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 vote.  No IRV 2nd round needed.  No issues.  Just like FPTP.

3. 3 or more candidates, no one got over 50% of the 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%, 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 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, in terms of 1st-choice votes it's 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 1st 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 1st 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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r b-j . _ . _ . _ . _ rbj@audioimagination.com

"Imagination is more important than knowledge."

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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 2nd 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 2nd 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.

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r b-j . _ . _ . _ . _ rbj@audioimagination.com

"Imagination is more important than knowledge."

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