User:BetterVotingAdvocacy/Big page of ideas

If you're going to use Category:Pairwise counting-based voting methods, I encourage considering using Pairwise counting#Negative vote-counting approach and Rated pairwise preference ballot.

This page is a loosely organized page for various ideas that might be of interest. I try to put more tangential ideas somewhere in the "Miscellaneous" section.

Images

I've made a number of images and GIFs on voting theory. See Special:ListFiles/BetterVotingAdvocacy.

Also see https://imgur.com/gallery/r8itfLp.

Description of some common Condorcet methods

Here are some how-to guides on using different voting methods for your own elections.

A slightly modified version of Schulze:

In general, any Condorcet method can be done with either rated or ranked ballots. That means you give every voter the ability to number all of the candidates either on a scale or from 1st to last.

When trying to find the result, start by, for each ballot, adding one vote in support of whichever candidate the voter preferred in every possible pairwise matchup. So, for example, a voter who voted A>B>C is treated as giving one vote to A>B, one to B>C, and one to A>C. This means that you'll have, for every pair of candidates, two values to store: the number of voters who prefer the first over the second, and vice versa.

Once you've done this, you'll want to find the margin of victory for every matchup. This tells you which candidate got more votes than the other in that matchup. Once you've found the margin, look for a single candidate who beats all others. If there is none, look for two candidates who beat all others (except possibly each other). Keep looking at more and more candidates until you find a beats-all group. This is the Smith set.

Now, eliminate everyone not in the Smith set. If there was only one candidate in the Smith set, they win, and if there are multiple candidates who tied (i.e. they don't beat each other), then they're tied. Otherwise, decide on your preferred measure of defeat strength (winning votes will be used for this example). Find the candidate who got the fewest votes in their favor in one of their victories against another candidate. Now, ignore this victory, treating it instead as a victory for both candidates. Now, repeat the process of finding the Smith set; you may be lucky and find that the group can be shrunken because some candidate who earlier lost to another now "wins" after ignoring their loss. Repeat this process until the winner is found.

Smith//Score:

Find the Smith set. Whoever has the most points in it wins.

Smith//Approval:

Find the Smith set. Whoever has the most approvals in it wins.

Finding the CW faster

Some examples of finding the Condorcet winner faster:

Take this example: http://web.math.princeton.edu/math_alive/6/Lab1/Condorcet.html

First, it can be observed that there are 55 voters, with all but 18 of them ranking Molson 5th (last), so ISDA allows us to say that Molson can be eliminated because a (mutual) majority prefer anyone but him. So the simplified preferences are:

18 S>MB>G>K
12: K>MB>S>G
10: G>K>MB>S
9: S>G>MB>K
4: MB>K>S>G
2: MB>G>S>K

It should be noted that Samuel Adams has 27 out of 55 1st choice votes, almost a majority, so it'd be prudent to check his matchups first.

S vs. MB: 27 vs. 28, a loss for Samuel Adams. So now it might be best to check MB's matchups.

MB vs. G: 30 in the top 2 lines alone, a majority, so MB is known to win just with that information alone.

MB vs. K: 18 + 9 + 4 = 31, a majority, so MB beats K. So Meister Braum is the Condorcet winner in this example.

A quick note: It can be seen that the top two lines rank only S or K above MB, and the top two lines are a majority. So one way to quickly figure out who won would've been to compare S and K to MB, and if MB pairwise beats them, then it is guaranteed that MB won, because when ignoring S and K, a majority prefer MB over all others. This is a demonstration of how Condorcet methods' attempts to make majority rule maximally comply with IIA helps in analyzing election scenarios.

Miscellaneous

One criterion that might be good for PR methods is the "Duplicated Quotas" criterion: if a PR method elects some candidate in the single-winner case, and the ballots are "duplicated" N times, then if N+1 seats are to be filled, the duplicated winner should win. Example for Condorcet PR:

2 A>B

1 B

2 C>B

2 D>E

1 E

2 F>E

B and E are the duplicated winners, since they're Condorcet winners when ignoring all of the voters who ranked the other.

