Chopping a 1m stick (statistics) [closed]

Question

What's the average of the longer part if I chop a 1m stick? I don't know how to solve this tricky question.

Answer 1

Assuming uniform splitting, the expected longer part can be described as

\begin{align*} \mathbb{E}[\max(X,1-X)]&=\int_0^1 max(x,1-x) dx\\ &=\int_0^{0.5} (1-x) dx + \int_{0.5}^{1} x dx \\ &= 0.5-\frac{0.5^2}{2}+\frac{1}{2}-\frac{0.5^2}{2}\\ &=\frac34 \end{align*}