# [EM] An odd case for Ranked-Pairs

Eric Gorr ericgorr at cox.net
Tue Jan 14 10:40:11 PST 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).

Now, the problem with this appears to be that E also does not have
any pairwise-victories, which would seem to indicate that it should
not have the opportunity to be selected as the victor.

Now, it would seem to me that, in the case of a tie, I should verify
that each option in the tie did have at least one pairwise-victory
and if they did not, eliminate them from the tie.

Would anyone else agree or have I made a mistake in here somewhere?

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