[EM] ER-Bucklin fails monotonicity

Kevin Venzke stepjak at yahoo.fr
Sun Jun 26 12:03:16 PDT 2005


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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