System chosen by the Wolf committee

After years of debate and simulations the " Equal Vote 0-5 Proportional Research Committee" (AKA the Wolf Committee for short) has come up with a system to endorse for producing results with high proportional representation. The system is Allocated Score. It is a compromise of dozens of conflicting desires. It passes all the important criteria while still being very simple. It is not totally official yet but this is likely to be the system chosen.

I know nearly everybody on this forum has a pet system. Allocated Score is not my system but it is a system I would endorse. I would like to ask for others to do the same. If you have comments or questions please ask but this is not the time to say “My system is better”. Comments, Tweaks and questions are more than welcome.

1 Like

This is very nearly what I have proposed previously as the proportional extension to “Instant Runoff Normalized Ratings”. I think there’s a bug in the implementation linked to in that it assumes that the maximal choice in each round is over the quota to win. I think you need to add loser disqualification rounds when that isn’t true.
Here’s an implementation in pure Python (no pandas)

This would break monotonicity. Anyway, since it is close to your system do you endorse it?

5 to elect
x: 5-0-0
2x: 0-4-5
2x: 0-3-0

You are not giving a lot of information here but I think I see what you are getting at. Allow me to alter your example a bit to take if off of the edge case.

Red = 21%: A5,B0,C0
Green = 41%: A0,B4,C5
blue = 38%: A0,B3,C0

The winner set is {‘B’, ‘B’, ‘B’, ‘A’, ‘B’} in that order. I think your complaint is that C is not elected for green who prefers C to B. If we check the stability of the winner set for the blocking set {C,C} we can see that {‘B’, ‘B’, ‘B’, ‘A’, ‘B’} gives 16 but {C,C} only gives 10. This means that {C,C} is not blocking and the example is no issue. The example

Red = 21%: A5,B0,C0
Green = 41%: A0,B2,C5
blue = 38%: A0,B3,C0

gives {‘C’, ‘B’, ‘A’, ‘C’, ‘B’} so there is a threshold of support that Green can give to B. Strategy is unavoidable. The more realistic

Red = 21%: A5,B0,C0
Green = 41%: A0,B2,C5
blue = 38%: A0,B5,C0

gives the same but in the order {‘B’, ‘C’, ‘B’, ‘A’, ‘C’}

Why would in the first example blue voters score their candidate 5 when they are better off if they score them 3?

They are not better off scoring them 3. The B candidates get elected earlier showing that in general their was more support. Since each individual does not know how others will vote it is always better to give your favourite a 5.

would give blue only 2 winners

Right now I see what you are getting at.

Red = 21%: A5,B0,C0
Green = 41%: A0,B4,C5
blue = 38%: A0,B3,C0

gives [‘B’, ‘B’, ‘B’, ‘A’, ‘B’]


Red = 21%: A5,B0,C0
Green = 41%: A0,B4,C5
blue = 38%: A0,B5,C0

gives [‘B’, ‘B’, ‘C’, ‘A’, ‘C’]

This looks like a monotonicity failure. Good spot on this. This is a problem.

@Jameson-Quinn and @Sara_Wolf. This comes from the sort and the selection being mismatched. This does not happen in SSS.

Only a little bit non-monotonic. I can show in KPY election space diagrams that the non-monotonic regions for IRNR are much smaller than for IRV.

Also I still think all ratings methods are better with normalization because without normalization I’m incentivized to vote MAX_RATING for everyone I like and MIN_RATING for everyone below my threshold and the system has degenerated to approval voting.

Normalization is a dirty trick. A good method would not need it.

Normalization means that everyone gets equal voting power. A vote [5,5,5,0,0,0] is bigger than a vote [5,4,3,2,1,0]. If I kinda like A,B,C, I should vote [5,5,5,0,0,0] to get the best chance of electing one of them over D,E,F. Normalization and disqualification rounds with re-normalization remove that incentive.

The method should handle that internally, without such an ugly extension.

  • For each candidate, map Droop quota of highest scoring ballots to that candidate
  • Elect a candidate whose lowest mapped score is the highest and use the number of higher-than-that scores as a tiebreaker
  • Remove all ballots that are mapped to the elected candidate

How would this work?

Getting this thread back on track, the issue is that the sort is done on the original score given. Switching this to the weighted score that that round does help. It gives [‘B’, ‘B’, ‘B’, ‘A’, ‘C’] but this is still a monotonicity failure. I think the system is sunk.

If you are committed to monotonicity!

It is not so much that failing monotonicity is bad but that the failure is normally a sign of a deeper flaw. In this case it provides free riding incentive. So these issues need to be added to the issue of nonmonotinicty.

Note that this isn’t, technically speaking, a violation of monotonicity. Candidates B₂, B₃, etc, are hurt by raising the score for candidate B₁, and the notation collapsed the difference between the different B candidates.

Essentially, the blue voters are successfully free-riding. The green voters could counter-strategize and rate B lower.

I don’t know of any proportional method where there’s never some form of free riding that’s possible. I agree that this is a problem scenario, but I think you’d find that most other methods have some possible effective strategy from the blue voters, especially if we assume the green voters are strategically naive. So I don’t think this is a show-stopper for this method.

Yes you are technically correct, the best kind of correct.

I suppose this just raises the issue of if this vulnerability to free riding is worse than other forms. This would be very hard to quantify. Typical free riding is to lower your second favourite to let another faction elect them. Since this effects the favourite my suspicion is that the consequences are higher. @Marylander do you have thoughts on this?

I think the scenario where the kind of strategizing demonstrated in the monotonicity example makes sense might actually be a problem in real life. Consider, for example, when a seat becomes open. Incumbents tend to get significantly more support than non-incumbents, particularly from voters who are inclined to “vote for what they know” and aren’t likely to try to maximize their voting power. So when there is an open seat, a party may well be able to count on receiving support from people who vote for the familiar incumbents. Since these votes aren’t necessarily loyal to that party, the party would probably prefer that these votes be spent before those of their loyal supporters. With a system such as Sequentially Subtracted Score, there is a fair bit of risk to doing something like this, since if you overestimate your support among nonpartisans, you could lose your seat. The people with the least appetite for taking this risk will be the incumbents themselves, as it is their seats that are being gambled, and they hold significant influence within the party. However, the flaw that has been brought up with Allocated Score makes this kind of strategy less risky, since party loyalists can still use midrange scores as a safety net in case the party overestimates its outside support.

So I would definitely recommend SSS over Allocated Score.