Jun 18, 2018 | science & tech

Coin Flip Probability with Python

distractions are welcome in any flavor

I'm writing up my dissertation...but occasionally I need a distraction. Don't get me wrong, I find studying hate speech very fascinating, but in all honesty, it gets to be a bit much sometimes. The world can be a depressing place. Plus, it's a dissertation; distractions are welcome in any flavor.

Today my distraction came in the form of a Tweet by David Robinson demonstrating how flipping a coin and getting a heads and then another heads takes 6 flips on average while a heads then a tails only takes 4.

He did this using tidyverse functions and then using base-R matrix operations. Well, I wanted to create the Python version because of said dissertation. I did so by creating a set of functions.

The first is simply a function to simulate flipping a fair coin...

python
import numpy as np
def flip_coin():
"""Simulate flipping a coin.
Returns
-------
str
"H" for heads/ "T" for tails.
"""
flip = np.random.binomial(1, .5, 1)
if flip[0] == 1:
side = "H"
else:
side = "T"
return side

Then I need a function to flip the coin multiple times and to stop only when a certain sequence of sides were met. In other words, stop when two heads were flipped in a row. From this, I want the number of times it took to achieve this sequence to be returned.

python
def flip_condition(stop_condition=['H', 'T'], print_opt=False):
"""Flip coin until flip pattern is met.
Parameters
----------
stop_condition: list
The sequence of flips to be matched before flipping stops.
print_opt: bool
Option to print the sequence of flips.
Returns
-------
int
The number of flips it took to match the pattern.
"""
flip_list = []
current_index = 0
current_condition = None
while current_condition != stop_condition:
flip_list.append(flip_coin())
if len(flip_list) >= len(stop_condition):
current_condition = [flip_list[i] for i in range(current_index - len(stop_condition) +1 , current_index + 1)]
else:
pass
current_index +=1
if print_opt:
print(flip_list)
return current_index

This also includes an option to print the sequence (in case you want to check my programming...I would 😜). And then I just run it many times and take the average for both conditions.

python
mean_ht = np.mean([flip_condition(['H', 'T']) for i in range(10000)])
mean_hh = np.mean([flip_condition(['H','H']) for i in range(10000)])
print("Average # of flips to achieve heads and then heads again: {}".format(mean_hh))
print("Average # of flips to achieve heads and then tails: {}".format(mean_ht))
# Average # of flips to achieve heads and then heads again: 6.0081
# Average # of flips to achieve heads and then tails: 3.9829

Looks like probability holds up in Python! Oh, but I also created the function to capture any pattern. How about heads, tails, heads?

python
np.mean([flip_condition(['H', 'T', 'H']) for i in range(10000)])
# 9.9231

or heads, heads, heads

python
np.mean([flip_condition(['H', 'H', 'H']) for i in range(10000)])
# 14.1544

Well that was neat! 🦖🦕

If you want to play around with the code, I've posted in this notebook here