[EM] An odd case for Ranked-Pairs
markus.schulze at alumni.tu-berlin.de
Tue Jan 14 11:16:02 PST 2003
you wrote (14 Jan 2003):
> This example was recently brought to my attention. Consider:
> The pairwise matrix is:
> 0 4 4 2 3 4
> 2 0 4 3 4 5
> 2 2 0 4 3 4
> 4 3 2 0 3 2
> 3 2 3 3 0 3
> 2 1 2 4 3 0
> Now, with Ranked-Pairs, the only kept-defeat will be B:F.
> However, once all of the defeats have been considered, there will be
> no kept-defeats for <someone>:E, which allows E to participate in a
> tie (according to my computations).
No oddities. Due to my computations, the defeat B:E=4:2 will be locked
in any case. Therefore, candidate E cannot be a (decisive or random)
Ranked Pairs winner.
Due to my computations, Ranked Pairs is indecisive between A, B, C,
and D. As I suggest to use Random Ballot as a tiebreaker, candidate A
wins with a probability of 1/3, candidate B wins with a probability of
1/6, candidate C wins with a probability of 1/6, and candidate D wins
with a probability of 1/3.
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