Some links:

Discussion on how to use Approval and Condorcet for legislative votes, including the ideas of "approval threshold" (how many people need to support an action being taken for it to happen; by default, it's a majority) and "concession threshold" (if an idea has a certain significant amount of support, you can indicate that you will switch from opposing to supporting it): https://www.reddit.com/r/EndFPTP/comments/futa9q/comment/fmmfulq?context=1

One way to think about how large the difference in required markings can be between pairwise counting and IRV is to consider an election scenario with a candidate who is a majority's 1st choice, where voters rank every candidate. In both Condorcet and RCV, the majority's 1st choice will win, but in RCV, only the 1st choices of each voter need be counted, resulting in one mark per ballot, while with pairwise counting, supposing there are 10 candidates, at a minimum 25 marks will need to be made per ballot (see Negative vote-counting approach for pairwise counting#Semi-negative counting procedure). However, pairwise counting will give more detailed information as to how much support each candidate has overall.

One way to explain normalization for scored ballots is that a voter attempts to put the maximal margin between every pair of candidates, while still preserving their relative strength of preference between each pair of candidates. Specifically, this means that you try to give your favorite the maximal margin (1 vote i.e. [max score - min score] points) against your least favorite.

Condorcet

It is arguable whether a voter can have maximal preferences between more than one transitive pair of candidates. Utilitarianism says if you maximally prefer A to B, then you must not prefer B to C, while Condorcet says you can for as many pairs as you like. An interesting method that goes one step away from utilitarianism towards Condorcet is "3-slot/tiered Smith//Approval": the voter may rank each candidate either 1st, 2nd, or last, and may approve either only their 1st choices, or also their 2nd choices. With only 1 tier, this would reduce to regular Approval voting.

Here are some facts about how various voting methods interact with Condorcet winners:

FPTP and any majority criterion-satisfying voting method will elect a CW with a majority of 1st choices
IRV and any dominant mutual third-passing system will elect a CW with over 1/3rd of 1st choices (or who can attain that after eliminations)
Any Category:Runoff-based voting methods will elect a CW who is in the runoff (supposing voters could express their preference for the CW)
Approval voting and Score voting (and most cardinal methods) have some kind of strategic equilibrium for the CW, especially if they are a Majority Condorcet winner.

Condorcet-cardinal hybrid methods

A basic reason to prefer Category:Condorcet-cardinal hybrid methods over most other Condorcet methods (or at least, over the Category:Defeat-dropping Condorcet methods) is that they allow the voters who prefer a CW to defend that candidate without needing to do Favorite Betrayal as much (though there may be errors with this analysis). As a general example, suppose there are two main candidates, with one being the CW, and there are some 3rd parties without about half as many pairwise votes in favor of them as the main candidates. The voters who prefer the losing main candidate can bury the CW under the minor candidates, and in the ensuing cycle, the non-CW main faction will win. There isn't anything that voters who prefer the CW as 1st choice can do to fix this, but the voters who rank a 3rd party 1st and the CW above the non-CW main candidate can do FB to prevent their favorite candidate from pairwise beating the CW. This ends the cycle and allows the CW's pairwise victory over the other main candidate to take precedence again. In rated Condorcet methods, however, FB isn't quite as necessary if the CW majority-beats the non-CW main candidate; this is because those who prefer the CW can do Min-max voting to give the CW maximal points by the majority and the non-CW no support by a majority; this will guaranteeably give CW enough points to win. More specific example of this at ^[1] and some explanation of how majorities can force their preference in rated methods in the Approval voting article. It is likely possible that the tied at the top rule can be made to work with something like Smith//Approval.

Note that when demonstrating the result of Smith//Score or similar methods, it isn't necessary to discover the entire Smith set in every case to find the winner. This is because if you confirm that some candidate is in the Smith set, and they have more points than all other candidates, then regardless of who else is in the Smith set, this candidate will win.

Rated pairwise preference ballot

It is possible to treat every pairwise matchup as being a Score voting match-up where the voter can score both candidates on a scale. In this sense, the traditional pairwise preference idea of a voter giving one vote to their preferred candidate in a match-up is akin to them giving the max score to that candidate, and a min score to the other candidate. This may make it easier to think about the rated support and pairwise support that are both allowed with the rated pairwise preference ballot.

Though the most important transitivity requirements for rated pairwise are likely the ones based on margins (if 1st choice is 4 points better than 2nd choice, then they must be at least 4 points better than 3rd choice, etc.), it's also possible to require transitivity of score for a candidate across matchups i.e. if 1st choice was a 3 against the 2nd choice's 2, they can't then become a 2 against 3rd choice, even if 3rd choice would be scored at 1 or lower (which preserves the minimum transitive margin of 3-2=1 point).

