Talk:Ranked Choice Including Pairwise Elimination

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Difference with Ranked Pairs
Hi User:VoteFair, this is the first time I've read this article, and I've only skimmed it. However, it seems somewhat unlikely that this method differs substantially from Tideman's Ranked Pairs (RP). Could you provide an example where this method chooses a different candidate than RP? -- RobLa (talk) 07:03, 2 February 2021 (UTC)


 * This is not a Condorcet method because it does not always yield the Condorcet winner. It's basically IRV with the addition of eliminating Condorcet losers when they occur. Currently I'm developing an example that compares it to IRV, which is the bigger competitor in the field of methods actually being considered for adoption. I have mathematically compared it to the Condorcet-Kemeny method and posted those preliminary results on the E-M forum. The similarity to the Arrow-Raynaud method was mentioned to me by Forest S. so that's something I mention here but have not yet explored in depth. (PS, recently I donated to the Miratze(?) site to help support this wiki, so I'll add that I greatly appreciate your work here on this wiki. Thanks RobLa!!) VoteFair (talk) 17:37, 4 February 2021 (UTC)

Clone independence
I don't think this method passes clone independence. Consider a Left, Center, Right election where we clone the Condorcet loser:

L is the Condorcet loser and is eliminated, then R is eliminated, then C wins.

Then clone L into a cycle:

Now there is no Condorcet loser, so C is eliminated due to having the smallest top-choice count. Then R beats L pairwise and wins. Hence we have a crowding failure. This example is fairly inelegant, but that's what came to mind right away; more effort might yield more realistic clone failures. Kristomun (talk) 17:00, 23 May 2022 (UTC)


 * I've moved clone independence back into the fail category. One failure, whether realistic or not, is all that matters. VoteFair (talk) 20:36, 23 May 2022 (UTC)

Summability
Now that I think about it, it probably isn't summable either. Suppose some candidates were eliminated as Condorcet losers, and then there's a cycle. Then the IRV step (determining the Plurality loser among the remaining candidates) breaks summability for the same reason it does in IRV. Kristomun (talk) 17:07, 23 May 2022 (UTC)


 * It makes sense that it isn't summable. I've made that change. Thanks! VoteFair (talk) 20:39, 23 May 2022 (UTC)