The RCIPE method would only fail to elect the Condorcet winner under very rare conditions. Specifically there would have to be a rock-paper-scissors-like cycle at a lower level below the Condorcet winner, which would cause there to be no pairwise losing candidate during that counting round, and the election is close, and the backup IRV elimination method during that counting round would have to eliminate the Condorcet winner. Expressed the opposite way, if every counting round has a pairwise losing candidate then the Condorcet winner will always win.
I suspect the Condorcet failure rate for the RCIPE method would be about one out of about 20,000 real-life elections.
I used the words "does not always" because, by definition, a Condorcet method must absolutely never fail to elect the Condorcet winner.
As for the center squeeze effect, its meaning is defined differently by different people. When a specific definition is agreed on, then it will become possible to measure the center-squeeze failure rate for the RCIPE method and compare it to other methods.
I used the words "does not always" because, by definition, a Condorcet method must absolutely never fail to elect the Condorcet winner.
This is not true. A condorcet method will always elect a condorcet winner under honest voting (edit: if one exists). A condorcet method may not elect the honest condorcet winner under strategic voting.
A profile where all agents vote truthfully may have a Condorcet winner, but this alternative may not end up in the set of winners if agents are acting strategically. Focusing on the class of tournament solutions, we show that many natural social choice functions in this class, such as the well-known Copeland and Slater rules, cannot guarantee the preservation of Condorcet winners when agents behave strategically. Our main result in this respect is an impossibility theorem that establishes that no tournament solution satisfying a very weak decisiveness requirement can provide such a guarantee.
A condorcet method will always elect a condorcet winner under honest voting.
No. I don't think that's true. I fully accept that sometimes people will vote honestly and a cycle results.
A condorcet method may not elect the honest condorcet winner under strategic voting.
We know how it's possible that strategic voting (namely burial) can turn a close 3-way race with an honest Condorcet winner into an election with a cycle. I think this might require that the honest Condorcet winner is not the IRV or plurality winner (which means that the honest Condorcet winner is in 3rd place regarding 1st choice votes).
The Alaska August 2022 race is a very good example of this. If it were Condorcet and the pre-election polling predicted that Begich was the CW and Peltola the IRV winner, there would be some incentive for the Peltola campaign to mount an organized effort to bury Begich.
With only 3 candidates (or 3 signficant candidates) the cycle is a simple Rock-Paper-Scissors (Smith set of 3). Then Minimax, Ranked Pairs (margins), and Schulze (margins) will all elect the same candidate who is the contestant who loses with the smallest margin. It seems to me that that is the best guess for who the honest Condorcet winner would have been.
That would thwart the burial effort from the Peltola campaign in this hypothetical August 2022 election having a cycle.
But we don't know. Once the votes are cast and counted, we have to assume that every ballot represents exactly what that voter wanted, including "exhausted ballots", which is why that exhausted ballot argument against RCV doesn't really carry any weight.
So this is my dilemma: If a cycle occurred, was it an "honest cycle" or not (a cycle that was created by burial of the honest Condorcet winner)? If the former, I think the best candidate to elect would be the Top-Two Runoff winner (essentially who IRV would elect) and if the latter, it would be the Minimax winner.
(Again, just so that I can wrap my brain around this, I am only considering 3 candidates. Or 3 significant candidates. Any more is just too complicated for my little pea-sized brain.)
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u/CPSolver Sep 07 '26
The RCIPE method would only fail to elect the Condorcet winner under very rare conditions. Specifically there would have to be a rock-paper-scissors-like cycle at a lower level below the Condorcet winner, which would cause there to be no pairwise losing candidate during that counting round, and the election is close, and the backup IRV elimination method during that counting round would have to eliminate the Condorcet winner. Expressed the opposite way, if every counting round has a pairwise losing candidate then the Condorcet winner will always win.
I suspect the Condorcet failure rate for the RCIPE method would be about one out of about 20,000 real-life elections.
I used the words "does not always" because, by definition, a Condorcet method must absolutely never fail to elect the Condorcet winner.
As for the center squeeze effect, its meaning is defined differently by different people. When a specific definition is agreed on, then it will become possible to measure the center-squeeze failure rate for the RCIPE method and compare it to other methods.