Two points are chosen, independently and at random, on a stick of unit length. What is the expected distance between them?
You may want to try to solve the problem, before you look at the solutions below.
Denote the positions of the two points by a and b, where a<b. The distance between them is b-a.
The sum, over all possible positions of the points, of the distance is
This integral was evaluated over the range 0<a<b<1. To find the mean value of b-a, we must divide by the extent of this range, given by
Thus the answer is the quotient ⅙ / ½ = ⅓.
We plot the 2-space in which the two points can be, as shown to the right. We are constrained by 0<a, a<b, b<1, giving the triangle shown in blue.
The value of the distance we are interested in is shown by the green lines. We want its mean, over the blue triangle. This will be the value of the distance at the centroid of the triangle, as marked by a black dot. The centroid of a triangle lies ⅓ of the way along any of its medians, so the answer is ⅓.
Consider instead the problem
Points p, q, r are chosen, independently and at random, on a ring of unit circumference. The ring is cut at p, straightened into a stick, and then cut at q and r. What is the expected length of piece qr?
We have cut a ring at three independently chosen random points, to form three pieces. The total length is 1, so the expected length of each piece must be ⅓. This problem is equivalent to the one originally given.
If you've understood how to use symmetry to solve the stick problem, you should be able to solve this one, without using a computer, a diagram, or written calculation:
I shuffle a pack of cards (52 cards including four aces), and draw cards from it, without replacement, until I find an ace. What is the expected number drawn?
This is one of several pages on using symmetry in mathematics.