[geoschmo]

Originally posted by BBegemott:
IMHO it can't be done with 3 player games with any number of players (except 3 of course).

Assume 9 players tournament. Nine players will be marked with numbers 1,2,...,9.
I could make only legal 10 triplets:

1) 123
2) 145
3) 167
4) 189
5) 246
6) 278
7) 259
8) 347
9) 369
10)358
11)48?
12)49?

There is no legal substitution for 11th & 12th triplet, because 4th player has already played 1,5,2,6,3,7. Putting any number instead of '?' contradicts to the rule 'everybody has to play everybody else once, but only once'.

Or have I missed some kind of hidden trick? I am really curious to see your solution

The problem is not every legal combination will work. For example, by using
1) 123
2) 145
3) 167
4) 189
you prevent the option of trying 146. This is fine for 1, 4, and 6 as long as they all play each other once, but might prevent others from being able to play each other, as you found out.

Instead if you do something like...
1) 123
2) 456
3) 789
4) 147
5) 158
6) 169
7) 249
8) 275
9) 268
10) 348
11) 359
12) 367

then it works.