Additivity of Shannon's entropy E.T. Jaynes writes, in "Probability Theory: the Logic of Science" the following in order to motivate the derivation of the entropy, $H$:

Suppose the robot perceives two alternatives, to which it assigns
  probabilities $p_1$ and $q := 1 - p_1$. Then the 'amount of
  uncertainty' represented by this distribution is $H_2(p, q)$. But now
  the robot learns that the second alternative really consists of two
  possibilities, and it assigns probabilities $p_2, p_3$ to them,
  satisfying $p_2 + p_3 = q$. What is now the robot's full uncertainty
  $H_3(p_1, p_2, p_3)$ as to all three possibilities? Well, the process
  of choosing one of the three can be broken down into two steps.
  Firstly, decide whether the first possibility is or is not true; the
  uncertainty removed by this decision is the original $H_2(p_1, q)$.
  Then, with probability $q$, the robot encounters an additional
  uncertainty as to events $2, 3$, leading to
$$H_3(p_1, p_2, p_3) = H_2(p, q) + q H_2\left(\frac{p_2}{q}, \frac{p_3}{q} \right)$$
as the condition that we shall obtain the same net uncertainty for
  either method of calculation.

I find this reasoning not compelling enough, especially the way $q$ appears  before $H_2$. Perhaps there's a better way to put what Jaynes' meant to say through an explicit example of a coin flip, where $T$ outcome turns out to be governed by another coin flip?
 A: Instead of Shannon entropy, let's talk about scoring rules. You assess a probability over a set of mutually exclusive and exhaustive outcomes, and then you win points based on the probability you assigned to the outcome that's correct.
Scoring rules, if they're any good, have a neat property called being "proper." That is, if I have some probability distribution $(p_1, 1-p_1)$ on two outcomes $1$ and $2$, under a proper scoring rule the best I can do is report my probability distribution as $(x_1,x_2)=(p_1, 1-p_1)$.
This isn't always the case! Suppose my scoring rule is linear: if $i$ happens, you get $x_i$ points. This is what would happen with, say, normal betting. Well, then if you think it's 60% likely that 1 will happen and 40% likely that 2 will happen, you win the most points by reporting $(1.0,0)$, not $(0.6,0.4)$. Calculate it: $1*0.6+0*0.4=0.6$, vs. $0.6*0.6+0.4*0.4=0.52$. If a roulette wheel lands on red 60% of the time, bet on red every time, not 60% of the time.
So there are many proper scoring rules. If we use the spherical score, then we get $x_i/\sqrt{\sum_jx_j^2}$ points, which we can easily prove is proper using geometric means.
All of the scoring rules work for sets with more than 2 alternatives. But are they consistent across multiple perspectives? That is, suppose I flip a coin, and if it's heads, I flip again. There are three possible sequences: T, HH, and HT. If the coin is fair, then the best probability distribution is $(0.5,0.25,0.25)$, and the expected spherical score is $\sqrt{3/8}$.
But what if we decompose this into two different flips, each of which is $(0.5,0.5)$, and only have the second flip half the time? Then the expected spherical score is $\sqrt{1/2}$ for the first flip, and $\sqrt{1/2}$ for the second flip, and so the total expected score is $3\sqrt{1/8}$, which is... different by a factor of $\sqrt{3}$.
So the spherical scoring rule is proper, in that it gets us to give the right probabilities. But it isn't additive, in that I can decompose a set of probabilities however I want and the score necessarily must work out to be the same. As it turns out, the only rule that has both properties is the log scoring rule, which is the same as measuring the entropy. 
