How do I simulate flip of biased coin?

Question:

In unbiased coin flip H or T occurs 50% of times.

But I want to simulate coin which gives H with probability ‘p’ and T with probability ‘(1-p)’.

something like this:

def flip(p):
   '''this function return H with probability p'''
   # do something
   return result

>> [flip(0.8) for i in xrange(10)]
[H,H,T,H,H,H,T,H,H,H]
Asked By: Pratik Deoghare

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Source

Answers:

import random
def flip(p):
    return (random.random() < p)

That returns a boolean which you can then use to choose H or T (or choose between any two values) you want. You could also include the choice in the method:

def flip(p):
    if random.random() < p:
        return 'H'
    else:
        return 'T'

but it’d be less generally useful that way.

Answered By: David Z

  • Import a random number between 0 – 1 (you can use randrange function)

  • If the number is above (1-p), return tails.

  • Else, return heads

Answered By: Paul

random.random() returns a uniformly distributed pseudo-random floating point number in the range [0, 1). This number is less than a given number p in the range [0,1) with probability p. Thus:

def flip(p):
    return 'H' if random.random() < p else 'T'

Some experiments:

>>> N = 100
>>> flips = [flip(0.2) for i in xrange(N)]
>>> float(flips.count('H'))/N
0.17999999999999999  # Approximately 20% of the coins are heads

>>> N = 10000
>>> flips = [flip(0.2) for i in xrange(N)]
>>> float(flips.count('H'))/N
0.20549999999999999  # Better approximation
Answered By: Federico A. Ramponi

Do you want the “bias” to be based in symmetric distribuition? Or maybe exponential distribution? Gaussian anyone?

Well, here are all the methods, extracted from random documentation itself.

First, an example of triangular distribution:

print random.triangular(0, 1, 0.7)

random.triangular(low, high, mode):

Return a random floating point number N such that low <= N < high and
with the specified mode between those
bounds. The low and high bounds
default to zero and one. The mode
argument defaults to the midpoint
between the bounds, giving a symmetric
distribution.

random.betavariate(alpha, beta):

Beta distribution. Conditions on the parameters are alpha > 0 and
beta > 0. Returned values range between 0 and 1.

random.expovariate(lambd):

Exponential distribution. lambd is 1.0
divided by the desired mean. It should
be nonzero. (The parameter would be
called “lambda”, but that is a
reserved word in Python.) Returned
values range from 0 to positive
infinity if lambd is positive, and
from negative infinity to 0 if lambd
is negative.

random.gammavariate(alpha, beta):

Gamma distribution. (Not the gamma
function!) Conditions on the
parameters are alpha > 0 and beta > 0.

random.gauss(mu, sigma):

Gaussian distribution. mu is the mean, and sigma is the standard
deviation. This is slightly faster
than the normalvariate() function
defined below.

random.lognormvariate(mu, sigma):

Log normal distribution. If you take
the natural logarithm of this
distribution, you’ll get a normal
distribution with mean mu and standard
deviation sigma. mu can have any
value, and sigma must be greater than
zero.

random.normalvariate(mu, sigma):

Normal distribution. mu is the mean,
and sigma is the standard deviation.

random.vonmisesvariate(mu, kappa):

mu is the mean angle, expressed in
radians between 0 and 2*pi, and kappa
is the concentration parameter, which
must be greater than or equal to zero.
If kappa is equal to zero, this
distribution reduces to a uniform
random angle over the range 0 to 2*pi.

random.paretovariate(alpha):

Pareto distribution. alpha is the
shape parameter.

random.weibullvariate(alpha, beta)

Weibull distribution. alpha is the
scale parameter and beta is the shape
parameter.

Answered By: nosklo

How about:

import numpy as np
n, p = 1, .33  # n = coins flipped, p = prob of success
s = np.random.binomial(n, p, 100)
Answered By: ticktock

One can sample from the X ~ Bernoulli(p) distribution nsamples times using sympy too:

from sympy.stats import Bernoulli, sample_iter
list(sample_iter(Bernoulli('X', 0.8), numsamples=10)) # p = 0.8 and nsamples=10
# [1, 1, 0, 1, 1, 0, 1, 1, 1, 1]

Return 'H' or 'T' instead using

def flip(p, n):
    return list(map(lambda x: 'H' if x==1 else 'T', sample_iter(Bernoulli('X', p), numsamples=n)))

print(flip(0.8, 10)) # p = 0.8 and nsamples=10
# ['H', 'H', 'T', 'H', 'H', 'T', 'H', 'H', 'H', 'H']
Answered By: Sandipan Dey
import random
def flip():
    return ["H" if random.randint(0,3) <= 2 else "T" for i in range(10)]

Right now probability of Head is 75% and tails is 25% (0,1,2 are all Heads and only 3 is Tails) . By using random.randint() you could have any probability of bias while still maintaining randomness.

Answered By: Rumi