Hint 4 for the 12 billiard balls maths prize (follows hint 3)

All the weighings must divide the possibilities into three more-or-less equal groups, so that the result of the weighing divides the number of possibilities remaining by three.

*So, each weighing must be one where “left scale heavier than right scale”, “right scale heavier than left scale”, and “scales equal”, are more or less equally likely.*

Now you can work out what to do at the first weighing. There are only six options for the first weighing – one randomly-chosen ball against another one, two randomly-chosen against another two, 3 against 3, 4 against 4, 5 against 5, 6 against 6 – so if it comes to it you can just check out all six to find the one which narrows the possibilities from 24 to 8.

Work out for yourself, for each first-weighing outcome, a second-weighing operation to narrow down from 8 to 3, 3, or 2.

Then work out for each second-weighing outcome what third-weighing method to use to narrow down from 3 (or 2) to one. And you’re done! Now write down your answer, hand it in, and claim your prize!

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