Probability of k wins out of n, given p?

Question

Given n trials, with a probability p of winning each trial, what is the probability of winning r or more trials?

My thinking goes as follows: Each combination of wins and losses has probability p^w * (p-1)^(nw) where w is number of wins.

Each number of wins can occur in nCr combinations, ex. winning 2 out 3 times means you might lose the first, the second or the third times, eg three combinations.

So the probability of winning 2 out of 3 times is 3C2 * p^2 * (1-p)^1. The probability of winning 2 or more times is just the sum of this calculation for 2 and 3 wins.

I have the following code:

import math

def nCr(n,r):
    f = math.factorial
    return f(n) / f(r) / f(n-r)

def prob_at_least(n, r, p):
    probs = [nCr(n,k)*pow(p,k)*pow(1.0-p,n-k) for k in range(r, n+1)]
    return sum(probs)

This code works, but is there a built-in function, or a shorter way to achieve the same?

Answer 1

From the scipy.stats module, you could use binom .

>>> import scipy.stats as scs
>>> print(scs.binom.pmf(2, 5, .5))
0.3125

Edit: to get r or more:

>>> trials = 0
>>> n = 5
>>> r = 2
>>> p = .5
>>> for i in range(r):
    trials += scs.binom.pmf(i, n, p)
 r_or_more = 1 - trials

Edit: the solution given by ljeabmreosn gives the cumulative distribution function which doesn't require my loop.

Answer 2

There are much faster ways to implement combinations :

import operator as op
from functools import reduce

def nCr(n, r):
    r = min(r, n-r)
    numerator = reduce(op.mul, range(n, n-r, -1), 1)
    denominator = reduce(op.mul, range(1, r+1), 1)
    return numerator / denominator

But if you're doing this a lot, you might want to consider a package like scipy, which has special.comb for efficient combination calculation