TITLE: Probability of getting a piece longer than $1/2$ on cutting a rope of length $1$ at two randomly chosen points
QUESTION [5 upvotes]: Here's a bootstrap approach I used to find the answer (python 3).
def num_pieces(num,lenght):
    ot = list(range(1,lenght+1))[::-1]
    nig = []
    for i in range(lenght-1):
        n = random.randint(1, num-ot[i])
        nig.append(n)
        num -= n
    nig.append(num)
    r = []
    for i in nig:
         r.append(i/10)
    return r

res = []
n1 = 1000
n2 = int(n1/10)

for i in range(n1):
    a = []
    for i in range(n2):
        r = num_pieces(10,3)
        if r[0] > 0.5 or r[1] > 0.5 or r[2] > 0.5:
            a.append(1)
        else:
            a.append(0)
    res.append(sum(a)/n2)

pd.DataFrame(res).hist();

And here's the result
If my code is correct (I believe it is), the probability is about 50%
Question - I do not know how to solve this task using plain math. In other words - what is the math base under the bootstrap result?

UPDATE
After receiving those awesome answers below, I checked the code and fixed the num_pieces generator. It worked incorrectly. All other code works just fine. So, here's the final solution
def get_random():
    a = [random.random(), random.random()]
    c = 1 - (max(a))
    b = max(a) - min(a)
    a = min(a)
    return [a,b,c]
res = []
n1 = 1000
n2 = int(n1/10)
for i in range(n1):
    a = []
    for i in range(n2):
        r = get_random()
        if r[0] > 0.5 or r[1] > 0.5 or r[2] > 0.5:
            a.append(1)
        else:
            a.append(0)
    res.append(sum(a)/n2)
pd.DataFrame(res).hist();

The probability is 75% indeed. Thank you all for answering!

REPLY [4 votes]: 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 ${1\over2}+{1\over2}\cdot{1\over2}={3\over4}$.