Re: math problem
I'm not sure that there's really a general principle at work here. One point is that after the initial bracket, one already knows the potential second and third place coins. Finding the second place coin (from 4) will take 3 trials, whether one does it bracket style or not; but with so few contenders, only one coin is eliminated from being the 3rd place one if the 2nd place guy is determined with a bracket comparison. It turns out that it's more efficient to spend those comparisons to eliminate other possible third place guys. One way to maybe understand this is to redo the problem but with 17 coins, or 32 coins.
By the way, I don't think that it is possible to 'intuitively see this as being optimal' -- I don't really know if there are good ways of proving that solutions to these sorts of problems are optimal, and would be curious to hear from others if there are.
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