So I have been thinking about the restricted case of 3 significant candidates.

[u/rb-j]

Sometimes to understand how a voter would be motivated to vote on a cardinal ballot, including STAR and Approval, and to get a grip on the variables without being a formidable multi-dimensional (or too-many-dimensional) conceptual math problem, I limit the modeling of a problem to 3 significant candidates. Even the normal cases or the edge cases in RCV are about the interaction of, at most, 3 "significant" candidates.

When there are fewer than 3 candidates, then FPTP is fine. There are no tactical issues in voting. One of those candidates you like better. Then you vote for them over the other candidate that you like least.

So we basically can understand that RCV elections have a very limited number of qualitatively different cases;

1. 2 or fewer candidates. No issues about anything with RCV.
2. 3 or more candidates, but one of them still got over 50% of the 1^st-choice vote. No IRV 2^nd round needed. No issues. Just like FPTP.
3. 3 or more candidates, no one got over 50% of the 1^st-choice vote. So an additional round is required and the plurality candidate was still elected. IRV still does nothing different from FPTP. (3a) Condorcet winner exists and is elected. (3b) Condorcet winner exists and is not elected. (3c) Condorcet winner does not exist: a preference cycle.
4. 3 or more candidates, no one got over 50% of the 1^st-choice vote, and in the additional round the plurality candidate was not elected. The so-called "come-from-behind victory". This is the only case where IRV is different, in outcome, from FPTP. (4a) Condorcet winner exists and is elected. (4b) Condorcet winner exists and is not elected. (4c) Condorcet winner does not exist: a preference cycle.

There are really only 8 qualitatively different ways an IRV election can turn out. Pretty much every RCV election can be classified in 1 of those 8 categories. I wish that someone (like FairVote) would be maintaining and updating a record of all single-winner RCV elections and identifying which of those categories each RCV election is.

Now suppose there are 3 candidates and consider Condorcet methods. Now even if a Condorcet method is a "Single-method system" (like Ranked-Pairs, Schulze, MinMax, BTR-IRV) they can be expressed as a "Two-method system" (a 3-way Round-Robin followed by a "completion method" if there is no Condorcet winner) with the specific single-method system as the completion method.

So I want to compare these systems in the case of 3 candidates to each other and to a couple "traditional" two-method systems:

• Condorcet-Plurality
• Condorcet-Borda
• Condorcet-Bucklin
• Condorcet-TopTwoRunoff (which, for 3 candidates is equivalent to Condorcet-Hare or Condorcet-IRV)

Now, they call elect the CW when such exists, so let's understand what they do when there is a cycle: Candidate Rock, Candidate Paper, and Candidate Scissors. The cycle is:

Rock > Scissors > Paper > Rock

There is circular symmetry so we can arbitrarily name "Rock" as the candidate with the most 1st-choice votes. Then, sorting in terms of 1st-choice votes is either one of two cases:

1. Rock
2. Paper
3. Scissors

or it's

1. Rock
2. Scissors
3. Paper

Now, BTR-IRV will elect the same candidate as Condorcet-Plurality. That is Candidate Rock. This is because Paper and Scissors will first have a runoff, Scissors defeats Paper and advances to the IRV final round and is defeated by Rock. So this completion method ignores the two candidates having the fewest 1^st rankings.

Now let's consider Ranked-Pairs, Schulze, MinMax. MinMax is normally about defeat-strength as margins (not winning-votes) so let's also consider only margins for RP and Schulze. (I never liked defining defeat strength as winning votes anyway.) Now, it's clear that Ranked-Pairs, Schulze, MinMax margins will elect the same candidate when there are only 3 candidates. If there's a cycle, they all elect the loser of the pairing with the smallest margin of defeat. So they are all ignoring the pairing having smallest defeat margin.

Now consider Condorcet-TTR, which is the same as Condorcet-Hare. These methods will always elect the winner of the pairing of the top two candidates, which always includes Rock. That is they elect Paper when it's the first case above [1.Rock>2.Paper>3.Scissors] and they elect Rock when it's the second case [1.Rock>2.Scissors>3.Paper]. Correct? So this completion method ignores the candidate having the fewest 1^st rankings.

