"Split" ballots - taking proportional representation to its logical conclusions?

[Toby Pereira]

Let's say we have a score ballot with a max of 10. It might be that I think candidate A is great in one policy area but they might have nothing to say in another area that's important to me. Candidate B might be the other way round. Candidates C and D might be quite good (but not brilliant) in both areas. My scores on a single ballot might be:

A=5

B=5

C=6

D=6

But this doesn't necessarily convey all the information I want to. For a single-winner election it's fine, but if this is a proportional election with two to elect, I'd prefer A & B to C & D, even though I gave A & B 5, and C & D 6. This information could be conveyed if I was allowed to submit two ballots worth half a vote each. Ballot 1:

A=10

B=0

C=6

D=6

Ballot 2:

A=0

B=10

C=6

D=6

If everyone voted like this, most non-sequential proportional score methods would elect A & B rather than C & D, and this seems right.

Different people are going to have different numbers of issues that are important to them and aren't always going to like candidates' policies in one area but not another, so with split ballots you'd allow the voters to choose how many parts they want their ballot to be split into (up to some limit). The ballots could be colour coded. For example, a single ballot could be white, two "half" ballots could each be red etc. So when you go into the polling station you tell them how many parts you want your ballot to be and they'd give you the relevant colour-coded ballot paper(s).

Now, I actually don't think that this would be realistic for most types of election (or even possibly any), but I think as a theoretical thing it's worth being out there.

This also has parallels with our real-life behaviour. You might have a favourite song, but if you're allowed a maximum of eight songs to listen to - e.g. https://en.wikipedia.org/wiki/Desert_Island_Discs - you probably won't pick your favourite and then the seven most similar to it. You'd probably pick a range for a variety of situations. Your top eight individual favourite songs are not necessarily the same as your single favourite set of eight songs, if you see what I mean. The point being that proportional representation, or something like it, works on the level of the individual as well as in groups.

[Brian Olson]

I am fascinated by this user interface concept of submitting multiple overlapping fractional-power ballots. I find it to be clever and elegant.

It also gives me an idea about testing election methods (assuming test software has been updated to allow ballot with vote-power [0..1.0]):

If voters are better off or worse off when submitting the same vote but split into N identical 1/N votes; something is wrong.

This could be a useful test for election methods where normalization comes into play (as I am fond of normalized rating vote methods).

It seems like a lot of work to put voters through, so I kinda think it'll be unlikely to find a real world application, but I hope we do find a use for it!

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[Kevin Baas]

both scenarios convey the same amount of information.

the scores are added together, so what you get is:

A= 10*0.5+0*0.5 = 5

B= 10*0.5+0*0.5 = 5

C= 6*0.5+6*0.5 = 6.

D= 6*0.5+6*0.5 = 6

Only actual difference is not you made the voter fill out twice as much for exactly the same result.

Well, that and now you've confused/misled the voter as to how the votes are tallied by implying that A = 10*0.5 + 0*0.5  is any different than A = 5*0.5 + 5*0.5, which it isn't.

furthermore, even 5/5/6/6 is wrong.  the voter would be better off voting 5/5/10/10.  but then his 2 5-point votes for a and b would hurt his c and d choices, so he'd be better off voting 0/0/10/10.   unless he knew a or b was more likely to win than c or d, in which case his best option would be to betray his favorite and vote 10/10/0/0.

so in sum really the best option for that voter is either 1 vote that's 10/10/0/0 or 1 vote that's 0/0/10/10, depending on whether a&b or c&d are favored by the majority.

[Toby Pereira]

I think you've missed the point of the post though. "Well, that and now you've confused/misled the voter as to how the votes are tallied by implying that A = 10*0.5 + 0*0.5  is any different than A = 5*0.5 + 5*0.5, which it isn't." - We're not just adding together the scores for the candidates because this is a proportional election. They're treated as separate ballots (well, separate half ballots). If two voters give a candidate a score of 5 out of 10, it's not the same as one voter giving them 10 and the other giving 0.

[Warren D Smith]

in most voting systems, being given the option of submitting
two half-votes, seems useless, because each voter does, strategically speaking,
as well or better by choosing one of them (the right one!) and giving
it full strength.



