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 ---- For more information about this list (subscribe, unsubscribe, FAQ, etc), please see http://www.eskimo.com/~robla/em
