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by wvlia5 2914 days ago
"that means you're not really getting any information if the beam balances"

Wrong. If it was true, that would imply that the probability of it balancing is 1, since the information you get is -log2(probability). Obviously the probability of it balancing is lower than 1, so you do get information.

1 comments

You don't get any information if the probability of the beam balancing is either 1 or 0. If you put everyone on the scale, the probability that it will be balanced is 0 - you don't learn anything. So the trick is to exclude some of the sample group in every measurement. You can use a binary search strategy where you divide the coins into "on the scale" and "not on the scale" groups, until you have a set of two that definitely contains the one you're searching for (and then one more measurement will tell you which one of those two it is) - but that strategy requires too many measurements, so you need to use more information at each stage, and you can't throw out coins you've determined to be non-counterfeit.