Cut until it becomes shorter than 0.5m

Let's double the sizes, so I've got a rope of length 2 and I cut it until its length is smaller than 1.

Let $E(a)$ be the expected number of cuts of a rope of length $a.$ So $E(a) = 0$ for $a < 1,$ and $E(1) = 1.$ After one cut, we have a rope of length $x$ for some random $x \in [0,a],$ so we get the equation $E(a) = 1 + \frac1{a} \int_0^a E(x) \, dx = 1 + \frac1{a} \int_1^a E(x) \, dx.$ Let $F(a) = \int_1^a E(x) \, dx.$ By the Fundamental Theorem of Calculus , $F'(x) = E(x).$ So we get $F'(a) = 1 + \frac{F(a)}{a}.$ This differential equation can be solved by standard techniques: rearrange and divide by $a$ to get $\begin{aligned} \frac{aF'(a)-F(a)}{a} &= 1 \\ \frac{aF'(a)-F(a)}{a^2} &= \frac1{a} \\ \left( \frac{F(a)}{a} \right)' &= \frac1{a} \\ \frac{F(a)}{a} &= \ln(a) + C \\ F(a) &= a\ln(a) + Ca \end{aligned}$ Since $F(1) = 0, C = 0,$ so $F(a) = a\ln(a),$ and $E(a) = F'(a) = 1 + \ln(a).$ In particular, $E(2) = 1 + \ln(2) = \fbox{1.6931}.$