[EM] Fwd: TACC (total approval chain climbing) example

[Jameson Quinn]

Oops; I mis-replied individually, instead of to the list. Here's what I
said:
----
This thread is interesting. However, it's also very heavy on the acronyms.
Can I ask that when you use an acronym, you make sure it's explained
adequately on the electorama wiki? At a quick check,
http://wiki.electorama.com/wiki/TACC is inadequate and
http://wiki.electorama.com/wiki/MAM is nonexistent.

As to the discussion: this is really the heart of the chicken dilemma. I
don't think there's any way to look at a set of ballots like

40 C
35 A>B
25 B

... and get an answer that's not going to be wrong some of the time. If
these ballots are honest, then B should win. If the B voters are truncating
an honest A second preference, than A would be the ideal winner, but
perhaps the system should choose C in order to discourage that strategy.
And if enough of the B voters are truncating C, you could make an argument
that C is the best winner.

The only system I know of where I wouldn't worry about this situation is
SODA. Under SODA, since C and B are required to declare second preferences,
the ballots above could not occur with lazy voters. So in order to get the
ballots above in a chicken dilemma situation, you'd have to have C voters
going out of their way to check "Do not delegate", in hopes that [B voters
would go out of their way to do so, in mistaken hopes that A voters would
not go out of their way to do so]; and then the C voters' hopes would have
to be right about the B voters but wrong about the A voters, while the B
voters were wrong about the A voters. In other words, as long as there's
common information (however complete or incomplete), the ballots above just
won't happen in SODA.

Short of SODA... I'm interested in the discussion of various preferential
methods, but I think that MAV is probably the best answer to the situation,
and it's not so great. It would elect B in the above scenario, possibly
rewarding a chicken strategy from B voters; but if 3/10 to 4/10 of the A
voters truncate, then it's very hard and risky for the B voters to get
their way.

Jameson


2014-04-23 10:44 GMT-04:00 C.Benham <cbenham at adam.com.au>:

>  Forest,
>
>
>  40 A>B>C
>  35 B>C>A
>  25 C>A>B
>
>
> In this example ballot set of yours (which you say is the result of the A
> faction burying against the sincere Condorcet Winner C) there is no reason
> to suppose that any of the ballots are more likely to be insincere than
> any of the others.
>
> In this sort of situation (with 3 candidates in a cycle and all the
> ballots containing the same amount of information) we should avoid electing
> a candidate
> X that is positionally dominated, especially by a candidate that pairwise
> beats X.   In this case that just says "not C".
>
>  35 A>B>C
>  40 B>C>A
>  25 C>A>B
>
>
> In this ballot set (which again you say is the result of the A faction
> burying against C)  TACC elects C.  But again there isn't any information
> on the ballots
> that suggests that some ballots are more likely to be insincere than any
> others.
>
> So the result must be justifiable on the assumption that all the votes are
> sincere. On these ballots C is positionally dominated and pairwise beaten
> by B.
> B also positionally dominates A.
>
> Approval Margins, Approval Margins Sort, IRV, Benham, Woodall and all
> other good  (and/or sane) methods elect B.
>
> Chris Benham
>
>
>
>
>
> On 4/23/2014 7:29 AM, Forest Simmons wrote:
>
>
>
>
>
>  On 04/21/2014 11:39 PM, C.Benham wrote:
>>
>>> Forest,
>>>
>>>  48 C
>>>> 27 A>B
>>>> 25 B
>>>>
>>>
>>>  Borda, TACC, and IRV based methods like Woodall and Benham elect C.
>>>>
>>>> But Borda is clone dependent, and the IRV style elimination based
>>>> methods fail monotonicity.  So TACC is a leading contender if we
>>>> really take the Chicken Dilemma seriously.
>>>>
>>>
>>> Benham and Woodall are a lot more resistant  to Burial than TACC (and
>>> other Condorcet methods that meet mono-raise, aka monotonicity) because
>>> they meet
>>> "Unburiable Mutual Dominant Third"
>>>
>>
>  How about the following set of sincere preferences?
>
>  40 A>C>B
>  35 B>C>A
>  25 C>A>B
>
>  Candidate C is the sincere winner under Benham, Woodall, TACC, or any
> other method meeting the Condorcet Criterion.
>
>  Suppose that the A faction decides to bury C:
>
> 40 A>B>C
>  35 B>C>A
>  25 C>A>B
>
>  Woodall, Benham, Condorcet (both wv and margins) reward this subterfuge
> by electing A, whereas under TACC  the tactic backfires by getting B
> elected.
>
>  Under Condorcet (wv) the C faction can defend itself by truncating to 25
> C.  This wouldn't help under Woodall/Benham:
>
> You get the same result if you replace the respective faction sizes 40,
> 35, 25 with any three positive numbers a, b, c that satisfy both a>b>c and
> b+c>a.
>
>  Now consider the following example where b>a>c:
>
> 35 A>B>C (sincere A>C>B)
>  40 B>C>A
>  25 C>A>B
>
>  Woodall/Benham still rewards the burial.  TACC elects C without any
> defensive move.  Condorcet elects B without any defensive move.
>
>  Under TACC or Condorcet it never hurts and sometimes helps for the CW
> supporters to truncate the rest of the candidates. But as you can see in
> the cases we are dealing with here, only much more drastic defensive action
> can save the CW under Woodall/Benham.
>
>  Conclusion: although in some cases the UMDT confers a superior defense
> against burial, that doesn't make Woodall and Benham uniformly more burial
> resistant than TACC.
>
>  Forest
>
>
>
> ----
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>
>
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