There's an alternate way to solve the original problem.
You can weigh them 3vs3. If the heavier ball is in one of those two groups in will tip the scale to that side, then you work from that group of 3. If it's not, you know it's one of the remaining 2 balls. So you can actually get it in two if it happens to be in the set of two, but no more than 3 if it isn't.
You can weigh them 3vs3. If the heavier ball is in one of those two groups in will tip the scale to that side, then you work from that group of 3. If it's not, you know it's one of the remaining 2 balls. So you can actually get it in two if it happens to be in the set of two, but no more than 3 if it isn't.
