Ranked Choice Including Pairwise Elimination
Ranked Choice Including Pairwise Elimination (abbreviated as RCIPE which is pronounced "recipe") is an election votecounting method that uses ranked ballots and eliminates pairwise losing candidates (Condorcet losers) when they occur, and otherwise eliminates the candidate who currently has the fewest topchoice counts.
This method modifies instant runoff voting (IRV) by adding the elimination of Condorcet losers. This addition would have prevented the failure of instantrunoff 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 (notyeteliminated) 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 leastpopular candidate. The Condorcet loser is the candidate who would lose every oneonone contest against each and every other candidate.
If an elimination round does not have a Condorcet loser, the candidate who has the fewest topchoice counts is eliminated. A candidate's topchoice count is the count of how many ballots rank that candidate highest compared to the other remaining candidates.
Unlike instantrunoff 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 runnerup 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 runnerup candidate would move to the runoff or general election along with the primaryelection winner. Small political parties would not qualify to move their runnerup candidate to the runoff or general election.
Importantly, the runnerup 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 multimember 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 two or more remaining highestranked candidates, the 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.
Also to avoid spoiled ballots, if a voter marks more than one ranking level for the same candidate, only the highestmarked 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 writein candidates whose names do not appear on the ballot being counted.
Tie breaker
If two or more candidates have the same smallest topchoice 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 notyeteliminated tied candidates who are ranked above that candidate, and adding those 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 notyeteliminated tied candidates who are ranked above that candidate, and adding those 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 nottied candidates is ignored when resolving ties.
If there is also a tie for the smallest pairwise support count, then another tiebreaking method is needed to identify which of the stilltied candidates to eliminate.
Example
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) 





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 topchoice 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 topchoice votes.
The only remaining candidate is Nashville, so it is declared the winner.
Chattanooga is the runnerup 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 ArrowRaynaud 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(N^{2}).