Ranked Robin

Ranked Robin is a Condorcet voting method focused on the presentation of the results such that everyday voters can understand them without extensive education. Ranked Robin uses a ranked ballot. Voters are free to rank multiple candidates equally on their ballots. The candidate who wins the most head-to-head matchups against other candidates is elected, much like a round-robin tournament. In the notation typically used on the EM-list, Ranked Robin is roughly "Copeland//Borda" with the addition of tiebreakers.

History

Ranked Robin was invented by Sass on 30 September 2021 and named by Sara Wolk on 7 November 2021. As an enthusiast of cardinal voting methods and a strong advocate for voter empowerment, Sass saw a timely need for a sufficiently-accurate ranked voting method that was on par with the simplicity of voting methods like STAR Voting and even Approval Voting, particularly in the United States. Ranked Robin is nearly identical to the earliest known Condorcet method, invented by Ramon Llull in his 1299 treatise Ars Electionis, which was similarly replicated by Marquis de Condorcet centuries later, and then again by Arthur Herbert Copeland. A mathematically identical method to Ranked Robin including the first tie-breaking mechanic was described by Partha Dasgupta and Eric Maskin in 2004. The primary innovation of Ranked Robin is the reduction and formatting of results in such a way that they are palatable to a general audience, as a full preference matrix can be overwhelming for most voters. This innovation can likely be adapted to simplify the results of other voting methods that use pairwise counting, particularly those that first restrict the set of winners such as Smith-efficient voting methods.

Balloting

Voters may rank as many candidates as they would like. Voters are free to rank multiple candidates equally. Skipped ranks are ignored and will neither hurt nor help a voter's vote. All candidates left unranked are considered tied for the last rank, below the lowest rank marked on a voter's ballot.

Local counting

Ranked Robin is precinct summable through the use of preference matrices. Full preference matrices can be created simply by hand if needed and then reported directly to the media and the public, allowing ballots and ballot data to remain local for recounts and risk-limiting audits without risking the threat of vote selling and voter coercion. This decentralization of tallying allows elections to remain robust against scaled election attacks, which is vital in jurisdictions that run geographically-spread or high-profile elections. In contrast, voting methods that are not precinct summable, like Ranked Choice (Instant Runoff) Voting and many expressive proportional voting methods, lose these benefits and can lead to distrust in election outcomes if fraud, attacks, or even simple mistakes happen under a centralized counting authority.

Tabulation

Elect the candidate who pairwise beats the greatest number of candidates.

Example election

8:Ava>Cedric>Deegan>Bianca>Eli

6:Ava=Bianca=Cedric>Eli>Deegan

6:Eli>Ava>Bianca=Cedric=Deegan

6:Deegan>Bianca=Cedric>Eli>Ava

4:Bianca>Ava>Eli>Deegan>Cedric

3:Eli>Deegan>Bianca=Cedric>Ava

2:Deegan=Eli>Bianca=Cedric>Ava

Create a preference matrix from the ballots.

Bold indicates wins, Italics indicates losses

# of voters who prefer	Ava	Bianca	Cedric	Deegan	Eli
Ava over	—	14	18	24	18
Bianca over	15	—	4	10	24
Cedric over	11	8	—	14	20
Deegan over	11	19	15	—	14
Eli over	17	11	15	19	—

Ava pairwise beats the greatest number of candidates, 3, so she is elected as the winner.

Tie-breaking mechanics

Frequency of ties

Almost all real-world elections using Ranked Robin will not have any ties for the winning candidate. However, ties under Ranked Robin may potentially be more common than ties under Choose-one Voting. While there are 4 degrees of tiebreakers defined, ties after the 1^st Degree tiebreaker are about as rare as ties under Choose-one Voting, and ties after the 2^nd Degree tiebreaker are much rarer than that.

Degrees of ties

If there is a tie (including Condorcet cycles), use the 1^st Degree tie-breaking method to resolve it. If there is still a tie, use the 2^nd Degree tiebreaker, and so on.

1^st Degree: Declare the tied candidates finalists. For each finalist, subtract the number of votes preferring each other finalist from the number of votes preferring them over each other finalist. The finalist with the greatest total difference is elected. For example, let $A_{w}$ be the number of votes preferring finalist ${\textstyle A}$ over each other finalist (equivalent to the sum of the values in ${\textstyle A}$'s row in a preference matrix consisting only of finalists) and let $A_{l}$ be the number of votes preferring each other finalist over ${\textstyle A}$ (equivalent to the sum of the values in ${\textstyle A}$'s column in a preference matrix consisting only of finalists); ${\textstyle A}$'s total difference is $A_{w}-A_{l}$. This is mathematically equivalent to the tournament-style of the Borda count (among only the finalists), where candidates get, per ballot, 1 point for each candidate they beat and ½ point for each candidate they tie. There are two alternate, more concise ways of describing the 1^st Degree tiebreaker.

