[EM] question about Schulze example (A,B,M1,M2)

[Richard Fobes]

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.
>