Lin Ma - 10 months ago 68

Python Question

Using Python 2.7. Suppose I have an unfair coin and I want to turn it into a fair coin using the following way,

- Probability of generating head is equal for unfair coin;
- Flip unfair coin and only accept head;
- When a head is appearing, treat it as 1 (head for virtual fair coin), when another head is appearing, treat it as 0 (tail for virtual fair coin), next time when head appears, treat it as 1, next time treat as 0, ..., and so on.

Not sure if this method works? Actually I am not quite confident about the method above and also how to use

`equalCoinHelper()`

If anyone have any good ideas, it will be great.

`from __future__ import print_function`

import random

counter = 0

# 0.3 probability return head as 1

# 0.7 probability return tail as 0

def unFairCoin():

if random.random() < 0.3:

return 1

else:

return 0

# probability of generating 1 is equal, so keep 1 only

def equalCoinHelper():

result = 0

while result == 0:

result = unFairCoin()

def equalDistribution():

global counter

# not think about how to leverage this better

equalCoinHelper()

counter += 1

if counter % 2 == 0:

return 1

else:

return 0

if __name__ == "__main__":

# generate 10 random 0/1 with equal probability

print ([equalDistribution() for _ in range(10)])

Answer Source

Getting a Fair Toss From a Biased Coin explains a simple algorithm for turning a biased coin into a fair coin:

- Flip the coin twice.
- If both tosses are the same (heads-heads or tails-tails), repeat step 1.
- If the tosses come up heads-tails, count the toss as heads. If the tosses come up tails-heads, count it as tails.

In Python this would be:

```
def fairCoin():
coin1 = unfairCoin()
coin2 = unfairCoin()
if coin1 == coin2:
return fairCoin() # both are the same, so repeat it
elif coin1 == 1 and coin2 == 0:
return 1
else:
return 0
```

The `elif`

and `else`

blocks can be simplified to just:

```
else:
return coin1
```