[ARIN-consult] Consultation on Implementing Single Transferrable Voting for ARIN Elections

[William Herrin]

On Fri, Jan 7, 2022 at 11:48 AM Richard Laager <rlaager at wiktel.com> wrote:
> On 1/7/22 1:24 PM, William Herrin wrote:
> > Refer to my prior post in which an example of "most opposed" is
> > readily inferred.
> > https://lists.arin.net/pipermail/arin-consult/2022-January/001507.html
>
> I'm aware of your example and that's exactly my point.
>
> Looking at the percentage numbers (I don't have time to dig up Condorcet
> voting software and run the example through it), candidates 2, 3, and 4
> could easily be clones. (Clones is a voting system term of art.)
>
> If they're clones, the electorate split across those candidates nearly
> evenly (21%, 20%, 19%), because they are pretty much the same. In that
> case, a good voting system should elect some of them. This a feature,
> not a bug:
> https://en.wikipedia.org/wiki/Independence_of_clones_criterion
>
>
> Take your vote counts, replace 2/3/4 with "C" (for clone), keep only the
> highest C in each line, and combine together. You get:
>
> 30% 1, C, 5
> 60% C, 1, 5
> 10% 5, C, 1
>
> The cloned candidate was clearly preferred

Hi Richard,

I don't buy that explanation but I'll run with it for a moment. In
such a case an equitable system should have selected candidate 1 plus
a random one of the clones slightly weighted towards candidate 2.
Instead STV selected two clones, excluding the more diverse voice from
candidate 1. STV shifted the election toward the presumptively
think-alike clones, reducing diversity of viewpoints among the
elected.

Now let's back up and move the numbers a bit further apart so that the
middle candidates aren't clones. Elect two of five candidates:

30% 1, 3, 2, 5, 4
24% 3, 1, 2, 5, 4
19% 4, 1, 2, 5, 3
8% 2, 3, 1, 5, 4
9% 2, 4, 1, 5, 3
6% 5, 4, 1, 2, 3
4% 5, 3, 1, 2, 4

Opposition to 4 (ranked 5th or unranked): 66%

Round A: 30% 1 24% 3 19% 4 17% 2 10% 5, eliminate 5
Round B: 30% 1 28% 3 25% 4 17% 2, eliminate 2
Round C: 30% 1 36% 3 34% 4, quotas met by 3 and 4.
3 and 4 elected.

FPTP: 73% 1, 66% 3, 34% 4, 17% 2, 10% 5, 1 and 3 elected.

Same problem. Candidate 1 who wins FPTP with a 73% majority is
defeated by STV while candidate 4 who is opposed by a 66% majority
gains election.

Regards,
Bill Herrin



-- 
William Herrin
bill at herrin.us
https://bill.herrin.us/