Problem Statement

A stick of length L drops and breaks at a random point distributed uniformly across the length.
What is the expected length of the smaller part?

Original Problem Link: Click here

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Solution

As given in the question, the point where the stick will break is uniformly random.

Now, lets consider, what is the probability, that a particular element (very small) point will be chosen (which is at a distance \(x\) from one end)?

Very small, right? Yes!

We must quantify this using calculus.

We can say that the probability will be:

\[\frac{dx}{L}\]

where \(L\) is the length of the rod.

Now, if this is the case, then the rod will break into 2 parts, one with length \(x\), other with \(L-x\).

Using the fundamental definition of Expectation, we can say that

\[E[l] = \sum (l.P(l)), \quad l \text{ varies from } 0 \text{ to } \frac{L}{2}\]

(as we are looking at the smaller edge)

Here, \(P(l)\) is:

\[\frac{dl}{L}, \quad \text{right? No!}\]

\(P(l)\) means what’s the probability that the smaller part will have a length \(l\).

Note that there are 2 points where a rod can break, for the smaller length to be a particular \(l\)
(One is at \(l\) distance from one end, the other is at \(l\) distance from the other end).

Thus,

\[P(l) = \frac{2 \, dl}{L}\]

Now, the sigma becomes an integration as we have elemental quantities involved.

Evaluating the integral, we get

\[E[l] = \int_0^{L/2} l \cdot \frac{2 \, dl}{L}\] \[= \frac{2}{L} \int_0^{L/2} l \, dl\] \[= \frac{2}{L} \times \left[ \frac{l^2}{2} \right]_0^{L/2}\] \[= \frac{2}{L} \times \frac{(L/2)^2}{2}\] \[= \frac{2}{L} \times \frac{L^2}{8}\] \[= \frac{L}{4}\]

Final Answer

Thus, the expected length of the smaller part is:

\[\boxed{\frac{L}{4}}\]

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