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Ranked Robin is a [[Condorcet method|Condorcet voting method]] focused on the presentation of the results such that everyday voters can understand them without extensive education. Ranked Robin uses a [https://electowiki.org/wiki/Ballot#Ranked_ballot 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 [[w:round-robin tournament|round-robin tournament]].
== History ==
Ranked Robin was invented by [[User:Sass|Sass]] on 30 September 2021 and named by [[Sara Wolk]] on 7 November 2021. As an enthusiast of [[Cardinal voting systems|cardinal voting methods]] and a strong advocate for voter empowerment, Sass saw a timely need for a sufficiently-accurate [[Ranked voting|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''<ref name="Hagele">{{cite journal |author1=G. Hägele |author2=F. Pukelsheim |lastauthoramp=yes | title=Llull's writings on electoral systems | journal=Studia Lulliana | year=2001 | volume=41 | pages=3–38 | url=http://www.math.uni-augsburg.de/stochastik/pukelsheim/2001a.html }}</ref>, which was similarly replicated by [[w:Marquis de Condorcet|Marquis de Condorcet]] centuries later, and then again by [[w:Arthur Herbert Copeland|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<ref>{{Cite journal|last=Maskin|first=Eric|last2=Dasgupta|first2=Partha|date=2004|title=The Fairest Vote of All|url=https://scholar.harvard.edu/maskin/publications/fairest-vote-all|journal=Scientific American|volume=|issue=290|pages=64-69|via=Harvard University}}</ref>. 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 [[Pairwise comparison matrix|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|Smith-efficient voting methods]].
== Balloting ==
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== Tabulation ==
{{Definition|Elect the candidate who pairwise beats the greatest number of
====
{{Ballots|1=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 [[Pairwise comparison matrix|preference matrix]] from the ballots.
{| class="wikitable"
|+
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 [[First Past the Post electoral system|Choose-one Voting]]. While there are 4 degrees of tiebreakers defined, ties after the '''1<sup>st</sup> Degree''' tiebreaker are about as rare as ties under Choose-one Voting, and ties after the '''2<sup>nd</sup> Degree''' tiebreaker are much rarer than that.
=== Degrees of ties ===
If there is a tie (including [[Condorcet paradox|Condorcet cycles]]), use the '''1<sup>st</sup> Degree''' tie-breaking method to resolve it. If there is still a tie, use the '''2<sup>nd</sup> Degree''' tiebreaker, and so on.
'''1<sup>st</sup> Degree:''' Declare the tied candidates finalists. For each finalist, subtract the number of
* ''If there is a tie, then for each tied candidate, subtract the number of
* ''Among the candidates who tie for winning the most head-to-head matchups, elect the tied candidate with the best average rank.''
'''2<sup>nd</sup> Degree:''' For each tied finalist, subtract the number of
'''3<sup>rd</sup> Degree:''' It is highly unlikely that there will still be a tie after the '''2<sup>nd</sup> Degree''' tiebreaker, but if there is, it is not recommended to use tie-breaking methods beyond the '''2<sup>nd</sup> Degree''' tiebreaker for government elections as voter trust may be shaken more by using the '''3<sup>rd</sup> Degree''' tiebreaker and beyond than drawing lots or hosting another election. In the event that there is a tie after the '''2<sup>nd</sup> 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 <math display="inline">A</math>, <math display="inline">B
</math>, and <math display="inline">C</math> are tied after the '''2<sup>nd</sup> Degree''' tiebreaker, then <math>A_w-A_l=B_w-B_l=C_w-C_l</math> (where wins and loses are calculated across the entire field of candidates), but it's likely that <math>A_w\neq B_w\neq C_w</math> (and by proxy that <math>A_l\neq B_l\neq C_l</math>). 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<sup>th</sup> Degree:''' If there is still a tie after the '''3<sup>rd</sup> Degree''' tiebreaker, it is unlikely that the '''4<sup>th</sup> 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
==== Example of a ballot set that requires all 4 tie-breaking degrees
{{Ballots|1=10: Eli>Deegan>Ava=Cedric>Fabio
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4: Ava>Bianca=Fabio
4: Ava=Bianca>Fabio
2: Bianca=Fabio>Ava=Eli}}Here's the preference matrix:
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|'''39'''
|'''42'''
|'''
|'''
|-
|'''Bianca over'''
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|''31''
|'''46'''
|'''
|'''
|-
|'''Cedric over'''
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|—
|'''30'''
|
|'''175'''
|-
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|''16''
|''24''
|''
|''
|''
|—
|'''
|-
|''Column total (votes against)''
|''149''
|''149''
|''
|''
|''
|''
|'''
|}
'''Ranked Robin:''' Ava and Bianca tie for pairwise beating the greatest number of other candidates, '''3'''.
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'''1<sup>st</sup> Degree:''' Ava and Bianca tie for the greatest total difference in votes for and against other tied finalists (both <math>29-29=0</math>).
'''2<sup>nd</sup> Degree:''' Ava and Bianca tie for the greatest total difference in votes for and against all other candidates (both <math>
'''3<sup>rd</sup> Degree:''' Ava and Bianca tie for the least ''losing'' (and '''winning''') votes between them, ''149'' (and '''
'''4<sup>th</sup> 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
== Presentation of results ==
If there is a [[Condorcet winner criterion|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
<blockquote>
Ava: 54%《》Bianca: 46%
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</blockquote><blockquote>
Ava vs. Bianca: +8% points
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Ava vs. Fabio: +51% points
</blockquote>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
<blockquote>
Ava won 4 matchups (against Cedric, Deegan, Eli, and Fabio)
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</blockquote><blockquote>
Ava vs. Bianca: -6% points
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Ava vs. Fabio: +51% points
</blockquote>These two scenarios will cover the vast majority of real-world elections.
=== If there's a tie ===
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'''Level 8:''' Show the full preference matrix.
==== Example of showing Level 4 with 3 finalists in a Condorcet cycle
<blockquote>
Ava, Bianca, and Cedric are finalists.
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</blockquote>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
<blockquote>
Ava, Bianca, and Cedric are finalists.
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Cedric is elected!
</blockquote>In the rare case of a '''2<sup>nd</sup> 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<sup>st</sup> Degree''' tiebreaker), Ranked Robin always elects a majority preferred winner, arguably including in cases of '''2<sup>nd</sup> Degree''' ties. This legal definition does not change the outcomes of Ranked Robin. Many municipalities in the [[United States]] are subject to a majority
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
If there is a '''2<sup>nd</sup> Degree''' tie, all of the finalists could potentially (but rarely) have a negative
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
* [https://en.wikipedia.org/wiki/Unrestricted_domain Unrestricted domain]
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* [[Independence of Irrelevant Ballots]]
=== Failed
* [[Participation criterion|Participation]]
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=== 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 [[w:Independence of clones criterion#Copeland|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:
{{ballots|
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.
{{ballots|
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 ==
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* [https://www.reddit.com/r/EndFPTP/comments/qkamzm/new_condorcet_method_that_doesnt_require_a/ Ranked Robin thread on r/EndFPTP] (starting November 1, 2021)
* [https://www.votingtheory.org/forum/topic/136/new-simple-condorcet-method-basically-copeland-margins Ranked Robin thread on Voting Theory Forum] (starting October 25, 2021)
* [https://www.equal.vote/ranked_robin Explanation of Ranked Robin from the Equal Vote Coalition]
== References ==
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[[Category:Condorcet methods]]
[[Category:Smith-efficient Condorcet methods]]
[[Category:Condorcet-related concepts]]
|