Hi Forest,

In 2005 I purported to show that Condorcet and FBC were incompatible, although
it does rely on a symmetric tie. I modified Woodall's proof regarding Condorcet
and LNHarm to get it.

You raise an interesting possibility of making them compatible by giving up UD.
But it seems to me the best we could do is find a format of voting under which
we can't determine how the definitions should apply.

I think you proposed two methods here, which are identical unless there are
pairwise ties.

The second one:

...seems to be the same as BTP, from Dec 2020.

This is a good FBC method but it's not compliant:

0.383: A>B>C
0.343: C=B>A  -->  C>A=B
0.179: A=C>B
0.092: B>C>A

A>B>C>A cycle, A wins, scoring off the two A-top factions.

But when the .343 lower B, C wins as CW with 100% score.

Kevin

(end)

Le vendredi 15 avril 2022, 19:51:08 UTC−5, Forest Simmons forest.simmons21@gmail.com a écrit :
The FBC (Favorite Betrayal Criterion) has long been thought (at least by me) to be incompatible with the Condorcet Criterion when restricted to Universal Domain election methods.

But today while contemplating how to propose DMC as it's own Condorcet completion method [lacking a CW, elect the most truncated candidate that pairwise beats every candidate with fewer truncations], my mind reverted back to a related DSV approval method that I had rejected because it was not precinct summable, sometimes requiring a second pass through the ballots to compactly summarize the necessary information:

Lacking an outright "True Majority Winner", elect the candidate that, on the most ballots, pairwise defeats every candidate ranked above it.

As I wracked my brain for a clever one-pass data compression idea, it suddenly hit me that this two pass DSV Approval method is both Condorcet and FBC compliant!

Suppose you raise your favorite F to equal rank with your compromise C on some ballot B. This move cannot decrease C's approval count, because C still pairwise defeats every candidate ranked above it on ballot B that it beat before. So the method passes the FBC.

How about the Condorcet Criterion? Well, the CW will always get a perfect 100 percent score, and will be ranked ahead of any other candidate X on at least one ballot, giving X a less than perfect Approval score.

Can a similar result be achieved by a one pass method?

For now let's call this method Two Pass FBC Condorcet (2PFBCC).

IRV routinely requires more than two passes thhrough the ballots, so 2PFBCC is better in this regard, since it only requires more than one pass when lacking a CW, i.e. extremely rarely, and never more than two ...soundly dominating IRV in summability ... not to mention monotonicity, Condorcet compliance and Compromise immunity (FBC) ... while of course retaining clone independence, etc.

And one more biggy ... simplicity and succinctness of definition: elect the candidate that, on the fewest ballots (if at all) is defeated head-to-head by any candidate ranked ahead of it.

Of course, for the lay person this definition must be supplemented by a definition of "head-to-head defeat" ... but that should not be too painful for a lover of democracy!

However, just for fun let's incorporate the head-to-head defeat definition into one complete definition for the entire method: 

Candidate X gets a point from ballot B if (and only if) every candidate Y ranked ahead of X on ballot B is merely an exception to the rule ...i.e more often than not X is ranked ahead of Y, even though on this particular ballot, candidate X is not ranked over Y.

It goes without saying that the candidate to be elected is the point winner.

This definition is self-contained including the heuristic that inspired it.

Heuristic: we can forgive X for being ranked below Y on ballot B, as long as that is more the exception than the rule when it comes to ballots in general.

A nagging question:

Should a point granted to X by ballot B be considered to be actual for X by the voter of ballot B even when B did not rank X at all, as long as X pairwise defeated all of the ranked candidates?

No, we withdraw the word "approval" originally used for this method in the DSV context ... but reserve the right to use the word consent:

Which is worse? ... that stretch of the word "consent" ? ... or the one that counts IRV voters as consenting to the IRV winner Y that they left unranked even though their favorite X defeated every other candidate pairwise, including Y.

In any case, here is my current proposal for 2PFBCC that skirts this issue: 

Lacking a CW ... for each ballot B, give a point to each candidate X that is ranked on ballot B, unless some candidate Y ranked above X on ballot B is also mostly (i.e. more often than not) ranked above X on other ballots, too. 

Finally, elect the point winner.

