New Thieletype proportional voting method

While thinking about how to keep some of the advantages of SPSV while ditching the KP transform, I came up with a new voting method that I'm currently calling sequential threshold average score voting, or STAS voting for short. It essentially operates the same way RRV does, but reweights ballots using a different formula (see the link above) that was inspired by the KP transform. This formula seems to preserve (and sometimes even strengthen) SPSV's tendency to avoid heavy deweighting when candidates are elected that were given low ratings on the ballot in question, while simultaneously using a simple "one weight per ballot" system instead of splitting ballots up like SPSV does. In my opinion this makes it a candidate for best Thieletype proportional voting method (though I doubt it's the best rated partyagnostic proportional voting method), but I'd like to see if anyone has any major objections or other comments.

@bternarytau If you are a political party and you tell your voters to give your party's candidates a score of 5/5 (and no one else anything), then your voters will have voting power 1 for the first seat, 1/2 for the second seat, 1/3 for the third seat, and so on.
If you tell your voters to give your party's candidates a score of 4/5 instead, then your voters will have voting power 4/5 for the first seat, 12/25 for the second seat, 28/75 for the third seat...

@marylander I do not see what you are getting at. Is that a free riding strategy?
@BTernaryTau I have looked at the system and understand your motivation for rejecting RRV and SPSV. I cannot see the justification for a simple average in STAS. I tackled this same problem and came up with Single Distributed Vote as a solution. If you are comparing RRV, SPSV and STAS I would suggest you add it to the mix. I only invented it to prove that such a system is better than RRV. I do not like Thiele systems in general.

@keithedmonds said in New Thieletype proportional voting method:
@marylander I do not see what you are getting at. Is that a free riding strategy?
28/75 > 1/3
If you expect to win lots of seats, you can increase your voting power by giving your preferred candidates less than a max score.
The optimal score is (1/2)*(s/(s1)) where s is the number of seats you think you can win.
Edit: This is a problem because it means that the method will favor larger parties even more so than either standard SPAV or RRV, since this is something that parties need a high seat count to be able to take advantage of. In fact it makes the method not really proportional. For example, with ballots:
4: A[1/2]
1: B[2/2]
Party A will win every seat no matter how many seats there are. 
I think the Thiele approach holds some water, but the sequential variants rarely approximate the optimal committee very closely.
That is... PAV good SPAV bad (at least in terms of outcomes. Obviously the computation complexity is a factor as well)

Party A will win every seat no matter how many seats there are.
Very good catch on this. These vulnerabilities would be tragic to find after it is implemented. Does Single Distributed Vote have such an issue?
@brozai Sequential is basically a requirement for it to be viable to the public. Also, it is much simpler to make a score version of SPAV than PAV. Here is one such attempt to make a score vesion of PAV https://www.rangevoting.org/QualityMulti.html

@marylander I see, thank you for catching this! I'm honestly quite surprised that this sort of behavior arises from what to me feels like a pretty natural extension of SPAV. Are there any other methods known to behave this way? I need to think more about what exactly went wrong here. If there aren't any other examples, I guess that would at least mean I discovered a voting method with a novel failure mode!
@KeithEdmonds SDV should not have this issue. If you give all candidates from your party the same initial score S and all others 0s, then your ballot will always contribute a score of SÂ²/(S + K Â· SUM) = SÂ²/nS = S/n, where n is a positive integer (assuming K=1 or K=2) that depends on how many candidates from your party have been elected. This means that a greater initial score will maximize your ballot power in this case.
More generally, the limit of SÂ²/(S + K Â· SUM) as SUM goes to infinity is 0, so no party should be able to win an arbitrarily large number of seats with a fixed number of voters.

Thought a bit about the results of applying other means in place of the arithmetic mean. The geometric mean seems to remove the issue where a party can win an arbitrarily large number of seats with a fixed number of voters (the relevant limit is 0), but it doesn't eliminate the more general problem of being able to win more seats by giving all preferred candidates lower scores. The harmonic mean doesn't have either problem, and in fact using it in place of the arithmetic mean yields RRV.

@bternarytau Thiele type methods have issues. Come join the cool kids and research Phragmen systems,
https://www.votingtheory.org/forum/topic/184/rulexextendedtoscoreballots

@keithedmonds Don't worry, I'm not planning to stick with Thieletype methods forever. For now I'm just trying to get a better feel for exactly what went wrong and what I need to look out for in the future.

@bternarytau Pretty sure it was that the average was not theoretically motivated