[EM] TACC (total approval chain climbing) example

[C.Benham]

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).

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?

> 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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