I'm working on a problem that reduces to the following:
I think this is a geometric probability, and to calculate answer we need to divide a total area of all circles containing all K points by area of square, which is obviously 1.
Let K = 1, then it's quite easy to calculate P, we take this point and kinda rotate circle around it, so it would draw a circle with radius 2R, which includes all circles of radius R containing this point; cut the parts lying outside the square, and calculate area of what left.
If K = 2, then first we check whether distance between points is not greater than 2R, so that circle containing both points exists. But I don't really understand how to calculate total area since final figure is going to be 4-leaf flower if that makes sense. And so on for greater K's...
I sense that might be a straightforward solution and wonder whether there is more elegant one.
Given the K points, a disk of radius R covers them if it covers their convex hull. If you try all positions of the disk, the most extreme will ones be such that the circumference goes through one or two hull vertices. If we consider all positions through a given vertex, we can rotate the disk inside an angle between the contacts with the two neighboring vertices on the hull. During this rotation, the center of the circle describes a circular arc of radius R.
Hence the locus of the center of the disk is a convex curvilinear polygon made of circular arcs of radius R. To get the desired probability, you intersect this polygon with the unit square and compute the common area.
On the figure, the convex hull of the K=6 points is in green. The red circles are the extreme positions with two contacts. The locus of the center of the disk is delimited in blue.
Once you have the convex hull, building the curvilinear polygon is not so difficult, and to get its area, you decompose in a standard polygon and a set of circular segments.
But clipping inside the unit square will make it a pain in the neck.