated around its four major cities, which are spread throughout the state. For this example, suppose that the entire electorate lives in these four cities, and that everyone wants to live as near the capital as possible.

The candidates for the capital are:

• Memphis, the state's largest city, with 42% of the voters, but located far from the other cities
• Nashville, with 26% of the voters, near the center of Tennessee
• Knoxville, with 17% of the voters
• Chattanooga, with 15% of the voters

The preferences of the voters would be divided like this:

42% of voters (close to Memphis)	26% of voters (close to Nashville)	15% of voters (close to Chattanooga)	17% of voters (close to Knoxville)
Memphis Nashville Chattanooga Knoxville	Nashville Chattanooga Knoxville Memphis	Chattanooga Knoxville Nashville Memphis	Knoxville Chattanooga Nashville Memphis

These ballot preferences are converted into pairwise counts and displayed in the following tally table.

All possible pairs of choice names	Number of votes with indicated preference
Prefer X over Y	Equal preference	Prefer Y over X
X = Memphis Y = Nashville	42%	0	58%
X = Memphis Y = Chattanooga	42%	0	58%
X = Memphis Y = Knoxville	42%	0	58%
X = Nashville Y = Chattanooga	68%	0	32%
X = Nashville Y = Knoxville	68%	0	32%
X = Chattanooga Y = Knoxville	83%	0	17%

In the first elimination round, Memphis is eliminated because it is the pairwise loser. This means Memphis loses the pairwise contest against Nashville (42% to 58%), and Memphis loses the pairwise contest against Chattanooga (42% to 58%), and Memphis loses the pairwise contest against Knoxville (42% to 58%).

If there had not been a pairwise loser, Chattanooga would have been eliminated (instead of Memphis) because it has the smallest number of ballots that rank Chattanooga as the first choice.

In the second elimination round, Knoxville is eliminated because it is the pairwise loser. This means Knoxville loses the pairwise contest against Nashville (32% to 68%), and Knoxville loses the pairwise contest against Chattanooga (32% to 68%).

If there had not been a Condorcet loser, Nashville would have been eliminated. The voters who marked Chattanooga as their first choice marked Knoxville as their second choice so in this elimination round Knoxville has 32% support (15% plus 17%). Memphis continues to have 42% support, and Nashville continues to have 26% support, which leaves Nashville with the smallest top-choice support.

In the third and final elimination round, Chattanooga is eliminated because it is the pairwise loser. Specifically Chattanooga loses its pairwise contest against Nashville (17% to 83%).

When there are only two candidates, and they are not tied, the Condorcet loser is always the same as the candidate who has the fewest top-choice votes.

The only remaining candidate is Nashville, so it is declared the winner.

Chattanooga is the runner-up candidate because it was the last to be eliminated.

Mathematical criteria

The frequency with which this method passes or fails each of the following criteria have not been estimated. This method is similar to the Arrow-Raynaud method so their frequency values are likely to be similar.

This method always passes the following criteria.

• Condorcet loser: pass
• Resolvable: pass
• Polytime: pass

This method sometimes fails the following criteria.

• Condorcet: fail
• Majority: fail
• Majority loser: fail
• Mutual majority: fail
• Smith/ISDA: fail
• LIIA: fail
• IIA: fail
• Cloneproof: fail
• Monotone: fail
• Consistency: fail
• Reversal symmetry: fail
• Later no harm: fail
• Later no help: fail
• Burying: fail
• Participation: fail
• No favorite betrayal: fail

It is summable with O(N²).