Hello,
Hopefully this is new information.
Consider these ballots:
40 A>B>D
35 D>B
25 C>D
B wins, with 75 votes in the second round.
Now let's raise B on some ballots:
40 A=B>D
35 D>B
25 C>D
Now candidate D wins, with 100 votes in the second round.
So, quite clearly, ER-Bucklin fails monotonicity. What's interesting is
that MCA *is* ER-Bucklin, but doesn't suffer from this problem, simply due
to only having two levels of approval.
However, if we use the "arbitrarily placed cutoffs" interpretation which
causes 3-slot methods to fail clone independence, then MCA would also fail
monotonicity, since nothing would prevent a formerly disapproved candidate
(such as D) from moving into the middle slot.
Kevin Venzke
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