FYI, the Condorcet-Kemeny method correctly ranks M1 as second-most
popular, and M2 as third-most popular. And it does so without the need
for a "tie-breaker" adjustment.
Richard Fobes
On 10/27/2011 8:46 PM, capologist wrote:
I recently conducted a vote under the Schwartz method. It produced a
result that is counterintuitive and that I don’t know how to justify.
Here’s a simplified version of the scenario:
*5x A > M1 = M2 > B*
*3x B > A > M1 = M2*
*2x M1 = M2 > B > A*
*2x M1 > M2 > B > A*
The partial ordering produced by the Schulze method has *A* beating
everybody else, *B* losing to everybody else, and *M1* and *M2* “tied”
in the middle:
The question regards the clone pair *(M1, M2)*. Why shouldn’t *M1* be a
winner over *M2*? Nobody would object to that. Some voters would prefer
it, and the rest don’t care one way or the other.
I don’t know how to explain to the voters who prefer *M1* over *M2* why
their preference shouldn’t be reflected in the results when nobody
disagrees with it.
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