Last Man Standing Tourney Still Open

[Thexder]

Right now I still only have 9 - so, I'm not sure how we'll work out the uneveness for a single elimination tourney. Even if I get one more player, then numbers will be 10 for round one, but will be 5 for round two...again uneven. 12 will be reduced to 6, which would be reduced to 3...again uneven (Math was never my strong suit).

You need to have 1 - 2 - 4 - 8 - 16 - 32 to make it straightforward. But if you have, say, 12 players, you can make special rule to end up to an "easy" number of players. When there's 12/2=6 pairs, you just make a rule that the two "best losers" will qualify to the second round. The you have 8 players and after the it goes smoothly: 8 --> 4 --> 2 --> WINNER.

,
Thexder