User:BetterVotingAdvocacy/Big page of ideas

See also User:BetterVotingAdvocacy/Article drafts.

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.


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

How strategic Approval voting can fail to elect CWs:

How Score voting and Condorcet can both be thought of in terms of pairwise preferences (as well as rated pairwise preference ballot):

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.


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


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:

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.

Score voting

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.

One way to understand why a voter should give maximal support to their lesser evil if their favorite isn't viable: if, whenever a candidate is very unlikely to win, the voter pretends they aren't in the election i.e. got eliminated, then eventually the voter will have a candidate who is their "favorite of the remaining candidates". If they are normalizing, then they ought to give this candidate the max score, and likewise give their least favorite of the remaining candidates (i.e. the greater evils) a 0.

It might help when thinking about using negative scores in Score voting to imagine a Score scale of -1 to 0. This would be equivalent to Approval voting, essentially. In addition, keep in mind that a vote on a scale of - 1 to 1 can be converted to a scale of 0 to 2 by adding 1 point to every candidate in each vote.

One way to understand a voter's absolute score for a candidate is that they are expressing their degree of support for that candidate pairwise against a candidate they don't support at all.

The expressiveness of a rated ballot, as measured by "number of possible votes (permutations) a voter can submit", is greatly reduced if removing non-normalized votes from contention (i.e. on the grounds that they reduce voter power). For example, with a scale of 0 to 3 and 3 candidates, there are 4^3=64 possible votes[1], but when removing non-normalized votes, there are only 18 possible votes.[2][3]


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):

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.

When designing a digital interface for the rated pairwise ballot, it's possible to use various software features to highlight the constraints on how a voter can vote. For example, if the voter submits an incorrect A>C preference, then their A>B and B>C preferences could be highlighted on the screen, with an indicator to show why the three preferences don't line up. Similarly, if the voter indicates a maximal margin for A>B and prefers B>=C, then the software can "fill out" the voter's ballot to put a maximal margin in favor of A in every matchup where the voter prefers B to the other candidate (For clarity purposes, the software should probably indicate, perhaps with colors or something, which matchups the voter filled out and which ones were auto-filled).

For rated pairwise, when there are 7 candidates, at least 21 lines are required on the ballot where the voters can score each matchup. When there are more than 7 candidates, it can be more practical to have the voters rank/score each candidate individually (like in Score voting) with only up to 6 slots offered, and then below that, allow the voter to indicate how they'd score the candidates in a matchup between their 1st choice and 2nd choice, 2nd choice and 3rd choice, etc. This is because with 8 candidates, at least 28 lines would be needed for each and every matchup, but with this compression trick, instead 8 lines are needed for each candidate's score/rank, and then 15 lines are offered to allow the voter to indicate their pairwise preference between candidates in each of the 6 gradations, for a total of 23 lines. From here, one line is added for each additional candidate i.e. 24 lines would be needed for 9 candidates, etc. As a bonus, voters who only fill out the ranked/scored section of the ballot and not the pairwise part can simply have their rankings/scores counted directly as their pairwise preferences (see the "Ranked or rated preference" implementation of the rated pairwise ballot). Also see [4].

With rated pairwise, if a candidate receives a score from the voter in some matchups but not all, the highest score they got in any matchup could be considered the voter's score for them in every matchup involving that candidate where the voter indicated no preference. Alternatively, it could be the highest score the candidate gets in any matchup where the voter prefers them in the matchup more than or at least equally as the other candidate in the matchup. In other words, if there are 4 candidates and the voter votes A:5 B:4, B:1 C:0, A:5 C:0, then it could be assumed the voter would vote B:1 D:0 rather than B:4.

Is it possible to do FPTP and IRV with rated pairwise? Fundamentally, you'd have to be limited to giving only your 1st choice(s) any rated pairwise support at a given point, and the main question is whether or not, with such a constraint, Condorcet cycles can still occur. Cycles can't occur when voters are assumed to give their 1st choices maximum pairwise preference over all other candidates (and no support to any other candidates), but could that change if voters give weak preferences in some matchups in favor of their 1st choice? Also, supposing no intransitivity, how should a voter's vote change, if at all, between rounds in IRV?

The answer to the above appears to be no. Here's an example:

1 voter: A>B 60%, A>C 70% (A>B>C)

1: B>A 40% B>C 100% (B>A>C)

1: C>A 80% C>B 60% (C>B>A)

The result is that A>B (A pairwise beats B) with 0.6 votes to 0.4, B>C 1 to 0.6, and C>A 0.8 to 0.7. So there's a cycle, with all of the voters each only supporting one candidate.

