is: 1 = NO/NO/NO, 2 = YES/NO/NO, The choice of code being arbitrary; we can use the binary representation, 1 = 001, 2 = 010, 3 = 011, 4 = 100, 5 = 101, 6 = 110, noting that the blocks 000 and 111 are unused. 3.4 Rényi's fake coin 3 = NO/NO/YES, 4 = YES/YES/NO, 5 = NO/YES/NO, 6 = YES/YES/YES, while the two messages YES/NO/YES and NO/YES/YES are unused (in ensemble lan- guage, U1 V2 n W = U2 V1 n W1 = 0, meaning that these ensembles are empty). Regardless of the code, we see that the symbol blocks require three symbols of binary value bits (YES/NO or 1/0), while the information carried by the message is only 2.584 bits. Why do we have an extra 0.416 bit of information in the message? The explanation for this mystery is in fact quite simple. Assume a source that has eight equiprobable symbols, i.e., Y = {1, 2, 3, 4, 5, 6, 7, 8}, which can also be written Y = X U {7, 8}. The information of Y is I(Y) = -log2(1/8) = log2 23 = 3 log2 2 = 3 bits. Since our previous messages describe the outcome of a die roll, the symbols (7, 8), which have a probability of zero, are never used. The consequence of never using these two symbols is that the actual probabilities of the six other symbols are raised by a factor of 8/6. This increase in likelihood corresponds to a relative information decrease (or less surprise!) of AI = -log(8/6) = -0.416 bit. Thus, the net information in the message is Inet = I(Y) + Al = 3 - 0.416 = 2.584 bits. One may justly argue that we get noninteger information because of the arbitrary choice of a base-2 logarithm. For the die roll, we could equally well choose a base- 6 logarithm, i.e., I (X) = -log6(x) = log6(6) = 1 sit (for "six-ary digit"). Then the information of the die roll can be given by a single sit symbol, instead of three bits. However, the sit is six-valued, which requires six different symbols per elementary information, as opposed to the two-valued bit, which only requires two symbols. Which of the two message blocks is better: one made of a single six-valued symbol, or one made of three t