Is it that easy? (Condorcet is almost incompatible with 1/n burial resistance for n>3)
Let a candidate be a dominant mutual nth candidate if he has more than
1/n of the share of first preferences and also pairwise beats everybody
else.
Let a method pass "strong" dominant mutual nth candidate burial
resistance if W is the dominant mutual nth candidate and voters who
prefer some A to W can't increase the probability of A winning by
burying W. (Ordinary DMnCBR requires that A becomes the sole winner.)
Then the maximum n compatible with this stronger form and Condorcet is
3, i.e. mutual third.
A very simple example that my linear programming solver found:
1: A>B>C
1: B>A>C
1: C>A>B
A is the CW and thus wins. Furthermore, A is a dominant nth candidate
for all n > 3 because he has exactly a third of the first preferences.
But then
1: A>B>C
1: B>C>A (burying A under C)
1: C>A>B
and every anonymous and neutral voting method must return a perfect tie,
which increased B's chance of winning.
Now I say "almost" because we don't know that the strategists prefer a
33% chance of either candidate to 100% of A, we only know that they
prefer B to A. So in that sense, no, it isn't that easy. But it seems
very close.
Maybe it's possible to get an actual proof by perturbing with some
epsilons and using resolvability and monotonicity to say that the method
must elect the candidate with the most first preferences, and then
letting B be it... e.g. something like
10000: A>B>C
10001: B>A>C
10000: C>A>B
and A wins by being a CW and nth dominant for n > 3.0001, i.e. 3 +
epsilon that can be made arbitrarily small; then
10000: A>B>C
10001: B>C>A
10000: C>A>B
by Tideman's resolvability almost suffices (since the perfect tie is
broken by one additional voter).
What do you think, is there any simple way to do the last step?
-km
On Fri, Jun 30, 2023, 12:30 PM Kristofer Munsterhjelm km_elmet@t-online.de
wrote:
Let a candidate be a dominant mutual nth candidate if he has more than
1/n of the share of first preferences and also pairwise beats everybody
else.
Let a method pass "strong" dominant mutual nth candidate burial
resistance if W is the dominant mutual nth candidate and voters who
prefer some A to W can't increase the probability of A winning by
burying W. (Ordinary DMnCBR requires that A becomes the sole winner.)
Then the maximum n compatible with this stronger form and Condorcet is
3, i.e. mutual third.
A very simple example that my linear programming solver found:
1: A>B>C
1: B>A>C
1: C>A>B
A is the CW and thus wins. Furthermore, A is a dominant nth candidate
for all n > 3 because he has exactly a third of the first preferences.
But then
1: A>B>C
1: B>C>A (burying A under C)
1: C>A>B
Because of the perfect symmetry, it is not possible to set up a sincere
runoff that excludes the candidate least likely to be the sincere CW.
But under standard game theoretic assumptions ... of perfect mutual
knowledge of sincere preferences, etc... any of the following ballots would
elicit "votes" to elect A:
A vs (B vs C)
B vs (A vs C)
C vs (A vs B)
In the first of these, since C is the sincere CL, rational voters know that
if the "runoff" reaches the B vs C decision, the winner will be B. But a
majority of voters sincerely prefer A to B .... therefore their rational
choice is for that majority to express A>(B vs C), etc.
For the second ballot ... if the second stage is reached, A will win.
Otherwise B will win. Since A is sincerely preferred over B, a majority of
the rational voters will vote B<(A vs C), etc.
For the third ballot, a majority of ballots will start C<(A vs B) because
no matter the result of A vs B, it will be preferred over C, etc.
and every anonymous and neutral voting method must return a perfect tie,
which increased B's chance of winning.
Now I say "almost" because we don't know that the strategists prefer a
33% chance of either candidate to 100% of A, we only know that they
prefer B to A. So in that sense, no, it isn't that easy. But it seems
very close.
Maybe it's possible to get an actual proof by perturbing with some
epsilons and using resolvability and monotonicity to say that the method
must elect the candidate with the most first preferences, and then
letting B be it... e.g. something like
10000: A>B>C
10001: B>A>C
10000: C>A>B
and A wins by being a CW and nth dominant for n > 3.0001, i.e. 3 +
epsilon that can be made arbitrarily small; then
10000: A>B>C
10001: B>C>A
10000: C>A>B
by Tideman's resolvability almost suffices (since the perfect tie is
broken by one additional voter).
What do you think, is there any simple way to do the last step?
-km
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info
On 7/1/23 03:05, Forest Simmons wrote:
On Fri, Jun 30, 2023, 12:30 PM Kristofer Munsterhjelm
<km_elmet@t-online.de mailto:km_elmet@t-online.de> wrote:
1: A>B>C
1: B>C>A (burying A under C)
1: C>A>B
Because of the perfect symmetry, it is not possible to set up a sincere
runoff that excludes the candidate least likely to be the sincere CW.
But under standard game theoretic assumptions ... of perfect mutual
knowledge of sincere preferences, etc... any of the following ballots
would elicit "votes" to elect A:
A vs (B vs C)
B vs (A vs C)
C vs (A vs B)
In the first of these, since C is the sincere CL, rational voters know
that if the "runoff" reaches the B vs C decision, the winner will be B.
But a majority of voters sincerely prefer A to B .... therefore their
rational choice is for that majority to express A>(B vs C), etc.
That's true. It's a Condorcet cycle, so it has the property that if a
majority could decide who to go for, it could get that candidate elected.
However, in this context, my point was more that suppose that the method
were to pass, say, the stronger form of mutual dominant quarter burial
resistance as well as Condorcet. Then the burial example must not
deprive A of his status as sole winner. But because the after-burial
election is symmetrical, the method has no way of knowing who was the
honest CW. Hence this is impossible.
Then I just tried to go from the (possibly too) strong criterion to what
we actually care about - that X>A buriers shouldn't be able to make X
win. But I can only do that with the additional assumption that in the
case of a perfect tie, any X-first ballot will make X the winner.
-km