a solution to the by-election problem?
Hi all,
The by-election problem is the following:
Starting with a Droop compliant solution for the M-winner problem,
exclude one of the M winners. Now determine a new set of M-winners
that includes the M-1 remaining previous winners and is also Droop
compliant. There is no STV method that does this. Either you preserve
Droop compliance but not all the M-1 previous winners are reelected,
or you protect the M-1 previous winners from exclusion and the new Mth
winner is not guaranteed to be Droop compliant.
However, I believe the method I describe in Section V of "A House
Monotone, Coherent, and Droop Proportional Ranked Candidate Voting
Method" https://arxiv.org/pdf/2506.12318 does this correctly when
setting the M-1 remaining winners as “previously elected." The proof
is in Section VI Claim 9, with N=M.
Can someone check this?
Best,
Ross
On 2026-04-20 23:39, Ross Hyman via Election-Methods wrote:
Hi all,
The by-election problem is the following:
Starting with a Droop compliant solution for the M-winner problem,
exclude one of the M winners. Now determine a new set of M-winners
that includes the M-1 remaining previous winners and is also Droop
compliant. There is no STV method that does this. Either you preserve
Droop compliance but not all the M-1 previous winners are reelected,
or you protect the M-1 previous winners from exclusion and the new Mth
winner is not guaranteed to be Droop compliant.
Don't combinatorial methods lke Schulze STV or CPO-STV do this?
Do the Condorcet election with every superset of the (M-1) winners that
have already been elected. Pick the winner according to the pairwise method.
Or can they still work themself into a corner they can't get out of?
-km
Perhaps I wrong about this. Elect M winners using a Droop
proportional STV method. Exclude one of the winners from all ballots
and rerun the election for M winners, protecting the M-1 previous
winners from exclusion. Can someone present a proof that it will
always be the case that the new M winners (which must include the
previous M-1 winners) satisfy Droop proportionality? Or can someone
present a counter-example?
On Tue, Apr 21, 2026 at 5:21 AM Kristofer Munsterhjelm
km-elmet@munsterhjelm.no wrote:
On 2026-04-20 23:39, Ross Hyman via Election-Methods wrote:
Hi all,
The by-election problem is the following:
Starting with a Droop compliant solution for the M-winner problem,
exclude one of the M winners. Now determine a new set of M-winners
that includes the M-1 remaining previous winners and is also Droop
compliant. There is no STV method that does this. Either you preserve
Droop compliance but not all the M-1 previous winners are reelected,
or you protect the M-1 previous winners from exclusion and the new Mth
winner is not guaranteed to be Droop compliant.
Don't combinatorial methods lke Schulze STV or CPO-STV do this?
Do the Condorcet election with every superset of the (M-1) winners that
have already been elected. Pick the winner according to the pairwise method.
Or can they still work themself into a corner they can't get out of?
-km
I think this example shows by-election failure for most STV methods
(including, sadly, my own).
35 C>A>D>B
34 C>B>D>A
31 D>C>A>B
2 winners Droop quota 33 1/3.
C and D win.
exclude C.
A and B must win. If you force D to be a winner Droop proportionality
is violated.
Is there an STV-like method that initially elects C and A?
On Tue, Apr 21, 2026 at 12:25 PM Ross Hyman rossahyman@gmail.com wrote:
Perhaps I wrong about this. Elect M winners using a Droop
proportional STV method. Exclude one of the winners from all ballots
and rerun the election for M winners, protecting the M-1 previous
winners from exclusion. Can someone present a proof that it will
always be the case that the new M winners (which must include the
previous M-1 winners) satisfy Droop proportionality? Or can someone
present a counter-example?
On Tue, Apr 21, 2026 at 5:21 AM Kristofer Munsterhjelm
km-elmet@munsterhjelm.no wrote:
On 2026-04-20 23:39, Ross Hyman via Election-Methods wrote:
Hi all,
The by-election problem is the following:
Starting with a Droop compliant solution for the M-winner problem,
exclude one of the M winners. Now determine a new set of M-winners
that includes the M-1 remaining previous winners and is also Droop
compliant. There is no STV method that does this. Either you preserve
Droop compliance but not all the M-1 previous winners are reelected,
or you protect the M-1 previous winners from exclusion and the new Mth
winner is not guaranteed to be Droop compliant.
Don't combinatorial methods lke Schulze STV or CPO-STV do this?
Do the Condorcet election with every superset of the (M-1) winners that
have already been elected. Pick the winner according to the pairwise method.
Or can they still work themself into a corner they can't get out of?
-km
On Tue, 21 Apr 2026 13:30:51 -0500
Ross Hyman via Election-Methods election-methods@lists.electorama.com wrote:
I think this example shows by-election failure for most STV methods
(including, sadly, my own).
35 C>A>D>B
34 C>B>D>A
31 D>C>A>B
2 winners Droop quota 33 1/3.
C and D win.
exclude C.
A and B must win. If you force D to be a winner Droop proportionality
is violated.
Is there an STV-like method that initially elects C and A?
This might be off topic, but in case its of interest,
why should C not given both wins/seats/voting power units?
If its because we must fill two litteral seats and
(possibly quite sensibly) refuse to use party lists
then we could just use something like "reserved seat numbers"
where representatives gains voting power units equal to
the number of seats they reserved in the election.
Depending on ones priorities, some people will find such an
implementation an improvement because the elected candidates
can then be directly held accountable for their votes,
rather then today where party mebers are simply doing their job
obeying the party line.
The individuals will also have more barganing power against lobbyists
increasing the expected cost of lobbying,
at the possible risk of slightly makeing the process easier to get in on
(good if you believe in competition, bad if you don't for this case).
Gustav