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

Wed Apr 23 14:49:37 PDT 2014

Michael,

You mentioned being more interested in strategy criteria than criteria like
monotonicity.  In that case you might find TACC preferable to
Woodall/Benham:

In fact, anytime IRV would squeeze out a Condorcet preference (from among
three candidates) sincere ballots will not be a Nash Equilibrium under
Woodall/Benham, because the supporters of the IRV winner will be fully
rewarded by a unilateral burial of of the Condorcet preference.

Of course, in the three candidate case if IRV does not squeeze out the
Condorcet Preference, then IRV itself will elect the Condorcet Preference
without any help from Woodall/Benham.

Therefore (at least in the three candidate case) only under zero
information conditions does Woodall/Benham have a better chance than IRV of
electing a true Condorcet candidate.

Of course under full information a majority that prefers the Condorcet
Preference over the IRV winner can take defensive action by appropriate
reversals, and thus reach a Nash Equilibrium under IRV (as well as under
Woodall/Benham).

But any method that requires that drastic action to arrive at a Nash
Equilibrium requires a lot of information, and therefore is highly
vulnerable to disinformation.  Lack of monotonicity only makes this
vulnerability worse.

[It is true that Approval requires a lot of information to reach a Nash
Equilibium, too, so it may not elect the Condorcet preference the first
time around, but at least no order reversals due to misinformation will
take place.  Monotonicity and FBC compliance mitigate the effect.]

In the case of TACC, most of the time when there is a Condorcet preference
the sincere ballot set will constitute a Nash Equilibrium position.  In the
few cases where this is not so, a Nash Equilibrium supporting the Condorcet
preference can be reached without order reversals, as in the case of
Condorcet(wv).

As for Clone Proofness  TACC is just as clone proof as Approval and Range.  In
Approval we don't expect "clones" to be voted on the same side of the
approval cutoff candidate on every ballot, but we do expect that they be
adjacent in the total approval order.  This may not be clear on small
ballot sets, but in any public election it would be statistically unlikely
for clones to come out unadjacent in the approval order.  But even if they
did, it would just mean that the worst that could happen under TACC  is for
a candidate with approval between the two clones of the previous winner to
get elected.

Forest

On Tue, Apr 22, 2014 at 8:13 PM, Michael Ossipoff <email9648742 at gmail.com>wrote:

> 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
>
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> 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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>> info
>>
>>
>
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