STAR-like method ("reverse STAR"?)

Toby Pereira

By going with most pairwise wins, it's a Copeland-based method, which fails independence of clones.

rob

@Toby-Pereira said in STAR-like method ("reverse STAR"?):

By going with most pairwise wins, it's a Copeland-based method, which fails independence of clones.

Gotcha.

So.... in a general sense, whether were talking about voting methods, seat belts, motorcycle helmets, or vaccines (sorry! political! ), I'm far more interested in knowing the degree of protection, as opposed to the simple boolean answer to whether or not it provides 100% protection. In my opinion, the latter is rarely useful, and often destructive to the goal of making any improvements at all.

Note that STAR voting also doesn't provide independence of clones (i.e. it is not 100% immune to the addition of similar candidates changing the outcome). I see a discussion on Reddit of STAR failing this criterion here: https://www.reddit.com/r/EndFPTP/comments/fp2b8r/example_of_star_failing_independence_of_clones/

For me, the main takeaway was:

"In order to deliberately game this flaw, you need to be able to "engineer" who the top-two candidates are. I would hope these would still be two very solid candidates if they were scored the highest."
Absolutely. STAR does have some serious concerns regarding overall manipulability (from the perspective of somebody who prefers IRV/RCV and Condorcet to anything cardinal these days) but this particular version of clone-proofness failure isn't really part of that.

I'm trying to visualize any scenario where this method ("Reverse STAR") is affected by the failure of that criterion. It seems to me that any scenario where independence of clones comes into play would be rare to the extreme, and certainly not something that could realistically be exploited, such as by strategic nomination. For starters, it needs to be a situation where there is no Condorcet winner, and where the score winner isn't even in the Smith set. I have attempted to contrive such a scenario at my Codepen ( https://codepen.io/karmatics/pen/ExKZVjM ) and it isn't easy.

Here are example results (actual ballots pasted at bottom, you can paste them back into the Codepen if you want) where there is a single "pairwise champion" that is not the Condorcet winner, where the Score winner is not that pairwise champion, where the pairwise champion has a fairly low total score (4th place), and where the STAR winner is neither the pairwise champion nor the Score winner. But let's be clear, this is an extremely close election where it took a lot of fiddling to get these results. And I really don't know if a clone-proofness failure is related to this, but it wouldn't surprise me if so.

****** Pairwise wins ******
d: 4
c: 3
e: 3
f: 3
a: 1
b: 1
****** Score ******
e: 2021 (2.5389)
f: 2014 (2.5302)
c: 1896 (2.3819)
d: 1827 (2.2952)
b: 1586 (1.9925)
a: 1513 (1.9008)
****** STAR ******
f: 240
e: 203

My current feeling on it is that the benefit this method has in simplicity and ease of explaining far outweighs any risk of clones causing a problem. I'm aware that Schultz and ranked pairs seem to solve this, but they do it at the cost of complexity which, in my opinion, is simply unmarketable. Meanwhile STAR seems easier to sell than Schultz or ranked pairs, but if there is ever an election where there is a Condorcet winner, and STAR chooses a different candidate, I would consider that a serious flaw... partly because it is directly observable that this happened, after the fact. (a la Burlington 2009 with IRV) There would be no end to people claiming it elected the wrong candidate.

Anyway, I'd be interested in seeing if you are able to clearly demonstrate this problem at the above Codepen. I'd be glad to dive deeper into analyzing the significance of issue, if you could at least show me a ballot set (or I guess, two ballot sets, one with the clone candidate and one without) that demonstrates it.

