AFB Ranked Pairs Attempt
So thanks to the feedback I got on my previous post,
I made an attempt at modifying Ranked Pairs
in hopes to create a voting system satisfying
Avoids Favorite Betrayal (AFB), Majority-Beat Condorcet critiera,
Monotonicity, Mutual Majority, and 1 of Later-No-Help/Harm.
But first I will do some definitions for clarification.
A candidate Alice Plurality-Beats another candidate Bob
if the number of votes strictly prefering Alice to Bob
is greater then the number of votes strictly prefering Bob to Alice,
or v(Alice > Bob) > v(Bob > Alice).
We futher define that Alice Majority-Beats Bob at the minimal majority
threshold if the number of votes strictly prefering Alice to Bob
is strictly greater then half the total number of votes.
I will use minimal majority of half as the majority threshold
unless otherwise specified.
If a voting method always without exception inferes complete
preference orders from the ballots,
and the total number of voter are predetermined and fixed
(spoiled ballots and noshows still counts to the total),
then Alice Majority-Beating Bob at the minimal majority threshold
conicides with v(Alice > Bob) > v(Alice = Bob) + v(Alice < Bob).
The classical Condorcet winner/loser candidate and criteria
as well as Smith and ISDA implicitly assumes Plurality-Beats,
but I will prefix the cirteria with PB and MB to avoid ambiguity.
A result of PB vs MB is that the PB-Smith set is a subset of the MB-Smith set,
and PB-Smith and PB-Condorcet criteras imply their MB counterparts,
but it is possible to be a PB-Condorcet winner without being a MB-Condorcet winner
as well as a MB-Smith set member without being a PB-Smith set winner.
The incompatability of PB-Condorcet cirteria with AFB and LN-Help/Harm
does not apply to the MB-Condorcet criteria, so I have been trying to create
a system satisfying AFB, MB-Condorcet critera, Monotonicity and 2 out of the
following 3 Mutual Majority, LN-Help, LN-Harm in hopes of getting a useful
baseline comparison.
Assuming the above assumptions for Majority-Beating coinciding
with v(A>B) > v(A=B)+v(A<B) holds, I propose the following system:
- Go through each pairwise matchup and check if it is a Majority-Beat,
if so add it to the list in the form of
Winner Majority-Beats Loser with Score
where Score is the number of votes where the pairwise winner
is ranked greater or equal to the pairwise loser,
or Score = v(Winner > Loser) + v(Winner = Loser).
This means pairwise matchups where neither Majority-Beats the other
won't end up on the list. - Sort the list according to the highest score first, lowest last.
- Go through the sorted list in order and lock in the pairwise matchup,
and marking the pairwise looser of each entry as defeated,
unless it would result in a cycle,
in which case we skip the entry. - If there is only 1 candidate which have not been marked defeated
after locking in the matchups, elect that candidate as the winner.
If there are more then 1, do tie-breaking among them.
There will always be at least 1 candidate which have not benn marked as defeated.
Compared to Rank Pairs, step (2), (3), and (4) are the same if I have correctly
understood whats on electowiki correctly.
https://electowiki.org/wiki/Ranked_pairs
https://electowiki.org/wiki/Maximize_Affirmed_Majoritues
https://electowiki.org/wiki/River
Step (1) differ in two ways, only pairwise matchups resulting in Majority-Beats
are considered, and how we score each matchup.
What the tie breaker actually is was unclear,
the links for random voter hierarchy went to the MAM page,
which description appears to be specific to MAM.
I will assume the tie-breaker used won't cause any confilict
with criteria compatability.
So far for criteria compatability I have only managed to prove
MB-ISDA which implies MB-Smith which implies Mutal Majority.
MB-Condorcet winner criteria, and MB-Condorcet loser criteria.
Every member of the MB-Smith set will have all their pairwise
matchups against non-members of the MB-Smith set locked in,
without risk of marking MB-Smith set members defeated,
resulting in MB-ISDA satisfaction.
Monotonicity should carry over from regular Ranked Pairs,
but I am unsure how to fully prove it.
I also believe it satisfy AFB and one of LN-Help/Harm based
on my understand of ICA and MMPO, but no proof so far.
I futher suspect that if set (2) is changed to sort the score in reverse,
that is lowest first to highest last, we get the other of LN-Help/Harm.
