Lopsided Round Robin format

[ianw]

Hi,

I would like to seek the forum's opinion a matter. Our pool league has 12 teams, but only 5 sets of tables. We play round robin with a bye during the season. All teams make play-offs, but since we are limited to 3 matches per day and need to play over the weekend, the proposed format is as follows:
Final round 4 teams play round robin
1st place gets a bye to the final on Day 2
The rest play on Day 1 divided into 3 houses. Winner of each house goes to Day 2
1 house would have 3 teams, the other 2 would have 4 teams. Again, round robin.

Question is, what arrangement of teams is a better/fairer distribution (numbers are in order of ranking):
House 1:   2     7  10
House 2:   3  5  8  11
House 3:   4  6  9  12
or:
House 1:   2     9  12
House 2:   3  5  7  11
House 3:   4  6  8  10

That's the simple assessment. team 2 gets an advantage of one less team to play. Question is make them play the weakest teams or balance it across all three?

The other catch is there is a handicap system in place. Not sure if it really matters, but I'll try to explain.
Each match is 5 players play each of 5 from other team per leg. 10 pts for a winner and 1pt per ball down to loser, so max 50 per leg. Team's winning legs, not points determine overall winner.
Each player has carries average from previous matches, the team's averages are added up and the rounded difference is the handicap added to the weaker team.
So team A, Average is 40 (8.0 per player), team B is 30 (6.0 pp).
In a leg, say score is A:44 to B:35. Add the handicap (10 pts) to B and they win the leg.

This could seriously affect the outcome if the spread is very wide since it could make it tough to beat a lucky-on-the-day weak team. Right now the top 3 teams all are in the 40-41 range, the rest between 40 and 36, except the last team with 31. So a 9 pt spread. During the season we have seen as high as a 12 pt spread, which is like playing 5 against 6+ players.

If it helps, when we had 10 teams, top 2 got a bye to Day 2. Then Day 1 was:
House 1: 3, 6, 8, 10
House 2: 4, 5, 7, 9

Thanks for everyone's feedback on this.

Ian.

[Ian Wakeling]

Ignoring the handicap system, I think you should think about the allocation to each house in terms of the most important match.  This is the match between the two highest ranked teams, since the team who wins this match is surely the favourite to win the house group.  With this in mind, I think both your examples give an advantage to team 2, not necessarily because there in one team less, but because there is a much larger rank difference between the top two teams.

Hope that helps.

[ianw]

Hi Ian (Wakeling),

This is the other Ian (the OP). Thanks you for your comment.

My thoughts were Team 2 (ie:2nd place) would have a major advantage by only playing 2 instead of 3 teams. Further, let the lower teams play first so Team 2 can decide their own fate against the other teams. Without a handicap system in place I would clearly have 2nd play 9 and 12 as that's the greatest advantage. Team 2 stand a very good chance of going through  7/10 as well. Playing 9/12 not only benefits their chances to get through but also makes it tougher for 4th place to get in, again giving an advantage in the finals to the 1st and 2nd place teams (assuming they advance)

However, with the handicap and the spread they might be seriously challenged to overcome the spread. Hence, playing 7/10 still gives an overall advantage and weakens the effect of the spread. That is my real concern.

Again, any further thoughts are appreciated.

Regards,
Ian