• If there is a tie, then for each tied candidate, subtract the number of votes preferring each other tied candidate from the number of votes with the opposite preference. Elect the tied candidate with the greatest total difference.
• Among the candidates who tie for winning the most head-to-head matchups, elect the tied candidate with the best average rank.

2^nd Degree: For each tied finalist, subtract the number of votes preferring each other candidate (including candidates who are not finalists) from the number of votes preferring them over each other candidate (including candidates who are not finalists). The tied finalist with the greatest total difference is elected.

3^rd Degree: It is highly unlikely that there will still be a tie after the 2^nd Degree tiebreaker, but if there is, it is not recommended to use tie-breaking methods beyond the 2^nd Degree tiebreaker for government elections as voter trust may be shaken more by using the 3^rd Degree tiebreaker and beyond than drawing lots or hosting another election. In the event that there is a tie after the 2^nd Degree tiebreaker, the differences for the tied candidates will be the same, but the values used to calculate them will likely be different. Elect the tied candidate whose values are closest to the tied differences. For example, if ${\textstyle A}$, ${\textstyle B}$, and ${\textstyle C}$ are tied after the 2^nd Degree tiebreaker, then $A_{w}-A_{l}=B_{w}-B_{l}=C_{w}-C_{l}$ (where wins and loses are calculated across the entire field of candidates), but it's likely that $A_{w}\neq B_{w}\neq C_{w}$ (and by proxy that $A_{l}\neq B_{l}\neq C_{l}$). The tied candidate with the greatest loss margin will also have the greatest win margin, and the tied candidate with the least loss margin will have the least win margin. Elect the tied candidate with the least loss and win margins as that is the least polarizing tied candidate.

4^th Degree: If there is still a tie after the 3^rd Degree tiebreaker, it is unlikely that the 4^th Degree tiebreaker will break that tie, as it will only work if the tied candidates have matchup losses against other candidates. Find the shortest beatpath from each tied candidate to each other tied candidate. For each tied candidate, for each shortest beatpath to another tied candidate, for each pairwise victory in the beatpath, subtract the number of votes preferring the losing candidate from the number of votes preferring the winning candidate. Sum these differences within each selected beatpath to get the total strength of each selected beatpath. Sum each tied candidate's total beatpath strengths over other tied candidates. Elect the tied candidate with the greatest sum of beatpath strengths. If there are multiple shortest beatpaths from one tied candidate to another, select the one with the lowest total strength.

Example of a ballot set that requires all 4 tie-breaking degrees

10: Eli>Deegan>Ava=Cedric>Fabio

9: Bianca=Deegan>Eli>Cedric

8: Deegan>Eli>Ava=Bianca=Cedric

8: Bianca>Ava>Fabio>Cedric

8: Fabio>Cedric>Ava>Deegan>Bianca

7: Ava>Eli>Bianca>Fabio

6: Fabio>Bianca=Cedric>Ava

6: Cedric>Deegan=Eli>Ava=Bianca>Fabio

5: Deegan>Ava=Bianca>Eli>Cedric

4: Cedric>Bianca>Ava

4: Ava>Bianca=Fabio

4: Ava=Bianca>Fabio

2: Bianca=Fabio>Ava=Eli

Here's the preference matrix:

Bold indicates wins, Italics indicates losses, Underline indicates ties

# of voters who prefer	Ava	Bianca	Cedric	Deegan	Eli	Fabio	Row total (votes for)
Ava over	—	29	26	39	42	56	204
Bianca over	29	—	35	31	46	51	204
Cedric over	33	28	—	32	32	42	167
Deegan over	38	37	32	—	30	38	175
Eli over	33	31	41	19	—	45	169
Fabio over	16	24	39	39	32	—	150
Column total (votes against)	149	149	181	168	190	232	1069 - 1069 = 0

Ranked Robin: Ava and Bianca tie for pairwise beating the greatest number of other candidates, 3.

1^st Degree: Ava and Bianca tie for the greatest total difference in votes for and against other tied finalists (both $29-29=0$).

2^nd Degree: Ava and Bianca tie for the greatest total difference in votes for and against all other candidates (both $204-149=55$).