Connection between Condorcet, Smith set, and Asset

One way to understand, Condorcet, Smith, and Asset voting: imagine you're having a discussion where you have to discuss one option at a time. So, one option, at any given point in time, "dominates the discussion". However, people can bring up other options one at a time, and then depending on the mood of the group, the group decides to discuss one option or the other. The group may discuss options for as long as they like, and can discuss the same options multiple times. If you start off with an option that is not in the Smith set, what you'll find is that if everyone is maximally intransigent i.e. doesn't yield any ground, then the majority-preferred option between any two options will always begin to dominate the discussion, resulting eventually in a matchup between a non-Smith option and Smith option where the Smith option must win. Now, because the Smith option is preferred by more people than any non-Smith option, a non-Smith option simply can't return to the discussion i.e. all of the Smith options will beat it any time a non-Smith option comes up for consideration. If people have weaker preferences, or don't vote in all of the matchups, this can bias the Smith set itself to being more utilitarian i.e. a minority can start to win certain matchups, resulting in the group trending closer to preferring the Score winner than the Condorcet winner or the Smith set candidates.

Condorcet PR

Here is an example illustrating the difficulty of creating a Condorcet multiwinner method along the lines of RRV:

34 A

35 B>C

31 C

B is the CW, so they'd win the first seat. If their supporters' ballots are reweighted by half, then C pairwise beats A 48.5 to 34 and wins, despite A being bullet voted by a Droop quota. One complicated way of possibly fixing this is to, after electing B, say that if B hadn't been in the election, C would have been the winner, and therefore both B and C voters' ballots should be reweighted by half since they both rank C above all candidates other than B (A), thus allowing A to beat C 34 to 33.

Vote-counting

Sequential Monroe Voting

SMV can be done by creating a weighted positional matrix (similar to that for Bucklin voting; see Summability criterion) which shows how many voters max-scored a candidate, one-point-below-max, and so on. This section will assume the use of a score scale of 0 to 5. A candidate's quota score can be reconstructed by looking at the number of voters who gave them a 5, and adding in all of the voters who scored them a 4 if there isn't a quota scoring them a 5, and repeating until a quota of voters has been found. The total points given by this group of voters is the candidate's quota score. Once the candidate with the highest quota score has been elected, reweighting occurs; if fractional reweighting is used, then at most there are 5 additional possibilities for every additional round for how voters can score the remaining candidates. For example, in the second round, either voters give a candidate an integer score between 0 and 5, or they give the candidate a fractional score whose strength depends on which of the 5 positive scores between 0 and 5 they gave to the winner of the first round.

This idea only requires two passes of all ballots for each winner elected; one to compute the round's matrix, and the second to isolate the ballots that are in a candidate's quota. It is possible to use the below-discussed idea for SPAV in SMV by, between rounds, only keeping track of the changes in scores for ballots that were reweighted. For example, it is possible to find the second round's matrix by noting that 5 voters are no longer giving a candidate a 4, with 3 of them now giving that candidate a 2.8 and 2 of them giving the candidate a 0. Thus, only changes would have to be made to the first round's matrix.

Negative pairwise counting approach

It's possible to fit both Negative vote-counting approach for pairwise counting info and rated info in a pairwise matrix by either creating a cell for each candidate that holds both pieces of info, or creating a separate column to contain one of the pieces of info. Example:


	Scores (Points totals)	A	B	C
A	50 points	(Ranked on) 2 ballots; 50 points	3 votes	5
B	33 points	12	3 ballots; 33 points	43
C	27 points	4	67	27 points

Because the Negative vote-counting approach for pairwise counting can involve some voters giving 1 vote to both candidates in a matchup, it can be worth thinking of the support a candidate has in a matchup as an approval rating i.e. both candidates might have majority-"approval". In these cases, if you're looking to get the tightest possible upper bound on number of voters that prefer one candidate or the other, then you can subtract the smallest x number of votes from both candidates in the matchup such that their combined number of votes is less than or equal to the total number of voters in the election.

When doing negative pairwise counting, it's possible to count 1st choices separately from other ranks, with voters who rank multiple candidates 1st potentially being counted separately too. See Pairwise counting#Uses for first choice information.

