Given a probability vector $\mathbf p\equiv (p_k)_{k=1}^n$ with $p_k\ge0$ and $\sum_k p_k=1$, Shannon's entropy is defined as $S(\mathbf p)\equiv -\sum_k p_k \log p_k$.
On the other hand, given two probability vectors $\mathbf p$ and $\mathbf q$, we say that $\mathbf q$ is majorized by $\mathbf p$, and write $\mathbf q\preceq\mathbf p$, if the sum of the $j$ largest elements of $\mathbf q$ is smaller than the sum of the $j$ largest elements of $\mathbf p$, for all $j$: $$\sum_{k=1}^j q_j^\downarrow \le \sum_{k=1}^j p_j^\downarrow, \quad \forall j=1,...,n.$$
Both of these quantities can be understood as quantifying how "disordered" a probability vector is, albeit obviously in different ways: the entropy assigns a number of each probability distribution, while majorization is a preorder in the set of probability distributions. Intuitively, $\mathbf p$ is "more disordered" than $\mathbf q$ when $S(\mathbf p)\ge S(\mathbf q)$ or $\mathbf p\preceq \mathbf q$.
It is known that $S$ is Schur-concave, i.e. that $\mathbf p\preceq\mathbf q$ implies $S(\mathbf p)\ge S(\mathbf q)$.
We also know that there are examples of $\mathbf p,\mathbf q$ which are not comparable via majorization but are such that $S(\mathbf p)\le S(\mathbf q)$. For example, $\mathbf p=(0.5,0.5,0)$ and $\mathbf q=(0.6,0.3,0.1)$.
Is there any intuition as to the different types of "disorder" quantified by entropy and majorization? Is either "superior" than the other for some applications?
Moreover, does $S(\mathbf p)\ge S(\mathbf q)$ imply $\mathbf p\preceq \mathbf q$ if we assume that $\mathbf q$ and $\mathbf p$ are comparable?
In my comment above I got the logic wrong, indeed at least it is true that $S(p) > S(q)$ and $p,q$ comparable imply $p \prec q$.
To see this, we use that $$p \prec q \iff \sum_i h(p_i) \leq \sum_i h(q_i)$$
for all convex $ h: \mathbb{R}_+ \to \mathbb{R}$, since $p_i,q_i \geq 0 \ \forall i$.
Now for a fixed $p$ this means that the set of probability vectors that are comparable to it can be described as
$$C_p =\{ q \in \Delta_n| q \prec p \lor p \prec q\}\\ = \left\{ q \in \Delta_n| \left(\sum_i h(p_i) \geq \sum_i h(q_i) \, \forall \ h \ \text{convex}\right)\right\} \cup \left\{q \in \Delta_n| \left(\sum_i h(p_i) \leq \sum_i h(q_i) \, \forall \ h \ \text{convex}\right)\right\} .$$
Since we know already one convex function ($ x \mapsto - x\ln x$) for which the requirement of the first set does not hold, $q$ has to be in the second set and thus $p \prec q$.