[EM] Favorite Betrayal and Condorcet

Forest Simmons forest.simmons21 at gmail.com
Sat Apr 16 11:22:34 PDT 2022


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 at gmail.com> escribió:

>
>
> El vie., 15 de abr. de 2022 10:10 p. m., Kevin Venzke <stepjak at yahoo.fr>
> escribió:
>
>> Hi Forest,
>>
>> > 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.
>>
>> 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:
>> > elect the candidate that, on the fewest ballots (if at all) is defeated
>> > head-to-head by any candidate ranked ahead of it.
>>
>> ...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.
>>
>
> 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
>>
>> (end)
>>
>>
>>
>> Le vendredi 15 avril 2022, 19:51:08 UTC−5, Forest Simmons <
>> forest.simmons21 at 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
>>
>>
>>
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://lists.electorama.com/pipermail/election-methods-electorama.com/attachments/20220416/59b17fac/attachment-0001.html>


More information about the Election-Methods mailing list