--
Warren D. Smith
http://RangeVoting.org <-- add your endorsement (by clicking
"endorse" as 1st step)

[Kevin Baas]

Unless you can transfer votes - if you can transfer votes that wasted half vote can be converted to a full vote.

That's the beauty of transferable votes - you dot have to know the propbabilities beforehand - it turns honest voting into (fair) strategic voting.

With some subtle exceptions such as favorite betrayal. But that's unavoidable due to Pareto efficiency.

Simply making a range voting system transferable vote adds all these benefits - eliminates the cost of half voting, so you don't have to vote 10-10-0-0. You can vote 10-9-4-3 and have your half votes proportionally redistributed intelligently. Obviously this isn't as helpful in a single winner election - there are no surplus votes.

[Clay Shentrup]

On Friday, July 22, 2016 at 4:43:34 PM UTC-7, Kevin Baas wrote:

Unless you can transfer votes - if you can transfer votes that wasted half vote can be converted to a full vote.

You don't want that because it's a distortion of actual preferences which produces worse results. See Warren's Bayesian Regret figures.

[Toby Pereira]

Arguably, the case for using score voting (and without any normalisation) is stronger for proportional representation than it is in a single winner election. I can understand the arguments for Condorcet single-winner methods - that people want to put their full force into every head-to-head, so whoever end up the two most popular candidates, everyone will get a proper and full vote on it. That doesn't necessarily work in practice with all the pitfalls of Condorcet methods and ranked-ballot methods in general, but it is a strong intuitive argument.

But with proportional representation, I don't have the same intuitive sense for ranked-ballot methods or normalisation. The point is that this is an election for several representatives, and single transferable vote methods work on the assumption that each voter has "their" representative and ignore what they think of the other elected candidates, whereas, done properly, a score system looks at how much each voter likes each candidate that is elected. For example, take these score ballots:

2 to elect

10 voters: A=10, B=0, C=9, D=9

10 voters: A=0, B=10, C=9, D=9

Single transferable vote methods would elect A and B, and presumably so would many that normalise (depending on exactly how it's done). But a score system that looks at every voter's rating of each candidate (rather than making the assumption that each voter has a single representative) would elect C and D, and this seems to be a much better result for society.

[Kevin Baas]

no you want it.  it's a better representation of actual preferences because voters have closer to equal effect on the outcome.

doesn't matter what warren's figures say.  logic trumps data.

[Kevin Baas]

No.  Transferable vote methods would elect C and D.

10.0 [10.0, 0.0, 9.0, 9.0, ]

10.0 [0.0, 10.0, 9.0, 9.0, ]

 totals: 100.0, 100.0, 180.0, 180.0, 

 quota: 280.0

 ignoring: 0, 

 totals: 0.0, 100.0, 230.0, 230.0, 

 quota: 280.0

 ignoring: 0, 1, 

 totals: 0.0, 0.0, 280.0, 280.0, 

 quota: 280.0

   elected: 2 280.0 280.0

10.0 [7.368421052631579, 0.0, 0.0, 6.63157894736842, ]

10.0 [0.0, 7.368421052631579, 0.0, 6.63157894736842, ]

 totals: 73.68421052631578, 73.68421052631578, 0.0, 132.6315789473684, 

 quota: 280.0

 ignoring: 0, 

 totals: 0.0, 73.68421052631578, 0.0, 206.31578947368416, 

 quota: 279.99999999999994

 ignoring: 0, 1, 

 totals: 0.0, 0.0, 0.0, 279.99999999999994, 

 quota: 279.99999999999994

   elected: 3 279.99999999999994 279.99999999999994

 winners:2, 3, 

[Toby Pereira]

Well, it depends how it's done. Your method might elect C and D, but ordinary single-transferable vote would just work off the ranks and elect A and B.

[Kevin Baas]

the analogy then would be if you convert to borda count:

5: 4,1,2,3

5: 4,1,3,2

5: 1,4,2,3

5: 1,4,3,2

score voting would produce a 4-way tie.

you said "and presumably so would many that normalise (depending on exactly how it's done)." here you're clearly talking allocation voting.  these ballots are already normalized.  the scores all sum to the same value: 28.  any reasonable multiple transferable vote method would elect c,d.