Here is an example of a situation where, if voters are assumed to normalize their scores, it is possible to justify a non-majoritarian winner even with only ranked preferences: suppose there are very many voters, with there being a majority faction only one voter larger than an opposing minority faction. The majority's preference is A>B>C>etc. while the minority bullet votes B. In this case, B would almost guaranteeably win in Score under the above assumptions, even if decimal scores were allowed, so long as the majority's preference for B was non-infinitesimal, since this would cut into their ability to express their A>B preference.

Regarding normalization, if a voter would end up having to choose between rounding down or rounding up their score for a candidate, then there ought to be a name to describe the voter rounding their score for that candidate closer to their original score i.e. if they originally scored A:2, and after normalization, ought to be scoring A:3.4, then they ought to round to 3 if forced to choose an integer score, because that's closer to 2, the original score.

One way to argue that voters should be able to use only some of their voting power is to consider this thought experiment: suppose every voter had 9 clones of themselves[5], and the voter could tell each of the clones whether or not to vote. In such a scenario, should it be required that the voter make all of their clones vote?

Approval voting can be thought of as multiple voters casting FPTP ballots (i.e. approving A and B is equivalent to one voter voting A and another voting B), Score can be thought of as fractional Approval, and FPTP can be thought of as a ranked method. Rated pairwise doesn't seem to easily fall within this framework.


This section is for things that I'm still working on; they may not be totally correct, but hopefully they have various insights in them.

(Work in progress, calculations and reasoning are off) Expressiveness of rated pairwise vs. other ballot types:

  • One way to evaluate the expressiveness of various ballot types is to look at how many possible votes a voter can cast with a certain number of candidates. [6] Rated pairwise, assuming a limited ballot design where voters can indicate their ranking of each candidate, and how they'd vote in a matchup between their 1st choice vs 2nd choice, 2nd choice vs 3rd choice, etc. (with matchups between candidates more than one rank apart calculated by adding up the margins of each intervening matchup i.e. the marginal preference a voter has for 2nd choice over 4th choice is calculated as the margin for 2nd>3rd + 3rd>4th), and if considering the "votes" a voter can cast as the possible rated pairwise preferences they can indicate on such a ballot, allows for at least 72 possible votes with 3 candidates, if allowing a scale of 0 to 3 in each matchup. This compares to a rated ballot with a scale of 0 to 3 itself, which allows for at most 62 possible votes.
    • 72 comes from 60 + 12. There are 6 possible rankings of the candidates if disallowing equal-rankings (A>B>C, A>C>B, B>A>C, B>C>A, C>A>B, C>B>A), with each of these having 10 possible preferences (for A>B>C, this can be demonstrated as (with "|" used to separate each possible vote): (A>B 0, B>C 0, A>C 0 | A>B 1, B>C 0, A>C 1 | A>B 0, B>C 1, A>C 1 | A>B 1, B>C 1, A>C 2 | A>B 1, B>C 2, A>C 3 | A>B 2, B>C 1, A>C 3 | A>B 2, B>C 2, A>C 3 | A>B 3, B>C 2, A>C 3 | A>B 2, B>C 3, A>C 3 | A>B 3, B>C 3, A>C 3). In addition, there are a number of rankings that involve equal-ranking (A=B>C, A=C>B, B=C>A, A=B=C, A>B=C, B>A=C, C>A=B), with each of these also allowing for an additional number of possible votes.
  • There are two variations of rated pairwise to consider here: one where the voter can only indicate their score for their more-preferred candidate in the matchup (i.e. they can only indicate marginal preference), and one where they can indicate their scores for both the candidates. The figures computed above are for the former variation, but the latter variation greatly increases the level of expressiveness, since, for example, a voter wishing to indicate they prefer A 1 point more than B can cast that as A:4 B:3, A:3 B:2, A:2 B:1, or A:1 B:0, while there is only one way to cast that preference using the margins-based variation.

I'm trying to evaluate whether is a way to essentially "track" a voter if voters are allowed to weaken their votes in Condorcet. By track I mean that you could figure out some voter's preferences by looking at the election result data; here is one example[7]. My guess for how you could create an example where such a thing is possible is to have an election with few voters, where only 1 of the voters weakens their vote at all. Keep in mind that this may be somewhat realistic when considering that each precinct releases its own vote totals, such that a very small precinct may be vulnerable to this type of thing, if it exists.

