STAR-RRV (no clone problem, more utilitarian)

STAR has two big problems:

  1. In cases like this
    Red 55%: A[5] B[4] C[0]
    Blue 45%: A[0] B[4] C[5]
    Makes A win instead of B.
  2. If the candidate with the highest sum has a clone, the method becomes equivalent to the SV.
    Example:
    If there are “far right” A, “center” B, “far left” C, each of which proposes at least 2 strong candidates, even if my real interests are these:
    A1 [10] A2 [5] B1 [2] B2 [1] C1 [0] C2 [0]
    in STAR I would definitely vote like this:
    A1 [10] A2 [10]…
    or like this:
    A1 [10] A2 [9]…
    If the supporters of B and C do the same thing, you get an SV (even if the system is STAR).

STARRV with range [0,5]:

  • found the candidate with the highest sum of points (eg A), who will be the first finalist.
  • A is removed from the votes and the score of each candidate, for each vote, is multiplied by the following value: 1 / (1 + sA / 5).
    sA = score that candidate A has in the vote.
    5 = maximum score of the range [0,5]
    This step removes the clone problem.
  • found the candidate with the highest sum (eg B), who will be the second finalist.
  • Runoff: considering the original votes, the score of the best candidate between A and B is set to 5 (leaving the score of the other unchanged).
    This makes the method more utilitarian, avoiding the problem 1) indicated at the beginning.
  • The candidate between A and B with the highest sum wins.

STARV (edit)
It’s STARRV where the Runoff takes place as in STAR: 1 point to the best of the 2 finalists, for each vote, and highest sum win.

This is precinct-summable, right? For every pair of candidates, you do RRV reweighting in case the first candidate is the first finalist, or the second candidate is the first finalist. And you also count the scores the voter would give those candidates with the “more-preferred candidate gets the max score” step.

For every pair of candidates

I didn’t understand this. In the method I have proposed I do not consider candidates in pairs.

You’re aware that for regular STAR, you can make it precinct-summable by doing pairwise counting, right? If you count the scores, and you also know the result of any possible runoff, then you can figure out who enters and wins the runoff with only one round of counting.

Ah ok. I don’t know if it’s precinct-summable.

On first pass this seems better than STAR, STLR or Score.

You use the simplified STLR normalization where you only change the higher. I suppose this is to reduce complexity? The system is actually already pretty complicated for the average person.

Speaking of simplicity, STAR tends to SV because of the clone problem, so SV would be the best.
However, if you consider the tactical votes of the SV where only very high (5,4) or very low (0,1) scores are used, then AV would be the best.

For me, the assessment of simplicity has to focus mainly on the shape of the paper ballot, also because almost all multi-winner methods are complex (more than AV or SV), and those are the real systems to aim for.
It makes sense to simplify a system, only if this doesn’t create real problems.

Not to go of on a tangent but if I was to start an advocacy campaign for a new parliament it would be for Single Member Approval voting. Simplicity is very important in advocacy. This does not mean that we should give up on finding better methods which are too complicated for current reform efforts.

In politics, it currently makes sense to support a Single Member Approval voting.
My point now is to spread surveys that use the range online, so that people get used to using the ranges.

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Do you see this as being a real problem if there are more than a few dozen voters? Like could a scenario that triggers this problem actually happen in the real world?

If C dropped out of the race before the election, you’d have a two candidate race and the majority choose A over B (even in Score voting A would win, since you can generally assume people will “normalize” their ballots by giving a 0 and 100 to at least one candidate). Would you consider that a problem?

There are various ways of explaining why this sort of example (which I see come up a lot) isn’t realistic, i.e. it is a contrived edge case. I’m working on a tool for helping to realistically look at this sort of thing (an “uncontriver”…?), but in the meantime:

Imagine that 100 people in an open office are voting for a temperature to set the thermostat (I know, I sound like a broken record with this example). Say A is 65 degrees, B is 70 degrees, and C is 75 degrees.

Wouldn’t it be odd that people are so universally favorable toward 70 degrees (every single person gives it a 4 out of 5), but not a single person picks it as a first choice?

Can you come up with any sort of real world narrative that would result in the electorate for human candidates A, B and C being so oddly clustered into two hard-edged groups? It’s not just that the electorate is polarized… but the voters are also open minded toward the middle ground in one sense (they all rate C as 4), but not at all in another sense (no one gives B a 5).

The example I have shown is in itself unrealistic, but it serves to clarify the problem linked to it, which is instead much more realistic.

