Like the birthday paradox, but for banking PINs. How many people should it take before you expect a duplicate PIN?

pin_paradox.py

import random

### Converging on 124.908224 after 125000 iterations

iterations = 0
total_pins = 0
max_pin = 9999

bucket_count = max_pin / 10
histogram = [0] * bucket_count
bucket_size = max_pin / bucket_count
max_bucket_count = 0


def stars(n):
    return "".join(['*'] * n)

def normalize(v, n, max_v):
    return int(((float(v) / float(max_v)) * n))


def display(h, bs, c):
    for i, b in enumerate(h):
        if b > 10:
            print("{:>10}: {}".format(i * bs, stars(normalize(b, 20, c))))

while True:
    iterations = iterations + 1
    pins = set()
    for endpoint in range(max_pin):
        next_pin = random.randint(0, max_pin)
        if next_pin in pins:
            #print "Found a double of {} after {} iterations.".format(next_pin, endpoint)
            break
        else: pins.add(next_pin)
    #print "endpoint was {}".format(endpoint)
    total_pins += endpoint
    histogram[endpoint / bucket_size] += 1
    if histogram[endpoint / bucket_size] > max_bucket_count:
        max_bucket_count = histogram[endpoint / bucket_size]
    if iterations % 5000 == 0:
        display(histogram, bucket_size, max_bucket_count)
        print "Converging on {} after {} iterations".format(float(total_pins) / float(iterations), iterations)