[EM] AFB Ranked Pairs Attempt

[glist at glas5.com]

On Mon, 20 Apr 2026 02:54:08 +0000 (UTC)
Kevin Venzke <stepjak at 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