Riddles

[Hypersmurf]

So, for the third weighing, I would normally weigh balls 2,7,8,12 against balls 5,6,10,11.
Since I know that balls 2,8,12,5,11 are all balanced, I can exclude them from the third weighing.
That leaves balls 7,6, and 10.

No - you know that ball 10 is balanced, because there's only one unbalanced ball. Your first weighing was unbalanced, and ball 10 wasn't in the first weighing.

If in that first weighing, the second pan was heavy... and since you know that either ball 6 or 7 is the odd ball (both of which were in the second pan), then the heavier of the two is the odd ball.

If in that first weighing, the second pan was light... and since you know that either ball 6 or 7 is the odd ball (both of which were in the second pan), then the lighter of the two is the odd ball.

But it's not necessary to leave any out of the weighing. You can, but you don't have to, since you know that 2, 8, and 12 in pan 1 are balanced, and 5, 10, and 11 in pan 2 are balanced - they won't affect the result.

-Hyp.