Csdrhrt

I have spent too long trying to figure this out and cannot seem to get it just right, so here I am asking the Math gods. This weekend I played in a volleyball tournament called "King of the Beach", where the tournament was structured with two pools of individual players. The format was in such a way that each player would play every other player in their own pool, with a player from the second pool as their teammate (ie pool A-F and pool 1-6). An example match would be the teams of A1 vs B2 or A1 vs C5.

Under this format, all letters (A-F) would play all other letters with a number (1-6) partner, and all numbers would player all other numbers with a letter partner.

I am trying to set up something similar for a team building exercise at work with 12 different people (needing two pools of 6). My issue is that I end up with all letters playing all other letters and all numbers playing all other numbers, yet player A may end up playing against number 4 three different times.

Maybe what I am trying to do is not even possible, but ideally I would take these two pools and make every letter play every other letter, while also only playing each number once. I have looked this up online and fail to find anything that is much help, so any help someone may have would be greatly appreciated. Thanks.

Pool 1

A

B

C

D

E

F

Pool 2

1

2

3

4

5

6

Try this: Let's pair each letter to each number for the first 5 games. i.e. $$begin{matrix} A & 1 \ B&2\C&3\D&4\E&5\F&6\end{matrix}$$
Why 5? Because there are $frac{n!}{k!(n-k)!}$ unique combinations of $k$ letters, so if we have 2, we have 15 ways of combining them. Since 6/2 =3, we have that we can have 5 games where each pairing is unique. Now for the last "round", each letter hasn't played the number it was on the team with. So we just swap the numbers for each pair like this: $$begin{matrix}A&2\B&1\C&4\D&3\E&6\F&5\end{matrix}$$ with the game pairs given by the letter $(A,B), (C,D), (E,F)$.

Hope this answers this question.

If you are plainly allowing to play every letter and number to play every other letter and number, without the obvious repetitions, we have:

• $6$ ways of choosing the first letter,


• $6$ ways of choosing the first number,


• $5$ ways of choosing the second letter and


• $5$ ways of choosing the second number.


• And because WX vs YZ is the same than YZ vs WX, we half everything.

Hence the total combinations are:
$$frac 12 6cdot 6 cdot 5 cdot 5= 450$$

A cycling strategy would be matching the 15 rounds in the group:

GROUP 1-1:
$$
A 1\
B 2 \
C 3\
D 4\
E 5\
F 6
$$

And then rematching from the new group, cycling the numbers:

GROUP 1-2:
$$
A 2\
B 3 \
C 4\
D 5\
E 6\
F 1
$$

These way you have the groups 1-1 to 1-6, each one with 15 matches.
After those $6cdot15=90$ games played, introduce a pair swap between 1 and 2 and rematch a new set of 90 games, cycling the number as already described, for a new pack of groups:

GROUP 2-1:
$$
A 1\
B 3\
C 6\
D 2\
E 4\
F cdot
$$

The $cdot$ indicates no match must be taken from that team.

Build that sequence of numbers from this list, which alternates each possible combinations:
$$
GROUP 1-X: 123456\
GROUP 2-X: 13624cdot\
GROUP 3-X: 1425cdot 3\
GROUP 4-X: 15463cdot\
GROUP 5-X: 16532cdot
$$

The last group comes from taking sequences unlisted in the previous cycling:
$$
GROUP 6: text{sequences from }{21,26,35,41,43,51,52,64}
$$

Hence you have all your $5cdot 90=450$ matches listed.

As you see, that is a lot of fun for a single summer...