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

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 . _ . _ . _ . _ [email protected]

"Imagination is more important than knowledge."

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