Random quads visible from a random point Although the MO question Limit of lights in rooms was quickly closed,
it suggests a related question:

Q0. What is the probability that a random quadrilateral $Q$
  is entirely illuminated from a random point $p \in Q$? 

What is a "random quadrilateral"?
What constitutes a random simple polygon is not easily answered.
See Jeff Erickson's webpage on this topic.
But perhaps we can take this
as the following definition of a random quadrilateral:
Let $p_i$ be four points uniformaly randomly distributed in a unit-radius disk. Let $Q$ be the quadrilateral formed by connecting the points in order: $(p_1,p_2,p_3,p_4,p_1)$.
A certain fraction of these will have crossing segments, and the others will be simple polygons. My simulations suggest that about half are simple, half have segment crossings.

Q1. What percentage of quadrilaterals (as described above) are simple (non-self-crossing)?
  Near $\tfrac{1}{2}$? Exactly $\tfrac{1}{2}$?

The answer to Q1 is likely known, but I could neither
find it nor derive it.
Among the random simple quadrilaterals $Q$, 

Q2. For what proportion does a random internal
  point $x$ insides $Q$ illuminates all of $Q$?


          


          

Any point in the kernel (yellow) will illuminate the quad.


I expect the answer to Q2 is well above a half, nearer to (but less than) $1$.
 A: Q1: assume that degenerate quadruplets do not exist (they have probability 0 of occuring). Observe that for any non-degenerate quadruple of points $P = (p_1, p_2, p_3, p_4)$ we have exactly one of three options:


*

*$P$ is a quadrilateral;

*$p_1 p_2$ intersects $p_3 p_4$;

*$p_1 p_4$ intersects $p_2 p_3$.


Let $B$ be the event of segments $p_1 p_2$ and $p_3 p_4$ crossing for random points $p_1, p_2, p_3, p_4$, then the probability of $A = \{P$ that form a quadrilateral$\}$ is $1 - 2 \Pr(B)$.
To find $\Pr(B)$, assign the unique intersection point $p$ to each $(p_{11}, p_{12}, p_{21}, p_{22}) \in B$. Introduce an injective change of coordinates among quadruplets $P$ for which $p$ is defined: 


*

*$(r, \alpha)$ = polar coordinates of $p$;

*$(l_{ij}, \alpha_{ij})$ = polar coordinates of $p_{ij}$ with respect to center $p$ and reference direction $\alpha$.


Since $p_{11} p_{12}, p_{21} p_{22} \ni p$, we must have $\alpha_i := \alpha_{i1} = \alpha_{i2} + \pi$ for $i = 1, 2$. $\Pr(B)$ can be found by integrating Jacobian $J = (l_{11} + l_{12})(l_{21} + l_{22})r \sin(\alpha_1 - \alpha_2)$ (up to a sign) over the region where all points lie within the circle (we only integrate for $0 \leq \alpha_1 \leq \alpha_2 \leq \pi$, but multiply by $8$ to account for different orders):
$\Pr(B) = 8\int_0^1 dr \int_0^{2\pi} d \alpha \int_0^{\pi} d \alpha_1 \int_{\alpha_1}^{\pi} d \alpha_2 \int_0^{\sqrt{1 - r^2 \sin^2 \alpha_1} - r\cos \alpha_1} dl_{11} \int_0^{\sqrt{1 - r^2 \sin^2 \alpha_1} + r\cos \alpha_1} dl_{12} \int_0^{\sqrt{1 - r^2 \sin^2 \alpha_2} - r\cos \alpha_2} dl_{21} \int_0^{\sqrt{1 - r^2 \sin^2 \alpha_2} + r\cos \alpha_2} dl_{22} \cdot (l_{11} + l_{12})(l_{21} + l_{22})r \sin(\alpha_2 - \alpha_1).$
Integrating $J$ within required bounds seems hard (probably no closed answer...), but numerical integration yields $\Pr(B) \approx 0.23482663$, and $\Pr(A) \approx 0.53034674$, which agrees well with my Monte Carlo computations.
Q2: let $C$ be the event "an unordered set of four random points is in convex position". Considering all $4!$ orderings in each case we can conclude that $\Pr(A) = \frac{1}{3}\Pr(C) + \Pr(\overline{C})$, which implies $\Pr(C) = \frac{3}{2}(1 - \Pr(A)) \approx 0.70448$. A convex quardilateral is lit by a random interior point with probability 1.
Consider a non-convex quadrilateral with convex hull $H = p_1 p_2 p_3$. Conditioned on vertices of $H$, the fourth point $p_4$ is uniformly distributed within the triangle $H$, and the side of $H$ to which $p_4$ is connected is equidistributed. Notice that affine transformations don't change uniformity of distribution nor that a particular interior point lights the quadrilateral. It follows that conditional probability "a random interior point lights a random non-convex quardilateral with convex hull $H$" does not depend on $H$. Let us find it for a standard triangle with vertices $(0, 0), (1, 0), (0, 1)$: $$\frac{\int_0^1 dx \int_0^{1 - x} dy \frac{1 - y(\frac{x}{1 - x} + \frac{1}{x + y})}{1 - y}}{1/2} = 7 - \frac{2\pi^2}{3} \approx 0.420264$$
Hence $$\Pr(\text{a random quadrilateral is lit by a random interior point}) \approx \frac{\frac{1}{3}\Pr(C) + 0.420264\Pr(\overline{C})}{\Pr{A}} \approx 0.676959,$$ which, again, seems to meet the Monte Carlo values pretty well.