Ex: there are 12 candidates, among which A,B,C are the strongest.
G1 25%: A [5] B [4] C [0] D …
G2 20%: A [0] B [4] C [5] D …
G3 15%: A [0] B [0] C [0] D [10] E …
G4 …
There are other smaller G groups (with lesser %) who support other candidates but in an irrelevant (or balanced) way A,B,C.
In the end, A and B have the highest sum and A wins.

If A and C are well defined and B is the middle way between the 2, it is not strange that the middle way is chosen more rarely in first place (with 5 points). It depends on how far A and C are.

The general problem is that in the STAR runoff, a vote like this [5,4] becomes [5,0] and if the candidate to lose the 4 points is more often one than the other, the described problem occurs.
STAR disadvantages the second choices a lot (too much), if this disadvantage is not equally distributed.

I agree that generally this problem is not easy to appear, but if it can appear then it is better to use a method that does not contain it.

I’m not sure what your more realistic example is supposed to be, because you left a lot of …'s in there. :slight_smile: If you care to fill it out more completely, I can give a better assessment of whether it is realistic.

Not trying to be difficult, but my guess is that, while you can come up with more complex examples that obscure the unrealisticness, they are just as unrealistic.

I mentioned a tool I’m building that helps with this (I’ll polish it up and put it in a JSFiddle so people can play around with it). In your simple case, it takes the 100 ballots you gave as examples, and produces some number of normalized “blends” of the ballots… averaging the numbers, then normalizing and then rounding so that there is a 0 and a 5 on each ballot. If I ask it to make 20 such ballots (each based on blending between 2 and 5 randomly selected ballots out of your set of 100), it produces these:

[A:0, B:5, C:4], [A:3, B:5, C:0], [A:5, B:5, C:0], [A:3, B:5, C:0],
[A:4, B:5, C:0], [A:5, B:4, C:0], [A:5, B:5, C:0], [A:3, B:5, C:0],
[A:5, B:5, C:0], [A:0, B:5, C:4], [A:0, B:5, C:5], [A:0, B:5, C:3],
[A:3, B:5, C:0], [A:0, B:5, C:4], [A:5, B:5, C:0], [A:4, B:5, C:0],
[A:5, B:5, C:0], [A:3, B:5, C:0], [A:0, B:5, C:4], [A:5, B:4, C:0]

The great majority of these (18 out of 20) have B as 5, with five having A tying for top position, and one having C tie for top position. If you added even a few of these blended ballots to the set of ballots, B would be the winner under STAR. (and would be the Condorcet winner as well)

Although there isn’t yet a lot of data on how people vote under STAR or Score, I bet we could learn a lot by looking at ranked choice ballot data from real elections. My hypothesis is that a) you can look at a set of ballots and mathematically detect the sort of contrived arrangements of voter preferences that trigger this supposed flaw, and b) these sort of things basically never happen in real world elections with more than a few voters. This should be testable.

It doesn’t seem so unrealistic to me that in some cases one candidate may be more disadvantaged (losing many more points due to the STAR runoff) than the other, to the point of losing even if he should win.
Many simulations often use random generations which prevent you from seeing this problem.

What we should do is consider reasons this disadvantage might unequally disadvantaged, or consider example data from the real world, that might show up in ranked choice ballot data, for instance.

Again, I’m arguing that this is highly unlikely. And it should be detectable mathematically, and to some degree, tested against real world data.

So again, your example (the only one that was complete), was this:

There are two separate issues that make it unrealistic. Only one of them could be detected if the ballots were reduced to being ranked rather than scored.

You could detect that lots of people ranked things A>B>C, and lots ranked C>B>A, while very few rated B on top (technically zero rated B on top in your example, but let’s generalize that into “very few”). So we should be able to see if this scenario ever happens with ranked choice, by looking at real world ballot data. I would guess that it would be very rare, even though it is easy enough to imagine a scenario where it would be expected, for instance a room full of people who either like cold temperatures or who like warm temperatures, with no one liking a middle temperature most.

The other thing that made your example unrealistic was that those who put B in the middle (everyone in your example) scored B in the highest middle position, with exactly zero scored B with 1, 2, or 3. We can’t tell this if the ballots were ranked choice, of course.

Note that this second situation is much harder to imagine a real world situation where that could happen. It’s one thing to imagine that all the people like either cold or warm best, but it is distinctly weird that middle temperatures are seen “almost as good” as their first preference (to every single person), while not a single person liked middle temperatures most. Why would that be? It makes no sense to me.

