Random Break of a Stick: Calculating the Average Length of the Smaller Piece

The problem of breaking a stick of length L into two pieces at a random point and finding the average length of the smaller piece is a classic example in probability theory. In this article, we'll delve into the mathematical steps to solve this problem, providing a clear understanding of the underlying concepts and formulas.

Introduction to the Problem

Consider a stick of length L. This stick is broken at a random point X, where X is uniformly distributed over the interval [0, L]. Our goal is to find the average length of the smaller piece resulting from this random break.

Break Point Selection and Piece Lengths

We start by denoting the position of the break point as X. Since X is uniformly distributed, it can be anywhere along the length of the stick. Consequently, we have two pieces:

Piece 1: of length X Piece 2: of length L-X

We are interested in the length of the smaller piece, which can be expressed as:

smaller piece min(X, L-X)

Calculation of the Expected Value

To find the average length of the smaller piece, we need to calculate the expected value, E[min(X, L-X)]. This involves integrating over the possible values of X in the interval [0, L].

Two Parts of the Integral

The smaller piece will be X when X is less than or equal to L/2 and L-X when X is greater than L/2. We can break the calculation into two parts:

Part 1: When X ≤ L/2

In this region, the smaller piece is X. The expected value of X is:

[E[min(X, L-X)] E[X] int_0^{L/2} x cdot frac{2}{L} , dx]

Evaluating the integral:

[E[X] frac{2}{L} int_0^{L/2} x , dx frac{2}{L} left[ frac{x^2}{2} right]_0^{L/2} frac{2}{L} cdot frac{(L/2)^2}{2} frac{2}{L} cdot frac{L^2}{8} frac{L}{4}]

Part 2: When X > L/2

In this region, the smaller piece is L-X. The expected value of L-X is:

[E[min(X, L-X)] E[L-X] int_{L/2}^L (L-x) cdot frac{2}{L} , dx]

Evaluating the integral:

[E[L-X] frac{2}{L} int_{L/2}^L (L-x) , dx frac{2}{L} left[ Lx - frac{x^2}{2} right]_{L/2}^L]

Evaluating the limits:

[ frac{2}{L} left[ L^2 - frac{L^2}{2} - left( frac{L^2}{2} - frac{(L/2)^2}{2} right) right] frac{2}{L} left[ L^2 - frac{L^2}{2} - frac{L^2}{2} frac{L^2}{8} right]]

Simplifying the expression:

[ frac{2}{L} left[ L^2 - L^2 frac{L^2}{8} right] frac{2}{L} cdot frac{L^2}{8} frac{L}{4}]

Combining the Results

Both regions contribute equally to the overall expected value. Therefore, the average length of the smaller piece is:

[E[min(X, L-X)] frac{L}{4}]

Thus, the average length of the smaller piece is boxed{L/4}.