Author: Ricardo Gibert
Date: 17:36:21 12/14/03
Go up one level in this thread
On December 14, 2003 at 20:08:22, David Dahlem wrote: >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 I would expect the only constraint is what the organizer thinks will produce a successful event. I would be surprised if you will find a such a rule. I overheard a major tournament organizer (Bill Goichberg?) say that a kind of magic number (in the US) for producing a large turn out is 7 rounds (a minimum). My guess is that going above 6 rounds makes an event something that is out of the ordinary and special, which makes the event more desirable and be taken more seriously. More common events have 5 or 6 rounds. Also, offering lots of prize money doesn't hurt either.
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