Probability that centre of the square lies inside the circle joining the two points inside the square Two points are uniformly and independently distributed (located) inside a square. A circle is drawn such that the segment joining the two points is a diameter. Find the probability that the center of the square lies inside that circle.
 A: Call the first point A, the second point B, and the center O. Join the line AO, and extend a line DOE through O, perpendicular to AO and continuing in both directions. If B is on the other side of DOE from A, then the circle joining A and B will contain the center. If B is on the same side of the line as A, then it will not. The line bisects the square's area, so the probability is 1/2. 
A: Although the answer has already been given, it's possible to check this via simulation.  In R:
n = 10^8
x1 = runif(n, 0, 1)
x2 = runif(n, 0, 1)
y1 = runif(n, 0, 1)
y2 = runif(n, 0, 1)
x.center = (x1+x2)/2
y.center = (y1+y2)/2
radius = sqrt((x.center-x1)^2+(y.center-y1)^2)
distance.to.center = sqrt((x.center-1/2)^2+(y.center-1/2)^2)
sum(radius > distance.to.center)

This simulates drawing $10^8$ pairs of points (x1, y1), (x2, y2), and finds the center and radius of the corresponding circles, and checks to see whether those circles contain the center of the square (1/2, 1/2).  When I ran this I got 49995062, meaning that the probability can be estimated as 0.49995062.  I'd say that's close to 1/2.
A: Another way to approach this problem would be to consider the angle made by the two points A and B at O (centre of the square).
In the case that O lies on the boundary of the circle with AB as diameter, the ∠AOB = 90 degrees.
If O lies inside the circle then the ∠AOB is greater than 90 degrees and if point O lies outside the circle then the ∠AOB is less than 90 degrees.
Since A and B are iid we can see the probability of ∠AOB being greater than 90 is 1/2
