Hey Tom, do you think it'd be possible to add two turns to the swiss system? Right now it seems that the default 3 rounds aren't enough in a tournament with 30 players.

Most of the tournaments have three rounds. To have four you need to have a huuuge number of players or either few players per game, and both are counterproductive...  \=

I think it'd be more fun to have some more round.

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Umm... this.

Yes. I was trying to figure out how to increase the number of rounds in a tourney I just made, and found I had to have over 60 players for this to happen! 3 rounds just isn't enough imo. 6 would be tons better! :)

Vataro wrote: Yes. I was trying to figure out how to increase the number of rounds in a tourney I just made, and found I had to have over 60 players for this to happen! 3 rounds just isn't enough imo. 6 would be tons better! :)

Probably 6 would be too much. Remember you have to consider number of players in the tournament as well as number of players per game.

To have 6 round Swiss tournament you need to have relatively few players per board, say 3 to 5, and a total number of players between 20-30 or 40-60 or more respectively

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I didn't even realize the # of rounds setting was dependent on the number of players. Why is that? This tournament: http://www.wargear.net/tournaments/view/68

Is a total waste of time now that we've all realized it was a 3-round tourney on a 4-player board. Maybe I don't understand the Swiss well enough, but why would the # of rounds and # of players be related at all?

From my limited research into the swiss system the number of rounds has to be at least log base 2 of the number of players for two player games. My guess is Tom extended that to log base players per game of the number of players.

That gives the floor for the number of rounds you need, but from what I read you can have more. And I think more would be good.

One of the basic problems I think we are struggling with in all this tournament stuff is much has been developed for 2 person or 2 team games. How to do tournaments with multi-player games seems to be much less established. So, open question - should their be at least as many rounds as players per game?

According to the Wikipedia article...
http://en.wikipedia.org/wiki/Swiss_tournament

"Determining a clear winner (and, incidentally, a clear loser) usually requires the same number of rounds as a knockout tournament, that is the binary logarithm of the number of players rounded up."

So, assuming a 32-player tournament on a 4-player board run knockout style (i.e. top two "advance"), the result would be akin to a 16-player tournament on a 2-player board, which would take four rounds to determine a clear "winner". Coincidentally, this is why the total number of rounds and number of player would be related - more players means more rounds to determine a clear winner, while fewer players means fewer rounds are needed.

However, the real fun part here is that the Swiss-system was designed for two player tournaments, not 4 player tournaments. It's possible for one pair to "advance" in one game, then one player to advance to the next but not the other player in that pair. Consequently, this muddies the waters a bit.

There is a maximum limit on the number of rounds - since the point of the Swiss system is to make sure the same player doesn't play another player twice, in a 32-player tournament on a 4-player board, that would be... crap, I used to know how to calculate this. I think it involves a factorial somewhere. Am I looking for a combination or a permutation? I think I want a k-combination (nCr-style, if I remember correctly), which would be... 35960? Okay, that can't be right. "Less than 31", let's put it that way.

(If somebody with a far more solid understanding of combinatorics could step in, that'd be fantastic.)

So, we have an upper bound that hopefully Hugh will find for us, and we have what seems to be a lower bound, which is the binary logarithm of the number of player-sets in the tournament. The question now is if there's a "fair" way to determine the number of rounds a Swiss-style tournament should take with more than two people on each board that prevents everyone from playing each other more than once, yet gives a realistic account of who "won" the tournament?

Tesctassa II wrote:

Vataro wrote: Yes. I was trying to figure out how to increase the number of rounds in a tourney I just made, and found I had to have over 60 players for this to happen! 3 rounds just isn't enough imo. 6 would be tons better! :)

Probably 6 would be too much. Remember you have to consider number of players in the tournament as well as number of players per game.

To have 6 round Swiss tournament you need to have relatively few players per board, say 3 to 5, and a total number of players between 20-30 or 40-60 or more respectively

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I was hoping for a 4 player board to have about 20-40 players in a tournament, so it actually sounds perfect to me :P.

I would be fine with 5 though... 4 or 3 just seems too little though.

Oatworm wrote:

There is a maximum limit on the number of rounds - since the point of the Swiss system is to make sure the same player doesn't play another player twice, in a 32-player tournament on a 4-player board, that would be... crap, I used to know how to calculate this. I think it involves a factorial somewhere. Am I looking for a combination or a permutation? I think I want a k-combination (nCr-style, if I remember correctly), which would be... 35960? Okay, that can't be right. "Less than 31", let's put it that way.

(If somebody with a far more solid understanding of combinatorics could step in, that'd be fantastic.)

So, we have an upper bound that hopefully Hugh will find for us, and we have what seems to be a lower bound, which is the binary logarithm of the number of player-sets in the tournament. The question now is if there's a "fair" way to determine the number of rounds a Swiss-style tournament should take with more than two people on each board that prevents everyone from playing each other more than once, yet gives a realistic account of who "won" the tournament?

I do not believe you need any combinatorics for what you want.  If there are 32 players in a tournament with 4 players per game, then you have 31 possible opponents you face 3 at a time, so there is a maximum of 31/3 = 10ish possible rounds. 

Unfornuately, from working on the 3-player round robin schedule, I can assure you that there is no easy way to make sure that you never face the same opponent twice if you want to have disjoint rounds (that is everyone plays in exactly one game each round).  Mongrel and I figured out the 3-player round robin scheduling problem without disjoint rounds and the 4-player problem is much more complicated.

Technically with up to 64 players, 3 rounds is enough to determine a 'true' winner on a 4 player board. After one round you have 16 undefeateds. After two rounds you have 4 undefeateds. They play the final game to determine the winner. If you lose in either the first or second round you can't win the tournament.