A realistic simulation wouldn’t use purely random ballots, of course, that would be meaningless. But it also shouldn’t use ones where they are completely contrived as in your example.

I have done simulations where I have created ballots with random input, but with structure to them. For instance, give every candidate a position in 2-dimensions (an x and a y, between 0 and 100), as well as a “universal appeal” value, also between 0 and 100. Then give every voter a random position in those same 2 dimensions. To make a ballot, consider the distance from the voter of each candidate, as well as that candidate’s universal appeal.

You can add extra sophistication by applying clustering algorithms to the locations of the voters. You can increase the number of dimensions beyond 2, and so on. You can also add varying degrees of random fuzz when determining each voter’s score for each candidate. You can put voters on a normal distribution rather than purely random. Etc.

While there is randomness, it isn’t purely random, and the set of ballots will have some structure to them.

It’s also vulnerable to tactical voting (optimal strategy never puts a gap of more than 1 point between two adjacent approved candidates, or between two adjacent disapproved candidates, and sometimes fakes indifference), a problem your second modification exacerbates by increasing the incentive to min (because giving points to disapproved candidates lessens influence in a runoff between an approved candidate and a disapproved candidate) and max (because preferences between approved candidates won’t be stretched very much in a runoff anyway).

STAR with clones, becomes equivalent to Score Voting so to maintain the positive properties of STAR it is necessary to avoid the problem of clones.
The modification I prefer is the third one, which avoids the problem of clones but is quite resistant to min-max (more than the second modification).

Isn’t STAR’s only positive property the misconception that it’s resistant to tactical voting? Even without clones, its first round is equivalent to Score’s (because even when voters don’t min-max, they minimize strength of preference) when the range is large enough (and when it isn’t, you have a rounding error problem), so the only consequence of the second round is to cause the utilitarian problem you identified (and that only happens in the absence of clones).

Thus, STAR’s clone problem is arguably a good thing, in that it negates its utilitarian problem. Your modification would eliminate the clone problem, thus returning the utilitarian problem (for instance, winning A in your example, just like STAR).

I say that the automatic runoff in STARV:

  • makes it more resistant to min-max than Score Voting.
  • makes it as resistant to min-max as STAR (which also uses a utilitarian procedure to choose the 2 finalists, so it too has the strategies linked to utilitarianism).

Why shouldn’t automatic runoff in STARV have these properties?

The question isn’t how resistant STAR is to any specific voting tactic; the question is how resistant it is to having the same first-round results as Score due to the voting tactics it’s vulnerable to. It’s vulnerable to a tactic, which I described in both my previous posts, that gets closer to min-maxing as the range increases (and shrinking the range is no solution, as that creates rounding errors and volatile tactics: “do I give A and B the same score to maximize influence in the first round or a different score to maximize influence in the second”?).

The whole idea of STAR is, if we sacrifice a little of Score’s utility under honest voting (by risking winning the second-best candidate in the runoff), we gain a lot of strategy resistance. If that strategy resistance is an illusion, which it is, all you have is downside, the utilitarian problem you identified in the OP.

I told you why in my first post.

I agree that STAR’s (and STARV’s) strategy resistance is an illusion but I wonder “how big is this illusion?”
The thing you seem not to consider is that in the real world there can easily be many voters who are honest no matter what, for more or less intelligent reasons (they don’t want to inquire about using an optimal strategy, they don’t trust mathematicians who tell them that a certain strategy works, morally they want to vote honestly in any case even if they could have a worse result, etc).
The question becomes, how many strategic votes (with optimal strategy indicated by you) are needed to generate a change in the outcome?

If Score voting needs for example about 60% strategic voters (on average) to get a change in the result, while STAR needs 80%, then if the honest voters are about 25%, STAR resists more than Score V, to that optimal strategy.

You told me that in STAR the utilitarian problem is caused by the second round when it is the first round (where the points are added) that causes it.
In the second round a vote like this: A[3] B[2] or like this: A[5] B[0] becomes in both cases like this:
A[5] B[0] or like this in head-to-head: A[1] B[0]
for this reason the second round doesn’t in any way require the voter to vote like this (strategic): A[5] B[0] instead of like this (honest): A[3] B[2].

About STARV, even intuitively, when you have to choose 2 winners (multiple winners) it is normal to reduce the weight of the votes based on who was the first winner; not doing so would be considered a serious mistake for multiple winner contexts but I don’t understand why it is not considered a problem in STAR.