I made some edits ( where I discussed some ways of figuring out what scores a voter would give to both candidates in a matchup (Ctrl+F "actual scores in the 1st vs 3rd matchup").

  • But I see a problem with the second way I described: it doesn't work properly with regular Score voting. Suppose a voter scores A:4 B:2 C:1. Their 1st vs 2nd preference is A:4 B:2, and 2nd vs 3rd is B:2 C:1. If adding these up as prescribed to obtain the A vs C preference, you get 6 points for A and 3 for C, which reduces to 5 and 2 respectively if fit within a 0 to 5 scale. But the actual A vs C preference was A:4 B:1, which has the same margin, but different absolute scores.
    • One way of solving this seems to be to instead do "take the lowest absolute score given to any candidate in any of the relevant matchups, and add the cumulative margin to that candidate's score to find the more-preferred candidate's score, moving this down as necessary to be capped at the max score." This always gives the appropriate result in regular Score voting. I wonder if it's equivalent or not to" take the highest absolute score, subtract cumulative margin, and move this up as necessary to be at least the min score."


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:

Note that the vote totals from a Condorcet election can be combined with the vote totals from FPTP, Approval, and Score. This can come in handy if doing a national popular vote for U.S. President. See Negative vote-counting approach for pairwise counting#Connection to cardinal methods and the image in that section for more details.

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 [8] 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 (i.e. are the Score winner), then regardless of who else is in the Smith set, this candidate will win.

I'd say Smith//Score can essentially be thought of as "Score except it passes the mutual majority criterion". This is because, when there isn't a majority faction, burying is likely to create cycles so large that the election essentially is resolved using Score voting on the major candidates; but when there is one, the minority's burying can't prevent one of the majority-preferred candidates from always winning. In order to reduce the vote-counting work for Smith//Score, the "Rated or ranked preference" implementation of Rated pairwise preference ballot might be useful.

Rated pairwise preference ballot

(See the rated pairwise preference ballot subsection of the vote-counting section below for more information on how to count the votes with this ballot type)

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.

If a voter scores the candidates, and indicates they want to put a maximal preference between all of then, then they could be allowed to only weaken their vote in some matchups manually, with all other matchups continuing to be assumed to be at maximal preference. See this discussion[9] for more about rated pairwise ballot designs and transitivity rules.

A way to measure how much voters are taking advantage of the ability to indicate maximal support in multiple transitive matchups in rated pairwise relative to ranking and scoring is to, for each voter, add up the margin they express in each transitive matchup from 1st choice to last i.e. add up 1st>2nd, 2nd>3rd+...+2nd to last>last. If, for a given voter, this cumulative margin exceeds 1 vote/100% support/(max score - min score) points, then that voter's rated pairwise preference can't be compressed into a scale. If this cumulative margin is added up for all voters, and is also calculated for each voter's ranked preference (this can be found from the data they submit on a rated pairwise ballot), then this also shows how much voters weakened their vote relative to if they had been forced to rank the candidates in Condorcet. One thing that confounds this analysis is that some voters might indicate a weak rated pairwise or rated preference between two candidates, but rank the two equally rather than differentiate the two, to avoid putting too much power in the matchup. There may be a way to do similar analysis simply by looking at the number of votes candidates have overall in rated pairwise versus ranked (compare the bottom 2 tables in [10] for an example).

Here is an example where the Score winner, Condorcet winner, and the "Condorcet winner based on rated pairwise preferences" (RPCW) are all different:

26 A:5 B:4

25 B:5 A:4

49 C:5

C is the Score winner. Pairwise, A beats B 26 to 25, and C 51 to 49, so A is the CW. But using rated pairwise preferences:

26 A:0.2 B:0, B:1 C:0 (by transitivity this implies A:1 C:0)

25 B:1 A:0, A:1 C:0

49 C:1 (this just means C:1.A/B:0)

Here, B beats A 25 to 5.2 (125 to 26 on a scale of 0 to 5), and C 51 to 49. So because A-top voters indicated weak preference between A and B, but maximal preference for them over C, they ensured one of the two won while still conceding to B.

Regarding Rated pairwise preference ballot#Preference threshold implementation, there is probably a way to allow a voter to indicate multiple preference thresholds, such that candidates within each threshold are weakly preferred at most to one another, but maximally preferred over all candidates in lower thresholds. The thresholds could even be allowed to overlap (i.e. by having the voter indicate the upper and lower bound scores for each threshold). See [11] for context; in the described situation, supposing there are two sides whose preferences are Favorite:5 2nd Choice:4 Compromise:1 Other Side:0, overlapping thresholds could allow each side to maximally boost their own candidates against the other side's candidates, while indicating their cardinal (weak) preference for their side's candidates over the compromise, and the compromise over the other side. This would only allow the compromise to win if they were scored highly enough by voters weakening their votes like this.


