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 (elimination-round-specific 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 pairwise losing candidates. This addition reduces the failure rate for the Independence of Irrelevant Alternatives (IIA), which is the kind of failure that occurred in the 2009 mayoral election in Burlington, Vermont. This check for pairwise losing candidates considers all the marks on all the ballots, which contrasts with IRV, which does not consider all the marks on all the ballots.
This method further modifies simple IRV by specifying how to handle ballots on which the voter has marked more than one candidate at the same ranking level.
The RCIPE STV method extends the single-winner RCIPE method to fill multiple equivalent seats.
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. Very 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. Instead, the RCIPE STV version should be used for elections that fill multiple seats, such as on a non-partisan city council or a dual-member legislative district.
Ballot Robustness
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.
The choice of how to handle a ballot on which a voter does not mark any ovals for a candidate depends on how write-in candidates are handled. If write-in candidates are not allowed, an unmarked candidate can be ranked at the ranking level below the lowest ranking level shown on the ballot. If write-in candidates are allowed, an unmarked candidate can be ranked at the lowest ranking level shown on the ballot, and that level also would be used for a write-in candidate whose name does not appear on that ballot.
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
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 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).
RCIPE STV
The RCIPE method can be extended to elect multiple candidates, such as when electing non-partisan members of a city council, or when electing two (or more) representatives from the same district.
The RCIPE STV method modifies the Single Transferable Vote (STV) method in the following ways:
- During each round of counting either one candidate is elected, or one candidate is eliminated, but not both in the same round.
- At the beginning of each counting round, the transfer count for each candidate is reset to zero.
- At the beginning of all the counting rounds, each ballot has an influence amount equal to one vote. During later counting rounds a ballot can have a reduced influence amount that is expressed as a decimal number that can range from zero up to one. If a law makes it illegal for a ballot to have an influence amount other than zero or one, the counting process can be modified as described below.
- When counting a ballot, the ballot's influence amount is added to the transfer count of the candidate who is ranked highest after ignoring the marks for already-elected and already-eliminated candidates.
- If a ballot ranks two or more candidates at the same preference level, and if there are not any remaining (not-yet-elected and not-yet-eliminated) candidates ranked higher on this ballot, then this ballot's influence amount is equally split among the remaining candidates who are ranked at that shared preference level. For example, if a ballot ranks candidate A highest, and ranks candidates B, C, and D at the next-highest level, and if candidates A and B have been elected or eliminated, then half of this ballot's influence amount transfers to candidate C and the other half transfers to candidate D.
- At the end of a counting round, the candidate with the highest transfer count is elected if that candidate's transfer count equals or exceeds the required quota count. The quota count is different for each counting round. Any reasonable formula for quota counts can be used.
- If all the candidates have transfer counts that are less than the quota number, and if there is a pairwise losing candidate during that round, the pairwise losing candidate is eliminated. During pairwise counting all the ballots are counted, but some ballots can have reduced influence.
- If a counting round does not elect a candidate and there is no pairwise losing candidate, then the candidate with the lowest transfer count is eliminated.
- When a candidate is elected, all the ballots that contribute full (not-reduced) influence to the candidate's transfer count have their influence reduced to the candidate's excess transfer count divided by the transfer count, where the excess transfer count equals the candidate's transfer count minus the quota count. Of course decimal calculation results are rounded down, not up.
- When a candidate is elected, all the ballots that contribute reduced influence (less than full influence) to that candidate's transfer count are reduced to have zero influence in future counting rounds.
If decimal influence amounts are not allowed, the following modifications can be used:
- Instead of giving a single reduced influence amount to ballots that support an elected candidate, some of the supporting ballots are given full influence and others are given zero influence.
- Selecting which ballots get full influence and which ballots get zero influence should make use of information about how many ballots share the same marking pattern. For example, if 100 ballots have the same marking pattern and the decimal calculation method would reduce their influence to 0.8 of a vote each, then selecting any 80 of these ballots to get full influence and giving the remaining 20 ballots zero influence produces the same result as the decimal approach.
- Ballots that have unique or uncommon marking patterns must be selected semi-randomly, yet the total number of supporting ballots getting zero influence must equal the quota count, and the total number of ballots getting full influence must equal the elected-candidate's transfer count minus the quota count.
- Instead of using the same semi-random selections from one counting round to the next, the semi-random selection process should be done for each counting round. This approach makes it unlikely that a specific ballot will get zero influence significantly more often than any other specific ballot that has similar markings.
- Ballots on which two or more remaining candidates share the same highest ranking level are distributed almost the same as in the decimal calculation method. The significant difference is that ballots with zero influence (during that counting round) are not distributed among the shared-level candidates.