Fun with Eyeglass wipes

wipes.py

"""
I order https://www.amazon.com/Zeiss-Pre-Moistened-Cleaning-Wipes-count/dp/B01E9NC2Z4/ref=sr_1_3_a_it?ie=UTF8&qid=1544464237&sr=8-3&keywords=zeiss+wipes
from Amazon pretty regularly.

The packets come with 2 attached together. When I pull one out that's still 2
attached, I tear off one of them to use and put the other back in the box.
I wanted to know how many times I have to pull a packet from the box before
I am more likely to pull a single packet instead of a double.

An example of how to run this is

python ./wipes.py 10000       # This will run the simulation 10000 times and average the results
Average is 90.9923214320655
"""

import random
import sys

from collections import namedtuple

Packet = namedtuple('Packet', ['num_of_wipes'])

def initialize_box():
    return [Packet(2) for x in range(200)]

def take_packet(box, packets_taken):
    choice = random.randint(1, len(box)) - 1
    chosen_packet = box.pop(choice)
    if chosen_packet.num_of_wipes == 2:
        box.append(Packet(1))

    packets_taken += 1
    return box, packets_taken

def chance_to_pull_double(box):
    num_double_packets = len([x for x in box if x.num_of_wipes == 2])
    chance_to_pull_double = num_double_packets / len(box)

    return chance_to_pull_double


def main():
    number_of_iterations = int(sys.argv[1])

    results = []
    for x in range(number_of_iterations):
        box = initialize_box()
        chances = 1
        packets_taken = 0
        while chances >= 0.5:
            current_box, packets_taken = take_packet(box, packets_taken)
            chances = chance_to_pull_double(current_box)
            results.append(packets_taken)

    print("Average is {0}".format(sum(results) / len(results)))

if __name__ == "__main__":
    main()