Hopefully of interest
Gustav
Hi Gustav,
I think I understand your definition, a Ranked Pairs based on majorities only,
ranking them essentially by losing side votes, fewest being strongest. So here are
some scenarios for you to consider. Let me know if you disagree on the winners.
Regarding favorite betrayal:
0.411: A>B>C
0.285: B=C>A
0.210: C>A>B
0.091: B>C>A
Defeat order is B>C, A>B, C>A. A wins.
0.411: A>B>C
0.285: C>A=B (i.e. B has been "betrayed")
0.210: C>A>B
0.091: B>C>A
This strengthens A>B and weakens B>C.
Defeat order is now A>B, C>A, B>C. C wins.
(MMPO elects C in both scenarios. ICA elects B in the first, A in the second.)
Regarding Later-no-harm:
0.297: C>A>B
0.294: A>B>C
0.289: B>C>A
0.117: A>B=C
Defeat order is A>B, B>C, C>A. A wins.
0.297: C>A>B
0.294: A>B>C
0.289: B>C>A
0.117: A>C>B (i.e. adding a C lower preference)
This weakens B>C without reversing it.
Defeat order is now A>B, C>A, B>C. C wins.
(MMPO elects C in both scenarios.)
Regarding Later-no-help:
0.327: C>A=B
0.269: A>C>B
0.262: B>A>C
0.110: C>A=B
0.030: A>B=C
Defeat order is C>B, A>C. B has no majorities. So A wins.
0.327: C>B>A (i.e. adding a B lower preference)
0.269: A>C>B
0.262: B>A>C
0.110: C>A=B
0.030: A>B=C
This creates a B>A defeat that is stronger than A>C.
Defeat order is now C>B, B>A, A>C. C wins.
I do think the method is monotone. But you have to be careful if you decide to have
a tiebreaker rule.
I hope I didn't make any errors in my math.
Kevin
votingmethods.net
So thanks to the feedback I got on my previous post,
I made an attempt at modifying Ranked Pairs
in hopes to create a voting system satisfying
Avoids Favorite Betrayal (AFB), Majority-Beat Condorcet critiera,
Monotonicity, Mutual Majority, and 1 of Later-No-Help/Harm.
But first I will do some definitions for clarification.
A candidate Alice Plurality-Beats another candidate Bob
if the number of votes strictly prefering Alice to Bob
is greater then the number of votes strictly prefering Bob to Alice,
or v(Alice > Bob) > v(Bob > Alice).
We futher define that Alice Majority-Beats Bob at the minimal majority
threshold if the number of votes strictly prefering Alice to Bob
is strictly greater then half the total number of votes.
I will use minimal majority of half as the majority threshold
unless otherwise specified.
If a voting method always without exception inferes complete
preference orders from the ballots,
and the total number of voter are predetermined and fixed
(spoiled ballots and noshows still counts to the total),
then Alice Majority-Beating Bob at the minimal majority threshold
conicides with v(Alice > Bob) > v(Alice = Bob) + v(Alice < Bob).
The classical Condorcet winner/loser candidate and criteria
as well as Smith and ISDA implicitly assumes Plurality-Beats,
but I will prefix the cirteria with PB and MB to avoid ambiguity.
A result of PB vs MB is that the PB-Smith set is a subset of the MB-Smith set,
and PB-Smith and PB-Condorcet criteras imply their MB counterparts,
but it is possible to be a PB-Condorcet winner without being a MB-Condorcet winner
as well as a MB-Smith set member without being a PB-Smith set winner.
The incompatability of PB-Condorcet cirteria with AFB and LN-Help/Harm
does not apply to the MB-Condorcet criteria, so I have been trying to create
a system satisfying AFB, MB-Condorcet critera, Monotonicity and 2 out of the
following 3 Mutual Majority, LN-Help, LN-Harm in hopes of getting a useful
baseline comparison.
Assuming the above assumptions for Majority-Beating coinciding
with v(A>B) > v(A=B)+v(A<B) holds, I propose the following system:
- Go through each pairwise matchup and check if it is a Majority-Beat,
if so add it to the list in the form of
Winner Majority-Beats Loser with Score
where Score is the number of votes where the pairwise winner
is ranked greater or equal to the pairwise loser,
or Score = v(Winner > Loser) + v(Winner = Loser).
