Negative Binomial Distribution

The negative binomial distribution is a Discrete Distribution of Random Variables

  • Geometric Distribution: probability of observing first success on $n$th trial
  • NBD: probability of observing $k$th success on $n$th trial
  • so NBD is a generic case of Geometric Distribution

A distribution is NBD if:

  • trials are independent
  • each trial is a Bernoulli Trial - i.e. has only two outcomes - success and failure
  • $p$ is the same for all the trials
  • the last trial must be success


  • $k$ - number of successes, $n$ - total number of trials
  • $p$ - probability of success, $q = 1 - p$ - probability of failure
  • pmf: $Pr(X = x) = C^{k-1}_{n-1} q^{n-k} p^{k}$


Example 1

A footballer can go home only after he scores 4th goal

  • $p$ - probability of success

Suppose he made 6 attempts

  • what's the probability that he scored 4 goals, and the last trial led to success?

Let's write down all possible sequences when the 4th kick is on the 6th attempt:

from itertools import permutations
p = set([x for x in permutations('SSSSFF')])
[x for x in p if x[-1] == 'S']
1 F S S F S S
2 S S F S F S
3 S F F S S S
4 F F S S S S
5 F S F S S S
6 S F S S F S
7 S F S F S S
8 S S F F S S
9 S S S F F S
10 F S S S F S

There are 10 sequences that lead to this outcome

  • note that Success is always last!

Let's calculate the probability of going home after 6 kicks (having 6th kick successful)

  • so $P(\text{go home}) = \sum_{i=1}^{10} P(\text{seq}_i)$
  • each sequence has the same probability of occurring:
    • $P(\text{seq}_i) = P(\text{seq}) = q^{n-k} p^{k}$
    • this is the probability of observing $n-k$ failures and $k$ successes
  • there are ${n - 1 \choose k - 1}$ ways to pick these elements ($C_{n - 1}^{k - 1}$)