I'll be working on an application that'll help me work out all the complex organization of a swiss-style tournament.
Swiss-style is where everyone (except possibly 1 person who gets a bye) gets to play in each round and the next round's pairings are based on the win-loss record of the participants at the end of the current round.
I've been working out a sample of a tournament and I'm trying to find a good way to do tie-breakers. Since we only keep one statistic (wins), it's really hard to sort players to decide the rankings for the elimination rounds.
Currently, I have:
Sort by # wins.
Sort by cumulative record (effectively, their record):
Quote from: http://en.wikipedia.org/wiki/Tie-breaking_in_Swiss_system_tournaments#CumulativeTo calculate this, sum the running score for each round. For example, if a player has (in order) a win, loss, win, draw, and a loss; his round-by-round score will be 1, 1, 2, 2½, 2½. The sum of these numbers is 9. This system places more weight on games won in the early rounds and the least weight on games won in the final rounds. The rationale for this system is that a player who scored well early in the tournament has most likely faced tougher opponents in later rounds and should therefore be favored over a player who scored poorly in the start before subsequently scoring points against weaker opponents (Just & Burg 2003:200–201).
After that, there's no other measure I can determine that would rank players by skill than by random chance (such that, if their pairings were swapped, the outcome would be different).
Here's some raw info that I've been playing with:
Round 1 pairings: 19|5, 7|13, 20|12, 10|3, 17|2, 14|4, 8|1, 6|15, 16|18, 9|11
Round 1 winners: 19 13 20 3 17 14 8 15 18 9
Round 2 pairings: 19|9, 13|18, 20|15, 17|14, 3|8, 5|11, 7|16, 12|6, 10|1, 2|4
Round 2 winners: 9 18 15 17 3 5 16 12 1 2
Round 3 pairings: 9|1, 18|3, 15|8, 2|19, 13|12, 20|16, 14|5, 1|7, 10|4, 6|11
Round 3 winners: 17 18 15 2 13 20 14 1 4 11
Round 4: 17|15, 18|2, 9|14, 1|3, 20|13, 8|12, 4|19, 11|16, 5|7, 10|6
Round 4 winners: 17 2 14 1 13 8 4 16 7 6
Person: <record> <cumulative total> <W-L>
1: LWWW 6 3-1
2: LWWW 6 3-1
3: WWLL 7 2-2
4: LLWW 3 2-2
5: LWLL 3 1-3
6: LLLW 1 1-3
7: LLLW 1 1-3
8: WLLW 5 2-2
9: WWLL 7 2-2
10: LLLL 0 0-4
11: LLWL 2 1-3
12: LWLL 3 1-3
13: WLWW 7 3-1
14: WLWW 7 3-1
15: WWWL 9 3-1
16: LWLW 4 2-2
17: WWWW 10 4-0
18: WWWL 9 3-1
19: WLLL 4 1-3
20: WLWL 6 2-2
Rankings:
#1: 17
#2-#3: 18, 15
#4-#5: 13, 14
#6-#7: 1, 2
#8-#9: 3, 9This is where I'm looking for some tie-breaking ideas. I'm hoping to use the statistics provided. With the cumulative record, there's no possible way for two people that have the same record to play each other.
Doing some research online, the metric would be based on opponent's record. But that is strictly based on the randomized pairings and the ticket to the elimination round was decided by a computer rather than skill.
BBH has suggested that there be a quick one-hand playoff to decide the tie-break and I think this may possibly be the fairest way to settle the tie.
it's a pickle all right
bamf defenitely got the short end of the stick at the last one, when i believe it was based on the records of people you lost to?
would basing the tie-breaker on the records of the people you defeated be better? a win vs someone 2-2 is surely more valuable than a win against someone 0-4.
i don't like the idea of a tie-breaking round, but it may be the best solution. we already use that method as a match tie-breaker if venture is equal when time is up.
The way that swiss works is that you should play with someone that has a similar record as you do in each round.
The problem is probably demonstrated at the extreme with #1 and #2 with the above info. Both are LWWW; 1 played 8, 10, 7 and 3; 2 played 17, 4, 19 and 18. Their opponent's respective points are: #1: 5/0/1/7 and #2: 10/3/4/9. If you do a straight adding up, #2 is clearly the winner of the tie break. But, if you go through the history of the games played, he was lucky to play #17 (1st overall) and #18 (tied for #2 overall) whereas #1 played #10 and #7 that ended up with losing records. Of course, this is the extreme case because #1 was also paired up with someone with a different record (playing against #7 [1 point] in round 3) and #2 being lucky and playing #18 [9 points] in round 4.
Now, another way to solve this would be to add an extra round of play because swiss-style tournaments are always best with a rounded-up value of log(players)/log(2). For our tournaments, 5 rounds of swiss and 3 rounds of elimination would take just about 6 hours. That might work out and it gives us a good envelope to expand to 32 players -- though with 32 players, elimination brackets would expand to a top 16...
The playoff might seem like a good idea but what if there's 3 people? That would require 3 mini-games to be played.
I'll simulate a round 5 and post the results.
EDIT:
One thing I might want to do is pair up people with exactly the same record in each round.
This is a horrible idea, you need to win the first round to make it into eliminations...
EDIT #2:
Same problem exists...