Is that a method most EM readers and their friends could live with?

How about the VoteFair and STAR vote people? 

How about RCV proponents in general? 

And how about Range/Score enthusiasts? 

[Among whom I count myself ... especially for Score Sorted Margins]

How about Majority Judgment supporters? ... to whom I am highly sympathetic, also.

I know we had our hearts set on a one pass method for Burlington, Vermont, But this method is de-facto one-pass (according to FairVote data) more than 99 percent of the time, and only 2-pass the rest of the time ... nothing compared to IRV's obligatorty multiple passes through the entire ballot set almost every election.

Try it, test it, and spread the word!

[or show me the simple bubble popping fact that I have over-looked]

-Forest

El vie., 15 de abr. de 2022 10:10 p. m., Kevin Venzke stepjak@yahoo.fr
escribió:

Or looking at it in reverse, when B is raised to equal top with C,  it
makes C lose because it no longer beats B pairwise, which makes it lose
points in the first and last factions ... which would have been OK had B
gained enough points to win.

Kevin,

It seems that my "proof" failed because I assumed that C's score could not
change by raising B equal to C ... but that's only true if we're talking
majority defeat: raising B to equal with C  cannot change a defeat of B by
C (or a non-majority defeat of C by B) to a majority defeat of C by B.

So let's try this fix:

Elect the candidate that on the fewest ballots is outranked by any
candidate that majority defeats it.

Or for lay person proposal completeness ...

Lacking a Condorcet winner, elect the candidate that on the fewest ballots
is outranked by any candidate that outranks it on a majority of ballots.

This is somewhat reminiscent of MDDA, so ties might happen with
non-negligible probability. Then, as in MDDA, why not break ties with
approval?

Implicit approval would keep it in the Universal Domain category. Explicit
approval could be a defense strategy lever.

Does that fix work?

Thanks,

-Forest

El sáb., 16 de abr. de 2022 12:05 a. m., Forest Simmons <
forest.simmons21@gmail.com> escribió:

Hi Forest,

Stated like this, this does satisfy FBC, however it is not a Condorcet method.

If I dub this method MajBTP, MajBTP is very close to MDDA as you note. It has
similar properties, including a Plurality failure risk with 4+ candidates. SFC
and SDSC/MD both seem to be preserved.

It's pretty interesting that this works, and to consider how this resolves
differently from MDDA. One example:

40: A>B>C
35: B>C>A
25: C>A>B

A>B>C>A majority cycle. MDDA elects B as the approval winner. MajBTP effectively
picks the first pref winner A, as every second-ranked candidate has a maj loss
to the first preference.

I might call this C//MajBTP. This can fail FBC in the same cases C//A does:

0.394: C=A>B
0.299: B=C>A  -->  B>A=C
0.179: B>A>C
0.126: A=B>C

A>C>B>A cycle, no majorities. A wins on approval.

When the .299 lower C, B becomes the CW.

Kevin

Kevin,

Thanks for your clarifications, insights and insightful examples.

-Forest

El sáb., 16 de abr. de 2022 3:54 p. m., Kevin Venzke stepjak@yahoo.fr
escribió:

It is well known that Range Voting, no matter its level of resolution, is
strategically equivalent to Approval. In particular, this means that under
perfect information conditions there always exists an optimal strategy that
makes no use of any intermediate ratings. [However, as in Linear
Programming, existence of an optimal "corner" solution in no way denies the
possible existence of other equally optimal non-corner solutions.]

Not so well known, but equally true, is that every Condorcet compliant,
Universal Domain (i.e. RCV) method reduces to  Approval when voters vote
only at the extremes.

Question 1. Does every perfect information UD Condorcet election have an
optimal strategy that makes no use of the intermediate rankings? This
certainly seems to be the tacit assumption of many Designated Strategy
Voting methods.

Question 2. Since Approval satisfies the Favorite Betrayal Criterion, does
it follow that any method that has an optimal strategy that makes no use of
the intermediate ranks in some sense satisfies the FBC? Could we call that
Strategic FBC?

And it seems possible that one DSV strategy for transforming a UD election
into an approval election might satisfy the FBC while another might not.