The problem we are seeing is that if the number is not exactly 64, then it is possible someone who lost one of their first two rounds will make the X-0's final game, can win that game, and cause multiple people to be at X-1 causing the ties we are seeing in just about every tournament.

A 'Swiss' tournament is supposed to be one in which although everyone gets to play in every round you never play the same opponent twice and you play opponents at or around you total score. To accomplish the first part of this you do need in excess of 64 players to have a fourth round.

What it sounds like it everyone wants is to do away with the first part and keep the second part. Allow additional rounds with people near your own score but allow people to play someone they have played before. With this criteria you can have unlimited rounds.

This can be done first creating groups of 4 by wins/loses, using the ranking score as the first tie breaker, and a random roll as the second tie breaker. While this will still have a possibility for a tie it grows less and less with each added round.

Seige07 wrote: The problem we are seeing is that if the number is not exactly 64, then it is possible someone who lost one of their first two rounds will make the X-0's final game, can win that game, and cause multiple people to be at X-1 causing the ties we are seeing in just about every tournament.

I think the problem we are seeing here is that aside from determining a clear winner, which of course matters (as it's possible to see from other threads about the scoring system), it should be possible to do it having fun! And this requires more turns to play in a tournament.

Furthermore, despite being 3 an optimal number of rounds to termine a 64-players tournament winner, it is also clear that playing that number of rounds doesn't allow mistakes. So it is equal to playing a knockout tournament. The swiss tournament instead should be half the way between the Round Robin, where you can lose lot's of games while still winning the tournament, and the Knockout, where you can't lose one game.

That's why usually you have more rounds than the "optimal number" required.

I think we should probably debate if it is necessary to change the Swiss system, or invent a new system from scratch, so that it fits better to a n-player game like Risk.

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Seige07 wrote: Technically with up to 64 players, 3 rounds is enough to determine a 'true' winner on a 4 player board. After one round you have 16 undefeateds. After two rounds you have 4 undefeateds. They play the final game to determine the winner. If you lose in either the first or second round you can't win the tournament.

As you alluded to at the end of your post, this goes back to the tiebreaker discussion being had in another thread. Again, in the 4-Play tournament I referenced above, it's over after 3 rounds with 6 of 32 players at the top of the board with 2-1 records. Hugh won on score but I think it's safe to say the result was unsatisfying for all.

asm wrote: Hugh won on score but I think it's safe to say the result was unsatisfying for all.

I can accept loosing on score, but only after a decent number of rounds (and a fair scoring system in case of RR tournaments)

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So this tournament just started it's final round:

http://www.wargear.net/tournaments/view/24

A number of the players went 1 and 1 (12 of them) and all have the same score. While 2 people went 2 and 0. 5 players per game.

So for the 3rd and final round 3 of those 12 got put in with the 2 leaders, 5 are in their own game, and 4 were put in with someone who went 0 and 2.

So now, if I am thinking it through right, the only people that have a chance of winning the tourney are the 2 who went 2-0 and 1 of the 3 in that game. Because even if the 2 who went 2-0 lose, the 1-1 guy that beats them will pick up more points from them then anyone else will get.

I don't know if this should be filed under "more rounds for swiss" or "breaking ties in a tourney", but in either case 3 people are given a title shot randomly, while 9 are not.

Good point Amidon. Tom, how are the players in the final game chosen? Seems like that's a definite reason for adding some more rounds in.

bengaltiger wrote: Good point Amidon. Tom, how are the players in the final game chosen? Seems like that's a definite reason for adding some more rounds in.

Players in the last round are chosen as in the previous rounds, that is players with the same score play against each other. In the tournament cited by amidon, since less that 5 players are tied at 2-0, some "randomly" chosen player with a 1-1 record is picked.

And this is not fair. Again some tie-breaking system should be used. In this case probably the "head-to-head" system could decide who deserves to play against the top players and a chance to win the tournament.

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The "randomly" is what I was wondering about. Is it actually random, or sorted somehow? Non-tourney player score, alphabetical name, etc?

It's a fair point which I think does need addressing. The Swiss system works well for 2 player games but when you extend it out to n player games it throws up problems like this.

Perhaps the final top table game should only include players who have gone unbeaten in previous rounds. The downside of this is you might end up with a 2 player game on a 10 player board or similar.

I've got two thoughts on things to try. I don't know if either will work well, but I don't think any of us do. We just need to try 'em and see how they work out.

Suggestion 1) Make the minimum number of rounds the larger of what is used currently or one more than the number of players per game.

Suggestion 2) Make swiss tourney's open ended on how many rounds and have them stop when one person has a better record than everyone else.

I'd like to try the 2nd one. Sure it could lead to long tourney's, but I think we would find a "natural" length for different sets of total players and players per game. Call it swiss+

Amidon37 wrote: I've got two thoughts on things to try. I don't know if either will work well, but I don't think any of us do. We just need to try 'em and see how they work out.

Suggestion 1) Make the minimum number of rounds the larger of what is used currently or one more than the number of players per game.

Suggestion 2) Make swiss tourney's open ended on how many rounds and have them stop when one person has a better record than everyone else.

I'd like to try the 2nd one. Sure it could lead to long tourney's, but I think we would find a "natural" length for different sets of total players and players per game. Call it swiss+

I like the 2nd idea better, but think that there should be a maximum on the number of rounds, say somewhere between double and triple what it is currently just so that the tournament ends (tie breaker here).  For 2 player swiss tournament this would not change anything as one player would still come out on top without any issues.

*edit: upon further thought, working out some examples, I think that for a small number of players per game (3,4,5), it would be highly unlikely for the tournament to take more than triple the current number of rounds before a winner is decided so it may be that no maximum on rounds is needed.