Colored arrangements of circles on the two sphere Let me define   a degree $n$ colored  arrangement of circles on $S^2$ to be a   collection $\mathcal{C}$  of $n$ disjoint, smoothly embedded circles $C_1,\dotsc, C_n\subset S^2$ together with a continuous  map (temperature)
$$  T: S^2 \setminus (C_1\cup\cdots \cup C_n)\to \lbrace 1,-1\rbrace $$ 
such that for any  circle $C_k$,  the temperatures  on opposite sides of $C_k$  are different.  In  other words, if $\gamma:(-\varepsilon,\varepsilon)\to S^2$, $t\mapsto \gamma(t)$,  is a  short smooth path  that intersects  exactly one circle  $C_k$, and it does so   transversally at $t=0$, then 
$$\lim_{t\searrow 0} \; T\bigl(\; \gamma(t)\;\bigr)= - \lim_{t\nearrow 0}\;    T\bigl(\;\gamma(t)\;\bigr). $$
Think of the components of $T^{-1}(-1)$ as icy regions and of the components of $T^{-1}(1)$ as liquid regions.  We will refer to $T$ as a coloring of the arrangement  of circles $\mathcal{C}$.
Two   arrangements of  circles $\mathcal{C}$ and $\mathcal{C}'$ are called equivalent  if there exists an orientation preserving  diffeomorphism of $S^2$ that maps one arrangement to the other. We denote by $\mathcal{A}_n$ the set  of equivalence classes of  arrangements on $n$ disjoint circles on $S^2$.
Two   degree $n$ colored arrangements  $(\mathcal{C}, T)$ and $(\mathcal{C}', T')$ are called  equivalent if  there exists an orientation preserving diffeomorphism    $\Psi: S^2\to S^2$  mapping the circles in $\mathcal{C}$ to the circles in $\mathcal{C}'$ and such that $T'\circ \Psi= T$.
Two colorings of a given arrangement $\mathcal{C}$ of $n$ disjoint circles on $S^2$  are called equivalent   if the colored arrangements  $(\mathcal{C}, T)$ and $(\mathcal{C}, T')$  are equivalent.   We get in this fashion a map
$$ K_n : \mathcal{A}_n\to\mathbb{Z}$$
that associates to an arrangement $\mathcal{C}$ the nummber of inequivalent colorings of $\mathcal{C}$.  Here are some questions I find interesting.
$\mathbf{Q_0}$ Find the cardinality of $\mathcal{A}_n$ ans its large $n$ asymptotics.
$\mathbf{Q_1}$    Find the cardinality of $\mathcal{S}_n$ and its large $n$ asymptotics. 
$\mathbf{Q_1^*}$ Investigate the map $K_n$.
$\mathbf{Q_2}$  We declare an colored arrangement $(\mathcal{C}, T)$ to be  selfdual if  it is equivalent  to $(\mathcal{C}, -T)$. Find the  number of selfdual colored arrangements of degree $n$ and  its  large $n$ asymptotics.
$\mathbf{Q_3}$  Equip $\mathcal{S}_n$ with the uniform probability density and denote by $X_n:\mathcal{S}_n\to \mathbb{R}$ the random variable o that associates to a colored arrangement $(\mathcal{C}, T)$ the  Euler characteristic of  the liquid region $T^{-1}(1)$.  Denote by $\mu_n$ the probability  distribution of $X_n$. Investigate the large $n$ behavior of $\mu_n$.
$\mathbf{Q_4}$ Can you think of a good notion of a random arrangement of $n$ disjoint circles on $S^2$?
 A: Consider the graph obtained by assigning to each region a vertex, and two vertices being joined by an edge iff the two regions are neighbors. It is easy to see that this is a tree, with the empty circles as terminal vertices: once you crossed a circle, the only way to return to the same region is by the same edge.
The tree isomorphism corresponds to the equivalence of arrangements of circles.
So the cardinality of $\mathcal A_n$ is the same as for the unlabeled trees. In the same wiki reference is mentioned he asymptotic estimate given by Otter (1948). For the cardinality of $\mathcal A_n$ is no known formula.
For the $\mathcal S_n$, there are only two ways to color a tree so that each edge connects two vertices of opposite colors. They may be not distinct (i.e. may be selfdual), since it is possible to have a tree automorphism which switches the colors. The asymptotics is similar to the uncolored case, because there are only one or two distinct colorings of the same tree.
The notion of randomness for trees depends on the purpose.
The answers are partial, mainly because the answers to the corresponding questions for the trees are partial known, but I hope that the connection with trees may help.
