foundations in probability

Let's let X = the position of the first cut. The distribution of X is uniform(0,1).

Let Y = the position of the second cut. The distribution of Y given x is uniform(0,x). Therefore the conditional density of Y given x is 1/x, for y between 0 and x. 

$E(Y) = \int_{0}^{1} \int_{0}^{x} \frac{y}{x} dy dx $

$= \int_{0}^{1} \frac{x}{2} dx$

$= \frac{1}{4}$