Scott Ritchie wrote:
Anyway, I'd like to prevent problems in the future, so:
1) Endorse an algorithm. I've already decided on Shulze, since that's
what Debian uses and they're our sister project.
2) Create a formal tie-breaking rule. My intuition says that we can
give Mark Shuttleworth (who already has special privileges) a second
vote that he only uses in the case of a tie, add that vote to the box,
and then rerun the election.
However, I'm not 100% sure that a regular vote cast in the CIVS sense
will actually resolve ties cleanly. So, I'm asking the list:
Suppose we have a tie among {A,B} for the marginal seat in an election.
Would a vote for A > B that expressed no other preferences always
result in an identical overall ranking except with a resolved tie?
If the tie is larger (or number of marginal seats larger), could he
similarly vote {A>B=C} or {A=B>C} to resolve them?
Could either vote affect the non-marginal seat order? Shuffling the
above winners, provided they all still win, isn't much of a problem,
however shuffling the losers might be an issue since we use the list
order for replacement candidates when someone steps down mid-term.
It might not resolve the actual tie itself, and it could produce further
ties. The Schulze method uses a strongest path calculation, which means
that (unless I'm mistaken) some contests can be changed without
affecting the result.
This is analogous to the minmax method, which Schulze is based on, and
which considers the worst defeat by any opponent, electing the candidate
whose worst defeat is the least. If you have two candidates, call them A
and B, and their worst defeat is against X and Y respectively, then no
ballot that ranks both X and Y above (or below) both A and B will
resolve the tie, because it alters both worst defeats by the same
amount. For Schulze, it's a bit more complex, but the same reasoning holds.
My educated guess is that Schulze is considerably less likely to produce
more ties in this case than it is likely to simply retain the ties that
do exist, and it is more likely still to actually resolve the tie while
not producing any more ties.
-
You could break the ties in a simple way by just using Mark's ballot
directly. If there's a tie between A and B, then you check if Mark voted
A above B or vice versa, and the candidate he voted highest wins. You
can do the same thing with multiple candidates: if there's a tie between
A, B, and C, and Mark's vote (over A, B, and C alone) is A>B>C, then A
wins, B comes second, and C third.
If that still doesn't resolve the tie, pick a random ballot and break
ties as shown above. Do so until the tie is broken or all the ballots
have been exhausted. If there's still a tie, that means nobody voted A>B
or B>A, so flip a coin. This is called the Random Voter Hierarchy
tiebreaker.
Also, I may be wrong about the subtleties of Schulze. I don't *think* I
am, but if you want to be sure, you might want to ask Markus Schulze
himself (if he doesn't see our posts and replies to either of them).
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