[EM] Weighted PR question
Toby Pereira
tdp201b at yahoo.co.uk
Thu Nov 13 16:52:21 PST 2014
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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