ories, namely, winning (x = 4, 2, 1) and losing (x = other), which led to the entropy H = 0.183 bit/symbol. Consider now all dice-roll possibilities, which are equiprobable. The total number of possibilities (regardless of degeneracy) is N = 6x6x6 : 216, each of which is associated with a probability p = 1/216. The corresponding informa- tion is, by definition, I(216) = log 216 = 7.7548 bits. We now make a partition between the winning and the losing events. This gives n(421) = 3! = 6, and ((other) = 216 - 6 = 210. According to Shannon's rule (3), and following Eq. (B17) of Appendix B, we have: I(N) = H (p1, p2) + p1I(n1) + p2I(n2) = H (421, other) + p(421)I[n(421)] + p(other)I[n(other)] = 0.1831 (1/36)I(6) + (35/36)I(210) = 0.1831 + 0.0718 + 7.4999 = 7.7548 bits, (4.10) which is the expected result. One could argue that there was no real point in making the above verification, since property (3) is ingrained in the entropy definition. This observation is correct: the exercise was simply meant to illustrate that the property works through a practical Entropy example. Yet, as we shall see, we can take advantage of property (3) to make the rules more complex and exciting, when introducing further "winning" subgroups of interest. For instance, we could keep "4, 2, 1" as the top winner of the 421 game, but attribute 10 bonus points for any different dice-roll result mnp in which m n + p = 7. Such a rule modification creates a new partition within the subgroup we initially called "other." Let's then decompose "other" into "bonus" and "null," which gives the source information decomposition: I(N) = H (421, bonus, null) + p(421)I[n(421)] (4.11) p(bonus)I[n(bonus)] + p(null)I[n(null)]. The reader can easily establish that there exists only n(bonus) = 9 possibilities for the "bonus" subgroup (which, to recall, excludes any of the "421" cases), thus p(bonus) = 9/216, and p(null) = 1 - p(421) - p(bonus) = 1 - 1/36-9/216 - = 0.9305 (this alter- native rule of the 421 game gives about 7% chances of winning something, which mo