Kristofer Munsterhjelm
20180331 21:57:00 UTC
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 Bfirst 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 threecandidate 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 4550% 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.

ElectionMethods mailing list  see http://electorama.com/em for list info
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 Bfirst 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 threecandidate 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 4550% 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.

ElectionMethods mailing list  see http://electorama.com/em for list info