Author: O. Veli
Date: 03:27:07 12/15/03
Go up one level in this thread
On December 14, 2003 at 21:51:23, Russell Reagan wrote:
>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.
Yes and no. Log2(n) where n=number of participants gives you the absolute
minimum number of rounds that you have to play to find a winner. The winner is
definetly not the best player at the moment, but rather a candidate. The more
rounds you play after Log2(n), the winner will be more reliable and the next
placements will be more reliable.
There are two formulas that are used:
i) 35*p=(5*r)-n, where p=number of placements, r=number of rounds
which is the formula you gave
ii) r=Log2(n)+2(p-1).
ie. You have 16 players and want to find out the best and worst (the swiss
results show both ends) 3 players. Then you will have log2(16) + 2(3-1) = 4+4=8
rounds. If we call Log2(n) first phase and 2(p-1) second phase, then we can
state that phase 1 is the absolute minimum to find a winner, and second phase
increases the probabilty that the winner is indeed the strongest player. The
longer the second phase, the more reliable the results will be.
Of course when you make the second phase really long, then you end up having
players playing with players from 2-3 levels down which effects the results. In
my experience, the second phase should be at most be as long as the first phase
and r<=n/2. Of course an IA would be more qualifed than I am but this has been
the case with me. If I were to organize IACG, I would keep it at most 8 round
SS, or better yet play a round-robin.
>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
I totally agree.
>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.
That may be fine for a weekend, or a fun tournament but not for any kind of
championship.
>"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).
You can play RR with up to 26 players, but they tend to be really long. There
were some tournaments in the 70s and 80s but are much less frequent now. The
last one I remember is the Indian Championship of 2001 or later. This of course
should be no problem for the computers.
>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
>
>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.
After this round, you have a clear winner but we are not sure that he/she/it
is the best player. The more round you play after this round is to make sure
that you find the strongest player but this also has a limit.
For example A beat B in the 4th round, so A is at 4 points and B is 3.
i) Let us assume that B is stronger than A, and had bad luck in their game. If
the tournament were to end here, you have an undeserved winner. If the
tournament were to last up to 4 more rounds, the B player (being the strongest
player he/she/it is) will win the remaining games, whereas player A may lose
some points. B has a chance to win the tournament if we continue.
ii) Let us assume that A is the stronger player. He/she/it will continue
winning the remaining games whereas B may lose some points. Than A will win the
tournament and we are statistically more sure that he/she/it is deserved winner
and not a lucky strong player.
Of course for a world championship, round-robin (preferably double RR) is the
best method. SS is not a good system for a championship, or a qualification
tournament since it has a luck factor in it. It is better used for open
tournaments, club tournaments where the aim is more geared towards gaining
experience.
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