STV with Condorcet for the final seat (among the non-exhausted ballots) to fairly represent majorities

The problem if 3 Ls win in the above example is that if the L party passes a law with their majority, and the C and R voters dislike that law, they would constitute a majority of voters but wouldn’t have a majority of representatives, so they couldn’t overturn the law. Here’s a possible fix: for the final STV seat, if there’s a Condorcet winner based on the non-exhausted ballots, elect them, otherwise just run STV as normal.

If run on the above example, after exhausting 4 Hare Quotas of votes and electing L1, L2, R1, and R2, you have:

7 L3 > C
6 C > L3
7 R3 > C

C pairwise beats L3 13 to 7 and beats R3 13 to 7, so they’re elected. (Under standard STV rules, C would have the fewest votes and be eliminated, then L3 would beat R3 13 to 7.) Now you have 2 Ls, 1 C, and 2 Rs, and the majority of voters are fairly represented.

Some points on this modification:

  • FairVote, if they were to accept a Condorcet method, would like it because in the single-winner case, it devolves into “elect the Condorcet winner, otherwise run IRV”, so they have a way to unify their voting method offerings in the single and multi-winner cases.
  • Cardinal advocates would prefer to guarantee electing Condorcet winners in single-winner races rather than IRV winners
  • FV won’t like it because it technically fails later-no-harm and all the other things Condorcet fails, but not in a very harmful way at least, since it should only really apply to the strategic voting for the final seat (plus they’re getting PR enacted, so it’s doubtful this would be the deal-breaker, but who knows)

P.S. It is possible to do “if there is no Condorcet winner on the final seat, eliminate the candidate with the fewest votes, transfer their voters’ votes to their next choice, and then check if there’s a Condorcet winner, if not, repeat” but that’s much more complicated for little gain.

Edit: C can fail to get elected if the votes are modified such that there are say, 5 L candidates and they each get 9 or 10 votes from the L faction. Then nobody on that side will have an outright quota and C will get eliminated before the final seat comes into play. It seems the only way to really ensure C wins is to either have voters Favorite Betray, equally rank, or have the voting method automatically elect a Condorcet winner before running STV on the remaining seats.

If a Condorcet winner is automatically elected before starting the STV count, maybe the ballot exhaustion could be done by prioritizing the ballots that ranked the Condorcet Winner highest (i.e. ballots ranking C 1st get exhausted first, then ballots ranking C 2nd, etc.) until a quota is exhausted.

Still just as vulnerable to free riding.

How often does free riding occur in STV? I’ve seen the examples you provided for Canada and I think one case in Ireland as well.

Often in Ireland (as has been discussed before), but not often in Australia (because you can just rank parties directly, and so people are too lazy to vote below-the-line, even though vote management would mean most voters should vote BTL.) So there is no one answer to this question; it depends on the conditions of the political environment where it is used.

How do voters themselves feel about these vote management strategies? What surprises me about the Irish case is that presumably, STV is in use there because it allows for candidate-oriented voting, so I’d guess maybe the Irish get angry that vote management is necessary there, but what discussions have they taken on fixing it, if any?

Here’s a method to consider … C still loses, but it could be combined with a Condorcet method for the first or last seat.

And it is not vulnerable to Woodall free riding

Edit: Skip to the edit at the end, this is probably wrong.

A Condorcet winner can be iteratively found by picking some candidate and testing whether anyone can be ranked higher than them by more voters. Here, it fundamentally seems that the issue is that STV isn’t a “Condorcet in each Hare Quota” type of method. Maybe it’s possible to look at all the top-ranked candidates, elect the ones with Hare Quotas, and for the remaining candidates, somehow see if they can assemble Hare Quota support (maybe by adding in lower ranks and testing if they can get a Hare Quota solely from those ranks that are being examined) while being a Condorcet winner within that Hare Quota.

In a sense, Asset PR would struggle to elect the right winners in situations where there isn’t a Condorcet winner in each Hare Quota, and Asset in the single-winner case is meant to be a non-deterministic attempt at a Condorcet method.

The reason I mention Hare Quotas rather than Droop Quotas is because Asset finds a Condorcet winner among all ballots (Hare Quota) rather than the Condorcet winner within the majority (Droop Quota, and equivalent to ignoring the minority’s ballots entirely whenever there’s a Smith Set, I believe.)

