Probability of getting a piece longer than $1/2$ on cutting a rope of length $1$ at two randomly chosen points

Think about it this way. Say $x$ is a number between $0$ and $1$, representing the position of the first cut. For instance, suppose $x = 0.25$. Now the second cut might occur on the short piece with probability $0.25$, and no matter where it is cut there, the long piece exceeds $0.5$. Or the long piece is cut with probability $0.75$, but if the second cut occurs too close to one of the ends of the long piece, the remaining length will still be greater than $0.5$; otherwise, if cut near the middle, the two pieces will be less than $0.5$. So we just have to figure out what we mean by being "too close" to one of the ends.

For the case $x = 0.25$, it is easy to see that if $y$ is the position of the second cut, if $0.25 < y < 0.5$, that is "too close" to an end--the remaining piece will be longer than $0.5$. And if $0.75 < y < 1$, then again, the second cut is too close to the other end, and the middle piece will be longer than $0.5$. So, given that $x = 0.25$, the probability that our second cut will give one section of rope that is longer than $0.5$ is the total length of all intervals satisfying $0 < y < 0.5$, or $0.75 < y < 1$; i.e., it is $0.5 + 0.25 = 0.75$.

Naturally, the next step is to generalize this to any choice of $x$. Without loss of generality, let $0 < x < 0.5$ (as the situation is symmetric about $x = 0.5$). Then as we saw above, every choice of $y$ between $0$ and $x$ will make the interval $[x, 1]$ longer than $0.5$, and every choice of $y$ between $x$ and $0.5$, or between $x + 0.5$ and $1$, will also make a segment longer than $0.5$. So the probability of the second cut occurring in the interval $[0, 0.5] \cup [x + 0.5, 1]$ is $1 - (x + 0.5) + (0.5 - 0) = 1 - x.$ That's a nice result!

Finally, we use calculus to integrate over all $x$ between $0$ and $0.5$, then multiply by $2$ because we need to account for symmetry. We have $$2 \int_{x=0}^{0.5} 1-x \, dx = \frac{3}{4}.$$ This is the desired probability.

The answer is 75%. It can be evaluated via $$ 2 \int_0^1 \int_0^y \begin{cases} 1, &\text{if } \max\{x,y-x,1-y\} > \frac12 \\ 0, &\text{if } \max\{x,y-x,1-y\} \le \frac12 \end{cases} \bigg\} \,dx \,dy $$ (where we use symmetry to declare $x$ the left cut point and $y$ the right cut point).

In the graph below, the black triangles are the points $(x,y)$ from the integral above that satisfy $\max\{x,y-x,1-y\} > \frac12$ (and the light gray triangles are their symmetric twins $(y,x)$), while the white triangles are the points that satisfy $\max\{x,y-x,1-y\} \le \frac12$.

We have a piece of length $>{1\over2}$ iff (a) both cuts are on the same half of the rope, or (b) are on different halfs, but the right cut is further right on its half than the left cut on its half. The probability that one of these happens is ${1\over2}+{1\over2}\cdot{1\over2}={3\over4}$.