Ballots for the above example results:

134: a[5] b[4] c[2] d[1] e[0] f[0]
64: a[5] b[4] c[3] d[1] e[0] f[1]
94: a[3] b[5] c[4] d[1] e[0] f[0]
40: a[2] b[2] c[5] d[2] e[0] f[0]
63: a[0] b[0] c[0] d[3] e[4] f[5]
55: a[0] b[0] c[1] d[3] e[5] f[4]
55: a[0] b[1] c[4] d[3] e[5] f[3]
48: a[0] b[0] c[5] d[3] e[5] f[5]
30: a[1] b[0] c[1] d[3] e[3] f[5]
30: a[0] b[1] c[2] d[5] e[5] f[4]
28: a[2] b[1] c[0] d[3] e[3] f[5]
27: a[0] b[0] c[0] d[3] e[5] f[5]
26: a[0] b[1] c[3] d[3] e[5] f[3]
22: a[1] b[0] c[0] d[3] e[4] f[5]
20: a[1] b[1] c[0] d[3] e[3] f[5]
4: a[1] b[3] c[5] d[2] e[2] f[0]
13: a[0] b[2] c[5] d[3] e[5] f[2]
13: a[0] b[0] c[0] d[3] e[3] f[5]
12: a[0] b[2] c[5] d[4] e[5] f[2]
6: a[0] b[0] c[1] d[3] e[5] f[5]
5: a[5] b[4] c[4] d[2] e[1] f[0]
4: a[1] b[0] c[0] d[3] e[5] f[5]
3: a[0] b[1] c[2] d[3] e[5] f[4]

Toby Pereira

I don't have any specific examples in mind to create a clone failure, but if you have an A>B>C>A cycle, then by cloning one of the candidates, you can award the win to the candidate that pairwise beats that candidate. Clone A and C wins; clone B and A wins; clone C and B wins. So to make something up on the fly (not necessarily realistic):

14: A>B>C
10: B>C>A
5: C>A>B

In this example, A beats B 19 to 10; B beats C 24 to 5 and C beats A 15 to 14. C has the worst pairwise win and the worst pairwise loss. But clone A and C will win. (Sorry for not going into Codepens.)

@rob said in STAR-like method ("reverse STAR"?):

So.... in a general sense, whether were talking about voting methods, seat belts, motorcycle helmets, or vaccines (sorry! political! ), I'm far more interested in knowing the degree of protection, as opposed to the simple boolean answer to whether or not it provides 100% protection. In my opinion, the latter is rarely useful, and often destructive to the goal of making any improvements at all.

My current feeling on it is that the benefit this method has in simplicity and ease of explaining far outweighs any risk of clones causing a problem. I'm aware that Schultz and ranked pairs seem to solve this, but they do it at the cost of complexity which, in my opinion, is simply unmarketable. Meanwhile STAR seems easier to sell than Schultz or ranked pairs, but if there is ever an election where there is a Condorcet winner, and STAR chooses a different candidate, I would consider that a serious flaw... partly because it is directly observable that this happened, after the fact. (a la Burlington 2009 with IRV) There would be no end to people claiming it elected the wrong candidate.

I agree with some of this. It seems to me that you want to find the simplest Condorcet method, since in terms of results, there won't be much between them.

So then the question becomes - is this the simplest Condorcet method? I don't know but it's probably quite good in that respect. I don't actually think ranked pairs is complex to understand, although I think Schulze is. Also there's the method that I think is called "Benham" where you sequentially eliminate the candidate with the fewest first places until a Condorcet winner exists. Though you don't have to use the word "Condorcet" in describing it. Just a candidate that head-to-head beats all the others.

There are of course cases where it's debatable whether you'd want the Condorcet winner to be elected. E.g.

49: A>>C>B
49: B>>C>A
2: C

Basically, two mainstream polarising candidates and a non-entity who is the Condorcet winner.

Jack Waugh

My concern with Reverse STAR is that I think that in many cases, it would reduce to an ordinal system. This throws away all the information about how strongly voters strategically support a candidate who falls in order between two others. For example, in Score, votes of Nader 1, Gore .01, Bush 0 and Nader 1, Gore .99, Bush 0 have very different effect, but a ranking system would treat those votes as the same, Nader > Gore > Bush.

rob

@Jack-Waugh said in STAR-like method ("reverse STAR"?):

This throws away all the information about how strongly voters strategically support a candidate who falls in order between two others

Yes I believe this information SHOULD be thrown away except to resolve a Condorcet cycle.

This is exactly why STAR throws away strength of preference (between the two front runners) as its final step.