Edit: This is wrong; the method I proposed here should really resemble Bucklin PR instead (elect the candidates who can get quotas with the fewest top ranks added in, and exhaust votes.)

@NoIRV Bucklin PR should handle your example with C losing in STV.

After L1 L2 R1 R2 win, we have 7 votes for L3, 6 for C, and 7 for R3. Nobody has a Hare Quota, so we add in the next rank and now C is ranked on 20 ballots and wins.

It seems Bucklin PR makes the most sense if done with Droop Quotas, because in the single-winner case that makes it regular Bucklin. Bucklin done with a Hare Quota in the single-winner case is equivalent to Approval where the candidate ranked on the most ballots wins, even if unanimously ranked last.

I’ve been thinking about this a bit more …

There may be some benefit to using Bucklin PR (that is, successively including lower ratings until a candidate achieves a quota of support) with either Droop or Hare quotas, then using a Condorcet method for the last seat.

In examples where I’ve tested this, such as the Schulze STV vote management case, the final Condorcet step achieves the same effect as Schulze STV, but at much less cost.

I’m using a method called Score Sorted Margins, a score-based variation of Approval Sorted Margins. It is fairly resistant to burial and chicken dilemma strategies, though not invulnerable.

This example actually seems like a bad one overall to me, since there are 2 Droop Quotas here, but whether or not they appear depends on whether they misorder their preferences solely within their coalition. So Vote Management actually yields the right result here.

A true optimal Condorcet PR method should always give the result an idealized Asset Voting PR negotiating equilibrium would give, when one exists. Here, looking at 1st preferences, we have 38 Andrea, 27 Brad, and 25 Carter. The two prospective winners are Andrea and Brad, and Brad voters bullet voted, so only consider Andrea’s voters: 26 of them prefer Carter to Brad, so if, say, 8 of them moved their votes from Andrea to Carter, then Andrea has 30 votes and Carter 33; Brad can’t win unless the 12 Andrea>Brad voters shift their votes to Brad, but then they’d be shifting the winner set to (Brad Carter) rather than (Andrea Carter), which is a strict worsening from their point of view. (Edit: I was wrong; an Asset Voting election, following certain algorithmic rules, would produce a tie between all 3 candidates here. If enough Andrea-top voters shift to Brad, both Andrea and Brad have a Droop Quota. At that point, enough Andrea>Carter>Brad voters can give Carter just enough votes that all 3 of the candidates have a Hagenbach-Bischoff Quota. Still, this winner set, where each of the 3 candidates has a 2/3rd chance of winning, differs from Schulze STV, which guarantees Andrea and Brad election.

Here’s a bit of a paradox to consider, actually: if we say that any of the three candidates could win, then there’s a chance Andrea won’t win. Yet Andrea has a Droop Quota of 1st choice votes; and yet, if we give Andrea priority in the event of a tie, we are then giving Andrea and Carter voters incentive to try to do Vote Management to ensure their two candidates will have priority in a tie, which makes doing all of this pointless.) So there is a negotiating equilibrium, therefore this is what an optimal Condorcet PR method should produce, since this result could be algorithmically found by just looking at the ranked ballots. If there wasn’t an equilibrium, there’s probably some cycle resolution method that could be tacked on, but that gets complicated. (Edit: We actually can determine a Condorcet cycle of winner sets based off of idealized Asset negotiations, and then use cycle resolution methods to pick a winner between these sets by comparing them, calculating pairwise comparisons of how many voters prefer one to the other, etc.)


27 A>B>C
24 C>D>E
26 A2>B2>C2
24 C2>D>E
3 D

As a 2-winner election, the candidates with the most votes are:
A A2) (27,26)
We can try “converging” on candidates outside of the most votes (leading) set to see if someone else can enter the set:
Converge on C no
CV (converge) on C2 no
We try converging on D, and find that 51 voters prefer D to be in the leading set over A2:
CV on D yes 51 votes
A D) (27,51)
Cv on A2 no
Cv on C no
50 voters prefer C2 to be in the leading set over A:
Cv on C2 yes 50 votes
C2 D) (50,3)
Cv on C yes 51 votes
C2 C) (50,51)
(Here I just skipped straight to comparing both A and A2, but it’s probably possible to go one by one and get leading sets of (C2, A), (A2, C), etc.)
Cv on A and A2 yes
A A2) (27,26)

Now we have a Condorcet cycle: we have (A,A2)>(C2,C)>(C2,D)>(A,D)>(A,A2). I’ll need to keep thinking about how ballots should be assigned to various candidates in pairwise comparisons here; maybe Schulze STV or CPO-STV could be used on only this “Smith Set” sequence of outcomes.

