Posts made by Marylander

@sarawolk I have indicated my availability.

@jack-waugh It is at the very bottom of the screen. It appears in the same place whether you are creating a new topic or editing the head post of an old one.

Edit: Send me a message if you are having trouble finding it.

@jack-waugh We could use tags to retain some of the information lost by cutting the number of categories.

@jack-waugh The amount of complexity in an IRV Yee diagram increases absurdly quickly as you add candidates, but I don't think there will be infinite complexity with a finite number of candidates the way fractals are infinitely complex.

@rob said in Why I'm now leaning toward Copeland//IRV as the best method to promote:

• the 440 elections monitored by FairVote that they had enough information to be able to determine whether there was a Condorcet winner.

How many of those elections reported the complete rankings of the ballots? In an election where the complete rankings are not reported, it will still be possible to determine the Condorcet winner if one candidate gets a majority of the first choices, and IRV will elect that candidate. However, if no candidate gets a majority of the first choices, then it will not be possible to determine the Condorcet winner. This could lead to selection bias in their dataset.

@rob said in Score Sorted Margins:

It sounds like an "iterate until it settles on an equilibrium" sort of thing. At first blush, I'm not 100% convinced it will always find an equilibrium. Has it been proven to?

Yes, it can be proven to always find an equilibrium.

Note that only adjacent candidates can be swapped in the ranking, and that when adjacent candidates are swapped, the only relation that changes is the one between those two candidates.

Thus if we start at a ranking in which A>B, and move to a ranking in which B>A, then at some step some point, A and B must have been adjacent, and then they must have swapped positions, implying that B beats A pairwise. So to move back to a ranking in which A>B, A would have to beat B pairwise. But A and B cannot both beat each other pairwise.

Therefore we cannot move from a newer ranking to an older one.

I'll grant that this system is probably too complex to be implemented in practice, but it's my favorite theoretically.

@jack-waugh Yes.

@jack-waugh How would the equal rankings be handled?

Also, you may want to look at the Nanson and Baldwin methods. They are Condorcet methods based on Borda that use an elimination process.

@jack-waugh I explained it in https://www.votingtheory.org/forum/topic/213/under-what-scenarios-will-the-smith-set-differ-from-copeland-set/5

First, find both the total scores for each candidate, and the pairwise matrix for each candidate.
Candidates are seeded by their total score. For each adjacent pair of candidates in the ranking, we check whether their order in the ranking agrees with their pairwise matchup. If each pair agrees, then stop and take this as the final ranking. If at least one pair does not agree, select the pair with the smallest difference in score that disagrees and switch the candidates. Repeat until you end up with a ranking that has no disagreements between adjacent candidates.

@rob The complexity of IRV has slowed its spread, even though there are far more complex electoral rules. At public hearings to implement it somewhere new, it gets criticized for being confusing. I think people in the voting reform space tend to underestimate the complexity of hybrid voting methods like this one because they are composed of building blocks we already are familiar with, but for people not already initiated, having to explain all the building blocks means more of a chance to lose them before you get to make an argument. If I were to try to promote some Condorcet method, I'd probably go with something like Minimax, since the cycle-breaker is fairly simple but also in practice very similar to the cycle-breakers of some of the more highly regarded Condorcet methods.

(I will note that in my opinion, we will almost never see a Smith set with more than 3 candidates, and furthermore, if there is a cycle, all of the pairwise matchups will be close; 51-49 all 3-ways will be far more likely than 60-40 all 3 ways.)

@rob said in Under what scenarios will the Smith set differ from Copeland set?:

I don't know about your favorite Score - Sorted Margins (I don't see an electowiki page). Would you be interested in helping/advising so I can work it into my Codepens where I am trying to do a bunch of methods that use score-type ballots. (if you are a JS coder, feel free to fork it and add it yourself)

I think it was invented by @Ted-Stern. It uses score ballots. How it works is:
First, find both the total scores for each candidate, and the pairwise matrix for each candidate.
Candidates are seeded by their total score. For each adjacent pair of candidates in the ranking, we check whether their order in the ranking agrees with their pairwise matchup. If each pair agrees, then stop and take this as the final ranking. If at least one pair does not agree, select the pair with the smallest difference in score that disagrees and switch the candidates. Repeat until you end up with a ranking that has no disagreements between adjacent candidates.

@rob

1: A>B>C>D
1: B>C>D>A
1: C>D>A>B

A beats B, B beats C and D, C beats A and D, and D beats A.

The Smith set is {A,B,C,D}, but the Copeland set is {B,C}.

This scenario is just the Condorcet triple, but with D being added as a clone of C.

I expect that in the real world, the most common Condorcet paradox would be the triple, since it is the simplest, so scenarios like this would be unlikely to occur, unless for some reason there was a reason to intentionally encourage clone-like candidates to run.