This in itself is probably a topic in its own, since your complaint applies to all Condorcet systems. There is good reason they throw away this information.

Remember, most of us consider that, if only two candidates are running, the one that has the majority should win. (right?) This also ignores strength of preference. That's on purpose, and is consistent with the idea of "one person one vote."

Generalizing further, let's look at a particularly "pure" example of why we want to do this. Say you are voting for the temperature to set the thermostat to (such as in an open office). Should the person who likes it at 95 degrees have more "pull" than the (also warm-preferring) person who likes it at 74 degrees? I say no. Strength of preference shouldn't matter, all that matters is whether their preference is above or below the median. See my demonstration of this at: https://pianop.ly/voting/median.html

Anyway, I will bring this up as a different topic where, instead of discussing it relative to one Condorcet method, we discuss it relative to all Condorcet methods or methods that tend to favor the Condorcet winner (or more generally, select the candidate that is "first choice of the median voter").

rob

@Toby-Pereira said in STAR-like method ("reverse STAR"?):

It seems to me that you want to find the simplest Condorcet method, ...

More or less. I want a more marketable Condorcet method. I'm not even 100% set on Condorcet, I'd be mostly happy with STAR if it had momentum but I think it is failing to gain traction. Simplicity is part of it, but simplicity isn't all that simple. For instance Approval voting is simple to explain the mechanics, but a lot harder to explain why it is good (such as how it avoids vote splitting, or why it conforms to the concept of "one person one vote"), or to explain how to strategically vote with imperfect information, etc.

I think cardinal ballots (0 - 5) are an improvement over ranked ballots, based on my long experience designing "cognitively efficient" UIs. Even if you are only using them for ranking, I think it is a better UI, especially for paper ballots as opposed to voting on a computer screen. In that sense, I think STAR is on the right track. I just think it gives too much emphasis to cardinal data when the ordinal is enough 99% of the time.

I am actually looking for a method that I think I can explain well, using some graphics and animation. Something I think I could explain to a voting theory newb in 20 seconds. I don't think I could do that with, for instance, ranked pairs.

I've looked at your example ballot sets, and have given them some thought. I think I'll address them soon in a different thread, since I think the issues are more general than applying to just this method.

Jack Waugh

Reverse STAR seems to compete well with Ranked Robin in regard to communicability to the public.

Jack Waugh

Can we show that reverse STAR never provides an incentive to invert ranks?

In fact, I have heard it repeated, including from WDS, I think, that Score never provides an incentive to invert ranks. But I don't actually know how to prove that.

rob

@jack-waugh I don't know but whenever people use words like "never" I find myself not caring. Too black and white.

Is Reverse-STAR likely to provide a significant incentive to such strategic voting behavior? By making it nuanced with words like "likely" and "significant", it becomes a far more meaningful question.

(And I doubt it is likely to provide a significant incentive)

Marylander

@jack-waugh said in STAR-like method ("reverse STAR"?):

Can we show that reverse STAR never provides an incentive to invert ranks?

Never? No, because it isn't true. The following example is effective at showing this for many Condorcet methods. (Not all, but many.)

40 A5 > B3 > C0
25 B5 > A3> C0
35 C5 > B3 > A0
A 275; B 350; C 175
B60-A40; B65-C35; A65-C35
B is a Condorcet winner and a score winner.
If 26 A voters lower B to 0, A will win the score vote, but B will still be a Condorcet winner. (At least 30 A voters must bury B to 0 to make B not a Condorcet winner just by burying B.)
However, if these 26 A voters also give C a point, then there is a Condorcet cycle, and so A wins.

26 A5 > C1 > B0
14 A5 > B3 > C0
25 B5 > A3 > C0
35 C5 > B3 > A0
A 275; B 272; C 201
B60-A40; C61-B39; A65-C35

However, I don't think this example is terribly damning because B voters only have to reciprocate a little to get back to winning. Although I do wish sorted margins was easier to explain because it is quite strong against this kind of strategy.

