PR spectrum

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Clay Shentrup

Jun 19, 2018, 12:36:02 AM6/19/18
to The Center for Election Science

Which three candidates should win? The hearts are the voter groups (all roughly equi-populous) and the candidates are the faces.

If you think PR is good, you say the three most outer. If you're more like me (skeptical of PR), you say the three center (just barely partisan) are best.

But what about splitting the difference and picking the halfway circle? How do we decide which is better apart from creating universes?

Brian Olson

Jun 19, 2018, 10:42:20 AM6/19/18
This illustrates a good point!

I like PR, and think it's perfectly natural for the R, B, and G groups to get their favorite representative, closest to their cultural center and causes and ideologies.

I hope that in practice it would then work out that on various issues there would be a coalition of two of the groups that would get together and do something. Maybe G and B like nature conservancy and B and R want to push through some tax changes and R and G do some school reform.

I'm skeptical of parties and don't want party-list proportional representation and I don't want any one party to be too powerful, so a PR legislature with shifting coalitions suits me just fine.

Also, the population probably isn't 3 poles, but a broader more general gaussian distribution over the political opinion space. Maybe it's bumpy, maybe it's bimodal or try modal, but really it's probably pretty broad with people all over the map. And there would be more than 3 seats. So even in a "three major parties" situation where the first 3 reps tightly align with party orthodoxy, by the time you get to the 50th representative we should be pretty well covering the map.

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Jun 19, 2018, 1:08:15 PM6/19/18
to The Center for Election Science
I picked the second biggest ring. Is there a voting method that would do that?

Ciaran Dougherty

Jun 19, 2018, 2:13:35 PM6/19/18
That's actually something I've been playing around with, but I haven't come up with something that works to my satisfaction.  I, too, am concerned with PR resulting in extremism of candidates, where the candidates are incapable of compromise without risking their seat next election.

...but isn't the alternative to effectively silence minority voices?  Your graphic is great, with 3 approximately equal groups, approximately evenly distributed throughout the space... but what if they aren't?  If you privilege the centroid for all seats, doesn't that merely give the largest faction (especially a true majority) disproportionate power?   

And as Brian pointed out, representatives need not compromise to get things done; if Blue & Green agree on some axis, they could co-sponsor legislation that expresses their views reasonably.  That would, however, require a change in parliamentary procedure, where you didn't have anyone playing gatekeeper in the discussion of what bills are to be discussed and/or voted on.

As to NoIRV's suggestion that the second biggest ring might be best... maybe?  Monroe's suggestion is that optimization of Proportional Representation is to consider only the score of ballots that are "represented" by the candidate in question.  Perhaps an alternative evaluation, where each person's opinion is considered twice?  Once at full weight for the seat representing them (per Monroe's suggestion), and 1/(Seats-1) for all of the other Seats. 
When I played around with that, it implied that it might trend towards that, (though the groups that had a full Hare quota got their hyper-partisan preference, which may be a good thing in that it mitigates the interest in bullet voting), while those who didn't have a full hare quota got their party's moderate (2nd biggest ring).

At the very least, it's worth looking into...

Jameson Quinn

Jun 19, 2018, 10:17:34 PM6/19/18
to electionsciencefoundation
Generally, candidates vary on dimensions other than ideology: intelligence, qualifications, honesty, acceptance of behavior norms, etc. Most voters would probably agree that, if ideology is held constant, they'd like all representatives to be smarter, more honest, etc.

I think that as ideological groups get smaller, the chance that their only candidates are low-quality goes up. I also think that diversity is good as long as candidates are high-quality, but adding low-quality representatives (even ones I agree with ideologically) tends to make outcomes worse. A toxic person can spoil a debate.

So in that sense, insofar as groups are small, it can be good to give them "compromise" representatives rather than their first choice; akin to the middle ring in Clay's diagram. Proportional methods with relatively high thresholds, but transfers of sub-threshold votes, tend to do this.

But other than that... I'm not sure it's possible, even if it were desirable. In the picture as drawn, with three equal-sized groups, each group would like the other two groups to compromise, while they themselves get their first choice. It's likely that in almost any method, voting strategy could affect how much your group compromises. So the more-strategic groups will tend to be represented relatively closely. The only way to be fair, then, is to represent any group above a certain size faithfully, without forcing compromise.

Toby Pereira

Jun 20, 2018, 1:33:36 PM6/20/18
to The Center for Election Science
With a ranked-ballot PR method, it's quite clear that the three candidates closest to the hearts would win. But I'm not sure that this is as obvious for all score-ballot PR methods, if the voters give honest scores at least. What would be a reasonable estimate for the scores each voter would give? I'd say something like 9, 7, 5, 4, 2, 0, for scores out of 9. Does this sound reasonable? I might have a look to see what different methods would achieve.


Toby Pereira

Jun 21, 2018, 5:47:22 AM6/21/18
to The Center for Election Science
Following on from this then, I'll call the outermost candidates R1, B1 and G1, with 2s and 3s as we go to the centre. For simplicity we can say there's just one of each colour voter. So the score votes out of 9 are as follows:

1 voter: R1=9, R2=7, R3=5, B1=0, B2=2, B3=4, G1=0, G2=2, G3=4
1 voter: R1=0, R2=2, R3=4, B1=9, B2=7, B3=5, G1=0, G2=2, G3=4
1 voter: R1=0, R2=2, R3=4, B1=0, B2=2, B3=4, G1=9, G2=7, G3=5

Using the KP transformation, we can split the voters into 9, giving us the following approvals:

2 voters: R1
2 voters: R1, R2
1 voter: R1, R2, R3
2 voters: R1, R2, R3, B3, G3
2 voters: R1, R2, R3, B3, G3, B2, G2

2 voters: B1
2 voters: B1, B2
1 voter: B1, B2, B3
2 voters: B1, B2, B3, R3, G3
2 voters: B1, B2, B3, R3, G3, R2, G2

2 voters: G1
2 voters: G1, G2
1 voter: G1, G2, G3
2 voters: G1, G2, G3, R3, B3
2 voters: G1, G2, G3, R3, B3, R2, B2

I'll come back to this later.


