Author: David Dahlem
Date: 17:08:22 12/14/03
Go up one level in this thread
On December 14, 2003 at 19:57:19, Ricardo Gibert wrote: >On December 14, 2003 at 12:39:34, David Dahlem wrote: > >>On December 14, 2003 at 05:29:04, George Tsavdaris wrote: >> >>>On December 13, 2003 at 22:56:28, Robert Hyatt wrote: >>> >>>>On December 13, 2003 at 21:09:13, David Dahlem wrote: >>>> >>>>>Is there a formula to calculate the number of rounds in Swiss tournaments with >>>>>various number of participants so as to optimize the odds for a clear winner? >>>>> >>>>>Regards >>>>>Dave >>>> >>>> >>>>log2(#players) rounded up to next whole number. If you want a more accurate >>>>2-3 place, add another round. >>> >>>I don't understand. This means that for 20 participants the number of round >>>must be Log2(20)= 4.32 ---> 5. Too little rounds i think. >> >>I don't understand this formula either. For example, the latest WCCC Tourney at >>Graz had 16 participants, and 11 rounds were played. How do tournament directors >>decide how many rounds will be played? >> >>Regards >>Dave > >You should probably be interpreting this formula as a minimum. People like more >rounds to make the result seem less random. In Graz, the tournament would have >been better if it had been played as a double round swiss. Ten rounds with 2 >games played played between the same programs instead of just one instead, but >for some reason this type of idea is not popular. > >What they did instead wound up having rather silly pairings towards the of the >tournament with the contestants at the top of the crosstable being paired >against the those near the bottom. This is not considered desirable, though it >does have the desirable effect of compelling the top contenders to play for a >decisive result in the last round. > >Actually there is no "formula." The formula above is ripped off from the one for >knockout tournaments where all the games are decisive. In chess, you have draws >and a swiss is not the same as a knockout tournament. In fact, swiss tournaments >use a number of different pairing systems, so no single formula is "best." > >In practice in human tournaments, a typical prize structure often militates >against the tournaments leaders seeking a decisive result in the last round. >Accepting a tied result is usually preferable leaving the relative rankings >among the top contenders unchanged. Call it "mutual fear" if you like, but >actually the money odds are dictating the drawn result. This makes the last >round often _seem_ redundant and avoiding a tied result difficult. BTW, this >does not seem to be a problem in computer events. > >Actually, by the way you phrased your question, the number of rounds that should >be played to "...optimize the odds for a clear winner," should be _infinity_. >Obviously that isn't helpful at all. Enough said. > >If you want to insist on having a formula, then I would suggest the same formula >with a slight modification. Add one extra round. This would ameliorate the >chances of tied result due to "mutual fear" in the last round. In any case, any >such formula should be taken with a grain of salt. Actually, i would think there should be some official Swiss tournament rules that cover this basic question. But i haven't been able to find anything. :-( Regards Dave
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