[EM] DH3 - error by me - should prefer BTR-IRV to Schulze beatpaths?

[Warren Smith]

As Benham pointed out (basically) my claim 
   http://RangeVoting.org/WinningVotes.html
that "all Condorcet methods" suffer from the DH3 pathology was
not quite right.  I should have said
"all Condorcet methods which handle a 3-cycle by eliminating the weakest defeat."
This includes Schulze-beatpaths, Heitzig-river, Tideman-ranked-pairs, 
least-reversal-classic-Condorcet, Simpson-Kramer-minmax, my method I call the
"maxtree" method...  but not ALL Condorcet methods.

(One also can consider emthods where a vote is a rank ordering AND 
an "approval threshold" but I will not do that here.)

An example of a Condorcet order-only method which is immune to DH3 is
Rob LeGrand's BTR-IRV method; another is the "IRV restricted to the Smith set" 
method;  another is the "plurality restricted to Smith set" method.

These are all immune in the sense that it does not make strategic sense 
for  the A- and B-voters to pretend D>C, because that does not work - C still gets
elected.  (They are all NOT immune in the perhaps-sillier sense that if the voters 
stupidly do it anyway, then you get the "dark horse" D winning.)

Of these three, which should we prefer?
None of the three are monotonic, all three suffer from "honest voting can hurt you"
paradoxes, and all three can suffer from embarrassing 
"winner=loser reversal paradoxes":

Plurality paradox:
2 A>B>C
2 A>C>B
3 B>C>A
7-voter scenario where 
plurality-winner = BTR-IRV-winner = IRV-winner = Condorcet-winner = A = plurality-loser

BTR-IRV paradox:
4 A>B>C
4 A>C>B
6 B>C>A
5 C>A>B
19-voter scenario where 
plurality-winner = BTR-IRV-winner = IRV-winner = A = plurality-loser = BTR-IRV-loser

IRV paradox:
9        B>C>A
8        A>B>C
7        C>A>B
24-voter scenario where 
IRV-winner = A = plurality-loser = IRV-loser = BTR-IRV-loser

Benham notes IRV enjoys (and I now note BTR-IRV also enjoys, but Plurality does not)
the "Dominant mutual third" property:
 If a more than a third of the voters rank (in any order) the members
 of a subset S of candidates above all others, and all the
 members of S pairwise beat all the non-members; then the
 winner must come from S.

That is because the IRV (or BTR-IRV)
election will, after eliminations and vote-transfers, reduce to
a 2-man contest between an S-member and somebody else.  (This is a weakened
form of the Condorcet or Smith Set property.)

Also BTR-IRV and IRV and Smith//IRV
are both clone-immune but plurality is not.  These facts cause
us to demerit Smith//Plurality.   

Now IRV is better than BTR-IRV in the sense it is immune to add-top failure
and enjoys "later-no-harm".  However, both these IRV advantages no longer hold
if we are speaking of Smith//IRV.  

So...  I guess BTR-IRV and Smith//IRV  both are good Condorcet methods and I have
uncovered no basis here for preferring one over the other.

But we HAVE uncovered a good reason (DH3 immunity) to prefer either to
Schulze-beatpaths, Heitzig-river, Tideman-ranked-pairs, 
least-reversal-classic-Condorcet, Simpson-Kramer-minmax, and my method I call MaxTree.

Benham pointed out there are two ways to go with  IRV restricted to the Smith set:
either eliminate the candidates not inthe Smith set beforehand, or do not
(i.e. just use regular IRV until have eliminated all but one Smith set member X
plus some perhaps-large number of non-Smith-set members also remain, then X wins).
I am not sure which of these two is better.
Ditto for BTR-IRV.  That complicates matters.

BTR-IRV also has the advantage over Schulze-beatpaths that it does not
matter whether you hew to the margins or winning-votes philosophies - BTR-IRV
is the same in either.  (With Schulze-beatpaths, you have to worry about that.)

In view of this, I should probably change CRV's recommendation
("if you insist on using a rank-order-ballot method") away from Schulze-beatpaths
and toward BTR-IRV or Smith//IRV - probably the former since it is simplest to describe?

Warren D Smith
http://RangeVoting.org