Author: José de Jesús García Ruvalcaba
Date: 15:02:15 06/16/99
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On June 16, 1999 at 16:32:55, Robert Pope wrote: >It has been mentioned that the Swiss Pairing System is designed to find an >absolute winner in a tournament, but if not enough rounds are played, the >ordering of the lower places is not obvious. > >Is it also true then that the pairing system also does the opposite at the same >time: finds the absolute worst-performing participant? Not neccesarily. I have played in five-rounds swiss players with only ten players, and the lower places are not facing each other in the last rounds; and not because they have already played but because they are paired against higher scoring opponents who have played among them. > I don't mean this >question as an insult to any of the participants. I was just wondering if this >is what the pairing system would do. > It can do that, but it may not. The pairings are done from top to bottom, i.e. top scorers are treated differently than bottom scorers. >Also, from my understanding, the pairings try to pair opponents with the same >number of points. What if most or all of the participants with the same point >score have already played each other? Then they face opponent with less points. Of course the lowest players have already played then they face higher scoring opponents, but the difference is that it is all accomodated to the pairing needs of the high scoring ones. > Can they be paired a second time? Never in a swiss-system tournament, as far as I know.
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