@jack-waugh Most good Condorcet systems tend not to satisfy later-no-harm, so there might be cases where it helps to bullet vote, although there also might be ways to specify a complete ranking that are also effective.

That said, I do not think it is much better to encourage voters to express later preferences that are purely strategic in comparison to having them be strategically left blank.

@andy-dienes Sorry I actually didn't mean to post that. I had it saved in drafts but hadn't fully worked out the example. I should probably check to make sure that other drafts of mine didn't get posted by mistake.

Edit: Yeah I think all of my drafts were posted because of some glitch with the mobile website.

@andy-dienes I see it as a "continuous" version of the AS reweighting procedure. One motivation for it is that the AS reweighting procedure being discontinuous can lead to chaotic outcomes.

25: A5 C3
25: A5
50: B5 C3
1: A5 B4
4 winners
A, B, C, A

This was a draft that got posted by mistake.

@cfrank said in Structure of a Voting System:

The fact that the best compression algorithms used in industry exploit dynamic information methods (machine pseudo-learning) to achieve state of the art results is suggestive, in my opinion. Any thoughts are welcome.

Voting methods have to deal with the fact that voters actively try to manipulate the decisions made by the method. This can be a challenge to ML models, for example, spammers design their spam to evade getting caught by fillters. However, spam filters can update themselves against the latest strategies, which doesn't really apply to voting methods. So there are unique challenges to voting method design that don't necessarily apply in ML.

@keith-edmonds

@keith-edmonds said in Proportionality Guarantees of Allocated Score (approvals):

Do you have thoughts on this idea? I can add code of you want.

That would be useful. How do you decide whom to give weight back to? Is it even across the ballots?

(Sorry I am getting to this so late. I haven't had the spare time I'd need to read the forum posts carefully.)

@andy-dienes said in Proportionality Guarantees of Allocated Score (approvals):

Although, may I ask why it is important to you that the price paid by a voter depends at all on their current ballot weight?

I'm not necessarily advocating this as a method. I just think that understanding this reweighting procedure is important to understanding what the change to allocated score to sort the ballots by weighted score rather than unweighted score actually does, because it is a "limit case" of the allocated score reweighted step. If instead of applying one allocated score reweighting step to spend the entire quota, we applied many reweighting steps that each spent a tiny portion of the quota (like partitioning an interval), then as the size of the largest reweighting step approached 0, the result would approach that of the procedure I described.

@andy-dienes said in Proportionality Guarantees of Allocated Score (approvals):

After all, the other voter might have another winner elected in a future round, and we do not go back and retroactively credit them some ballot weight.

I did make a method that did just that. (It's an SSS variant.) However, similar to what you said about Meek STV,

@andy-dienes said in Proportionality Guarantees of Allocated Score (approvals):

it relies on iterating an optimization program until an arbitrary tolerance level is reached

@keith-edmonds said in Proportionality Guarantees of Allocated Score (approvals):

Correct but not a monotonicity failure on the score. It was more like a vote management vulnerability.

Looking back on the debate about Allocated Score at the end of the Wolk committee, I am thinking that sorting by weighted score was effectively a janky way of trying to make the reweighting step of AS spend midrange scores for a party before all of the higher-level scores are spent. I call it jank because in my opinion, the following reweighting step is more consistent with the underlying logic of sorting by weighted score (but it is also more complex; essentially it continuously recalculates weighted score as it spends weight from the ballots):

• Sort the ballots by the weighted score they gave the elected candidate. B_(1) is the ballot with the highest weighted score, B_(2) second highest, and so on.
• If B_(1) has a higher weighted score than B_(2), then spend enough weight from B_(1) such that B_(1) and B_(2) have the same weighted score.
• After that, if B_(1) and B_(2) have higher weighted scores than B_(3), then spend enough weight from B_(1) and B_(2) so that they will have the same weighted score as B_(3).
• Keep repeating this process until a quota of weight is spent. If B_(i) is the last ballot to require any of its weight to be spent, then only spend enough weight from ballots B_(1),...,B_(i) so that a full quota will be spent, and proportion the spend such that their weighted score for the winning candidate will be equal.

For example, the case that motivated the shift:
Red = 21%: A5,B0,C0
Green = 41%: A0,B4,C5
Blue = 38%: A0,B3,C0
(5 winners)

The first winner is from party B. Standard allocated score allocates 20% worth of green ballots to the first winner.

This alternative reweighting step would first spend 10.25% from the green ballots. After that, the weighted score for party B that the green and blue ballots have become equal, but there is still 9.75% that needs to be spent. Approximately 4.36% of this comes from green, and 5.39% from blue.

So in total, ~14.61% is spent from green, and ~5.39% is spent from blue to elect the first candidate.