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).

Regarding transitivity, there are two ways it can be formulated. If a voter votes A>B at a 30% margin and B>C at 40%:

  • The strongest matchup in the A>B>C chain is 40%, so the margin of A>C must be at least 40%.
  • The matchups from A to C in A>B>C add up to 30%+40%=70%. So therefore, A>C must be at least 70%.

The justification for the first type of transitivity can be seen as follows: if the voter had indicated no preference between A and B, then logically A should have the same margin against every other candidate that B has against those candidates. Therefore, if B>C is 40%, then A>C has to be at least 40%. And since the voter evaluated A as being better than B, this can only strengthen this logic.

The justification for the second type of transitivity (which is based on Score voting) can be seen colloquially: if the voter indicates they "somewhat" prefer A to B, and "strongly" prefer B to C, then it'd seem that they ought to "very strongly" prefer A to C.

If using Score-based transitivity, there are two situations where constraints must be applied, supposing the voter's preference is A>B>C: if the voter votes A>B 30% and A>C 35%, then they can at most vote B>C 5%. This is because A>B + B>C ought to provide the minimum margin for A>C. And if the voter votes A>B 40% and B>C 20%, then they must give A>C at least 60% support. Also note that with Score transitivity, there is an implication the voter's marginal strength of preference for a candidate must increase as they are pitted in matchups against other candidates, the less and less preferred they are i.e. a voter's marginal support for 1st choice>3rd choice must be greater than 1st>2nd. This is because in order to indicate a 2nd>3rd preference, the voter has to put some kind of margin there, which then factors into the minimum margin required for 1st>3rd. To some extent, this makes this type of transitivity easier to work with than the other type of transitivity, because there will be more matchups where the voter is required to give 100% support, which means less matchups where the ratings can change the result. For a scale of 0 to 5, for example, the voter is guaranteed to give 1st>6th 100% support, as well as to 1st>7th, etc. In general, if there are n possible scores (evenly spaced apart, and including the min score), then a candidate in the j-th rank will have 100% support against someone at the (j+n-1)-th slot and below. To generalize further, if there is a certain % of spacing between two consecutive evenly spaced scores, then two candidates who are n ranks apart will have the more-preferred candidate getting at least n*the % support, and at most 100% support. For example, 2nd>4th must be at least 40% strong with a scale of 0 to 5. Thus, if printing a ballot where the voter is asked to give rated pairwise support to candidates between each rank, it would make sense to actually omit certain scoring options in some cases which would be prohibited by this constraint i.e. for the 2nd>6th matchup, the voter shouldn't be allowed to give either candidate a 3 out of 5, since that would prevent them from creating the minimal 80% margin required for the matchup (3 out of 5 is 60% support; if you add or subtract 80%, you get 140% and - 20% respectively, which are not allowed vote values).

The use of the rated pairwise ballot may increase the likelihood of a transitive result (i.e. having a Condorcet winner) relative to the use of a ranked ballot in Condorcet methods.

  • This is because, when every voter casts a Score voting-style ballot (i.e. they score each candidate the same in all of their matchups), the Score winner will be the CW. In fact, it is possible to mathematically determine how secure a "rated CW" (the CW according to rated pairwise) is in avoiding being in a cycle (having a pairwise defeat) by looking at their pairwise margins against every other candidate. In other words, if using a score scale of 0 to 5, if the rated CW's weakest pairwise victory has a 6-point margin when every voter votes Score-style, then even if one of the voters now switches to cast a non-Score-style ballot (i.e. they increase their pairwise power in at least one matchup beyond what Score voting would allow), the most this can do is make the rated CW have only a 1-point margin victory.
  • It's also worth considering who can viably be in the Smith set, depending on how voters pairwise-rate the candidates. If only a few of the voters casts a non-Score ballot, then this can only plausibly mean candidates whose utilities are very close to the Score winner can be in the Smith set along with the Score winner, since candidates whose utilities are significantly lower can't make up their large margin of loss to the higher-utility candidates with the help of only a few voters maximizing their pairwise support for these candidates. So there can be some value to analyzing the rated utilities/Score voting result or ranking (order of finish) of the candidates when analyzing how the Smith set ranking produced by rated pairwise ballots will look like.

