[EM] Election-Methods Digest, Vol 118, Issue 8

Tue Apr 22 20:13:29 PDT 2014

About the Monotonicty and Clone-Independence issues regarding TACC:

I've never been very swayed by the Monotonicity criterion. Yes, when I was
only interested in current-conditions, I accordingly oppsed IRV, at that
time, because current conditions need FBC. And, I'll admit, at that time, I
used Monotnicity against IRV, even though FBC was my only only reason for
opposing IRV. But, I always made it clear that Monotonicity wasn't my
reason for opposing IRV, in those days when I opposed IRV because I was
only intereted in current conditions.

So I've never considered Monotonicity important. I've only been interested
in strategy criteria.

As someone who's been advocating Benham and Woodall all this time, and
continues to, clearly I don't evaluate voting systems by Monotonicity, and
clearly I don't consider nonmonotonicity a disqualification, even a little.

TACC's compliances can be debated, because the conclusion about any
critrerion-compliance depends on the full, detailed definition of the
criterion.

In this discussion of Benham, Woodall, and TACC, what's important?

I say it's MMC, CD, and, to a lesser-extent, Condorcet.

>From what I've heard (and haven't heard challenged), Benham, Woodall, and
TACC all meet MCC, CD, and Condorcet.

Clone-Independence is a desirable nice too, of course. TACC can have its
winner changed if a new candidate is added, whom everyone ranks  adjacently
to an old candidate, but across the ballots' approval cutoff.  . However we
define Clone-Independence, it can't be denied that TACC's clone-indepence
has a weak-spot. Whether or not that describes something that will happen
significantly often is a separate question that i won't try to answer.

Undeniably, Benham and Woodall  have the advantage that a new candidate
unniversally ranked adjacent to an old candidate can never change the
winner...period.

Of course in a comparison of one method versus another, there are various
different advantages to compare.

As Forest agreed, IRV's great familiarity and popularity, and the great
familiaity of runoffs, makes IRV, and therefore, Benham and Woodall, easy
to propose. That's an advantage for those methods.

But, in most comparisons of 2 methods, typically each has one or more
advantages over the other.

UMDT and Invulnerability to 2nd-Place Complaints are two such criteria in
this comparison.

If I there's a significant chance that a party will run two similar
parties, and that they'll straddle some voters' approval-cutoff, then
that would count against TACC.

I have to say that I'd be pleased for either one to be enacted in the Green
scenario, or be offered in a progressive party platform.

Say, that some progressive party is elected to office, and is the new
government, and that there's going to be public initiative or referendum
vote on choosing a voting system. (Say IRV is the initial voting system,
and the one in which the voting-system vote will be conducted).

I'd rank Benham, Woodall, and TACC adjacently, because they all comply with
MMC, CD, and Condorcet.  I trust that advocates of any one of those methods
would do the same.

In that initiative or referendum, there'd be no reason for advocates or
preferrers of any of those methods to propose or nominate their favorite.
...and may the most publicly-liked one win.

Michael Ossipoff

On Mon, Apr 21, 2014 at 10:29 PM, Forest Simmons <fsimmons at pcc.edu> wrote:

> Questions about compliances of Implicit Approval Chain Climbing
>
>
>
>
>> ---------- Forwarded message ----------
>> From: Steve Eppley <SEppley at alumni.caltech.edu>
>> Date: Sun, Apr 20, 2014 at 7:48 PM
>> Subject: Is Chain Climbing really monotonic??
>> To: Michael Ossipoff <email9648742 at gmail.com>
>>
>>
>> Begin with:
>>      7: A B C
>>      6: B C A
>>      5: C A B
>> First add C to S since C is bottom-ranked by the most (7).
>> Then add B to S since only B is unbeaten pairwise by a candidate in S.
>> Then elect B.
>>
>> Suppose two voters raise B from the bottom:
>>      7: A B C
>>      6: B C A
>>      3: C A B
>>      2: C B A
>> First add A to S since A is bottom-ranked by the most (8).
>> Now B can never be added to S since B is beaten pairwise by a candidate in
>> S.
>> So now B can't win.
>>
>> Have I made a mistake?
>>
>
> No, but since the approval order changed from B>A>C to B>C>A this is not a
> test of mono raise winner; i.e. since C was also raised (in the approval
> order) we shouldn't be surprised that C became the new winner.
>
>>
>>
>> ------------------------------
>>
>> Message: 2
>> Date: Mon, 21 Apr 2014 14:21:45 -0400
>> From: Michael Ossipoff <email9648742 at gmail.com>
>> To: "election-methods at electorama.com"
>>         <election-methods at electorama.com>
>> Subject: [EM] Fwd: Is Chain Climbing really independent of clones??
>>
>> ---------- Forwarded message ----------
>> From: Steve Eppley <SEppley at alumni.caltech.edu>
>> Date: Sun, Apr 20, 2014 at 7:22 PM
>> Subject: Is Chain Climbing really independent of clones??
>> To: Michael Ossipoff <email9648742 at gmail.com>
>>
>>
>> The "least implicit approval" score used in Chain Climbing doesn't look
>> cloneproof.
>>
>> Under Plurality Rule, two "top" clones can split their vote, allowing a
>> third candidate to win by spoiling.  Chain Climbing seems very similar:
>> Two
>> "bottom" clones can split their "implicit disapproval" so that a third
>> candidate will instead be added to S.
>>
>> Begin with the example from my recent email:
>>      5: A B C
>>      4: B C A
>>      3: C A B
>> First add C to S since C is bottom-ranked by the most (5).
>> Then add B to S since only B is unbeaten pairwise by a candidate in S.
>> Then elect B.
>>
>> Suppose we add a clone of C:
>>      3: A B C C'
>>      2: A B C' C
>> 4: B C C' A
>>      3: C C' A B
>> First add A to S since A is bottom-ranked by the most (4).
>> Now B can never be added to S since B is beaten pairwise by a candidate in
>> S.
>>
>> Have I made a mistake?
>>
>> In the context of approval, two candidates that are ranked or rated  on
>> opposite sides of the approval cutoff are not considered clones.
>
>
>
>> Some Range devotees go even further and require true clones to have equal
>> scores on all ballots.  The idea is that they could differ infinitesimally,
>> but then they would necessarily round to the same standard score when the
>> set of allowable ratings is a standard finite set, as in any standard
>> public election.
>>  ------------------------------
>>
>> Message: 3
>> Date: Mon, 21 Apr 2014 14:34:01 -0400
>> From: Michael Ossipoff <email9648742 at gmail.com>
>> To: "election-methods at electorama.com"
>>         <election-methods at electorama.com>
>> Subject: [EM] Fwd: Does Chain Climbing fail Resolvability?? (was Re:
>>         Is Chain Climbing really monotonic??)
>>
>>
>> Steve questioned TACC's compliance with Condocet, in situations where two
>> candidates are unbeaten from S, and have equal implict-approval totals, if
>> TACC adds them both simultaneously to set S. I replied that Jobst and
>> Forest probably intended a random choice, by some means, between those two
>> candidates, to determine which to first add to S. In that way, there
>> doesn't remain any difficulty with Condorcet compliance.
>>
>> Below is a reply from Steve. Because he requested that I forward his other
>> TACC comments, I assume that he'd like me to forward this one as well:
>>
>>
>> ---------- Forwarded message ----------
>> From: Steve Eppley <SEppley at alumni.caltech.edu>
>> Date: Mon, Apr 21, 2014 at 12:38 PM
>> Subject: Does Chain Climbing fail Resolvability?? (was Re: Is Chain
>> Climbing really monotonic??)
>> To: Michael Ossipoff <email9648742 at gmail.com>
>>
>> I wrote a few days ago about Chain Climbing's lack of decisiveness. In the
>> same vein, I suspect Resolvability is failed by TACC with the random
>> tiebreaker (assuming 'implicit disapproval' is defined by Bottom(A) rather
>> than Bottom(Z)... 'absolute implicit disapproval' rather than 'relative
>> implicit disapproval').  Suppose one terrible candidate is unanimously
>> ranked bottom, and a bunch of other candidates cycle.  Call the cyclic
>> candidates C.  Given the random tiebreaker to add one at a time to S, many
>> candidates in C have a non-zero chance to win.  To try to make the winner
>> deterministic, a vote added to the collection of votes can rank a subset
>> of
>> C at the bottom.  Call that subset Cb, and let Ct denote the rest of C.
>>  If
>> C contains many candidates, then at least one of Cb & Ct must contain two
>> or more candidates.  If at some point TACC needs to add one of those
>> "still-tied" candidates to S, the randomness of that pick may cause the
>> winner to still be non-deterministic.
>>
> Steve,
>
> you seem to overlook the possibility of equal rankings and truncations.  I
> think that most voters would truncate at least one of the three cyclic
> candidates along with the despised candidate.  Since the despised candidate
> is covered there is no possibility of his election.
>
> There are many deterministic ways of breaking ties in addition to the
> random ones.
>
> For example if two candidates are in bottom position on the same number of
> ballots, the one ranked equal top on the most ballots comes out ahead of
> the other one.
>
> As I mentioned to Michael Ossipoff, all of these worries disappear if we
> stick to score ballots, for example ballots with allowable scores 0, 1, ...
> 10.  If two candidates have the same number of zeroes, the one with the
> fewest ones comes out ahead in the approval order. If that doesn't resolve
> it, then the one with the fewest number of twos, etc. In the very rare case
> that two candidates have identical distributions of scores, random ballot
> can be used.
>
> Forest
>
>
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>
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