[EM] Burial examples
Chris Benham
cbenhamau at yahoo.com.au
Sun Apr 1 05:10:45 PDT 2018
On 1/04/2018 7:27 AM, Kristofer Munsterhjelm wrote:
> Passing dominant mutual quarter burial resistance is really hard, if
> even possible:
I am sure that for a Condorcet method it isn't possible.
Meeting dominant mutual third burial resistance isn't so easy. I don't
know of any Condorcet method
that does that is better than the "Hare-Condorcet hybrid".
The version I like: voters strictly rank from the top as many or as few
candidates as they wish. If there is
a CW (or single pairwise-undefeated candidate?) x then elect x.
Otherwise eliminate the candidate that (among remaining candidates) is
voted top-most on the fewest
ballots and if among remaining candidates there is pairwise-beats all
candidate (or single pairwise-undefeated
candidate?) x then elect x. If not, repeat and so on.
But it fails mono-raise and can fail to elect a positionally dominant
uncovered candidate, unlike my other
favourite Condorcet methods.
It comfortably handles your four burial scenario examples. In the first
three the burying has no effect and in
the fourth it backfires.
My other favourite Chicken Dilemma criterion complying Condorcet method,
MMLV(erw) Margins-Sort Elimination,
also easily elects the sincere CW in the first 3 examples but the burial
succeeds in the fourth.
https://wiki.electorama.com/wiki/Chicken_Dilemma_Criterion
I first suggested that method in a post to EM on 18 April 2015.
http://lists.electorama.com/pipermail/election-methods-electorama.com/2015-April/000007.html
*Voters rank the however many candidates they wish, equal-ranking
allowed. In determining pairwise scores
a ballot that equal-ranks above bottom candidates A and B gives a whole
vote to each, a ballot that votes
them equal-bottom (i.e. above no other candidate/s) give no vote to
either and a ballot that votes A above B
gives 1 vote to A and nothing to B.
Give every candidate a score equal to her/his smallest losing pairwise
score. List them in order of score.
If the list is in "pairwise order" (i.e. the candidate with the highest
score pairwise beats the candidate with
the second highest score who in turn pairwise beats the candidate with
the third highest and so on) then
elect the candidate at the top of the list.
If not, then swap the places in the list between the two adjacent
candidates who are pairwise out of order by
the smallest margin. If two or more are out of order by the same
margin, then swap the pair that is lowest on
the list. Repeat until the whole list is in pairwise order.
If no more than 3 candidates are on the list, elect the candidate on top
of the list. Otherwise eliminate the
candidate on the bottom of the list and repeat (ignoring eliminated
candidates) as many times as needed.*
Taking your first example:
10: A>B
11: B>C (sincere is B>A)
03: C>A
C>A 14-10, A>B 13-11, B>C 21-3.
Scores: B11 A10 C3
Both BA and AC are pairwise out of order. The score margin between B
and A is 1 (11-10) and the score margin
between A and C is 7 (10-3). So the pair with the smallest margin swap
places and now no adjacent pair is pairwise
out of order and only 3 candidates are on the list (A>B>C) so A (the
sincere CW) now being on top of the list, wins.
Chris Benham
> First of all, I'm not sure if it's "dominant mutual third" or "mutual
> dominant third". If I got it wrong, I apologize :-)
>
> I was reading through some past posts here and got the impression that
> some were considering Smith,Plurality to be resistant to burial. Then
> I found a counterexample. In case it's of use to others, I'll put it
> here:
>
> 10: A>B>C
> 11: B>A>C
> 3: C>A>B
>
> A is the CW. Now let the B voters bury A:
>
> 10: A>B>C
> 11: B>C>A
> 3: C>A>B
>
> There's now an ABCA cycle and B wins.
>
> If I've understood the criterion correctly, this example shows that
> Smith,Plurality fails weak burial resistance/unburiable DMT, because
> if we let the DMT set be {A}, then
>
> A is ranked above A_comp = {B, C} on more than a third = 8 ballots
> A beats B and C pairwise (since A is the CW)
> but the example shows we can alter B-first ballots to change the
> winner from A to B.
>
> ==
>
> I then thought I'd find other burial examples.
>
> Smith,Plurality and Smith, Antiplurality both fail weak burial
> resistance:
>
> 10: A>B>C
> 6: B>A>C
> 5: B>C>A
> 3: C>A>B
>
> A is the CW. Bury A on 5 BAC ballots:
>
> 10: A>B>C
> 1: B>A>C
> 10: B>C>A
> 3: C>A>B
>
> we have an ABCA cycle and B is both the Plurality and Antiplurality
> winner.
>
> -
>
> Smith, Coombs:
>
> 6: A>B>C
> 5: A>C>B
> 2: B>A>C
> 10: B>C>A
> 2: C>A>B
>
> A is the CW. Bury A on the two B>A>C ballots:
>
> 6: A>B>C
> 5: A>C>B
> 12: B>C>A
> 2: C>A>B
>
> ABCA cycle. A has the greatest number of last place votes and so is
> eliminated first by Coombs, then B beats C pairwise, so B wins.
>
> -
>
> Baldwin:
>
> 10: A>B>C
> 4: B>A>C
> 3: C>A>B
> 7: C>B>A
>
> Bury A on the BAC ballots:
>
> 10: A>B>C
> 4: B>C>A
> 3: C>A>B
> 7: C>B>A
>
> A is eliminated first, then B beats C pairwise. According to
> http://www.cs.wustl.edu/~legrand/rbvote/calc.html, Raynaud also elects
> B with positive probability.
>
> -
>
> Passing dominant mutual quarter burial resistance is really hard, if
> even possible:
>
> 7: A>B>C
> 9: B>A>C
> 6: C>A>B
> 2: C>B>A
>
> A is the CW. Now bury A on the BAC ballots:
>
> 7: A>B>C
> 9: B>C>A
> 6: C>A>B
> 2: C>B>A
>
> Pretty much every method elects B here.
>
> ==
>
> Some thoughts: I've been investigating a particular type of Condorcet
> methods on and off. These are three-candidate methods where the CW
> wins if there is one, otherwise I relabel all the candidates so
> there's an ABCA cycle and the candidate I want to find the score of is
> A. Then that candidate's score is f(ABC, ACB, ..., CBA) for a custom
> function f. I've run some tests (using a combination of IC and
> Gaussian models) to see what sort of strategic susceptibility the
> different choices of f give.
>
> I've been noticing that methods that pass both monotonicity and
> reversal symmetry can't seem to go below 45-50% strategy
> susceptibility (meaning half of the randomly chosen elections let one
> of the losers alter their ballots so that the loser wins). In
> contrast, the best methods that pass only one of the two can get down
> to around 10%. (Minmax gets around 70%, and the best I've seen so far
> from methods that are neither is 5%)
>
> Perhaps DMTBR/UDMT is the thing that separates the good performers of
> the second category from those of the first; and perhaps it's
> impossible to have all of Condorcet, monotonicity, reversal symmetry,
> and DMTBR/UDMT. If I find more time, I might try to devise a proof of
> that... if the "perhaps" is true.
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