[EM] TACC (total approval chain climbing) example

[Forest Simmons]

On Mon, Apr 21, 2014 at 2:39 PM, C.Benham <cbenham at adam.com.au> 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" that says that if a set S of candidates
> are all voted together on top of more than a third of the ballots (i.e. on
> more than a third of the ballots no other candidates are voted between or
> above them) and all the members of S pairwise beat all the non-member
> candidates, then the winner X must come from S  (so far the MDT criterion);
> and if some of the ballots that vote any Y above X are altered by further
> lowering X then if the winner is changed it can't be to Y.
>
> 34 A>B
> 17 C>A
> 16 B>C
> 33 B
>
> A>B 51-49,  A>C 34-33,  B>C 83-17
>
> A is the DMT winner, but if  2 of the 33 B ballots are changed to B>C then
> TACC  (like some better methods)  elects B.
>
> Also, for what it's worth, Benham and Woodall have no zero-info.
> truncation incentive, and little informed truncation incentive. IRV has
> none.
>
> Whenever there are 3-candidates in a  cycle, TACC always elects the
> candidate that pairwise beats the least approved candidate (which strikes
> me as weird and arbitrary).
> If the approval order is A>B>C and the pairwise results go  A>B>C>A, then
> TACC  will elect B.
>
> That is a violation of a criterion (I like) that says that the winner
> can't be pairwise-beaten by a more approved candidate. (You may have had a
> name for it).
>

We talked about this before, and I agreed with you on that.  But that was
before I started taking the Chicken Dilemma seriously.

Now look at this cycle  A > B > C > A with the approval order A>B>C.

On this information alone we might say that A is strong because it doubly
beats B, both in approval and pairwise.  However, when we realize that
A is pairwise beaten by the weakest candidate C, and is on the losing end
of a short beat path from B, some of the wind is taken from her sails.

>
> A much better method that meets Chicken Dilemma, Condorcet and Mono-raise
> is  Approval Margins. We determined that it is equivalent to suggestion of
> yours, "Approval Margins Sort".
> From early (probably March) 2005, does this ring a bell?
>

Yes, it rings a loud bell.  I'll have to revisit it in the context of the
chicken dilemma.

>
>  I wonder if the following Approval Margins Sort
>
> (AMS) is equivalent to your Approval Margins method:
>
> 1. List the alternatives in order of approval with
> highest approval at the top of the list.
>
> 2. While any adjacent pair of alternatives is out of
> order pairwise ... among all such pairs swap the
> members of the pair that differ the least in approval.
>
> This method is clone independent and monotonic, and
> yields a social order that reverses exactly when the
> ballots are reversed. If AMS and AM are the same, it
> might be useful to have this alternative description.
>
>
> AMS is monotonic in a strong sense: if ballots are
> changed so as to increase alternative X's approval or
> to give X a victory that it didn't have before, while
> leaving all of the other approvals and pairwise
> defeats the same, then X cannot move down in the
> social order produced by this AMS method. In other
> words, AMS is monotonic with respect to the entire
> social order it produces.
>
>
> After one example it is pretty obvious that AM and
> AMS are equivalent when there are only three
> alternatives, since they both yield the CW when there
> is one, and they both preserve the approval order if
> the only upward defeat arrow is from the bottom to the
> top, and they both reverse the closest approval margin
> pair, otherwise.
>
>
>
> Chris Benham
>
>
>
>
> On 4/20/2014 8:58 AM, Forest Simmons wrote:
>
>        Let's consider the following ballot set:
>
>  48 C
>  27 A>B
>  25 B
>
>  Plurality says that A cannot win, because C has more top votes than A is
> ranked.
>
>  The Chicken Dilemma Criterion says that B cannot win, because there is a
> possibility that the B faction is defecting from a true preference of B>A.
>
>  That leaves C as the only acceptable winner for this ballot set.
>
>  How do various methods stack up on this ballot set?
>
>  MinMaxPairwiseOpposition (MMPO) elects A.
>
>  Condorcet(wv) and Condorcet(margins) both elect B.
>
>  Implicit Approval elects 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.
>
>  But what if the ballot set is sincere?
>
>  In that case it seems like B should be the winner.
>
>  The problem is that standard election methods have no way of detecting
> ballot sincerity.
>
>  Therefore there is no strategy free method for covering both scenarios.
>
>  The question now becomes which requires more drastic strategy?
>
>  Note that if the above ballot set represents sincere preferences, the A
> faction can help elect B by voting A=B, which requires no order reversal.
>
>  This move will work in any of the above mentioned methods!
>
>  Now suppose that the true preferences were
>
>  48 C>B
>  27 A>B
>  25 B>C
>
>  Then B is the Condorcet Winner.
>
>  But under both Condorcet wv and margins, there is a burial incentive for
> the C>B faction to change votes to 48 C>A .  If B takes no defensive
> action, this burial strategy will succeed.
>
>  Under TACC that burial strategy will backfire by electing electing A.
>
>  However, if the C>B faction may still be tempted to take the a more
> sincere approach of merely truncating B.  When the B faction responds to
> this threat by truncating C, we are led back to the original ballot set
> given above:
>
> 48 C
> 27 A>B
> 25 B
>
>  Which must have C as the winner if we want to honor Plurality and CD.
>
>  But then the A>B faction has a strong incentive to raise B to 27 A=B.
>
>  This solves the problem without any order reversals.
>
>  Forest
>
>
>
>
>
>
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>
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