Let $\cal{A}_n$ be a simple arrangement of $n$ lines in $\mathbb{R}^2$.
(*Simple*: each pair of lines meet in a distinct point, i.e.,
no three lines pass through the same point.)
Start a random walk at an arrangement vertex $v_0$.
Choose to walk along one of the lines $L_0$ passing through $v_0$,
and turn left or right with equal probability at the next vertex $v_1$.
Then walk along the line $L_1$, where $v_1 = L_1 \cap L_0$. Turn at
$v_2 = L_1 \cap L_2$, and so on.
(Never proceed straight through; never back-up.)

^{ An arrangement of $25$ lines. Random walk starting at $v_0$. }

. As a function of $n$, which arrangements $\cal{A}_n$ of $n$ lines have the highest probability that the walk will revisit some $v_0$?Q1

If the four cells incident to $v_0$ are two triangles and two quadrilaterals, then there should be at least a $$ \tfrac{1}{16} + \tfrac{1}{16} + \tfrac{1}{32} + \tfrac{1}{32} = \tfrac{3}{16} $$ chance of returning to $v_0$. For example, for one triangle, choose $L_0$ and a direction along $L_0$—$\frac{1}{4}$, then turn left twice—$\frac{1}{2} \frac{1}{2}$; so return to $v_0$ by traversing that triangle with probability $\frac{1}{16}$.

Eventually, any walk will escape to $\infty$.

. As a function of $n$, which arrangements $\cal{A}_n$ of $n$ lines lead to the quickest escapes to $\infty$, in the sense that every starting vertex $v_0$ leads to a quick escape?Q2

In other words, let $E(v)$ be the expected number of steps for a walk starting at $v$ to escape to $\infty$. Then I seek the arrangements $\cal{A}_n$ that minimize the maximum of $E(v)$ over all $v \in \cal{A}_n$. So no $v$ is "deep" inside the arrangement.

A possible extremal configuration: Each line is tangent to a central circle.

Related MO question: "A random walk on random lines."

twoquads (not four). There is an example in the figure, at about 11 o'clock. $\endgroup$ – Joseph O'Rourke Oct 9 '18 at 15:48