# The Little End of the Stick

## Problem

(a) If a stick is broken in \(2\) at random, what is the average length of the smaller piece?

(b) What is the average ratio of the smaller length to the larger?

## Part (a) Solution

Clearly, the smaller piece will be no more than \(\frac{1}{2}\) the length of the stick. Let \(S\) be the length of the smaller piece, which follows a continuous uniform distribution over \([0, \frac{1}{2}]\). Thus, the expected length is:

\[E[S] = \frac{0 + \frac{1}{2}}{2} = \frac{1}{4}\]

## Part (b) Solution

To calculate the average ratio, we apply the definition of the expected value by integrating over the possible lengths of the smaller piece:

\[\int_0^{\frac{1}{2}} \left(\frac{x}{1-x}\right) \frac{dx}{1/2}
= \ln{4} - 1 \approx 0.3863\]

The above integral is leasily solved with the substitution \(u=1-x\), giving an average ratio of about \(0.3863\).