Re: False coin, math problem.
I think it's max 4, but we always have a 50% shot at 3, so stochastically it's 3.5.
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Firstly there's no point weighing 6 vs. 6 (tells us nothing)
TRY1: Take any 6, and balance them. We then have 6 normal coins (if the balance is off, then it's the 6 we didn't select.)
TRY 2: Take 3 normal coins, and weigh them against three unknowns. If the balance is off (50% chance), then we know that these three unknowns are candidates, and also whether the answer is lighter of heavier. Goto 3a. If they're measured equal, then the other three unknowns are the only candidates. Goto 3b.
TRY 3a: Weigh two of the candidates against each other. If they're equal, it's the other one. If they're not equal, then we know which one of those on the balance it is. (3 steps, but 50% of not getting this route)
TRY 3b+4: We do TRY2 on this group to gauge heavier/lighter, followed by 3a. (4 tries, 100%)
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