This does hint at a very interesting use case for cardinal ratings in a Condorcet PR method; with rankings, there’s no way to know if a voter would “negotiate” to get both their second and third choice elected, or only their first choice. As previously discussed though, in a 100-winner election, a voter may be more satisfied by a candidate they rated a 100/100 rather than 100 candidates they rated a 1/100, so even with utilities indicated on the ballot, there can be issues.

I think this modification to address Vote Management for cardinal methods could be adjusted to be used for ordinal methods as well. Then this becomes Condorcet in the single-winner case and is maximally (probably only somewhat) resistant to Hylland free-riding while probably being easier to compute than something like CPO-STV.

It would probably be an improvement if, when finding the Condorcet method’s winner for the final seat of STV, all previously eliminated candidates were un-eliminated (with the ballots that rank them allowed to give them only so much support as they have weight).
One advantage of this improvement is that

wouldn’t matter because C would be re-included. Though it might make a difference in another example if C-top voters had some of their votes spent on a lower choice and thus didn’t have enough weight to elect C in the final round.

Edit: The following idea has the same issue that you showed in your deficit handling non-monotonicity example:

How well does the following proposal address free riding: “elect at least the first N - 1 seats using STV, using surplus distribution to fill the final seat with a quota if possible, but otherwise, the final seat is elected with a Condorcet method, with all previously eliminated candidates back in contention. However, ballot weight is restored to make sure the weight paid for each of the first N - 1 seats was equal before starting the Condorcet count.”

To take your example of a majority being flipped:

Modifying this to the ranked context, suppose the 51 B voters vote B1>B2>B3. So in standard STV, 2 B’s win, then B3 is eliminated for being furthest from quota, letting 3 A’s win. But with my modification, we do STV as normal up until the 4th seat, so 2 A’s and 2 B’s win, then, taking note that the least weight paid for any seat was the 4th seat for 16 votes, we restore ballot weight such that if the Hare Quota of 20 votes was used, the B voters get 4*2=8 votes back for a total of 11+8=19 votes and the A voters 1. Then, in the Condorcet count, B3 wins with a majority of 19 to 16 for A3 and 1 for A1.

A few notes unrelated to free riding, so you can skip this part. The Condorcet method (or any single-winner method really) has to pass mutual majority to retain STV’s Droop proportionality, and even then it might still force a failure.

More than one seat can be elected without a full quota in an STV election, and vote management often causes this to happen. When this occurs, the last seats are not filled one by one, but by an elimination leaving as many candidates in contention as there are seats. Consider the Dail elections held last Sunday. In 3-seat Dublin Rathdown, Fine Gael took the final two seats without either of their candidates meeting the quota. At the end of count 8, the results were:
Quota: 10601
Catherine Martin (Green Party): 11444, elected after count 5
Neale Richmond (Fine Gael): 9704 (elected)
Josepha Madigan (Fine Gael): 8677 (elected)
Shay Brennan (Fianna Fail): 8277 (eliminated)

The point being that the last two seats were filled at the same time. How do you deal with that?

Fill the second to last seat with the candidate with the most votes to fill it, and then initiate the Condorcet count with the vote restoration. So Neale Richmond would’ve won the second to last seat under my suggestion.

I should probably mention that this can be generalized as “when there are as many uneliminated candidates as seats to be filled, elect all of them except the one with the fewest votes.”

Is it possible to deal with STV free-riding by using a quota smaller than a Droop quota (perhaps in between Imperiali and Droop), and then ramping up the quota to a Droop quota and restarting the count if too many candidates would be elected (I.e. for a 5-seat election, if you kept the count going, and found you could elect 6 different candidates with Imperiali quotas, then you restart the count using Droop quotas)? This seems like an easy solution which can probably deal with the majority of free-riding if it works. The most important thing for deciding how to make it work would likely be to observe how small the “free-riding quota” (baliwick size) is in practice in relation to the Droop quota, and then pick a quota size closest to that as the starting point.