[EM] Best Ranked Preference Deterministic Method?

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[EM] Best Ranked Preference Deterministic Method?

Forest Simmons
Elect the alternative that on the greatest number of ballets ballots, pairwise beats or ties the top choice.
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Re: [EM] Best Ranked Preference Deterministic Method?

Kristofer Munsterhjelm-3
On 22/12/2020 22.01, Forest Simmons wrote:
> Elect the alternative that on the greatest number of ballets ballots,
> pairwise beats or ties the top choice.

Isn't that Stensholt's BPW? And isn't BPW nonmonotone?

(I'm not sure what you mean by "top choice" if it isn't the Plurality
winner.)

-km
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Re: [EM] Best Ranked Preference Deterministic Method?

Kevin Venzke
Le mardi 22 décembre 2020 à 16:29:19 UTC−6, Kristofer Munsterhjelm <[hidden email]> a écrit :

>On 22/12/2020 22.01, Forest Simmons wrote:
>
>> Elect the alternative that on the greatest number of ballets ballots,
>> pairwise beats or ties the top choice.
>
>Isn't that Stensholt's BPW? And isn't BPW nonmonotone?
>
>(I'm not sure what you mean by "top choice" if it isn't the Plurality
>winner.)

What you do is compare a given candidate to the top candidate on each ballot.

By the way, you can extend BPW to more candidates by using a variation on TACC. I call it FPCC. Sort the candidates by descending first preference order. Start with the first candidate placed into a bucket. For each other candidate, add them to the bucket if they pairwise beat every candidate currently in the bucket. Elect the last candidate added to the bucket. But yes, it's not monotone.

I'll call Forest's method BTT (beat/tie top). With four candidates, it seems most similar in results to the decloned Copeland, and then TACC and IRV. For truncation and burial incentive, it seems pretty good. Worse than IRV, better than TACC, comparable to dcCopeland. For compromise, BTT is slightly worse than TACC and dcCopeland but all are better than IRV. None are that great though. BTT showed no Plurality issues.

(Trying with three candidates: BTT becomes closest to fpA-fpC, and is also close to Stensholt SV. And BTT's burial incentive seems greatly reduced for some reason.)

BTT seems to satisfy Condorcet except possibly in cases where the CW has no first preferences and another candidate beats every candidate except the CW. In that case, multiple candidates have a perfect score.

The mono-raise seems pretty good but is not perfect:
0.400: D>C>A>B
0.275: B>D
0.243: C>B>A>D
0.080: C>D   ----> changes to D>C

Original scenario elects D. The revised scenario elects B. The revision provides a boost to B, since B can count D-top ballots but not C-top ones.

Kevin
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[EM] Best Ranked Preference Deterministic Method?

Forest Simmons
In reply to this post by Kristofer Munsterhjelm-3
I'm not familiar with Stensholt, but here's my attempted proof of mono-raise:

Raise the winner W on ballot B. Then W pairwise beats or ties a superset of the same candidates as before, so its score does not decrease.

Suppose some other candidate W' increases it's score without W also increasing its score. This can only happen if W (by moving to top) becomes the only top ranked candidate on B that does not beat W' and is not itself beaten by every other top ranked candidate on B ...

I suppose that is possible ...

Is there a simple fix?

On Tuesday, December 22, 2020, Kristofer Munsterhjelm <[hidden email]> wrote:
On 22/12/2020 22.01, Forest Simmons wrote:
> Elect the alternative that on the greatest number of ballets ballots,
> pairwise beats or ties the top choice.

Isn't that Stensholt's BPW? And isn't BPW nonmonotone?

(I'm not sure what you mean by "top choice" if it isn't the Plurality
winner.)

-km

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