[EM] RCIPE version 2
Richard, the VoteFair guy
electionmethods at votefair.org
Mon Jul 26 15:16:52 PDT 2021
On 7/26/2021 6:20 AM, Kristofer Munsterhjelm wrote:
> ...
> ... Is that right? ...
Your calculations are correct.
> And if that's right, what would A's count be given the ranking
>
> 1: A>B>C=D=E=F
>
> is it 5 or 2?
> 1: A>B>C=D=E=F
For this ballot, candidate A would get 5 up-arrows. Or a score of 5 if
each up-arrow is regarded as a point toward a score.
If this voter just marked:
A>B
... that would still be a score of 5 for A because un-marked rankings
are virtually marked at the oval for the lowest ranking. (Here in
Oregon everyone votes by mail.)
And of course the A>B ballot gives B a score of 4 because the other four
candidates are ranked lower than B.
> Before I respond properly, ...
I look forward to your response!
Richard Fobes
The VoteFair guy
On 7/26/2021 6:20 AM, Kristofer Munsterhjelm wrote:
> Before I respond properly, I'd like to check that I've got your proposed
> "sum of candidates ranked lower" elimination mechanism right.
>
> In this election:
>
> 13: D>A>F>B>C>E
> 11: C>F>A>E>B>D
> 9: B>C>F>A>E>D
> 7: A>B>E>D>F>C
> 7: E>B>F>A>C>D
>
> I calculate the candidate counts as:
>
> A has 152 arrows
> B has 138 arrows
> C has 111 arrows
> D has 79 arrows
> E has 87 arrows
> F has 138 arrows
>
> so D would be eliminated if there were no Condorcet loser. Is that
> right? (D happens to be the Condorcet loser anyway; I'm just asking
> about your particular fallback mechanism.)
>
> And if that's right, what would A's count be given the ranking
>
> 1: A>B>C=D=E=F
>
> is it 5 or 2?
>
> -km
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