Couldn't find an accurate estimate of π anywhere, so I decided to Monte Carlo it

π.py

#!/usr/bin/python

import random

N_REP = 9999999
c = 0
for i in range(N_REP):
    x, y = (random.random()-.5, random.random()-.5)
    if (x*x+y*y)**.5 < .5:
        c += 1
print("pi is approximately {0}".format(4.0 * float(c) / float(N_REP)))