Forums
Most Recent
 Hand falling asleep at mouse
 beginers to play with?
 Should IM's be commenting on GM's games?
 4/24/2017  All In The Name Of Winning
 Do chess.com blitz 1900's drop queens in the opening?
 African chess players
 Stalemate
 How to play against the french defence?

Opening theories not for under 1600 players?
FM chuddog
 Which "Old" SuperGM's Games Should I Study First?
Forum Legend
 Following
 New Comments
 Locked Topic
 Pinned Topic
2530 players should be 56 rounds or so.
That's my guess, according to my experiences.
In 9 players, every round will have a person getting a 'bye', but not the same person twice.
In general, the more rounds the better. As you increase the number of rounds in a Swiss, the more it resembles a round robin. I think the time available, rather than the Swiss system itself dictates the number of rounds in a tounament.
Yes, the more rounds the better; the minimum for 2530 players is 5 rounds. The general rule is that x rounds are usually enough to have a crear winner out of 2^x players: 5 rounds is enough for up to 2^5=32 players. 4 rounds is enough for up to 16 players.
Usually it is normal to calculate no. of players divide by 3 to get optimal round number. So for 25  30 players it is normal to get 9 rounds in swiss system and all top plyers should play each other.
Usually it is normal to calculate no. of players divide by 3 to get optimal round number. So for 25  30 players it is normal to get 9 rounds in swiss system and all top plyers should play each other.
Think how many rounds would be needed for a tournament of 498 players. 166 rounds?
http://www.cappellechess.fr/fr2/default.php?page=3967 or nine?
9 is a normal no. of rounds in a tournament, or 13 (i think ) on chess olimpiad. We ware here talking about round number for small tournament how to calculate.
One way to think of it is that for every round, up to half the players can get 1 point. So for 30 players, after rd 1, there can be 15 with 1 point, then 8 after rd 2, 4 after rd 3, 2 after rd 4 and finally 1 after rd 5.
If you have a calulator with a ln (logarithm) function you can calculate:
rounds = ln(players)/ln(2), then "round" up. (no pun intended!)
or, rounds = log2(players) (log2 = base 2 logarithm).
Or use a table like this where for each round, you double the maximum number of players:
1 rd, 2 players
2 rds, 4 players
3 rds, 8 players
4 rds, 16 players
5 rds, 32 players
6 rds, 64 players
and so on...
Of course, if you have time for an extra round or two, more of the top players will have a chance to play each other.
Usually it is normal to calculate no. of players divide by 3 to get optimal round number. So for 25  30 players it is normal to get 9 rounds in swiss system and all top plyers should play each other.
Think how many rounds would be needed for a tournament of 498 players. 166 rounds?
http://www.cappellechess.fr/fr2/default.php?page=3967 or nine?
Your post seems to indicate that you thought 9 rounds was adequate for 498 players. Your link shows that places 17 had identical scores of 7. The next 15 places (621) were half a game off the pace with 6.5 wins. I imagine a good number of the top 21 didn't lose a game. How many of the top 21 got to play each other in 9 rounds?
Usually it is normal to calculate no. of players divide by 3 to get optimal round number. So for 25  30 players it is normal to get 9 rounds in swiss system and all top plyers should play each other.
Think how many rounds would be needed for a tournament of 498 players. 166 rounds?
http://www.cappellechess.fr/fr2/default.php?page=3967 or nine?
Your post seems to indicate that you thought 9 rounds was adequate for 498 players. Your link shows that places 17 had identical scores of 7. The next 15 places (621) were half a game off the pace with 6.5 wins. I imagine a good number of the top 21 didn't lose a game. How many of the top 21 got to play each other in 9 rounds?
You could have a 166 rounds and everyone could be tied at 83 points. A minimum number of rounds in a swiss style does not guarantee no ties. It guarantees there are no tied players who won all their games.
2 to the power of 8 is 256, 2 to the power of 9 is 512 so nine rounds is the minimum to determine a winner of a swiss style tournament of 257 to 512 players if two players dominate the others.
Round 1  512 players tied with 0 points
Round 2  Maximum 256 players tied with 1 point
Round 3  Maximum 128 players tied with 2 points
Round 4  Maximum 64 players tied with 3 points
Round 5  Maximum 32 players tied with 4 points
Round 6  Maximum 16 players tied with 5 points
Round 7  Maximum 8 players tied with 6 points
Round 8  Maximum 4 players tied with 7 points
Round 9  Maximum 2 players tied with 8 points meet in the last round.
If there were only eight rounds the top two players would not have met.
As indicated by previous posters you could have more rounds.
