This is the hardest problem of Euler project: https://projecteuler.net/problem=597
Before start, please do not share any clue about the answer because this would be a spoiler to participants.
In the question, the case with 3 boats with 160 meters long river, the probability of no boats bumping into other is given as $\dfrac{4}{15}$.
I prepared a draft scheme to make it easier to understand:
However, I could not understand why. To my calculation, the condition of no boat bumping is:
$$\frac{80}{\vec{c}} \leq \frac{120}{\vec{b}} \leq \frac{160}{\vec{a}}.$$
How do we extract the given probabilty from this condition?
$\def\e{\mathrm{e}}\def\d{\mathrm{d}}$Using the notation in the original description of the question, let $d = 40\text{m}$, $s = L - 2d = 80\text{m}$. Also, denote by $V_1, V_2, V_3$ the velocities of $C, B, A$, respectively.
The question says that the velocities of each boat are i.i.d as $V = -\ln X$, where $X \sim U(0, 1)$, thus $V \sim \mathrm{Exp}(1)$. Because the equivalent condition on which no bumps occurs is$$ \frac{s + 2d}{V_3} > \frac{s + d}{V_2} > \frac{s}{V_1}, $$ then probability of no bumps is\begin{align*} &\mathrel{\phantom{=}}{} \int_0^{+\infty} \e^{-v_3} \,\d v_3 \int_{\frac{s + d}{s + 2d} v_3}^{+\infty} \e^{-v_2} \,\d v_2 \int_{\frac{s}{s + d} v_2}^{+\infty} \e^{-v_1} \,\d v_1\\ &= \int_0^{+\infty} \e^{-v_3} \,\d v_3 \int_{\frac{s + d}{s + 2d} v_3}^{+\infty} \e^{-v_2} · \exp\left( -\frac{s}{s + d} v_2 \right) \,\d v_2\\ &= \int_0^{+\infty} \e^{-v_3} \,\d v_3 \int_{\frac{s + d}{s + 2d} v_3}^{+\infty} \exp\left( -\frac{2s + d}{s + d} v_2 \right) \,\d v_2\\ &= \int_0^{+\infty} \e^{-v_3} · \frac{s + d}{2s + d} \exp\left( -\frac{2s + d}{s + d} · \frac{s + d}{s + 2d} v_3 \right) \,\d v_3\\ &= \frac{s + d}{2s + d} \int_0^{+\infty} \exp\left( -\frac{3s + 3d}{s + 2d} \right) \,\d v_3\\ &= \frac{s + d}{2s + d} · \frac{s + 2d}{3s + 3d} = \frac{s + 2d}{3(2s + d)} = \frac{4}{15}. \end{align*}
P.S.: The probability of no bumps with $n$ boats is