A voter's rated pairwise preference can be partially expressed as a ranked preference augmented with margins. So, for example, A 30%>B 50%>C 100%>D 100% (which could also be written as A 0.3>B..., or A 40% to 10%>B 80% to 30%, etc.) would indicate the voter gave a 30% margin to A>B, a 50% margin to B>C (implying, by transitivity, that they gave at least a 50% margin to A>C as well), etc. This doesn't tell you how the voter scored matchups involving candidates who are more than one rank apart, however.

One of the reasons it would be difficult to allow a voter to directly express their preference for their 1st choice>3rd choice is because they could theoretically give 1st>2nd and/or 2nd>3rd large margins, but then give 1st>3rd a small margin, which would violate rated pairwise transitivity. One way of solving this would be to allow a voter to express their 1st>3rd preference by indicating how much larger they want the margin in that matchup to be than what would be required by transitivity. So for example, if 1st>2nd: 30% and 2nd>3rd: 40%, and the voter indicates that 1st>3rd should be 20% higher than usual, then that could be interpreted as "add up 1st>2nd and 2nd>3rd to find the minimum margin transitively required for 1st>3rd (i.e. 30%+40%=70%), and then add in the voter's additional 20% preference to yield a margin of 90% for 1st>3rd."

One way to understand why, regardless of transitivity, a voter may indicate their strength of preference independently in pairwise matchups between candidates one rank apart (i.e. how you score 1st choice vs 2nd choice is totally separate from 2nd vs 3rd) is because if your 1st choice dropped out of the race, then your 2nd choice would become your new 1st choice, so you'd then want to have accurately expressed the strength of your 2nd>3rd preference.

Criterion compliances

If using rated pairwise in a Condorcet method:

  • The majority criterion and mutual majority criterion are passed if the majority indicates maximal preference for all of their preferred candidates in matchups against all candidates they don't prefer.
  • Condorcet and Smith criteria are passed if the voters who prefer a candidate in the Smith set over a candidate not in the Smith set always maximize their preference.

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.


It's possible to do pairwise counting on only a subset of the candidates. In other words, you can choose to only know the matchups between certain candidates i.e. because you've eliminated all of the other candidates.

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.

Here are the first two rounds of a 5-seat SMV election done using this idea:

Hare quota is 400
Number of voters that

scored a candidate a

certain way

(Cumulative voters in


Scored the candidate a 5 305 380 375 250
Scored the candidate a 4 270 (575) 220 (600) 80 (455) 100 (350)
Scored the candidate a 3 300 400 500 200 (550)
Scored the candidate a 2 500 600 300 20
Scored the candidate a 1 150 100 200 100
Hare quota scores: 1905













The small, italicized cells are irrelevant to determining each candidate's Hare quota score.

With a Hare quota of 400 voters, candidate B is the first winner. Spending 400 of those ballots yields:

Number of voters that

scored a candidate a

certain way

Scored the candidate a 5 200 --- 150 250
Scored the candidate a 4.54 35 (235) --- 70 (220) 0
Scored the candidate a 4 100 (335) --- 60 (280) 80 (330)
Scored the candidate a 3.63 20 (355) --- 20 (300) 10 (340)
Scored the candidate a 3 40 (395) --- 100 (400) 100 (440)
Scored the candidate a 2.72 5 (400) --- --- ---
Scored the candidate a 2 --- --- --- ---
Scored the candidate a 1.81 --- --- --- ---
Scored the candidate a 1 --- --- --- ---
Scored the candidate a 0.90 --- --- --- ---
Hare quota scores: 1765.1 --- 1680.4 1786.3

Because fractional reweighting was used here, we have to now take into account that every voter who scored candidate B a 4 has 10/11th of their ballot weight remaining, since 1/11th of them (20 out of 220) were necessary to fill out B's quota. This is why there are 5 additional positive scoring possibilities.

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:


(Points totals)

A 50 points (Ranked on) 2 ballots;

50 points

3 votes 5
B 33 points 12 3 ballots;

33 points

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.

