[EMAIL PROTECTED] wrote:
Chris Benham wrote:
I'm happy with its performance in this old example:
101: A
001: B>A
101: C>B
It easily elects A. Schulze (like the other Winning Votes "defeat
dropper" methods) elects B.
It meets my "No Zero-Information Strategy" criterion, which means that
the voter with no idea how others will vote does best to simply rank
sincerely.
This is an interesting example, partly because it seems to me that C would
be a better winner. I ran through some possibilities on the Ranked-ballot
voting calculator at http://cec.wustl.edu/~rhl1/rbvote/calc.html, and got
the following:
101: A
001: B>A
101: C>B
A: Baldwin, Nanson, Raynaud
C: Black, Borda
(Other methods required a random tiebreaker).
That on-line voting calculator you used automatically symmetrically
completes the ballots, and then doesn't
do methods that normally allow truncation but not above-bottom equal
rankings (like Hare aka IRV).
So for pairwise algorithms it only does Margins.
This is an interesting example, partly because it seems to me that C would
be a better winner.
To me electing C just looks like a massive massive failure of
Later-no-Help. It looks like there are two
serious antagonistic factions, the A supporters versus the C supporters.
The A faction don't properly
understand the voting method...maybe there are normally only two
candidates so they had no incentive
to bother. Then one of A's supporters (B) decides as a joke to stand as
an independent (wrongly assuming
that in a preferential system with supposedly no split-vote problem it
can do no harm), and then the C
supporters all give their second preference to B (maybe hoping or
knowing that doing so will help
C) and then C wins.
More than half the voters prefer A to C, and B is solidly supported
by 1 voter versus 101 for each of the
others. In my book this is an election that Hare gets resoundingly right.
Strangely, if you reverse all the rankings, you get:
101: C=B>A
1: C>A>B
101: A>B>C
A Baldwin, Nanson, Raynaud
B Black, Borda
Which means it made no difference to Baldwin, Nanson, or Raynaud. Black
and Borda gave different answers for the reverse order, which seems
logical.
Reverse Symmetry is purely a "mathematical elegance" property that I
would trade nothing of any use to
have.
Now let's look at some possibilities for the second and third choice for
those who picked A.
101: A>B>C
1: B>A>C
101: C>B>A
A: Carey, Hare
B: Baldwin, Black, Borda, Bucklin, Coombs, Copeland, Dodgson, Nanson,
Schulze, Simpson, Small
B looks like a good choice. Carey and Hare give a rather bizarre result,...
My Australian eyes see nothing "bizarre" about electing A, but I agree
that electing B is at least reasonable.
I'm happy with Simmons again easily electing A.
(Of couse, right now I'm kind of punchy from pain pills, so I could be
missing something. :D )
Michael, get well soon.
Chris Benham
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