Simulating Pi using Monte Carlo

Question

This code estimates the value of pi and it then compares it to the real pi value by a certain accuracy which is defined as 'c'. Then it decreases 'c' to a smaller number and does the calculation again.

The values of c are .01,0.001,0.0001,0.00001.

What I am trying to do is the whole process 10 times and and find the average for the amount of 'd' which the amount of times it runs the code to get to the accuracy level I want.

import math
import random
pi = math.pi

n = 0
d = 0
ratios = []
xs = []
ys = []
c = 0.1
simulating = True

while c >= 0.0001:

    while simulating:
        x=random.random()
        y=random.random()
        xs.append(x)
        ys.append(y)
        if x**2 + y**2 <= 1.0:
            n += 1
        d += 1
        ratio = 4*n*1./d
        ratios.append(ratio)
        if abs(ratio-pi) / pi <= c:
            print "Draws Needed: ", d
            break

    c = c*.1
    print c

Could you reduce the scope of your question? We would love to help you. — Alexander Ventura, Oct 06 '13 at 00:42
What is your question? I have been trying to figure that out for 5 minutes. — Groditz, Oct 06 '13 at 00:46
Well, nobody can figure out what you're asking - but I can confirm that this is a decent approach to estimating pi via a Monte Carlo method. — Tim Peters, Oct 06 '13 at 00:58
Your code has several problems. The corrected one is given below. — Developer, Oct 06 '13 at 03:47
See a [previous question](http://stackoverflow.com/q/18139934/2647279) about calculating `pi` Monte-Carlo style, it shows some of the drawbacks of this method. — Bas Swinckels, Oct 13 '13 at 18:51

score 0 · Accepted Answer · answered Oct 06 '13 at 03:26

Here are our corrections:

from __future__ import division
import random

pi = random._pi
error = 0.1
inCircle, Total = 0,0
while (error >= 0.0001):
    print '%g ...'%error
    while True:
        x,y = random.random(), random.random()
        if (0.5-x)**2+(0.5-y)**2 <= 0.25: inCircle += 1
        Total += 1
        estimate = 4*inCircle/Total
        if abs(estimate/pi-1) <= error:
            print '{est.} %g vs. {pi} %g after %d trials, {err} %g\n'%( \
                   estimate,pi,Total,error)
            break
    error *= 0.1

results:

0.1 ...
{est.} 3.33333 vs. {pi} 3.14159 after 6 trials, {err} 0.1

0.01 ...
{est.} 3.11765 vs. {pi} 3.14159 after 68 trials, {err} 0.01

0.001 ...
{est.} 3.14286 vs. {pi} 3.14159 after 70 trials, {err} 0.001

0.0001 ...
{est.} 3.1417 vs. {pi} 3.14159 after 247 trials, {err} 0.0001

Some explanation would help as to what you're doing and why, otherwise you may as well have been given the code in your dream by a divine power. — Abhijit Sarkar, Mar 04 '19 at 06:11

Simulating Pi using Monte Carlo

1 Answers1