An interesting thing to note is that vote-counters may already be doing negative counting in their heads when doing the regular counting approach. This is because when, say, a counter marks a voter's preferences for their 2nd choice, they have to remember not to mark any preference for 2nd choice>1st choice or 2nd choice>2nd choice. This may seem easy, but consider that only a moment ago, that same counter was likely marking the voter's 1st>2nd preference, which in a horizontally oriented matrix is a cell that is directly above or below 2nd>2nd. More generally, the counter starts off by marking willy-nilly the voter's preference for 1st choice against each and every other candidate, but then has to remember to reduce the number of marks they make for every sequentially ranked candidate by one, while in the negative approach they increase it by one.

It might be useful to do negative counting by sequentially counting the ranks of a voter's ballot by starting at the last rank and going upwards. Note that, when a voter doesn't skip ranks, for them to have ranked a candidate last, they must have given that candidate a ranking number equivalent to the number of candidates (if equal ranking isn't allowed) or less than that (i.e. a "higher" rank).

Technically, the markings required for the negative counting approach can be reduced almost by half in the following manner: when a voter ranks a candidate last, make no marks for them. When a voter ranks a candidate one rank above last, the only mark made is that the voter prefers this candidate over the last-place candidate; this way, rather than marking negative votes in almost all of this candidate's matchups, only one mark has to be made. And so on. For ballots that rank all candidates, the top-ranked half of the candidates would be counted negatively, while the bottom-ranked half would be counted in this way. But this could potentially be more confusing and/or require more data storage (i.e. separately counting the negative and positive pairwise votes for each candidate).

When doing the "semi-negative" counting procedure mentioned in the previous paragraph, some voters will be able to contribute votes to both candidates in a matchup, while other voters won't, purely based on how highly or lowly they ranked them. If this creates legal or procedural issues, it is possible to have each precinct only submit the margin they found in every pairwise matchup, rather than the votes on both sides as well. In other words, if, for the A vs B matchup, in Precinct 1 A has 15 votes and B 10, while in P2 A has 7 and B 8, then it is possible for P1 to submit that A has 5 votes more than B, and P2 to submit that B has 1 vote more than A. This can be used to find that A has 4 votes more than B in the combined electorate of the two precincts.

Examples

Here's a theoretical example to compare negative pairwise counting with the regular approach (taken from ^[2]): a 2018 election in Washington state involved 28 candidates ^[3]. For simplicity's sake, suppose there had only been 100 voters in that election, with 80% of them ranking 5 candidates and 20% ranking all 28 candidates. (See the "Formula for number of marks that need to be made" section in the negative counting article for info on the following calculations.)

Under negative counting, at least 8,760 marks would be made.
- Calculation: 378 * 20 + 15 * 80
  - The 378 comes from (1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+21+22+23+24+25+26+27). The 15 is from adding the first 5 values in this series, 1+2+3+4+5.
Regular counting would require 17,560 marks.
- Calculation: 378*20+125*80
  - 125 comes from 27+26+25+24+23.
Miscellaneous:
- FPTP would've required 100 markings.
- Supposing voters scored every and only every candidate they ranked, Score voting would have required 960 marks (28*20 + 5*80).

Some caveats are that the work in the regular approach could have been reduced significantly because many of these candidates could have potentially been left off of the ballot if there had been tighter ballot access requirements, and the number of rankings could have been limited (i.e. voters could have been given only 5 ranks to put the candidates into), which would make some voters have to equally rank more candidates. These factors also reduce the amount of work for the negative counting approach to some extent.

Negative counting in non-pairwise methods

Negative counting approaches can be applied to various voting methods. For example, it's possible to reserve a special mark in Score voting that indicates that a voter gave every non-write-in candidate the max score, and then also count negative scores for the voter in such a way as to reproduce their actual scores. The practicality of this would likely be limited to ballots that max-score nearly all of the candidates, though. This type of special mark actually changes the worst-case number of marks needed to count Approval; if every voter approves every candidate (which is unlikely, since the voter is indicating no preference between any of the non-write-in candidates), then only 1 mark needs to be made per ballot, rather than [number of candidates] marks. In fact, such a mark reduces the worst-case number of marks from [number of candidates] down to roughly [0.5*(number of candidates)], because when, say, a voter approves one more than half of the candidates, that can be counted with the special mark along with negative marks for the one less than half of the candidates disapproved by the voter, for a total of [0.5*(number of candidates)] marks, rather than this same number plus one. Of course, reducing the number of marks that need to be made doesn't always result in less work (manual or cognitive) overall.