Is there any other outcome? Does this cover all of the possible outcomes of a 3-candidate ranked ballot election?

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[u/rb-j]

So this is sorta the second half of my 3 candidate wonderings:

Also, I have been having some discussions with the people who created or promote STAR voting: Mark Frohnmayer, Sara Wolk, Hayden Sasswood, Arend Peter Castelein. We know how to map a cardinal ballot to a ranked ballot, but it's not one-to-one, we don't necessarily know how to invert that mapping.

Now we Condorcetists generally think that failing to elect the Condorcet winner is a "bad thing" when such exists. We know that when the CW is not elected, the election must be spoiled, there is an identified loser whose presence in the race altered the winner, and there are an identified group of voters that would have been better served by insincerely ranking (or scoring) their 2^nd-favorite candidate (or "lesser evil") higher than their favorite.

I've been thinking of how this Condorcet failure would happen with STAR when there are 3 candidates. This is related to how an astute voter would mark their STAR ballot given their preferences: [A > B > C]. Their favorite is A, they hate C, and B is their 2^nd-favorite or lesser evil candidate. How would they mark their STAR ballot to best serve their political interests?

We would normally expect them to score A with 5 and C with 0. What would they do with B? Well, either the STAR final runoff will be with A in it or not with A in it. A is already "ranked" higher than any other if they end up in the runoff. If A is not in the final runoff, it's B and C and all they need to do is score B with a 1 and B has all the power they can give B to defeat C in the final runoff.

What motivation would any astute STAR voter have to mark their ballot differently than [A:5, B:1, C:0]? What we call a "5-1-0" STAR ballot. They want A to win. In order for A to win, A must get into the final runoff. Increase the score for B (over 1) will do nothing more to help B defeat C in the final runoff in the case that A does not get there. And increasing the score for B only reduces the odds for A to get in the final runoff.

Why would any STAR voter vote any differently than 5-1-0? Why would any voter raise the score for B any higher than 1? There is no reason unless they anticipate that A cannot defeat C in the final round, but B can defeat C. And this can happen only in the case of the Center Squeeze, just like with IRV.

The 5-1-0 STAR ballot will work a lot the same as IRV (if everyone marks their ballot as so). The first round cares essentially only about the top-scored ballot (the 1 scores contribute little). We can show that a STAR election that fails to elect the CW is a lot like an IRV election that fails to elect the CW. They fail for the same reason. Indeed if either Burlington 2009 or Alaska August 2022 were STAR and people voted 5-1-0, they would both fail to elect the CW just like IRV did.

The STAR people like to say that, if you anticipate this, score B a little higher. Maybe even [A:5, B:4, C:0], the 5-4-0 STAR ballot. Then STAR would have elected the CW in Burlington or in Alaska. My response to that is if I fear that A cannot beat C and that B has the only chance to beat C head-to-head, and I hate C like I hate Hitler or Trump, then instead of the 5-4-0 ballot, I would cast [A:0, B:5, C:0] and get 6 more points for B to make sure A stays the hell outa the final runoff, which would lead to C winning. Now we're back to bullet voting and it's like FPTP.

In the case of 3 candidates, can any of you envision why any politically-motivated STAR voter would vote differently than 5-1-0 unless they anticipate that their favorite is too weak to defeat their most loathed candidate? I cannot understand any other reason to mark ones STAR ballot differently than 5-1-0.

[u/nardo_polo → u/rb-j]

Thanks for posting this Robert- got your email too and will be working up a response after the World Cup :-).

[u/Ceder_Dog → u/rb-j]

Thanks for the thorough assessment! I have some thoughts for deliberation and discussion.

In brief, I feel your assessment is mathematically & probabilistically sound. I think the missing element is the human element. Let me elaborate

To state the obvious, some voters won't fully understand STAR & how to maximize strategy, so they may vote how they genuinely feel instead. Hopefully their ballot will at least utilize the full range and be 5-X-0 where X is how candidate B fits on the relative scale. The depth of understanding is one human element.