3^rd Degree: Ava and Bianca tie for the least losing (and winning) votes between them, 149 (and 204).

4^th Degree: The shortest beatpath from Ava to Bianca is Ava→Deegan→Bianca and the shortest beatpath from Bianca to Ava is Bianca→Cedric→Ava. The difference between the number of votes preferring Ava over Deegan and the number of votes preferring Deegan over Ava is $39-38=1$. From Deegan to Bianca, the difference is $37-31=6$. The sum of the differences in the beatpath from Ava to Bianca (the total beatpath strength) is $1+6=7$. From Bianca to Cedric, the difference is $35-28=7$. From Cedric to Ava, the difference is $33-26=7$. The total beatpath strength from Bianca to Ava is $7+7=14$. Bianca has the greatest (sum of) total beatpath strength(s) among tied candidates, so Bianca is elected.

Presentation of results

If there is a Condorcet Winner, then simply show each of the winner's pairwise matchups against other candidates. This can either be shown as percentage of total votes for each candidate in a given pairwise matchup, or as the percentage point difference in favor of the winner if there's a desire to show less information.

Two different ways to present the results of the same election with Condorcet Winner Ava

Ava: 54%《》Bianca: 46%

Ava: 59%《》Cedric: 39%

Ava: 63%《》Deegan: 31%

Ava: 64%《》Eli: 22%

Ava: 64%《》Fabio: 13%

Ava vs. Bianca: +8% points

Ava vs. Cedric: +20% points

Ava vs. Deegan: +32% points

Ava vs. Eli: +42% points

Ava vs. Fabio: +51% points

If there is no Condorcet Winner but a single candidate wins without any tiebreaker, show how many matchups each candidate won in addition to each of the winner's pairwise matchups.

Example of how to present the results of an election where the winner Ava is not a Condorcet Winner

Ava won 4 matchups (against Cedric, Deegan, Eli, and Fabio)

Bianca won 3 matchups (against Ava, Eli, and Fabio)

Cedric won 3 matchups (against Bianca, Eli, and Fabio)

Deegan won 3 matchups (against Bianca, Cedric, and Fabio)

Eli won 2 matchups (against Deegan and Fabio)

Fabio lost all matchups

Ava vs. Bianca: -6% points

Ava vs. Cedric: +20% points

Ava vs. Deegan: +32% points

Ava vs. Eli: +42% points

Ava vs. Fabio: +51% points

These two scenarios will cover the vast majority of real-world elections.

If there's a tie

When there is a 1^st Degree tie, it's often a Condorcet cycle (rock-paper-scissors-style tie) with more than two candidates (now finalists). In the case that there are only two finalists, present results as the above example without a Condorcet Winner, but highlight the matchup between the two finalists, which alone breaks the tie. Otherwise, present results according to the level desired as described below.

Level 1: Simply state who the winner is.

Level 2: Show which candidates are finalists (and optionally which candidates were eliminated).

Level 3: Show how many matchups each candidate won as shown above.

Level 4: Show each finalist's total difference in votes for and against other finalists. These values can be given a name like "Total Advantage".

Level 5: Show the breakdown of each finalist's total difference in votes for and against other finalists by showing the difference within each finalist matchup. These values can be given a name like "Matchup Advantage"

Level 6: Show a preference matrix that's just wins and losses (and ties).

Level 7: Show a preference matrix using percentages.

Level 8: Show the full preference matrix.

Example of showing Level 4 with 3 finalists in a Condorcet cycle

Ava, Bianca, and Cedric are finalists.

Ava's Total Advantage: -23.5% points

Bianca's Total Advantage: -3.3% points

Cedric's Total Advantage: +26.8% points

Cedric is elected!

Note that all of the Total Advantages sum to 0. This can be used to check the math performed.

Example of showing Level 5 with 3 finalists in a Condorcet cycle

Ava, Bianca, and Cedric are finalists.

Ava vs. Bianca: +22.1% points

Ava vs. Cedric: -45.6% points

Ava's Total Advantage: -23.5% points

Bianca vs. Cedric: +18.8% points

Bianca vs. Ava: -22.1% points

Bianca's Total Advantage: -3.3% points

Cedric vs. Ava: +45.6% points

Cedric vs. Bianca: -18.8% points

Cedric's Total Advantage: +26.8% points

Cedric is elected!

In the rare case of a 2^nd Degree tie, if there are many candidates, it is recommended to focus on who the tied finalists are and their Total Advantages over all other candidates (which likely will not sum to 0).

