Throw a Two

Without the loss of generality, we can assume the centre of one wooden stick lands between two given lines, and we can also assume the centre lies on some given vertical line. If we denote the distance along the vertical line from the bottom horizontal line to the centre of the stick as $x$ , then it is clear the stick cannot cross both lines if $0<x<\frac14$ or $\frac34<x<1$ . Given that $\frac12 <x<\frac34$ , it is evident by considering the associated right-angled triangle that the probability $P_x$ that the stick will cross both lines for some $x$ is $P_x=\dfrac{2\arccos \left(\frac{4x}{3}\right)}{\pi }.$ Thus, since $P=\frac12\mathbb{E}[P_x]$ , we have $\begin{aligned} P=\dfrac{1}{2}\times \dfrac{1}{\left(\frac34 -\frac12 \right)}\int_{\frac12}^{\frac34}\dfrac{2\arccos \left(\frac{4x}{3}\right)}{\pi }\; dx=\dfrac{\sqrt{5}-2\arccos \left(\frac23 \right)}{\pi }&= 0.176321597814...\\ \Longrightarrow 10^{10}\times P&\approx \boxed{1763215978}. \end{aligned}$

Problem is attached to Geometric Probability .