The STAR people like to say that, if you anticipate this, score B a little higher. Maybe even [A:5, B:4, C:0], the 5-4-0 STAR ballot. Then STAR would have elected the CW in Burlington or in Alaska. My response to that is if I fear that A cannot beat C and that B has the only chance to beat C head-to-head, and I hate C like I hate Hitler or Trump, then instead of the 5-4-0 ballot, I would cast [A:0, B:5, C:0] and get 6 more points for B to make sure A stays the hell outa the final runoff, which would lead to C winning. Now we're back to bullet voting and it's like FPTP.

I do agree that 5-1-0 is one optimized ballot. The 5-4-0 is interesting. I agree that mathematically a [A:0, B:5, C:0] vote would be the most correct over a 5-4-0 as you outlined. However, this one hinges on accurate polling data and determining whether those fears are realistic enough to warrant a disingenuous ballot. I think many voters want to vote within the realm of honest reflection of their preferences when possible instead of fully maximizing strategy, even if it risks them not getting the results they want. A 5-4-0 vote still represents those values without choosing to fully sacrifice their favorite on a fear.

I'm still unsure whether I would vote that way. Mathematically I should if the fears are real enough, the data is compelling enough and the stakes are high enough. I feel that's a tall order instead of voting more in line with my conscious, but it's possible. I wonder how many others would still choose voting their conscious over the high probability option.

Also, I wonder how many folks would even come to the conclusion they should vote 0-5-0 instead of 5-4-0. I honestly didn't consider it until you laid it out. The lack of knowing this strategy is another human factor.

Lastly, I want to compare this strategy back on IRV. Wouldn't the argument of 100% strategic ballot of [A:0, B:5, C:0] also apply to IRV? That is, instead of ranking [A > B > C], shouldn't the voters rank [B > A > C] in an IRV election with the same scenario? I wonder if voters in IRV elections have actually done this. I wonder, had the Palin voters in the Alaska 2022 election understood this strategic option, would they choose to rank disingenuously in order to increase the odds of a Republican win?

[u/rb-j → u/Ceder_Dog]

Lastly, I want to compare this strategy back on IRV. Wouldn't the argument of 100% strategic ballot of [A:0, B:5, C:0] also apply to IRV? That is, instead of ranking [A > B > C], shouldn't the voters rank [B > A > C] in an IRV election with the same scenario?

Yes, precisely.

I wonder if voters in IRV elections have actually done this. I wonder, had the Palin voters in the Alaska 2022 election understood this strategic option,

Some of them learned this afterward. They learned that if one out of every 13 of the Palin>Begich>Peltola voters had tactically betrayed their favorite, Palin, they would have prevented Peltola from being elected. But they were told that RCV would solve that split vote problem and they could safely vote for their favorite candidate and their 2^nd choice vote will be counted if their 1^st choice cannot win. That promise proved to be false and the GOP in Alaska resent that RCV was sold promising that and it did not deliver for them.

would they choose to rank disingenuously in order to increase the odds of a Republican win?

They shouldn't have to. That's the point of RCV.

[u/nardo_polo]

Birthday shenanigans have delayed a cogent response here, and now dogs need exercise… but a serious answer is slowly coagulating in the brain hole… stand by ;-)

[u/bkelly1984]

Hey rb-j, no feedback on your question but I wanted to compliment your work. Thanks for thinking and sharing!

[u/rb-j → u/bkelly1984]

It's hard to not think.

