Ranked Robin

Ranked Robin is a Condorcet method focused on the presentation of the results such that everyday voters can understand them without extensive education. 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. A strict series of tie-breaking mechanics are defined.

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 ranked 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 other candidates.

Tie-breaking methods

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 and eliminate all other candidates. For each finalist, subtract the number of ballots on which they lost to each other finalist from the number of ballots on which they beat each other finalist. The finalist with the greatest total difference is elected. For example, let $A_{w}$ be the number of ballots on which finalist ${\textstyle A}$ beats 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 ballots on which ${\textstyle A}$ loses to each other finalist (equivalent to the sum of the values in '${\textstyle A}$s column in 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.

2^nd Degree: For each tied finalist, subtract the number of ballots on which they lost to each other candidate (including eliminated candidates) from the number of ballots on which they beat each other candidate (including eliminated candidates). 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 Level 2, 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 ballots on which the losing candidate was preferred over the winning candidate from the number of ballots on which the winning candidate was preferred over the losing 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

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	52	188
Bianca over	29	—	35	31	46	47	188
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	35	35	28	—	138
Column total (votes against)	149	149	169	156	178	224	1025 - 1025 = 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 $188-149=39$).

3^rd Degree: Ava and Bianca tie for the least losing (and winning) votes between them, 149 (and 188).

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 voters who prefer Ava over Deegan and the number of voters who prefer 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. If there is no Condorcet Winner but a single candidate wins without any tie-breaker, show how many matchups each candidate won in addition to each of the winner's pairwise matchups. These two scenarios will cover the vast majority of real-world elections.

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

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: -8% points

Ava vs. Cedric: +20% points

Ava vs. Deegan: +32% points

Ava vs. Eli: +42% points

Ava vs. Fabio: +51% points

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: +33.8% 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: +33.8% points

Cedric is elected!