Updated 12 September 2026

Negative binomial distribution

Flip a biased coin with success probability $p$ until you accumulate exactly $r$ successes. The number of failures along the way is negative-binomial distributed.

What to notice

• Raising $r$ shifts mass to the right (you’re waiting for more successes, so more failures pile up) and smooths the shape toward a Normal, since the total is a sum of $r$ independent geometric waits.
• Lowering $p$ stretches the whole distribution — rare successes mean long waits.
• Variance exceeds the mean by a factor of $1/p$. Compared to Poisson (where variance equals the mean), the negative binomial is overdispersed — which is why it’s the go-to for count data that clumps.

Relationship to other distributions

• Geometric. With $r=1$, you’re waiting for the first success, which is the failure-count geometric distribution. Our geometric page instead counts the successful trial too, so its values are one larger.
• Sum of geometrics. More generally, a negative binomial with parameter $r$ is the sum of $r$ independent geometrics — using the failure-count convention. This parallels a positive-integer-shape gamma distribution, built from independent common-rate exponential waits.
• Poisson-gamma mixture. A Poisson whose rate is itself gamma-distributed integrates out to a negative binomial. This relationship requires gamma variation in the Poisson mean; arbitrary variation need not give a negative binomial.

Two counters for one experiment

At r=3 and p=0.4, suppose the sequence is failure, success, failure, failure, success, success. There are three failures and six total trials. Switching the counter adds r to every result; it changes the mean from r(1−p)/p=4.5 to r/p=7.5 but leaves the variance at r(1−p)/p²=11.25.

Exactly two failures before the third success requires the fifth trial to succeed and exactly two of the first four to succeed. There are six such arrangements, each with probability 0.4³×0.6², giving 0.13824. The compute panel adds these masses to answer an at-most question under either convention. A total-trial threshold below r is impossible.

Why varying Poisson means produce overdispersion

Suppose a count conditional on Λ is Poisson(Λ), and Λ has a gamma distribution with shape r and rate β. Integrating over Λ gives a failure-count negative binomial with p=β/(β+1). Conversely β=p/(1−p). Its mean is r/β and variance is r/β+r/β²: variation in Λ adds the second term. This is an exact model relationship, not a proof that any overdispersed dataset was generated this way.

At r=3 and p=0.4, β=2/3. The average conditional Poisson mean is 4.5, but the marginal variance is 11.25. A fixed-rate Poisson with mean 4.5 has variance 4.5 instead. The stopping experiment uses integer r; the gamma-mixture family also permits positive noninteger shapes.

References

Random Services: negative binomial uses the total-trial convention; Stanford: mixture models derives the gamma–Poisson mixture and its parameter conversion.