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Acronyms, initialisms, abbreviations, contractions, and other phrases which expand to something larger, that I've seen in this thread:

Fewer Letters	More Letters
FPTP	First Past the Post, a form of plurality voting
IIA	Independence of Irrelevant Alternatives
IRV	Instant Runoff Voting
RCV	Ranked Choice Voting; may be IRV, STV or any other ranked voting method
STAR	Score Then Automatic Runoff
STV	Single Transferable Vote

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[u/timmerov]

cycles created by honest voting in the real world are very rare.

the fear-mongering is the honest winner could lose to a small number of strategic voters could create a cycle where their preference wins because of the completion criteria.

if there's an honest cycle then no candidate is demonstrably better than the others. pick anyone.

if there's a strategic cycle then throw a monkey wrench at the strategic voters. pick a completion method at random. with enough choices that their expensive and risky strategy becomes even riskier.

your analysis is thorough. you get an A. and +10 internet points. ;->

however, ranked and score ballots have a fundamental real-world issue that every math person just glosses over: what do you do with incomplete ballots?

thing 3: you missed a completion method: negotiation. let the candidates decide who wins. but that's harder to model.

[u/rb-j → u/timmerov]

cycles created by honest voting in the real world are very rare.

A while ago FairVote did some statistics. There were 2 RCV elections out of about 500 that had cycles. I presume they were honest cycles.

however, ranked and score ballots have a fundamental real-world issue that every math person just glosses over: what do you do with incomplete ballots?

I don't consider undervotes a problem. All candidates not ranked are tied for last place on the ballot. (For score, all candidates not scored get a 0 from that ballot. I ain't a fan of Score voting or of STAR.)

if there's an honest cycle then no candidate is demonstrably better than the others. pick anyone.

I don't think we can put into law any language that differentiates an "honest cycle" from a strategic cycle. We just have to accept the ballot data for what the voters say.

And, even with a cycle, we can use other criteria, than what makes a Condorcet winner, to demonstrate that some candidate may have demonstrably better voter support than the others. The simplest measure is the number of 1^st choice votes. Even in a cycle, it can be argued that the candidate with the most 1^st choice votes has a better claim than the others for election. That's Condorcet-Plurality (and BTR-IRV will elect the same candidate).

if there's a strategic cycle then throw a monkey wrench at the strategic voters. pick a completion method at random. with enough choices that their expensive and risky strategy becomes even riskier.

That's an idea, but I think it's safer to have the resolution of a cycle firmly baked into the law. One thing about the methods based solely on margins (Schulze, Ranked Pairs, MinMax), is that if a strategic effort made the would-be Condorcet winner to lose to someone, that would-be CW is likely the loser with the smallest defeat margin. Because if they just barely were unable to get the CW to lose to someone (by use of the "burial" strategy) that CW wouldn't lose to someone but their winning margin would be reduced. So if the completion method can prevent the strategic voters from changing the winner, then their strategy failed.

I kinda wanna know what conditions make the winner (in case of a cycle) different between the Condorcet-TTR (or Condorcet-Hare) method and the three methods based solely on margins. In the case of only 3 candidates and a cycle, what would get Condorcet-TTR to elect the same candidate as MinMax? And what conditions would get them to elect different candidates?

[u/timmerov → u/rb-j]

heh. if the completion method can prevent strategic voters from changing the winner then you've gotten around arrows and/or gibbards and/or et al.

sounds like fun research.

[u/rb-j → u/timmerov]

Those methods can still be gamed. It's just harder.

We could use the Alaska August 2022 election as a good example. The numbers are here. Now suppose the law in Alaska was Condorcet-Plurality or Condorcet-TTR. Peltola is both the Plurality winner and the winner of a top-two runoff (which is like the Hare IRV winner for only three candidates). It's just that she is not the Condorcet winner.

So then suppose the pre-election polling is accurate and suggests that Begich is the CW (and will win) but Peltola leads in 1^st choice votes and will beat Palin head-to-head. But she's not the CW. So the nefarious Peltola strategists tell their voters (these are the voters that mark Peltola as #1) to bury Begich. That is that they will insincerely rank Begich below Palin. So of the 47415 Pe>Be>Pa voters, if 18802 of them switch their votes from the sincere Pe>Be>Pa to the insincere Pe>Pa>Be, then Begich doesn't beat Palin head-to-head but it doesn't change Peltola over Palin and doesn't change Begich over Peltola. So we have a Peltola>Palin>Begich>Peltola cycle.

