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.
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.