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<a class="moz-txt-link-abbreviated" href="mailto:mrouse1@mrouse.com">mrouse1@mrouse.com</a> wrote:<br>
<blockquote type="cite">
<pre wrap="">Chris Benham wrote:
</pre>
<blockquote type="cite">
<pre wrap="">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
</pre>
</blockquote>
<pre wrap=""><!---->dropper" methods) elects B.
</pre>
<blockquote type="cite">
<pre wrap="">It meets my "No Zero-Information Strategy" criterion, which means that
</pre>
</blockquote>
<pre wrap=""><!---->the voter with no idea how others will vote does best to simply rank
sincerely.
</pre>
<pre wrap=""><!---->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 <a class="moz-txt-link-freetext" href="http://cec.wustl.edu/~rhl1/rbvote/calc.html">http://cec.wustl.edu/~rhl1/rbvote/calc.html</a>, 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).</pre>
</blockquote>
That on-line voting calculator you used automatically symmetrically
completes the ballots, and then doesn't <br>
do methods that normally allow truncation but not above-bottom equal
rankings (like Hare aka IRV).<br>
So for pairwise algorithms it only does Margins.<br>
<br>
<blockquote type="cite">
<pre wrap="">This is an interesting example, partly because it seems to me that C would
be a better winner.</pre>
</blockquote>
To me electing C just looks like a massive massive failure of
Later-no-Help. It looks like there are two<br>
serious antagonistic factions, the A supporters versus the C
supporters. The A faction don't properly<br>
understand the voting method...maybe there are normally only two
candidates so they had no incentive <br>
to bother. Then one of A's supporters (B) decides as a joke to stand
as an independent (wrongly assuming<br>
that in a preferential system with supposedly no split-vote problem it
can do no harm), and then the C<br>
supporters all give their second preference to B (maybe hoping
or knowing that doing so will help<br>
C) and then C wins.<br>
<br>
More than half the voters prefer A to C, and B is solidly supported
by 1 voter versus 101 for each of the<br>
others. In my book this is an election that Hare gets resoundingly
right.<br>
<br>
<blockquote type="cite">
<pre wrap="">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.</pre>
</blockquote>
Reverse Symmetry is purely a "mathematical elegance" property that I
would trade nothing of any use to<br>
have. <br>
<br>
<blockquote type="cite">
<pre wrap="">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,...</pre>
</blockquote>
My Australian eyes see nothing "bizarre" about electing A, but I agree
that electing B is at least reasonable.<br>
I'm happy with Simmons again easily electing A.<br>
<br>
<blockquote type="cite">
<pre wrap="">(Of couse, right now I'm kind of punchy from pain pills, so I could be
missing something. :D )
</pre>
</blockquote>
Michael, get well soon.<br>
<br>
Chris Benham<br>
<br>
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