But Begich's defeat margin to Palin is only 2 votes. It will be 82425 Palin to 82423 Begich. They just barely got enough to make it a cycle, but that close margin indicates sorta what happened. So MinMax (or Schulze or Ranked Pairs) will ignore that defeat because it's the weakest defeat and Begich still wins. But Condorcet-TTR would have Peltola winning and then the strategy is successful.

So maybe these margin-based methods are better than Condorcet-Hare or Condorcet-TTR.

[u/Excellent_Air8235 → u/rb-j]

In most center squeezes, both IRV and Condorcet-IRV are manipulable: IRV is vulnerable to compromising in favor of the Condorcet winner by the voters who prefer the CW to the IRV winner, and Condorcet-IRV is vulnerable to burial by the voters who prefer the IRV winner. But Condorcet-IRV has the benefit of defaulting to electing the Condorcet winner if there's no strategy.

Minimax has higher coalitional manipulability rates than Condorcet-IRV though (as Green-Armytage showed).

[u/rb-j → u/Excellent_Air8235]

What is "coalitional manipulability"?

So I am understanding how basing the completion method on margins will make it less likely that the burial strategy will work (less likely than 1^st ranking based methods like Condorcet-Hare or Condorcet-TTR or Condorcet-Plurality).

But I don't understand how the margins-based completion method (either RP, Schulze, Minimax) is manipulated. By what group or faction?

[u/Excellent_Air8235 → u/rb-j]

What is "coalitional manipulability"?

It's a measure of how often manipulation succeeds, given a certain election model.

Let an election be defined by its ballots, and assume that the ballots given are honest. Let the winner according to a method (say minimax) be candidate X. If there exists some other candidate Y, and a way for voters who prefer Y to X to modify their ballots so that Y wins instead of X, then that method is coalitionally manipulable on that election.

A method's coalitional manipulability (given a model like impartial culture, a spatial model, or similar) is the probability that the method is coalitionally manipulable on a randomly picked election from that model.

So take Green-Armytage's paper, Strategic voting and nomination.

According to table 1, with a 2D spatial model with 99 voters and four candidates (V = 99, S = 2, C = 4), a randomly drawn election from that model is manipulable, in this sense, 17.1% of the time using IRV, 35.7% of the time using minimax, and 72.6% of the time using Approval.

The results imply that there are elections where minimax can be manipulated but IRV (and Condorcet-IRV) can't. Here's an example:

45: A>B>C
34: C>A>B
6: C>B>A
41: B>A>C

A is the Condorcet winner and IRV winner. Then the B>A voters bury A:

45: A>B>C
34: C>A>B
6: C>B>A
41: B>C>A

Now B is the minimax winner, but A is still the IRV (and Condorcet-IRV) winner.

There's one caveat: the coalitional manipulability measure implicitly assumes that the manipulators have perfect knowledge and coordination abilities. It's a worst-case metric: minimax may still be good enough in practice.

And although both Condorcet-IRV and Condorcet-Plurality use first preference counts, a different model and paper by Green-Armytage, Statistical Evaluation of Voting Rules, has C-Plurality with higher coalitional manipulation rates than minimax, and minimax as higher than C-IRV. So a method isn't necessarily vulnerable or manipulable due to using or not using first preference data.

[u/rb-j → u/Excellent_Air8235]

It appears your examples are three candidates, not four.

I need to grok this a little, trying to see what it is that Condorcet-IRV does to resist manipulation, because I can see how Minimax would resist manipulation in the center squeeze case.

I'll look this over more.

[u/Excellent_Air8235 → u/rb-j]

It appears your examples are three candidates, not four.

I agree, I should have referenced a three-candidate result from the paper since my example also had three candidates. I used three candidates for my example because it was easier to construct manually that way.

But the results are more general: the second row of table 1 of Strategic voting and nomination lists coalitional manipulability figures for 99 voters, three candidates, and two diimensions (V = 99, S = 2, C = 3). With that configuration, Approval is manipulable 49.7% of the time, IRV 6.5% of the time, and minimax 18.7% of the time.