This means pairwise matchups where neither Majority-Beats the other
won't end up on the list. - Sort the list according to the highest score first, lowest last.
- Go through the sorted list in order and lock in the pairwise matchup,
and marking the pairwise looser of each entry as defeated,
unless it would result in a cycle,
in which case we skip the entry. - If there is only 1 candidate which have not been marked defeated
after locking in the matchups, elect that candidate as the winner.
If there are more then 1, do tie-breaking among them.
There will always be at least 1 candidate which have not benn marked as defeated.
Compared to Rank Pairs, step (2), (3), and (4) are the same if I have correctly
understood whats on electowiki correctly.
https://electowiki.org/wiki/Ranked_pairs
https://electowiki.org/wiki/Maximize_Affirmed_Majoritues
https://electowiki.org/wiki/River
Step (1) differ in two ways, only pairwise matchups resulting in Majority-Beats
are considered, and how we score each matchup.
What the tie breaker actually is was unclear,
the links for random voter hierarchy went to the MAM page,
which description appears to be specific to MAM.
I will assume the tie-breaker used won't cause any confilict
with criteria compatability.
So far for criteria compatability I have only managed to prove
MB-ISDA which implies MB-Smith which implies Mutal Majority.
MB-Condorcet winner criteria, and MB-Condorcet loser criteria.
Every member of the MB-Smith set will have all their pairwise
matchups against non-members of the MB-Smith set locked in,
without risk of marking MB-Smith set members defeated,
resulting in MB-ISDA satisfaction.
Monotonicity should carry over from regular Ranked Pairs,
but I am unsure how to fully prove it.
I also believe it satisfy AFB and one of LN-Help/Harm based
on my understand of ICA and MMPO, but no proof so far.
I futher suspect that if set (2) is changed to sort the score in reverse,
that is lowest first to highest last, we get the other of LN-Help/Harm.
On Mon, 20 Apr 2026 02:54:08 +0000 (UTC)
Kevin Venzke stepjak@yahoo.fr wrote:
Hi Gustav,
I think I understand your definition, a Ranked Pairs based on majorities only,
ranking them essentially by losing side votes, fewest being strongest. So here are
some scenarios for you to consider. Let me know if you disagree on the winners.
Regarding favorite betrayal:
0.411: A>B>C
0.285: B=C>A
0.210: C>A>B
0.091: B>C>A
Defeat order is B>C, A>B, C>A. A wins.
0.411: A>B>C
0.285: C>A=B (i.e. B has been "betrayed")
0.210: C>A>B
0.091: B>C>A
This strengthens A>B and weakens B>C.
Defeat order is now A>B, C>A, B>C. C wins.
(MMPO elects C in both scenarios. ICA elects B in the first, A in the second.)
That is quite the counterexample,
but mind if I ask why you did not mention the monotonicity failure
for the reversed version where we go from C winning to B winning
by lowering B?
Regarding Later-no-harm:
0.297: C>A>B
0.294: A>B>C
0.289: B>C>A
0.117: A>B=C
Defeat order is A>B, B>C, C>A. A wins.
0.297: C>A>B
0.294: A>B>C
0.289: B>C>A
0.117: A>C>B (i.e. adding a C lower preference)
This weakens B>C without reversing it.
Defeat order is now A>B, C>A, B>C. C wins.
(MMPO elects C in both scenarios.)
Everything looks correct here (except my hope for LN-Harm compatability).
Regarding Later-no-help:
0.327: C>A=B
0.269: A>C>B
0.262: B>A>C
0.110: C>A=B
0.030: A>B=C
Defeat order is C>B, A>C. B has no majorities. So A wins.
0.327: C>B>A (i.e. adding a B lower preference)
0.269: A>C>B
0.262: B>A>C
0.110: C>A=B
0.030: A>B=C
This creates a B>A defeat that is stronger than A>C.
Defeat order is now C>B, B>A, A>C. C wins.
Everything look correct here (except my hope for LN-Help compatability).
So at this point I decided to play around with similar versions to the proposed method,
since my guess that sorting by greater or equal would be the key.
specifically I found replacing the scoring of each entry from
v(Winner > Loser) + v(Winner = Loser) to strictly v(Winner > Loser),
while keeping sorting largest first to lowest last in place,
makes things a little more hopefull
In the first example the order becomes
0.621 A>B
0.585 C>A
0.502 B>C
in the first part and
0.621 A>B
0.586 C>A
0.501 B>C
in the second with
C winning in both (and reversed sorting/locking order B winning both)
In the LN-Harm counterexample, this modification to the scores gives
C winning both (and reversed sorting/locking order B winning both).
