I want it to be three, winnowing down from four balls on each side, to two, to one.

But there is no requirement to weigh all the balls at once, so if you only weigh two on each side, you have a 50% chance of being able to find the heaviest ball in only two weighings and 50% chance of finding it in three.

So by further application, you have a one in four chance of finding the heaviest ball with only one weighing, and a maximum of four, by placing only one ball on each side at a time. Thus the minimum turns out to be one.

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