My approach is maybe more naive than the others posted.
Break the unit interval at $x$ and $y$ where $x < y$. Our lengths are then $x$, $y - x$, and $1 - y$. It's not hard to show that they all have probability $1/3$ of being the shortest. In any case, our joint PDF is given by $f(x,y) = 6$ (since $x$ and $y$ remain uniform random variables on $1/6$th of the square $[0,1] \times [0,1]$). Each triangle in the diagram below corresponds to the domain of the PDF for one of the three cases.
I'll take care of the case when $x$ is shortest, that is, $x \leq y - x$ and $x \leq 1 - y$. This is the leftmost triangle. Since we're assuming $x$ is least, we are looking for
$$E[x] = \int_0^{1/3} \int_{2x}^{1 - x} 6x \;dy \;dx = 1/9$$
The cases when $y - x$ and $1 - y$ are shortest are similar.
$U \sim U(0,1)$
Define $X = \text{max}\{U, 1-U\}$
so that
$$X =
\begin{cases}
1-U, & 0 < U < 1/2 \\
U, & 1/2 \leq U < 1
\end{cases}
$$
It follows that
$$X = \frac{1 + |2U - 1|}{2}$$
$$F_X(x) = P (X \leq x), \:\:\:\: 1/2 < x < 1$$
$$=P\left(\frac{1 + |2U - 1|}{2} \leq x \right)$$
$$=P(|2U - 1| \leq 2x-1)$$
$$=P(-2x+1 \leq 2U - 1 \leq 2x-1)$$
$$=P(1-x \leq U \leq x)$$
$$=P(1-x < U \leq x)$$
$$=F_U(x) - F_U(1-x)$$
Differentiating with respect to $x$:
$$f_X(x) = f_U(x) + f_U(1-x), \:\:\:\:\: \text{for}\:\:\: 1/2 < x < 1$$
It follows that
$$f_X(x) =
\begin{cases}
2, & \text{if}\:\:\:1/2 < x < 1, \\
0, & \text{otherwise}
\end{cases}
$$
and since we have for a uniform variable
$$f_Z(z) =
\begin{cases}
\frac{1}{b-a}, & \text{if}\:\:\:a < z < b, \\
0, & \text{otherwise}
\end{cases}
$$
$$E[Z] = \frac{b+a}{2} $$
It follows that
$$E[X] = \frac{1 + 1/2}{2} = \frac{3}{4} $$
Best Answer
Imagine your rope as one unit long. Putting it on the number line, the endpoints will be at 0 and 1. Cutting at $1/4$, the ratio is $1:3$, and any cuts left of that also give smaller ratios. By symmetry, the cuts at greater or equal to $3/4$ also give $3:1$ ratios. Thus, we have two quarter length segments that work, in a segment of unit length, giving the probability $1/2$.