[EM] ART Single Winner Method

Kevin Venzke stepjak at yahoo.fr
Wed Dec 2 20:23:14 PST 2020

Hi Nate,

Using Approval to select two finalists has been proposed many times over the years, but what has never sat well with me is the theoretical Clone-Loser issue. A candidate who wins by approval, but who can't beat the #2 candidate by approval, could in theory run with a partner try to claim both of the top approval spots. It might not help very often, I would admit. It would probably be a more prevalent strategy in a race where voters are primarily concerned about which party wins and not which candidate.

I'm not sure I've heard before, the specific idea to take these approval finalists and do the head-to-head comparison instantly. (With other ways of selecting finalists, yes... For instance staples of mine are pitting the top-ranking winner against the approval winner, or the top-ranking winner against the candidate who is most approved on the ballots not approving the top-ranking winner.)

>From a Condorcet standpoint, you're only using one pairwise contest. IRV, too, will use one, once you're at two candidates remaining. So I don't think the Condorcet element is so evident.


Le mercredi 2 décembre 2020 à 21:31:16 UTC−6, Nathaniel Allen <npwning at gmail.com> a écrit : 
>Hey all,
>Rob asked me to send out this method I've created to see what everyone thinks. I call it 'ART' for 
>Approval, Runoff Tally (in the name of a good acronym). The method works as follows: Voters rank 
>each candidate 'Good', 'Acceptable', or 'Bad' (like 321). To find the winner you simply add up the 
>'Acceptable' and 'Good' scores for each candidate and the top two advance, ensuring whoever wins 
>is amongst the most favorable. Then the winner is selected based on a head-to-head automatic
>runoff, higher score on more ballots wins. The condorcet winner wins every time in this method except
>when they are not among the most approved. This is the biggest question I have however, I think that
>the weight of Approval in this method is stronger than condorcet winning (especially in a method where
>head-to-head tie scores are possible). The condorcet loser never wins.

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