[EM] An odd case for Ranked-Pairs

[Markus Schulze]

Dear Eric,

you wrote (14 Jan 2003):
> This example was recently brought to my attention.  Consider:
>
>    A>C>B>F>D>E
>    B>C>E>F>D>A
>    D>B>A>F>E>C
>    E>A>B>C>F>D
>    E>D>A>B>C>F
>    F>C>D>A>B>E
>
> 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.

Markus Schulze

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