I don&#x27;t think that page explains it very well, but have poor math background. I imagined notebook paper with horizontal lines spaced L apart and then the needle dropping at any angle. When the needle is vertical the probability it cross a line is 1, when horizontal it is zero.  The length of the needle L is the hypotenuse of a triangle. If we call the angle from horizontal x, the &quot;height&quot; of the needle can be anywhere within h=Lsin(x) for x between 0 and pi&#x2F;2.<p>The &quot;lines&quot; are like a sample of a point from a uniform distribution U with width L, and h is an interval inside U. The probability a number sampled from a distribution of width L will fall within interval h is h&#x2F;L. Substituting for h gives p(cross|x) = sin(x).<p>Then assuming the needle is equally likely to drop at any angle, for any one angle theta we get probability density p(theta=x) = 1&#x2F;(pi&#x2F;2-0)= 2&#x2F;pi.<p>The probability the needle drops at angle x AND crosses a line is the product of p(theta=x)p(cross|x)= (2&#x2F;pi)sin(x). As mentioned, x can range between 0 and pi&#x2F;2. To get the probability the needle drops at angle x1 OR x2 OR x3, etc and cross we need to sum all these. So take the integral of (2&#x2F;pi)sin(x) from 0:pi&#x2F;2. This gives 2&#x2F;pi.