Edit:
@rob said in STAR-like method ("reverse STAR"?):

(BTW you made an error on the second line, you meant C0 not A0)

Fixed.

rob

@marylander It's an interesting example, although I agree it isn't "terribly damning" because, well, I just can't imagine anyone in a real world election trying to use that strategy because they'd have to be extremely knowledgeable about how others are going to vote, to not risk doing more harm than good. (i.e. giving the election to C, who they like the least)

In theory, we could alter the method to (after narrowing the field to the ones that have the most pairwise wins) then narrow it to the top two score winners, then choose the pairwise winner from those two. So then it would be even more like STAR, but with the pairwise step at the beginning.

26: A[5] C[1] B[0]
14: A[5] B[3] C[0]
25: B[5] A[3] C[0]
35: C[5] B[3] A[0]

****** Pairwise wins ******
A: 1
B: 1
C: 1
****** Score ******
A: 275 (2.7500)
B: 272 (2.7200)
C: 201 (2.0100)
****** STAR ******
B: 60
A: 40

This would give it to B, which would be the Condorcet winner if everyone voted sincerely. In this case it is basically STAR (since the pairwise step didn't actually narrow it down at all), but of course most of the time it's just gonna pick the Condorcet winner.

(BTW you made an error on the second line, you meant C0 not A0)

culi

@rob Very late to this conversation but what you're describing is exactly the Dasgupta-Maskin method[1] which is just Copeland with a Borda tie breaker. This method has actually been used in figure skating competitions under the name of "one by one"[2]

There are different version of Copeland depending on how you wanna score wins, ties, and losses. Most commonly used is probably the 1/½/0 (1 point for wins, 0.5 points for ties, 0 for losses) method but Lull proposed a 1/1/0 method and 3/1/0 is commonly used in sports. But it seems like you're just using a 1/0/0 Copeland here

This system can also be compared to Black's which is essentially just Condorcet with Borda tiebreaker

[1] https://scholar.harvard.edu/files/maskin/files/voting.pdf
[2] https://sci-hub.se/10.1287/opre.2014.1269

culi

Perhaps we can differentiate them like so:

star: top two runoff based on borda score and pairwise winner for the second round
dasgupta_maskin/one_by_one: 1/0.5/0 copeland with borda tiebreaker
reverse_star: 1/0/0 copeland with borda tiebreaker
black: simple condorcet with borda tiebreaker

Jack Waugh

@culi In Borda, you can't skip ranks, but in reverse STAR (which could be called RATS), you can.

rob

@culi Well it isn't a Borda tie breaker, it is a Score tiebreaker. This method uses Cardinal ballots.

Jack Waugh

This may be the fairest system I have ever heard of for a single winner.

rob

@jack-waugh Cool I'm not sure whether it is the absolute fairest but I like its balance of fairness (i.e. "one person one vote" or more accurately described as "everyone has equal voting power") and simplicity.

Jack Waugh

@rob OK, I'll bite -- what do you think is fairer, and on what grounds?

rob

@jack-waugh My primary criteria for fairness is that each voter has equal pull. As you know, I've said the nearest to perfect example of this is voting for a number (such as temperature) and picking the median. If you translate that to single winner elections with a finite number of candidates, you want to make it "as Condorcet as possible." That is, you want to -- as much as possible -- ignore strength of preference. (because factoring in strength of preference incentivizes exaggeration, etc)

I have a suspicion, but I haven't confirmed it, that the "most Condorcet" method is the recursive IRV one I proposed. (https://www.votingtheory.org/forum/topic/276/recursive-irv) The deeper you recurse, the better. If you only do it one level deep, it is plain old IRV. Go one level deeper and it will make it Condorcet compliant. Go 3 or 4 levels deep, and, well, it will just keep getting more and more Condorcet-ish.

That said, this reverse STAR one is plenty good, simply by virtue of being Condorcet. A simple improvement (based on what I said above) would be to normalize the ballots of the members of the Copeland set prior to calculating scores, but that makes it more complicated and I think it is unnecessary.

Toby Pereira

@jack-waugh said in STAR-like method ("reverse STAR"?):

@rob OK, I'll bite -- what do you think is fairer, and on what grounds?

You're saying it's the fairest. I think the onus is on you to justify that!