In the LN-Help counterexample, this modification to the scores gives
A winning first, and then C, keeping the LN-Help failure,
but the reversed sorting/locking order first elects A but gives
B the win in the second part.
So thanks for the great conter examples,
this will save my much otherwise wasted time and effort.
Since the modification to scoring each Majority-Beat appeared to
hold up to the counterexamples, I will repropose the similar method in case
it is of any interest to anyone.
(1) Go through each pairwise matchup and check if it is a Majority-Beat,
if so add it to the list in the form of
Winner Majority-Beats Loser with Score
where Score = v(Winner > Loser) (no differences).
(2) Sort the list according to highest score first (top), lowest (last).
(3) Go thourgh the sorted list in order and lock in the pairwise matchups,
and marking the pairwise loser of each entry as defeated,
unless it would result in a cycle, in which case we skip the entry.
(4) If there is exactly 1 candidate no marked as defeated, elect that candidate.
If there are more the 1, do tie-breaking.
Since this is in practice 2 method, I will refere to them as
Top to Bottom (T2B) and Bottom to Top (B2T) for locking in highest first
to lowest score last and lowest first to highest last respectively.
(So something like Ordered Majority-Beats? OMB(T2B), OBM(B2T)?
I could use some naming suggestions if this second attempt succedes.)
Based on the counterexamples above hopefully this means
OMB(T2B) satisfies AFB + Mono + LN-Harm (as it fails the LN-Help example)
OMB(B2T) satisfies AFB + Mono + LN-Help
but I could only prove MB-ISDA in general for both of them.
I do think the method is monotone. But you have to be careful if you decide to have
a tiebreaker rule.
I fully agree here on the tiebreaker.
I hope I didn't make any errors in my math.
I could not find any.
It looks like you understood everything perfectly.
Gustav
Hi Gustav,
Le lundi 20 avril 2026 à 13:00:19 UTC−5, Gustav Thorzen via Election-Methods election-methods@lists.electorama.com a écrit :
That is quite the counterexample,
but mind if I ask why you did not mention the monotonicity failure
for the reversed version where we go from C winning to B winning
by lowering B?
I wasn't checking the reversed method, so it wasn't on my mind. From your original
post I wasn't totally sure whether you thought the reversed method was still
monotone, but it surely isn't, because it penalizes candidates for getting more
votes in their losing contests.
The reversed method of the new suggestion has a similar issue: Being penalized for
getting more votes when on the winning side.
So at this point I decided to play around with similar versions to the proposed method,
since my guess that sorting by greater or equal would be the key.
specifically I found replacing the scoring of each entry from
v(Winner > Loser) + v(Winner = Loser) to strictly v(Winner > Loser),
while keeping sorting largest first to lowest last in place,
makes things a little more hopefull
This territory is a little better explored, but I think four candidates are needed
for some of the demonstrations.
The issue is not how you count the defeat strengths, but that the locking mechanism
permits chaotic effects. I tried to warn about this earlier, but I didn't mention
locking per se, maybe.
Favorite betrayal:
0.411: C=A>B>D -> A>B>C=D (i.e. C is betrayed)
0.377: D>A>C>B
0.158: B>D>C>A
0.052: A>D>C>B
Win moves from D to A.
(First the B>D win is rejected, then D>A is.)
Later-no-harm:
0.377: C>A>B>D
0.277: D>A>B=C -> D>A>C>B (i.e. add lower preference for C)
0.199: A>B>D>C
0.145: B>D>C>A
Win moves from A to C.
(First the C>A win is rejected, then D>C is.)
Later-no-help:
0.350: B>A>C=D -> B>A>D>C (i.e add lower preference for D)
0.331: D>C>B>A
0.240: C>A>D>B
0.077: A>B=D=C
Win moves from C to B.
(First C has no majority losses; then it does, to D, and C places last in the ranking.)
Again, hopefully no errors.
A pattern is that which defeats get rejected or locked seems to lack a clear
connection to how we modified the ballots. This makes it extremely difficult to
offer guarantees about the effects of vote changes.
