Our annual city championship is coming and I would like to know the opinions of this group of TDs how to make the groupings.
We will have 12 players that have qualified throughout the year to play.
We break this into 2 groups of 6 players. Each group plays a round robin to determine a winner of that group. Then the 2 group winners play a small match to determine the city champion for the year.
All the players are USCF rated.
How would you make the groupings?
I would seed the players by rating, and group as follows:
Group A
1
4
5
8
9
12
Group B
2
3
6
7
10
11
This is not the only way to do it; it’s just the way I’d do it.
Presumably, the idea is to split the 12 players in such a way as to make the two groups relatively equal in strength at the top and in rating spread throughout the field. If that is the aim here, then I would think Mr. Price’s proposed distribution (or some fairly small variation therein) would be the way to go.
I agree with Messrs. Price and Reed with the caveat that the average rating should be as close as possible.
Alex Relyea
Since the sections are going to be RR, I’d split them 1,3,5,7,9,11 and 2,4,6,8,10,12 unless there were some major gaps it the ratings distribution.
The draw for colors and pairings in the RR sections takes care of the rest of random allocation issues.
The way WinTD handles this if you want sections of roughly equal strength is to start with 1-4 and 2-3. Then put 5 in the section with the lowest average, and 6 in the other one. Then 7 in the section with the lowest average for the first three players and 8 in the other, etc.
Another way would be to look it as a 12 player (corrected from 16) seeded event.
1 and 12 go in group 1.
2 and 11 go in group 2
3 and 10 go in group 1
4 and 19 go in group 2
etc.
(I suspect if you asked a dozen TDs this question, you’d get at least 15 answers.)
But Mike, only MY way is the right way (I’ll hold off from giving a written answer because it might change from event to event). 
Years ago we had a club championship in Lake County that ran this way, except there were often more than 2 sections to “normalize”. I’d use an “alternating method” like some of those described above to construct the groups, and then check BOTH the average rating of the groups and the standard deviation of the groups. (The argument being that two groups of roughly the same average rating - but one where they were all rated about the same and the other where they were all outliers - were not really equivalent-ish groups.) I would then use “guess and fix” to try to normalize the groups into a mode where there were as alike as reasonably possible.
How about taking two participants and making them Section Captains, picking the other participants in their section? Might be interesting to see who they’d pick!
12 person unrated blitz tournament. 1st, 3rd, 5th, 7th, 9th, and 11th places go into one group for the main tournament, and the rest go into the other group.
At first, I thought that would be how I would divide the players. But then I realized that guarantees that the average rating in the second group is lower than that of the first group. (Well, in a degenerate case, if the ratings of every pair of players 1-2, 3-4, 5-6, 7-8, 9-10, and 11-12 are the same, the average ratings would be equal.)
I think alternately assigning the higher rated player of the pair to the first or second group would help keep the average rating of the two groups closer.
Then again, with 12 players, there are only 332,640 ways to divide them into two groups of six, so it should not take too many computrons to determine the optimal division for a given collection of players. 
The way it has been done in the past is
1, 4, 5, 8, 9 and 12 in the first group.
2,3,6,7,10 and 11 in the second group.
I like the way Tom says WinTD does it with the average rating of the groups making a difference which player goes into it. I will make that suggestion tonight to the other officers.
I was figuring 924 variations, or 12! / (6!*6!). That is 12/6 * 11/2 * 10/5 * 8/4 * 9/3 * 7/1.
It has been a lo-o-o-ong time since I took statistics, so what did I miss?
Actually, you are a lot closer than I am – I calculated permutations, not combinations.
However, you calculated C(12,6), which is the number of ways to choose 6 items of 12. However, you double counted. For every way you could choose six players, the way in which you choose the other six players gives the same division of players. For instance, if you choose {1,2,3,4,5,6}, that gives you the same division as choosing {7,8,9,10,11,12} – it’s just that the two groups are “swapped.”
I actually calculated P(12,6)/2, which was a sad mistake. 
If you do not want to do rating as all of the previous suggestions trend towards, then why not by when the individuals qualified. First to qualify #1 in group 1, second to qualify #1 in group 2, third to qualify #2 in group 1, etc. to fill up both groups.
Also there is simply the option of randomly drawing what players are in what group, along with the seedings in each group.
Larry S. Cohen
Another vote for 1,4,5,8,… vs 2,3,6,7,…
If you play fantasy sports you have probably already seen this concept as a “snake draft” order, in which team owners draft the first round in a given order, then reverse the order for the next round, then reverse it again, etc.
In this case there are two “owners” drafting chess players for two teams. A chooses 1 and B chooses 2, then B chooses 3 and A chooses 4, and so forth.
I think this is a issue almost like tiebreak choices. Someone will always be unhappy with how the
groups are divided. I understand the reasoning for the 1,4,5,8,… vs 2,3,6,7,… concept in making the
two groups as close as possible in average rating. I’m just not sure that that in itself is the best goal.
I realize that splitting 1,3,5,7,… guareentees that (call it group A) group A will be stronger than
Group B.
Having said all that I still lean towards the 1,3,5,7 split for these reasons.
With the 1,4,5, splitt it seems like the higher rated player has a tremendous advantage over the
number 2 player. In order to win his section he plays a weaker field. While #2 has to be able to win
against #3.
I view the fairness factor of (all things being equal) for example #1-2000 vs #3-1800 being equal to
#2-1900 vs #4-1700 as a fairer contest.
And while it is true that Group A on average is Stronger than Group B the reality is that in theory if
#1 and #2 win their groups then they both have to play each other. But that is just theory. What
happens in reality is that some player gets hot in each group and wins that group and then they have to
face each other.
I do like Mike Nolans suggestion of having team captains choose the groups. Doing that I would
have 1 and 2 choose, but they would choose not for themselves but for the opposing group, with #2
getting first go.
Even random selection seems somewhat interesting.
An alternative is to assign to equalize averages from the bottom up. That is, start with 9 & 12 in one group and 10 & 11 in the other. Put 8 in the group with the highest average, and 7 in the other. Then 6 in the group with the highest average and 5 in the other, etc. That would tend to make the average of the 2nd through 6th spots in each section similar, so the #1 and #2 seeds would face relatively similar strength fields.
A slight refinement of this idea might be to give the first Section Captain one choice, the second two, the first two more, the second two more, etc. 1-2-2-2-2-2-1.
Bill Smythe