Suppose a DSV method M converts  UD elections into approval elections in a
CW preserving way, i.e. if X is the CW of some UD ballot set beta, then X
will also be the CW of the approval ballot set M(beta) and therefore the
approval winner. It seems like a voter voting through that DSV method M
would not be highly tempted to rank Favorite under Compromise, especially
if under M, candidates ranked top or equal top always get approved on
M(beta).

Here is an example of just such a DSV method M closely related to (but
better than) the flash in the pan method 2PFBCC:

First, for the ballot set beta, find and summarize the pairwise defeats and
ties in some convenient form.

Then convert each ballot B of beta into approval form by use of an
inclusive approval cutoff K defined by the lowest ranked candidate of B
that is not pairwise defeated by any candidate ranked ahead of it on ballot
B.

Since the inclusive cutoff is defined by a "ranked candidate" it cannot be
a truncated candidate. So a ballot that truncates all of the candidates
that are not ranked equal top makes approval the same as equal top. This
feature gives voters that don't trust M the ability to specify their own
approvals.

Note that if there is a CW it will define the cutoff on every ballot that
ranks it, a d since the cutoff is inclusive, the CW will be approved by
every ballot that ranks it.

It follows that the CW will be the approval winner, since it is ranked
(hence approved) above any rival on more ballots than not.

So here's my proposal ...

Given an RCV ballot set beta, elect the CW of beta if there is one, else
elect the CW of M(beta).

-Forest

El sáb., 16 de abr. de 2022 4:33 p. m., Forest Simmons <
forest.simmons21@gmail.com> escribió:

On 18.04.2022 02:21, Forest Simmons wrote:

As I understand it, in proper Condorcet methods, it's sometimes useful
to use intermediate rankings because you can both express a preference
for A over B and one for B over C at the same time.

There are modifications of Condorcet that pass the FBC, e.g. Kevin's ICA
and Mike Ossipoff's ICT and Symmetrical ICT. These work by making
Approval strategy equally-optimal so that if (for the purpose of
contradiction) your favorite is F and compromise is C, then F=C>... can
be no worse then C>... But in so doing, they lose Condorcet efficiency.[1]

Kevin's simulations of
http://lists.electorama.com/pipermail/election-methods-electorama.com/2005-June/114476.html
seem to indicate that Condorcet methods (at least "advanced" ones like
Schulze) have a low rate of FBC failure. The "Improved Condorcet"
methods would presumably be the flipside of this coin, passing FBC
absolutely but having some (low?) rate of Condorcet failure.

I also seem to recall that MMPO doesn't reduce to Approval despite
passing FBC, but my memory is not the clearest, so I could be mistaken.

-km

[1] A similar trick applied to Borda is Mike's Summed Ranks method.

Kris,

Your comments remind me that (if I remember correctly) there is supposed to
always exists a Nash equilibrium approval ballot set which elects the
sincere CW candidate when one exists.

But a DSV method that finds such an equilibrium (along with its concomitant
candidate) would have to satisfy the FBC, since any one voter defecting
from that equilibrium to approve her favorite F would get away with it ...
if the winner changed at all it would have to change to F.

So all we need is a constructive proof of the alleged Nash Equilibrium
existence.

Can someone clear up this mystery?

El lun., 18 de abr. de 2022 2:40 a. m., Kristofer Munsterhjelm <
km_elmet@t-online.de> escribió:

It doesn't, in fact Kevin's first EM post twenty years ago (plus or minus a
few months) was MMPO in the context of Approval ballots. His example showed
that MMPO does not reduce to approval and (unlike Approval) fails both
Plurality and the ballot Condorcet criterion.

My example DSV method M below is a conversion of any RCV style ballot set
beta into an approval ballot set M(beta) such that the approval winner of
M(beta) is the only possibility for a CW of beta.

Kevin gave a  example to show that method M fails the FBC, and I showed
that changing the phrase "is not defeated by" to " is not majority defeated
by " makes the method FBC compliant and even preserves the property of the
CW getting maximum DSV approval ... but is not decisive because (analogous
to Copeland) there is an appreciable chance that the argmaxapproval set
will not be a singleton. If you break this DSV approval tie with something
like Plurality or Implicit approval, it seems that the CW is not guaranteed
to win. So the situation is very similar to ICT and MDDA.

-Forest