(Apologies for the lack of latex) I was thinking today about simple ways of proving that the area of a circle is pi*r^2, and I came up with the following argument using probability. I googled around for similar arguments, and I found nothing. I am curious if there is a deeper and more general result this could be a special case of, or if there if anyone else knows of something relating to this. We know that the area of a square of side length r is r^2. Consider the unit circle centered at the origin, and a circle of radius R centered at the origin. We randomly select two numbers N and M from [-1,1], with uniform distribution. The probability that N^2+M^2<=1 is thus the ratio of the area of the unit circle to the area of a square of side length 2. Now we randomly select two numbers N' and M' from [-R,R]. The probability that N'^2+M'^2<=R^2 is the ratio of the area of the circle of radius R to the area of a square of side length 2R. However, our random selection from [-R,R] is the same as a random selection from [-1,1] and then multiplying by R. Thus, we can calculate that N'^2+M'^2<=r^2 has the same probability as N^2+M^2<=1. Thus, the area of a circle of radius R is proportional to the area of a square of side length 2R. It follows that the area of a circle of radius R has area pi*R^2, for some constant pi.