So thanks for the great conter examples,
this will save my much otherwise wasted time and effort.
I hope I could help, all in all.
Kevin
votingmethods.net
On Mon, 20 Apr 2026 19:54:43 +0000 (UTC)
Kevin Venzke stepjak@yahoo.fr wrote:
Hi Gustav,
Le lundi 20 avril 2026 à 13:00:19 UTC−5, Gustav Thorzen via Election-Methods election-methods@lists.electorama.com a écrit :
That is quite the counterexample,
but mind if I ask why you did not mention the monotonicity failure
for the reversed version where we go from C winning to B winning
by lowering B?
I wasn't checking the reversed method, so it wasn't on my mind. From your original
post I wasn't totally sure whether you thought the reversed method was still
monotone, but it surely isn't, because it penalizes candidates for getting more
votes in their losing contests.
The reversed method of the new suggestion has a similar issue: Being penalized for
getting more votes when on the winning side.
I initially put these ideas aside thinking just that,
but I got hopefull when I discovered that MB-ISDA
was obtainable independently of locking order,
since that threw out a conjecture I used in an attempt to
disprove Monotonicity. Not that it mattered in the end.
So at this point I decided to play around with similar versions to the proposed method,
since my guess that sorting by greater or equal would be the key.
specifically I found replacing the scoring of each entry from
v(Winner > Loser) + v(Winner = Loser) to strictly v(Winner > Loser),
while keeping sorting largest first to lowest last in place,
makes things a little more hopefull
This territory is a little better explored, but I think four candidates are needed
for some of the demonstrations.
The issue is not how you count the defeat strengths, but that the locking mechanism
permits chaotic effects. I tried to warn about this earlier, but I didn't mention
locking per se, maybe.
Favorite betrayal:
0.411: C=A>B>D -> A>B>C=D (i.e. C is betrayed)
0.377: D>A>C>B
0.158: B>D>C>A
0.052: A>D>C>B
Win moves from D to A.
(First the B>D win is rejected, then D>A is.)
I just noticed the ratios not summing to 1,
since it looks like rounding errors I will assume
this won't affect anything relevant.
First round
0.830 A>B
0.788 C>B
0.587 D>C
0.569 B>D
0.535 D>A
OMB(T2B) winner D
OMB(B2T) winner B
Second round
0.842 A>C
0.830 A>B
0.569 B>C
0.569 B>D
0.535 D>A
0.535 D>C
OMB(T2B) winner A
OMB(B2T) winner A
So the AFB failure applies to both.
Later-no-harm:
0.377: C>A>B>D
0.277: D>A>B=C -> D>A>C>B (i.e. add lower preference for C)
0.199: A>B>D>C
0.145: B>D>C>A
Win moves from A to C.
(First the C>A win is rejected, then D>C is.)
Later-no-help:
0.350: B>A>C=D -> B>A>D>C (i.e add lower preference for D)
0.331: D>C>B>A
0.240: C>A>D>B
0.077: A>B=D=C
Win moves from C to B.
(First C has no majority losses; then it does, to D, and C places last in the ranking.)
Again, hopefully no errors.
I did not find any errors,
but I did not check in detail for LN-H* counterexamples
since the AFB failures leaves the methods moot for baseline comparison,
and I am out of Ranked Pair variant ideas with potential.
(Actually there is MMPO locking order, but that have
edge cases I need to figure out definitions to deal with,
before it becomes an actual method.)
In any case MB-ISDA follow trivially from only
considering Majority-Beats and refusing to lock cycles,
regardless of locking order,
in case anyone have other ideas to try.
A pattern is that which defeats get rejected or locked seems to lack a clear
connection to how we modified the ballots. This makes it extremely difficult to
offer guarantees about the effects of vote changes.
I agree on this, I initially made good progress using
conservation laws resulting from only considering Majority-Beats,
but failed to cover some cases (including every equally scored matchups),
not that it would have worked knowing these counterexamples.
So thanks for the great conter examples,
this will save my much otherwise wasted time and effort.
I hope I could help, all in all.
Well, until I get decent at finding nontrivial counterexamples on demand,
simply knowing what is and is not compatible makes learning by reverse enginering
theorems and proof a lot easier (let alone time and effort saving on dead ends),
and by extension every attempt to prove something.
Gustav