# Fair coin from biased coin

Marton Trencseni - Tue 22 March 2022 - Math

## Introduction

Recently I saw a fun puzzle on Hacker News: given a biased coin, construct a fair coin. Ie. you're allowed to toss a biased coin repeatedly and record the sequence of results (like 001...1, and the goal is to return 0/1 with 50%-50% probability, so use the biased coin to simulate a fair coin. I quickly scanned the article and comments to get some ideas, but I wanted to work through solution(s) myself. The code is up on Github.

Let's assume the biased coin comes up 1 with $p$ probability and 0 with $q=1-p$ probability.

First, some helper functions:

def coin_flip(bias=0.5):
if random() < bias:
return 1
else:
return 0

def make_biased_coin(bias=None):
if bias is None:
bias = random()
print(f'Bias = {bias:.3f}')
return lambda: coin_flip(bias)


## von Neumann's probabilistic solution

The most elegant solution is from von Neumann: toss the biased coin twice. The 4 possible outcomes:

• 00: probability $P=p^2$
• 11: probability $P=q^2$
• 01: probability $P=pq$
• 10: probability $P=pq$

Observe that the last 2 probabilities are the same. Algorithm: if the result is 00 or 11, repeat the procedure. If the result is 01, return 0, if the result is 10, return 1.

In code:

def probabilistic(biased_coin):
while True:
s, w = biased_coin(), biased_coin()
if s == 0 and w == 1:
return 1
elif s == 1 and w == 0:
return 0


## Statistical solution

I wanted to come up with a solution of my own for this puzzle. I thought, first let's measure (estimate) the bias of the biased coin, and then somehow correct for it. Let's flip the coin $N/2$ times first, take the mean of the flips, let's call this $\hat{p_1}$, our first estimate of $p$. Then, do another $N/2$ flips (for a total of $N$ flips) to get $\hat{p_2}$, our second estimate of $p$. We can pretend that the first measurement gets the true value of $p$, and then the second measurement varies around $p$ in a symmetrical guassian way, so $P(\hat{p_2} > \hat{p_1}) = P(\hat{p_2} < \hat{p_1})$.

What if get unlucky, and $\hat{p_2} = \hat{p_1}$? We repeat the process and try again. In code:

def statistical(biased_coin, N):
while True:
num_flips = int(N/2.0)
s = sum([biased_coin() for _ in range(num_flips)])/num_flips
num_flips = N - num_flips
w = sum([biased_coin() for _ in range(num_flips)])/num_flips
if s < w:
return 1
elif s > w:
return 0


## Permutations solution

This is based on a comment I saw on Hacker News. Flip the biased coin $N$ times to get an $N$ long sequence of bits. Count the number of 1s in the sequence, let's say it's $k$. The probability of getting this specific sequence is $P=p^k q^{N-k}$. We notice that all sequences with $k$ 1s have this probability. So if we can somehow enumerate all such sequences, we can order them lexicographically, and we can divide them into two sets of equal cardinality: pick the middle bit sequence, and take the ones smaller than the middle and the ones bigger than the middle bit sequence.

Implementing this is trivial with one more restriction: let's only consider bit sequences such that the bit sequence is not equal to its reverse (ie. if the bit sequence is 01011 then it's reverse is 11010). In this case each bit sequence has exactly one pair, its reverse, and we can create a trivial ordering between these pairs. If we get unlucky, and the bit sequence is the same as its reverse, we repeat the process and try again. In code:

def permutations(biased_coin, N):
while True:
inst = ''.join([str(biased_coin()) for _ in range(N)])
revr = inst[::-1]
if inst > revr:
return 1
elif inst < revr:
return 0


## All 3 solutions are the same

For the second and third solution, what should $N$ be? For the statistical solution, first I thought that the higher $N$, the more accurately it approximates a fair coin. But, this is not the case. All of the solutions are perfectly accurate at all $N$:

def experiment(num_flips, num_experiments, coin_type, coin_func):
start = time()
means = [mean([coin_func() for _ in range(num_flips)]) for _ in range(num_experiments)]
m, s = mean(means), stdev(means)
elapsed = (time() - start) * 1000
print(f'{coin_type}: mean = {m:.3f}, stdev = {s:.3f}, elapsed = {elapsed:.0f} msec')

num_experiments = 1000
num_flips = 1000
biased_coin = make_biased_coin(0.2) # fix bias at 0.2 for measurement

experiment(num_flips=num_flips, num_experiments=num_experiments, coin_type='Fair coin',
coin_func=lambda: coin_flip())

experiment(num_flips=num_flips, num_experiments=num_experiments,
coin_type=f'Probabilistic coin',
coin_func=lambda: probabilistic(biased_coin))

for N in [2, 50, 100]:
experiment(num_flips=num_flips, num_experiments=num_experiments,
coin_type=f'Statistical coin (N={N})',
coin_func=lambda: statistical(biased_coin, N=N))

for N in [2, 50, 100]:
experiment(num_flips=num_flips, num_experiments=num_experiments,
coin_type=f'Permutations coin (N={N})',
coin_func=lambda: permutations(biased_coin, N=N))


Yields something like:

Bias = 0.200
Fair coin:                 mean = 0.501, stdev = 0.016, elapsed = 629   msec
Probabilistic coin:        mean = 0.500, stdev = 0.016, elapsed = 1815  msec
Statistical coin (N=2):    mean = 0.500, stdev = 0.016, elapsed = 5293  msec
Statistical coin (N=50):   mean = 0.500, stdev = 0.016, elapsed = 13198 msec
Statistical coin (N=100):  mean = 0.500, stdev = 0.015, elapsed = 24275 msec
Permutations coin (N=2):   mean = 0.500, stdev = 0.016, elapsed = 4569  msec
Permutations coin (N=50):  mean = 0.499, stdev = 0.016, elapsed = 18403 msec
Permutations coin (N=100): mean = 0.499, stdev = 0.016, elapsed = 35645 msec


Given that, we should run them at the minimal $N$, which is $N=2$ coin flips. Then, if we think about it, the statistical and permutations based solutions trivially reduce to the von Neumann solution at $N=2$ coin flips:

• statistical: flip the coin once, it's either 0 or 1, so the first mean $\hat{p_1}$ is either 0 or 1. Flip it a second time to get $\hat{p_2}$. The conditions of $\hat{p_1} < \hat{p_2}$ and $\hat{p_1} > \hat{p_2}$ reduce to the 01 and 10 cases (probability $P=pq$), and we flip again in the 00 and 11 case.
• permuations: there are 4 possible outcomes of 2 flips. 00 and 11 are their own reverse, so we flip again, in the other two cases of 01 and 10 we return 0 or 1 (probability $P=pq$).

## Conclusion

I had great fun thinking through this trivial puzzle.