On Wednesday, 20 June 2018 18:33:36 UTC+1, Toby Pereira wrote:


Jun 21, 2018, 2:16:18 PM6/21/18
to The Center for Election Science
So the centrists tie with 13 approvals each. The moderate partisans get 11, and the extremists get 9.

This is centrist biased, at least for now.

Toby Pereira

Jun 24, 2018, 7:04:23 AM6/24/18
to The Center for Election Science
That's only in terms of raw approvals though, which is what you'd probably expect. I might look at what different approval-based PR methods would do with these ballots. 

Ciaran Dougherty

Jun 24, 2018, 8:25:28 PM6/24/18
Using Toby's Inputs, the results would be as follows

Apportioned Range:
Under straight apportioned range, the first seat would be awarded to one party's centrist (by tie-breaking procedure), trigger the back-off, then seat that party's extremist.
The next seat would be the next party's centrist-->extremist.
The final seat would be the final party's extremist.

Monroe Score (maximum possible score of the hare quotas for the seat that represents them): 27

Apportioned Range, without backoff:  (This is mostly for Clay, who asked why I included the Backoff step)
Without the backoff, the first seat would go to some party's centrist.
The other two parties have all of their candidates in a 6 way tie, so within that set, the winner is entirely random.
The last seat would then go to the final party's Extremist.

Monroe Score: 21, 19, or 17, depending on whether the second party gets their Extremist, Moderate, or Centrist, respectively

Apportioned Range/KP Transformed:
After the KP Transformation, the first seat would be awarded to some party's centrist (by tie-breaking procedure), having 9 approvals, and apportions them the partisan ballots for their party
R1-3, all 6 "partisan plus centrists" ballots, and 2 drawn from "anybody but the opposing extremist" ballots.
The second seat would be B2/G2, by tie breaker, having 7 approvals: their own partisan (1), non-centrist partisans (2), the remaining non-opposing extremists (4), plus two other ballots, chosen at random from the extremists (r1, b1, g1).
The final seat would be the extremist from the party not yet represented

Monroe Score: 19

Reweighted Range:
The first party's Centrist would be seated (party chosen by tie breaker).
The second party's Centrist would then be seated (party chosen by tiebreaker)
The final party's Centrist would then be seated.

Monroe Score: 15

Under KP transform RRV/PAV, the first seat would be awarded to one party's centrist (by tie-breaking procedure) .
The other two parties have all of their candidates in a 6 way tie, so within that set the winner is entirely random.
The last seat goes to the final party's extremist.

Monroe Score: 21, 19, or 17, depending on whether the second party gets their Extremist, Moderate, or Centrist, respectively

Conclusion: RRV and ARV (as written) are the more fair (where order of your party getting a seat doesn't impact which of your party's candidates gets seated), than ARV-without-backoff, KP-PAV, and KP-ARV.  Worse, if there isn't a tie, if instead of 1/1/1 voters, there were 1001/1000/999 voters, the party that got more votes would be more likely to be forced into their compromise candidate, and the party that gets a seat last has to compromise least.  That sounds like the recipe for  bizarre combination of Favorite Betrayal and Free-Riding (though, not quite either Woodall Free-Riding nor Hylland Free-Riding)

Between RRV and ARV, the higher satisfaction of the voters (according to their ballots) is found with ARV. And while both KP Transformed algorithms return a higher Monroe Score than RRV, I am greatly discomforted by the significance of Order to KP versions.


Toby Pereira

Jun 25, 2018, 6:36:48 PM6/25/18
to The Center for Election Science
Thanks for doing this. I do think though that sequential methods are likely to give odd results here based on how the thing is set up. The centrist candidates have a higher total score, so one of them would be elected first under most sequential methods, and then it's likely that the two other centrists will be elected for balance. Well, this is what happens in RRV anyway.

But I'm more interested in methods that give a score to the whole slate of candidates, rather than electing them one at a time. I've just looked at non-sequential PAV (on the KP transformed ballots). I haven't compared everything but I've compared R1, B1, G1 with R2, B2 G2 and R3, B3, G3 with both D'Hondt and Sainte-Laguë divisors. The satisfaction scores are as follows:

R1, B1, G1 (extremists) - 27 with D'Hondt or Sainte-Laguë
R2, B2, G2 (mid-level) - 26 with D'Hondt, 24.2 with Sainte-Laguë
R3, B3, G3 (centrists) - 25 with D'Hondt, 21.4 with Sainte-Laguë

So the extremists take it in this case. I might have another look later with mixing up the candidates - e.g. R1, B2, G3. But I suspect that nothing will beat 27.

I also want to work out the proportions in which the candidates would be elected with the following algorithm that I've been playing around with for a while:

"Work out the probability of each candidate being picked given 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 that ballot and continue. If any tie cannot be broken, then elect the remaining candidates with equal probability."

It's a bit fiddly and might be easier to just simulate it.

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Toby Pereira

Jun 26, 2018, 6:37:00 PM6/26/18
to The Center for Election Science
I've had another look at this with the other combinations of candidates. R1, B1, G2 and R1, B1, G3 (and equivalent) also score 27 points using D'Hondt divisors (all other combinations score lower). This means that electing two extremists and any candidate from the other party maximises the score. But using Sainte-Laguë the scores drop to 26.33 and 25.67 respectively, whereas for R1, B1, G1 it stays at 27. So overall PAV seems to favour the extremists.

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