[EM] A possible low-strategy PR concept when picking n-1 winners from n candidates

[Gustav Thorzen]

On Fri, 1 May 2026 23:38:43 +0200
Kristofer Munsterhjelm via Election-Methods <election-methods at lists.electorama.com> wrote:

> When picking a winner from two candidates, majority rule is 
> strategyproof. I was playing with ways to generalize this to multiwinner 
> to find at least *some* domain where a method can be both Droop 
> proportional and strategy-proof, and found this for elections with n 
> candidates and (n-1) seats:

Unless I misunderstood what is ment here,
it seems like you claim to have found a counterexample
to Duggan-Schwartz impossibility theorem (reducing to Gibbard-Satterthwaite when single winner)
https://en.wikipedia.org/wiki/Duggan-Schwartz_theorem (wikipedia summary only)
and there is also a corresponding generalization to Gibbards theorem without
the rankorder ballots requirement (though I don't know any common name for that one).

> Consider a set of n-1 winners {A_1, ..., A_(n-1)}, and call the loser B.
> If there exists strictly more than a Droop quota of voters who prefers 
> A_k to B for any k, (but not necessarily the same voters for each A_k) 
> then say that this set is a stable set. (That is, some subset of the 
> voters prefers A_1 to B, another subset of the voters prefers A_2 to B 
> and so on, and all of these subsets contain more than a Droop quota's 
> woth of voters.)
> 
> If there is only a single stable set, and the method elects that set, 
> then it passes the DPC for that election and it's also strategyproof.
> 
> Suppose that some group of voters wants to kick out A_k and replace him 
> with B. These voters must necessarily (honestly) rank B over A_k. But 
> we've established that there exists some Droop quota of voters who rank 
> A_k over B. The B>A_k voters can do nothing to stop the existence of 
> such a contingent of voters, because they are already maximally 
> contributing to B over A_k, so they can't break the stable set's 
> condition of a Droop quota or more ranking A_k over B, and thus they 
> can't undo this stable set.
> 
> So the only thing they can try to do is to establish another stable set 
> and hope that the method switches over. Their objective is to kick out 
> A_k, so this other candidate set must be {A_1, ..., A_(k-1), A_(k+1), 
> ..., A_(n-1), B}. But for such a set to exist, there must exist at least 
> a Droop quota's worth of voters who rank B over A_k. Since the B>A_k 
> voters are already ranking B over A_k, they can't contribute to the 
> creation of such a subset either. Hence if they create another stable 
> set by altering their ballots, it is not the one they want.
> 
> Since this argument holds for all A_k, there should be nothing anybody 
> who has an incentive to get B elected can do to get B elected.
> 
> (I'd imagine there's also a Condorcet extension for this concept for 
> electing k out of n candidates, but I can't quite see it yet. From a DSV 
> perspective, such a generalization would be "automatic vote management" 
> if single-winner Condorcet is "automatic compromising".
> 
> The argument above also resembles the reasoning I used for the resistant 
> set, so maybe there's a generalization that connects it with that.)
> 
> The problem is that there may be multiple stable sets. Impartial culture 
> elections are almost certain to have more than one. Here's an example:
> 
>   50: A>B>C
> 306: A>C>B
>   65: B>A>C
> 426: B>C>A
> 260: C>A>B
>   93: C>B>A
> 1200 voters in total, and the quota for two seats is 400.
> 
> {A,B} is a stable set because we can assign
> 	306 A>C>B voters,
> 	 50 A>B>C voters, and
> 	 65 B>A>C voters
> to the group that prefers A to C. This is more than a Droop quota in 
> total. Similarly, we can assign 426 B>C>A voters to the group that 
> prefers B to C.
> 
> But {B, C} is also a stable set:
> 	 65 B>A>C
> 	378 B>C>A
> 	= 443 voters preferring B to A, and
> 	260 C>A>B
> 	 93 C>B>A
> 	 48 B>C>A
> 	= 401 voters preferring C to A.
> 
> So what we might wonder is: does there exist a way for a voting method 
> to pick a particular stable set when more than one exists, and to stick 
> to that stable set no matter what the voters who prefer the other stable 
> set may do?
> 
> Unless my simulations are wrong, Schulze STV is almost always 
> strategyproof for impartial culture and (n-1) out of n, so that suggests 
> that it's possible at least some of the time. But Schulze STV has a 
> nonzero coalitional manipulability rate for (n-1) out of n with other 
> models, say a spatial model. Is that because the method it uses to pick 
> a particular stable set is strategyproof almost always for IC but not in 
> a spatial model? What's special about it in that case? Is it something 
> like "for three candidates and two to be elected, it's strategyproof 
> whenever there are only two stable sets" and then spatial model, but not 
> IC, sometimes has three?
> 
> For three candidates, at least, there should always be at least one 
> stable set, unless there's an exact tie. I don't know if that holds for 
> every (n-1) of n election.
> 
> In any case, it might be something to investigate further at some point.

If I have understood correctly so far then you have not subverted the assumptions
of Duggan-Schwartz impossibility theorems and choosen:
Voter symmetry implicitly +
Candidate symmetry implicitly +
Explicitly placed a bound on the maximum number of winners
strictly lower then the number of candidates,

and therefore must be vulnerable to some form of strategic dishonesty,
in some cases.

Hopefully of interest
Gustav