I was wrong. My initial approach appears to be the ninja approach like sampo said above. I got 50% -_- import numpy as np import random as rnd impo

I was wrong. My initial approach appears to be the ninja approach like sampo said above. I got 50% -_- import numpy as np import random as rnd import math as m chords = [] samp = 10000 for c in range(samp): #random point on infinite plane is described by a single coordinate dist = rnd.uniform(1., 1000.) #DO NOT KNOW HOW TO DO INTERNAL SLICES #ignore all possible lines that do not intesect circle xsec_half = m.atan(1./dist) angle = rnd.uniform(-xsec_half, xsec_half) #make 2 points on the randomly chosen line x1, y1 = -dist, 0 x2, y2 = dist, 2*dist*m.tan(angle) #copy and paste some shit from wolfram alpha dx = x2-x1 dy = y2-y1 dr = (dx**2 + dy**2)**0.5 D = x1*y2 - x2*y1 sqr = (dr**2 - D**2)**0.5 xa = (D*dy + np.sign(dy)*dx*sqr) / (dr**2) xb = (D*dy - np.sign(dy)*dx*sqr) / (dr**2) ya = (-D*dx + np.abs(dy)*sqr) / (dr**2) yb = (-D*dx - np.abs(dy)*sqr) / (dr**2) chords.append(((xa-xb)**2 + (ya-yb)**2)**0.5) print len([a for a in chords if a > 3**0.5])/float(samp) Basically make the unit circle at 0,0. Pick any radius up to infinity (infinite plane). By symmetry, we don't need 2 coordinates, just a distance from the circle. Choose any angle from that point which will draw a vector that intersects the circle (to not waste CPU). Calculate the length of the intersecting chord. I was really surprised by the distribution of chord lengths, it peaks at chord lengths of 2.0, which is the maximum chord length, and falls exponentially with smaller chord lengths. I don't know what I expected, but not that.