One issue that can pop up in negative pairwise counting is that, for a voter who ranks every candidate, the vote-counter might accidentally count negative votes for the last-ranked candidates. This doesn't hurt the accuracy of the count, but it can make things significantly slower (specifically, it makes negative counting slower than regular counting when voters rank every candidate); there is no need to count negative votes for last-ranked candidates, since they are not ranked above anyone, and thus can be ignored (One way to think about it is that if the voter hadn't ranked those last-rank candidates at all, they'd have the exact same effect on the election, ignoring write-ins, and no marks would be counted for those last-rank candidates in that case). Here are some ways vote-counters can mentally avoid making that mistake:

  • 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).
  • The ballot itself can be made to have one rank less than the number of candidates. For example, if there are 6 candidates, then only allow voters to rank candidates from 1st to 5th. This doesn't functionally change anything, since a voter wishing to rank some candidate last simply needs to rank the other 5 candidates in any order, and then not rank that particular candidate; because they aren't ranked, they will be treated as if they were ranked last. The reason this would help is because the last-ranked candidate(s) wouldn't be marked on the voter's ballot.
    • Note that this trick doesn't always work if equal-ranking is allowed; this is because, for example, with 6 candidates, the voter could rank one of them 1st and 5 of them 2nd; the 2nd-ranked candidates would all functionally be ranked last.

A basic justification for using some kind of pairwise counting procedure where every candidate a voter ranks 1st can be counted with only one mark each: suppose you use the "rated or ranked preference" implementation of Rated pairwise preference ballot, and a voter does min-max voting with their scores and casts a rated preference. This voter would only need one mark to count each candidate they gave a max score to, and no marks for the min-scored candidates. And in effect, this voter is giving one set of candidates maximal support against all candidates not in the set, while casting no preference between the candidates in the set, which is equivalent to ranking them 1st and all other candidates last. But, if this voter were to switch to now casting a ranked preference, the vote-counters would have to increase the number of marks they count for the voter's ballot, while not essentially capturing any different information (except that the voter would now be essentially treated as giving 0 votes to both candidates in the matchup between two equally-ranked candidates, rather than potentially giving both of them 1 vote i.e. because they might have max-scored both).

Semi-negative counting procedure

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.

Rated pairwise preference ballot

For the rated pairwise ballot, semi-negative counting can sometimes be beneficial, but there is some nuance in how to apply it. Take the following two rated pairwise votes for example (on a scale of 0 to 5):

A:5 B:0, B:5 C:0 (so A:5 C:0)

A:5 B:4, B:5 C:3 (transitivity implies something like A:5 C:2)

For the former vote, it'd be easier to assume the voter gave the max score to every candidate in each matchup, and then do negative counting i.e. count it as

A 5
B -5 5
C -5 -5 5

Note that with a scale of 0 to 1, this particular example looks exactly like negative pairwise counting as applied to a ranked ballot.

For the latter vote, it'd be easier to assume the voter is somehow constantly scoring the candidates in each matchup, and then only add or subtract points in as necessary i.e.

A 5
B 4 +1
C -1 3

So it's as if the voter had filled out a rated ballot, with the only difference being that they gave B 1 more point in the B vs C matchup than would normally be allowed, which then forced the voter to give C 1 less point against A (depending on how you implement the ballot design and transitivity requirements).


Here's a theoretical example to compare negative pairwise counting with the regular approach (taken from [12]): a 2018 election in Washington state involved 28 candidates [13]. 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.

See also [14]


  1. This is derived as follows: pretend there are infinite candidates in the election. When there is only 1 candidate in the election, we can pretend everyone except that 1 candidate is scored the same no matter what (let's say they are all scored at the min score, which is usually 0, for this example), and that that 1 candidate can be scored [number of scores in the scale] ways. So with 1 candidate and a scale of 0 to 4, there are 4 possibilities. With 2 candidates, the 2nd candidate is scored a 0 for every possible score the 1st candidate is scored at. But the 2nd candidate could also be scored a 1 for every possible score the 1st candidate is scored at, etc. So the result is 4*4=16 possibilities. For 3 candidates, there are 4 possible scores the 3rd candidate could be at when the first 2 candidates are moving through all of their possible scores, so that is 4*4*4 or 4^3=64 possibilities.
  2. There are 20 votes ignored in that analysis, so that's 20 + 26 = 46 non-normalized votes total to ignore.
  3. (This is because a normalized vote requires one candidate to be at the max score and one candidate to be at the min score, so there are only 4 possibilities for how the 3rd candidate can be scored.) A:3 B:2 C:0 (=2*3=6, since you could swap B and C here, and you can also swap B or C for A) A:3 B:1 C:0 (=2*3=6) A:3 B:0 C:0 (=3, since you can't swap B and C here) A:3 B:3 C:0 (=3)
  5. This can be thought of as similar to the KP transform and how, in some implementations, it treats each voter as casting multiple ballots.
  6. See Critique #3.