[EM] SFC, condorcet, participation, full support of A>B and B>C at same time

[Warren Smith]

ok, sorry for my dimness here.

Thanks to Benham and Venzke we now have two readable definitions of SFC:
Benham-SFC: "If  more than half the voters vote X over Y and it is possible to 
complete truncated ballots in a way to make X the CW, then Y must not win."

Venzke-SFC: "If more than half of the voters 
rank A over B, but there is no majority of the voters ranking some third 
candidate over A, then B doesn't win."

I'm taking their words for it that these are equivalent and to Ossipoff's intended
SFC definition.   

I don't terribly see why these rather obscure properties should exactly be
the foundation of voting theory or why they terribly address the issue that started this,
which was the question of whether it can really be said to be true that
in Condorcet you can "fully support A>B and B>C at the same time" whereas in range voting
you cannot.

An issue that seems more relevant to that question is the participation criterion.
Benham has argued that IRV fails participation in a severe way (3-candidate elections)
whereas (he hoped) some or all Condorcet systems fail it only in a less-severe
way (4-or-more candidate elections needed).  That hope, if true, sounds interesting
an souns like a tangible advantage for some or all Condorcet systems over IRV.

But is it true?  

Certainly SOME Condorcet systems fail participation in 3-candidate scenarios.
Proof:
Say C is the Condorcet winner.   If you add your honest vote
B>C>A then C is no longer the Condorcet winner and A wins; you are
better off not voting.
before you vote (C is the Condorcet Winner):
C:B  5:4
A:B  8:1
C:A  6:3
after you vote:
C:B  5:5
A:B  8:2
C:A  7:3
after you and your co-feeling friend vote:
B:C  6:5
A:B  8:3
C:A  8:3
now in Condorcet systems that "drop the weakest defeat" C would still win,
but in other Condorcet systems like BTR-IRV, SmithIRV, and Simmons
it seems that A could indeed now  be the winner.  (All these margins
are achievable by actual ballots, at least if a sufficiently
large constant is aded to all numbers.)
E.g. (most simply) in particular with Copeland with random tiebreak,
C wins before you vote; after you+friend vote  it is a 3-way tie,
i.e. your two votes made it worse if your utilities were  B=9, C=8, A=0.
Therefore, at least one of your & your friend's votes were a bad move
versus not voting, in Copeland's Condorcet system.
Q.E.D.

However, weakest-defeat-drop Condorcet systems (pure ordinal rank orderings only)
which feature positive tie-responsiveness always do satisfy participation
in 3-or-fewer-candidate scenarios.  Proof sketch:  the 5 cases are
1. If you vote for the CW top, your vote did not hurt you.
2. If you vote for the CW bottom, your vote did not hurt you.
3. If you vote for the CW middle, your vote can only hurt you if
it stops him from being CW, i.e creates a cycle, A>B>C>A,
and then causes your bottom-ranked to win.
However in that case the former-CW now necessarily suffers the weakest defeat
and hence still wins.  [Or there is a tie for weakest defeat, but
going thru the cases it seems these ties always constitute improvements if so, e.g. if
A>B>C (as is, A wins) and if
you vote C>B>A ==> 3way tie.
you vote B>A>C ==> BA tie
you vote B>C>A ==> A=B>C=A cycle, AB tie.
you vote C>A>B ==> C=A>B=C cycle, AC tie.]
4. If there was no CW and your vote created one, it helped.
5. If there was no CW and you vote did not create one, then
if your vote changed the winner it did so by leaving the cycle the same but weakening
a defeat (which helped).
Q.E.D.

Returning to the issue of "full support of A>B and B>C at the same time,"
it seems to me that NO rank-order voting system (equalities allowed, or not)
allows doing that.
That is, there is always an election situation, with 3 candidates,
where you, by voting A>B>C, cause a worse result than if you had voted
in some other (dishonest) way.  
Proof: this is the Gibbard-Satterthwaite theorem. 
Q.E.D.

Warren D Smith
http://rangevoting.org