Pick two points at random on a circle and draw the chord connecting them.
Pick two more points and connect them with a second chord.
What is the probability that these chords intersect?
If the first chord is a diameter, the chances that the chords cross is 50%. If the first chord is less than a diameter, the chances that the chords cross is less than 50%. So, averaging all possible chord lengths, the probability must be less than 50%.
/*********************************************/
I suggest solving by transforming the problem out. Pick a circle with unit circumference, break it at one of the two points, and straighten it into a unit line. Now the problem becomes:
Pick one point on a unit length.
Pick two more points.
What is the probability that they lie on opposite sides
of the first point?
Well, if the first point is at x,
then the probability is 2x*(1-x)
And the puzzle solution
is the integral from 0 to 1 of (2x*(1-x))dx
Unfortunately, my calculus is a little rusty. I'll do the integration later, unless somebody does it first.
Edited on May 11, 2009, 8:34 pm
That can't be right (Hint, Spoiler)