Notice a knockout style tournament would have the same nine rounds.
Notice that a knock out tournament would need more games because of draws, even if you don't call replaying after a draw a new round. Notice what 9 rounds did to Petrosian who probably didn't lose a game. If there was a draw after 166 rounds, I would say it was a legitimate result. After 9 rounds, when few of the top players have played each other I think a draw is unsatsfactory, and going to tie breakers doesn't prove anything. 166 rounds would be impactable because of time restrictions. 13 rounds would have been a vast improvement. I still say the more rounds the better. Your example tends to show why 9 rounds is not optimal.
but what if we have 9 players?How many round in that case?Or is it better ,in that case play round robin?
but what if we have 9 players?How many round in that case?Or is it better ,in that case play round robin?
If you have time for 8 rounds then a roundrobin. If you have time for 16 rounds then a double round robin.
If you do not have the time then see the post from RoyLupez
...use a table like this where for each round, you double the maximum number of players:
1 rd, 2 players
2 rds, 4 players (3 to 4 players)
3 rds, 8 players (5 to 8 players)
4 rds, 16 players (9 to 16 players)
5 rds, 32 players (17 to 32 players)
6 rds, 64 players (33 to 64 players)
and so on...
The calculation is this:
2 to the power of 3 = 8, so 3 rounds for 8 players.
2 to the power of 4 = 16, so 4 rounds for 9 to 16 players.
If you do not have time for 8 rounds then a Swiss with a minimum 4 rounds works for you.
One way to think of it is that for every round, up to half the players can get 1 point. So for 30 players, after rd 1, there can be 15 with 1 point, then 8 after rd 2, 4 after rd 3, 2 after rd 4 and finally 1 after rd 5.
If you have a calulator with a ln (logarithm) function you can calculate:
rounds = ln(players)/ln(2), then "round" up. (no pun intended!)
or, rounds = log2(players) (log2 = base 2 logarithm).
Or use a table like this where for each round, you double the maximum number of players:
1 rd, 2 players
2 rds, 4 players
3 rds, 8 players
4 rds, 16 players
5 rds, 32 players
6 rds, 64 players
and so on...
Of course, if you have time for an extra round or two, more of the top players will have a chance to play each other.
Also to be considered is how many places need to be accurately determined. Once first is determined you may need two more rounds to get 2nd, four more to get 3rd. For team selection tournaments in the late 60's, a round robin was the way to go. This is what I put together to create round robin schedules a few years ago: http://home.comcast.net/~wporter211/realsite/chess_etc/rrpair.htm
One way to think of it is that for every round, up to half the players can get 1 point. So for 30 players, after rd 1, there can be 15 with 1 point, then 8 after rd 2, 4 after rd 3, 2 after rd 4 and finally 1 after rd 5.
If you have a calulator with a ln (logarithm) function you can calculate:
rounds = ln(players)/ln(2), then "round" up. (no pun intended!)
or, rounds = log2(players) (log2 = base 2 logarithm).
Or use a table like this where for each round, you double the maximum number of players:
1 rd, 2 players
2 rds, 4 players
3 rds, 8 players
4 rds, 16 players
5 rds, 32 players
6 rds, 64 players
and so on...
Of course, if you have time for an extra round or two, more of the top players will have a chance to play each other.
ok 16 players 4 rounds... but what if the "tournament" is meant for qualifying reasons and you want to get the best 2 players, then 3 rounds would make sense for 16 players since you have virtually 2 swiss with 8 players? or actually making two parallel 3rd swiss with 8 players is better?
It would still make sense to play 4 rounds otherwise the result might be a bit unfair (there would be an high chance of having one player qualifying with much weaker pairings for example). A mere 3 rounds tournament can have really random results.
For 16 players and two places to be accurately determined, you need six rounds; four to determine 1st place and two more for 2nd place.
Isn't it possible though, to overshoot the optimal number of rounds? E.g. 16 players would ideally have 4 rounds to convincingly choose a winner, with pairings getting closer with each round, til the top two players face each other in the last round (assuming an ideal situation where higher rated players always beat lower rated opponents). The final sorting ends up perfect. But if you have 5 rounds, the top players have already faced each other and can't play each other again. One person that's vying for maybe fourth place ends up having to face the #1 player, an unfair matchup compared with the pairings that he is otherwise tied with. Perhaps it would be best if the top player (out of 16) would get a fullpoint bye after 4 rounds. That way he'd get the point that he'd probably otherwise win anyway, while allowing an extra round for sorting the rest of the places without one of them facing an unfairly tougher opponent.