Author: David Dahlem
Date: 19:40:03 12/14/03
Go up one level in this thread
On December 14, 2003 at 21:51:23, Russell Reagan wrote: >On December 14, 2003 at 05:29:04, George Tsavdaris wrote: > >>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. > >Here is how I understand this. > >This depends upon what the tournament is supposed to produce. Is the tournament >supposed to produce one clear winner (a "champion"), or do you care about the >accuracy of the rankings of lower finishers? > >A rough formula for determining the number of top finishers that will be >accurate is: > > (5 x number of rounds) - number of players >number of rounds = ------------------------------------------ > 7 > >So for the WCCC, there were 16 participants and 11 rounds. This means that ((5 x >11) - 16) / 7) = 5.57. So approximately the top 5 places will be accurate. This >means that you can't accurately say that the person who finished 8th is really >better than the person who finished 9th or 10th. You can only make statements >like that for the top 5 finishers in this case. > >In an event where you only care about crowning one champion, as in a "world >champion," you only want the first place to be accurate ideally. When you add >more rounds beyond log2(participants), you get a more accurate 2nd place >finisher, and then 3rd place, and so on. > >This would make you think that the more rounds, the better the results, but I >believe that the problem is that when you add more rounds beyond the "optimal" >number of rounds, then you create a log jam at the top. If you are just holding >a tournament for fun, that's fine. If you are holding one for the "world >championship," that isn't so good, because you really only care about who wins, >not who finishes 2nd or 3rd. A log jam for places 2-5 is fine as long as we have >one clear winner. > >As for how the swiss system works and why 5 rounds is enough for 20 >participants, it would probably be easier to understand if we used a game that >didn't allow for draws, because draws change the math a little bit. If you are >using the swiss system in a game that allows for draws, then you actually will >need less than log2(participants) rounds, so even 4 would be acceptable. > >Let's assume for a minute that there would be no draws, just for the sake of >explaining why ths swiss system works. Let's say we have 16 participants. In a >swiss tournament, you play other participants with the same number of points as >you (if possible), and you don't play the same person twice. So each round every >undefeated participant will play another undefeated participant, and after the >game, we will have eliminated half of the undefeated participants. > >Before round 1: 16 undefeated participants >After round 1: 8 undefeated participants >After round 2: 4 undefeated participants >After round 3: 2 undefeated participants >After round 4: 1 undefeated participant, the winner > >This is intended to make the good participants play other good participants and >"weed out" people as quickly as possible. For only 16 participants, it would >probably make more sense to just have a round robin. The swiss system was >created for something like a national championship where there may be many >thousands of participants, and you want to have one clear winner (obviously a >round robin is not possible). > >Now for an example of why playing more rounds than optimal can mess things up. >If there are 16 participants, then the optimal number of rounds is 4. Assuming >no draws, then after 4 rounds we have the following standings: > >Rank Points >1. 4 >2. 3 >3. 3 >4. 3 >5. 3 >6. 2 >7. 2 >8. 2 >9. 2 >10. 2 >11. 2 >12. 1 >13. 1 >14. 1 >15. 1 >16. 0 > >After this round, we have a clear winner. Notice that after 1st place there is a >log jam. If we play another round we could have three participants with 4 >points, then we either have to have tie breaker rounds or use tie breakers. I'd >rather have 4 rounds and a clear winner than three people tied for first and the >"champion" be determined by a tie breaker. When our goal is to determine a world >champion, we are willing to accept a log jam between places 2-5 to get one clear >winner. > >This is how I understand it anyway. Hi Russell Thanks, very informative info. The more i read, the more confused i am. :-) Here is something to add to my confusion, it's some text i found on the web, i don't remember exactly where. I think it's from a TD at a chess club. <quote> It is usual when running a Swiss event to aim for n+2 rounds, where 2n is the next highest power of two above the number of participants. This is said to optimise the probability of a single winner. 23 = 8 24 = 16 25 = 32 26 = 64 27 = 128 28 = 256 To test this I have written a computer simulation of a Swiss tournament for up to 32 players. It shows, for instance, that in a 20-player competition run over 7 rounds (n+2 where 2n=32), there is a 62% chance of an outright top and a 26% chance of a 'resolved tie' - that is two or more players on the same number of wins but one has beaten the other(s). So, this gives an 88% probability of a single winner. The other 12% are 'unresolved ties' - typically three players equal on wins and each having beaten one of the other two. In this example n+2 rounds are superior to n+1, at only 60%, and n+3, at 68%. However, nine rounds (n+4) proves to be slightly better for a 20-player event, at an 89% probability of a single winner. Running the simulation over several different player numbers and rounds, we find that up to 24 players there is generally a double peak optimum, at about 90%, for n+2 and n+4. Above 24 players interestingly the double peak moves to n+3 and n+5. For instance, for 32 players 10 rounds are optimum, at 92%, just ahead of 8 rounds at 88%, with 7 rounds (n+2) down at only 69% and 9 rounds (n+4) at 74%. The above results all assume each player in every game has a 50% chance of winning. In reality this will not be so. In a handicap tournament we would generally expect a shallow normal distribution of abilities and in a level event a skewed distribution. The simulation was re-run for both of these distributions. Surprisingly the results did not vary a great deal - an improvement of 2-4% being typical. So, what should be done about the one in ten of all Swiss events that do not yield an outright winner? I have seen counting net points and shooting at the peg, but both are terrible solutions. Mathematically the fairest would seem to be trying to ascertain which of the tied players has faced the strongest opposition, for instance by totalling the number of wins by the opponents of each. <unquote> Regards Dave
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