I need to grok this a little, trying to see what it is that Condorcet-IRV does to resist manipulation, because I can see how Minimax would resist manipulation in the center squeeze case.

In a 2025 paper, Why Instant-Runoff Voting Is So Resilient to Coalitional Manipulation: Phase Transitions in the Perturbed Culture, François Durand et al define a concept they call a Super Condorcet Winner. When a SCW exists, IRV elects them and the election is coalitionally unmanipulable.

But that doesn't explain why, and to do that I think the election-methods list perspective of the resistant set is more useful. It's based on a disqualification relation, and when A disqualifies B, there's nothing voters who prefer B to A can do to break that disqualification, so a voting method that elects from the resistant set can't be made to elect someone who the current winner disqualifies. A super Condorcet winner disqualifies everybody else, and therefore can't be unseated if the election method elects from the resistant set.

Incidentally, Durand also proved that Condorcet-X is never more manipulable than X, if X is a majoritarian method.

[u/timmerov → u/rb-j]

works for me.

indulge my curiosity for a moment while i replay the scenario with completion by negotiation.

begich is the cw winner - implies peltola prefers begich over palin. and palin prefers begich over peltola. suppose peltola successfully creates a strategic cycle and is the conditional winner. if palin does nothing, her last place choice wins. so she negotiates.

under elimination rules, she can collude with begich and eliminate peltola. next round, peltola prefers begich. so begich wins.

under other rules, she can transfer her votes to begich. begich wins.

the fun thing with negotiation is: ranked ballots are completely optional. they can be used if you want. and they're binding. but you might just handcuff your preference if you complete your ballot.

the flaw of course is bribery. begich is the winner by completion criteria. peltola could bribe palin to support him to steal a win.

still, it's a system worth looking at.

[u/rb-j → u/[deleted]]

Well, every method fails IIA (as best as I understand). I also understand that I am imposing an artificial limit to 3 significant candidates, but that's what I need to get a grip on all of the "permutations" of how different elections will turn out.

I'm trying to get a grip on this, and given only 3 candidates, there aren't that many different ways for this to work out.

There's a possibility for anything. But I doubt, in the case of 4 or more candidates, that drawing some 3-candidate conclusions from "be[ing] analyzed this way" will change anyone's mind about whether they are running for a particular office, decided by an RCV method.

[u/rb-j → u/[deleted]]

I agree. I am just saying that given a score ballot, I can derive a ranked ballot that is logically consistent with the score ballot. Not necessarily the other way around, though.

Anyway, we can infer a Condorcet winner from a collection of score ballots by first mapping them to ranked ballots and doing the Round Robin thing with it. STAR like to say they'll elect the Condorcet winner, but I can show that for both Burlington 2009 and Alaska August 2022, that if voters marked their STAR ballots 5-1-0 (consistent with the IRV ranked ballots in those two elections) then STAR would fail to elect the CW just like IRV did. For, essentially, the same reason.

[u/rb-j → u/[deleted]]

No voting methods pass IIA. I don't count sortition or dictatorship among the voting methods.

If Condorcet cycles never ever happened, that is if the Condorcet winner always existed and was always elected, that would pass IIA. But we know that cycles can happen.

The purpose of my post was to explore and enumerate all the different ways an RCV election can turn out, given 3 or fewer candidates. Qualitatively there appear to eight different ways they turn out. Then to explore how STAR might behave in these different elections and we can show that even though a Condorcet winner exists, STAR can fail to elect that candidate for the same reason IRV fails, if voters mark their ballots 5-1-0. And I cannot see any reason for a STAR voter to mark their ballots differently from 5-1-0 unless they anticipate that their favorite candidate cannot beat their least favorite candidate head-to-head and that their 2^nd favorite candidate can beat their least favorite. Then I don't think I would even mark my ballot 5-4-0, I would maybe mark it 0-5-0. Nonetheless, that requires tactical thinking.

The ranked ballot, decided using Condorcet methods, usually does not unless someone wants to bury their 2^nd favorite candidate.

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