acquire the information I through a minimum number of trials or weighting measurements. A painstaking approach but not the smartest, as we shall see, consists in comparing the weights of all coins with respect to a same reference coin, which the jeweler would initially pick up at random. If he happens to pick up for reference the fake coin, he gets the answer right away with the first measurement. The probability for this lucky conclusion is 1/27, or 3.7%. In the worst case, where all of his successive measurements balance out, it would take 25 operations to conclude the test. 10 Yet, we may observe that this is not a bad method if the scale is too delicate and cannot take more than one coin in each pan. Let us assume that the scale is robust and sufficiently accurate to weigh groups of ten coins or SO. Consider then the outcome of any scale measurement with groups of several coins with equal numbers, as selected at random. There are three possible, and equiprobable outcomes: (a) Left pan heavier; (b) Right pan heavier; (c) Pans balanced. The outcome is a ternary answer (YES/NO/NEUTRAL) with equiprobable outcomes. Each of these single measurements (a), (b), (c), thus, yields an information of I'm = -log2(1/3) = log2 3 bit, and n such independent measurements yield the information nIm = n log2 3 = nI/3. This result shows that the minimum number of measurements required to sort the fake coin from the group is n = 3, since 31m = I. This is a second hunch towards the problem solution. However, we don't know anything yet about how to proceed, in order to get the answer through this "theoretical" minimum of three operations. The solution provided by Rényi is illustrated in Fig. 3.1. It consists in making succession of selective measurements, first with two groups of nine coins, then two groups of three coins, then with two groups of one coin, each one identifying where the fake coin is located. This solution proves that three measurements indeed are sufficient to get the answer! Indeed, if the first 25 measurements wit