Negative Binomial Point Probabilities

Why Parameterization Matters

A negative binomial point probability gives the probability of observing exactly a specified count. In claim-frequency work, the count might be the number of claims over a period. The calculation is often direct once the model is specified, but the phrase “negative binomial” can refer to different ways of defining the count and its parameters. Before substituting values, identify what is being counted and which probability belongs to that count.

This tutorial uses the convention that $N$ counts failures before a fixed number of successes. It also shows how to translate a model specified by its mean and dispersion, and how to read the alternative convention that counts successes before a fixed number of failures. These translations are important because using the right formula with the wrong parameter meaning still gives the wrong answer.

Definition: Suppose independent trials have the same success probability $p$, where $0<p<1$, and let $r$ be a positive integer. The negative binomial random variable $N$ counts the number of failures before the $r$th success. Its support is $0,1,2,\ldots$, and $$ P(N=k)=\binom{k+r-1}{k}p^r(1-p)^k,\qquad k=0,1,2,\ldots. $$ Here $p$ is the probability of a success on each trial, and $1-p$ is the probability of a failure. A success need not mean a claim; its meaning is set by the problem.

Where the Point Probability Comes From

To have exactly $k$ failures before the $r$th success, the final trial must be a success. In the first $r+k-1$ trials, there must be $r-1$ successes and $k$ failures, in any order. The number of such orders is $\binom{r+k-1}{k}$. Each order has probability $p^{r-1}(1-p)^k$, and the final success contributes another factor $p$. Multiplying the number of orders by the probability of each sequence gives the point-probability formula.

\binom{r+k-1}{k}\,p^{r-1}(1-p)^k\,p = \binom{r+k-1}{k}p^r(1-p)^k

This construction also explains the support: any nonnegative integer number of failures is possible before the $r$th success. Unlike a fixed-trial binomial count, there is no fixed maximum count. The formula applies to independent trials with a constant success probability and a fixed target of $r$ successes. If the trial probabilities vary or the stopping rule is different, this derivation does not apply.

Worked Example: Exactly Two Failures Before the Third Success

In a hypothetical model, each independent exposure is classified as a success with probability $0.8$, and the number of failures is counted until the third success occurs. Find the probability of exactly two failures. State the method and check that the result uses the requested count.

Method. This is the failures-before-$r$-successes convention, with $r=3$, $p=0.8$, and $k=2$. The complementary failure probability is $1-p=0.2$.

Substitute into the point-probability formula:

\begin{aligned} P(N=2) &=\binom{2+3-1}{2}(0.8)^3(0.2)^2\\ &=\binom{4}{2}(0.512)(0.04)\\ &=6(0.02048)\\ &=0.12288. \end{aligned}

The probability is $0.1229$ to four decimal places. As a check, the sequence must contain three successes and two failures, and the final trial must be a success. The formula includes three factors of $0.8$, two factors of $0.2$, and six possible arrangements of the first four trials. The common mistake is to count two successes instead of two failures; here the requested value $k=2$ is the number of failures.

Mean and Dispersion Parameterization

A negative binomial model may be specified by its mean $\mu$ and a positive dispersion or size parameter $r$, rather than directly by $p$. Under the failures-before-$r$-successes convention, the corresponding success probability is $p=r/(r+\mu)$. This follows from the negative binomial mean $\mu=r(1-p)/p$: solving for $p$ gives $p=r/(r+\mu)$. Thus the probability of a failure is $1-p=\mu/(r+\mu)$.

Parameter conversion: For a negative binomial model with mean $\mu>0$ and size parameter $r>0$, use $$ p=\frac{r}{r+\mu}, \qquad 1-p=\frac{\mu}{r+\mu}. $$ Then the point probability is $$ P(N=k)=\binom{k+r-1}{k} \left(\frac{r}{r+\mu}\right)^r \left(\frac{\mu}{r+\mu}\right)^k. $$ In the trial interpretation, $r$ is a positive integer. In the mean-dispersion extension, $r$ may be any positive real number, with the binomial coefficient interpreted using the gamma function.

One common dispersion convention uses $\alpha=1/r$. With this convention, the negative binomial variance is $\mu+\alpha\mu^2$; larger $\alpha$ means greater variance relative to the mean. Equivalently, larger $r$ means less extra dispersion for a fixed mean. Always check how a problem defines its dispersion parameter: some descriptions report $r$, while others report its reciprocal $\alpha$. A symbol alone does not identify the convention.

Worked Example: Convert a Mean and Size to a Point Probability

A hypothetical portfolio's claim count $N$ follows a negative binomial model with mean $\mu=6$ and size $r=3$. Find the probability of exactly two claims. Give the answer to four decimal places.