Legal and economic viability

When legally defined as always reducing to a finalist set first and then electing the finalist with the greatest total difference (Total Advantage) among finalists (as described in the 1^st Degree tiebreaker), Ranked Robin always elects a majority preferred winner, arguably including in cases of 2^nd Degree ties. This legal definition does not change the outcomes of Ranked Robin. Many municipalities in the United States are subject to a majority clause in their respective state's election code, often requiring those jurisdictions to run two or more elections for certain races. Ranked Robin can satisfy many of these majority clauses in a single election, allowing municipalities to eliminate an election if so desired, helping to offset the costs of implementing Ranked Robin, typically entirely within one election cycle.

If there is only 1 finalist, then they are voted for by a majority of voters who had a preference among finalists.

If there are multiple finalists, at least 1 finalist will have a positive Total Advantage and at least 1 finalist will have a negative Total Advantage because the sum of advantages will always equal 0. Because the finalist with the greatest Total Advantage is elected, that winner is guaranteed to have a positive Total Advantage, demonstrating that among finalists, they are a majority preferred winner.

If there is a 2^nd Degree tie, all of the finalists could potentially (but rarely) have a negative Total Advantage when compared to all candidates, but it could be argued that because the finalist with the greatest Total Advantage is elected, the winner was voted for by a majority of voters who had a preference among finalists. This argument is further strengthened in the case that exactly 2 finalists experience a 2^nd Degree tie, which covers almost all cases of 2^nd Degree ties. If this argument is found not to satisfy a particular majority clause, it may be desirable to leave the 2^nd Degree tiebreaker out of the legislation and legally declare a tie in the equivalent case of a 2^nd Degree tie, which is about as rare as a tie under Choose-one Voting.

Furthermore, in most cases with only 1 finalist, including all elections with a Condorcet Winner, the winner will be majority preferred over all other candidates because the winner’s Total Advantage is positive; however, there are rare theoretical cases in which the only finalist has a negative Total Advantage over all other candidates. If the “majority preferred among finalists” argument doesn’t legally hold when there’s only 1 finalist, then this rare case could either explicitly be denoted as not electing a majority winner (thus requiring an extra election to be run), or an alternative winner could be calculated by selecting the candidate with the greatest Total Advantage among all candidates (completely ignoring any reduction to a set of finalists).

Criteria

Passed

• Unrestricted domain
• Non-imposition (a.k.a. citizen sovereignty)
• Non-dictatorship
• Homogeneity
• Condorcet criterion
• Majority criterion
• Pareto criterion (a.k.a. unanimity)
• Monotonicity criterion
• Majority loser criterion
• Condorcet loser criterion
• Reversal symmetry
• Resolvability criterion
• Polynomial time
• Mutual majority criterion
• Smith criterion
• Independence of Smith-dominated alternatives
• Independence of Irrelevant Ballots

Failed

• Participation
• Independence of irrelevant alternatives
• Consistency
• Independence of clones
• Sincere Favorite

A note on cloneproofness

Ranked Robin can fail clone independence in one of two ways: either by its Copeland component or by its Borda component.

The Copeland component fails clone independence by crowding and teaming. It can be argued that a party stands nothing to gain (or lose) by running clones as far as the crowding vulnerability is concerned, because all a candidate A can achieve by triggering a clone failure is to change the candidate from some B to some other C, which doesn't help A since A lost anyway -- unless C just happens to be closer aligned with A's position than does B. However, the teaming incentive may be more conventionally exploitable, since it directly benefits a candidate who runs clones.

The Borda component fails clone independence by teaming. If the Copeland set consists of more than one candidate, as can happen with some Condorcet cycles, then this could expose the Borda component and allow teaming to succeed. For instance, consider this pre-cloning election:

12: A>B>C>D>E>F
11: B>C>A>D>E>F
10: C>A>B>D>E>F

The Copeland set is {A,B,C}. A and B tie for Borda score, but this can be shifted in favor of A by teaming, e.g.

12: A1>A2>B>C>D>E>F
11: B>C>A1>A2>D>E>F
10: C>A1>A2>B>D>E>F

after which A wins.

Ranked Robin passes vote-splitting clone independence: cloning a candidate can't make that candidate lose.

External links

• Ranked Robin thread on r/EndFPTP (starting November 1, 2021)
• Ranked Robin thread on Voting Theory Forum (starting October 25, 2021)
• Explanation of Ranked Robin from the Equal Vote Coalition