[EM] Weighted PR question

Tue Jun 14 11:09:44 PDT 2016

I've been looking at a potential new method and thought I would bump this discussion to see how the new method would work with this. Forest's original approval ballots were:
40: AB30: AC20: C10: BC
This new method would elect candidates in proportion to the probability that they would be picked by the following algorithm:
Pick a ballot at random and note the candidates approved on this ballot. Pick another ballot at random, and strike off from the list all candidates not also approved on this ballot. Continue until one candidate is left. If the number of candidates goes from >1 to 0 in one go, ignore the last ballot and continue. If any tie cannot be broken, then elect the remaining candidates with equal probability.
The proportions would be as follows (percentages in brackets):
A: 33/70 (47.14)B: 13/90 (14.44)C: 121/315 (38.41)

      From: Toby Pereira <tdp201b at yahoo.co.uk>
 To: Toby Pereira <tdp201b at yahoo.co.uk>; Juho Laatu <juho4880 at yahoo.co.uk>; EM <election-methods at lists.electorama.com> 
 Sent: Friday, 14 November 2014, 20:15
 Subject: Re: [EM] Weighted PR question

I think that the optimum proportions using the measure I described would be A(56), B(60), C(93), so adding up to 100 it would be A(26.79), B(28.71), C(44.50). The total of the squared representations would be 102.87, which is indeed the lowest of those I've measured.
You could probably make a case for then making these the probability weights in a single-winner lottery method as well. But again, this is purely considering proportionality and not positive support for candidates generally. For example:
2 to elect, approval voting
10: A, B10: A, C
This measure would find BC to be the most proportional, even though A has universal support, because AB and AC are unbalanced.
Toby

      From: Toby Pereira <tdp201b at yahoo.co.uk>
 To: Juho Laatu <juho4880 at yahoo.co.uk>; EM <election-methods at lists.electorama.com> 
 Sent: Friday, 14 November 2014, 0:52
 Subject: Re: [EM] Weighted PR question

Looking at this, no intuitive answer sprung out at me.
Leaving aside considerations of positive support for the moment, I think the best measure of actual proportionality is to minimise the sum of the squares of the individual voters' "total representation". If v voters vote for a particular candidate, then if that candidate is elected then voters who voted for that candidate get a representation of 1/v from that candidate and everyone else gets 0. Assuming I've made no mistakes:
For A(70), C(20), B(10), the total would be
40*(70/70+10/50)^2 + 30*(70/70+20/60)^2 + 20*(20/60)^2 + 10*(10/50+20/60)^2 = 116.
For A(70), C(30) it is 115.
For B(50), C(50) it is 108.3.
For Juho's A(35), B(25), C(40) it is 103.3.
This would make Juho's ratios the most proportional of those suggested. But that doesn't rule out others being even better. However, this proportionality criterion is not monotonic and can violate Pareto in some cases, so most proportional doesn't necessarily mean best even in an election that's supposed to be proportional. That being said, I'm not aware of a monotonic proportional approval system that I find to be reasonable enough for use, and this isn't a case where I currently see an intuitively obvious answer.
With a single winner lottery, I'd probably go for Juho's answer. I am aware of a website that uses random ballot approval voting for a choice that has to be made daily. Essentially, a random ballot is picked and a random candidate approved on that ballot is picked as the winner. It seems to work well and there's no incentive to vote dishonestly. And it's equivalent to what Juho has described.
Toby

  From: Juho Laatu <juho4880 at yahoo.co.uk>
 To: EM <election-methods at lists.electorama.com> 
 Sent: Thursday, 13 November 2014, 23:39
 Subject: Re: [EM] Weighted PR question

On 14 Nov 2014, at 00:26, Forest Simmons <fsimmons at pcc.edu> wrote:

> Consider the following approval ballot set:
> 
> 40 AB
> 30 AC
> 20 C
> 10 BC
> 
> Of the following weighted representations, which would be better?
> 
> (I)  A(70), C(20), B(10)
> 
> (II) A(70), C(30)
> 
> (III) B(50), C(50)
> 
> or some other?

How about
A 35 = 40 * 1/2 + 30 * 1/2
B 25 = 40 * 1/2 + 10 * 1/2
C 40 = 30 * 1/2 + 20 + 10 * 1/2

> 
> How about if the weights stood for probabilities in a single winner lottery.  Which would be best?

Same 35/25/40 approach could do also here. Or maybe 100% to A (the most approved candidate). Or maybe 70/50/60 (based on the number of approvals). I guess different single winner lotteries could have different targets.

Juho

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