Method. Convert the mean-size specification to the failures-before-$r$-successes parameters, then evaluate the point probability. The given $r=3$ is the size parameter, not the requested number of claims.

First calculate the per-trial success and failure probabilities:

p=\frac{3}{3+6}=\frac{1}{3}, \qquad 1-p=\frac{6}{3+6}=\frac{2}{3}.

The requested count is $k=2$. Therefore,

\begin{aligned} P(N=2) &=\binom{2+3-1}{2}\left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right)^2\\ &=6\left(\frac{1}{27}\right)\left(\frac{4}{9}\right)\\ &=\frac{24}{243} =\frac{8}{81}\\ &\approx 0.0987654. \end{aligned}

Thus $P(N=2)=0.0988$ to four decimal places. The parameter conversion is consistent with the stated mean because $r(1-p)/p=3(2/3)/(1/3)=6$. A common error is to use $p=\mu/(r+\mu)=2/3$ as the success probability. That reverses success and failure probabilities; the correct success probability for this convention is $r/(r+\mu)=1/3$.

Recognizing the Alternative Convention

Some sources define a negative binomial count as the number of successes before the $r$th failure. This is a different description of the count, so the probability attached to the exponent $r$ is now the probability of failure. If each trial has success probability $p$, the number $M$ of successes before the $r$th failure has $P(M=k)=\binom{k+r-1}{k}(1-p)^r p^k$. This is the same general form with the roles of success and failure exchanged.

Worked Example: Count Claims Before the Second Claim-Free Period

In a hypothetical sequence of independent periods, the probability of a period with a claim is $0.25$, and the probability of a claim-free period is $0.75$. Let $M$ count the claim periods before the second claim-free period. Find the probability of exactly three claim periods.

Method. The count is successes (claim periods) before a fixed number of failures (claim-free periods). Treat claim-free periods as the stopping outcome, with probability $0.75$, and claim periods as the counted outcome, with probability $0.25$. Here $r=2$ and $k=3$.

\begin{aligned} P(M=3) &=\binom{3+2-1}{3}(0.75)^2(0.25)^3\\ &=\binom{4}{3}(0.5625)(0.015625)\\ &=4(0.0087890625)\\ &=0.03515625. \end{aligned}

The answer is $0.0352$ to four decimal places. As a check, every qualifying sequence has two claim-free periods and three claim periods, and the second claim-free period is last. There are $\binom{4}{3}=4$ possible orders for the first four periods. The common mistake is to put $0.25$, the probability of a claim, to the power $r=2$. The fixed stopping outcome here is claim-free, so its probability $0.75$ is raised to the power $r$.

Calculation Checks and Exam Traps

A point-probability calculation is easier to audit when the count convention is written down before the formula. Use the following sequence:

Name the counted outcome.
Decide whether the random variable counts failures before successes or successes before failures. Do not infer this from the words “negative binomial” alone.

Identify the stopping number and target count.
Write $r$ for the fixed number of stopping outcomes and $k$ for the requested number of counted outcomes. The support for the count is $k=0,1,2,\ldots$.

Assign probabilities to outcomes.
For failures before $r$ successes, the fixed stopping outcome has probability $p$, and the counted failure has probability $1-p$. For successes before $r$ failures, these roles reverse.

Evaluate and check the point event.
Use $\binom{k+r-1}{k}$ for the arrangement count, verify the powers match the numbers of each outcome, and round only after the full calculation.

The coefficient can be written either as $\binom{k+r-1}{k}$ or as $\binom{k+r-1}{r-1}$; these are equal. It counts arrangements of the first $k+r-1$ trials, not arrangements including the final stopping trial. The point-probability formula itself already accounts for that last trial through the factor for the stopping outcome raised to $r$.

Keep the request specific. “Exactly $k$” calls for one point probability $P(N=k)$. A request such as “at most $k$” or “more than $k$” involves several point probabilities or a complement, not just the single formula evaluated at $k$. The calculations here are for exact counts only.

Key takeaway: First determine what the random variable counts and which outcome ends the sequence. For failures before $r$ successes, use $p^r(1-p)^k$; for successes before $r$ failures, reverse those probability roles. When given a mean and size, convert using $p=r/(r+\mu)$ under the failures-before-successes convention.

Check Your Understanding

For each problem, identify the convention before evaluating the exact probability.

A sequence of independent trials has success probability $0.6$. Let $N$ count failures before the second success. Find $P(N=1)$. (Answer: 0.2880)
A negative binomial count $N$ counts failures before the fourth success, with success probability $0.5$. Find $P(N=2)$. (Answer: 0.1563)
A claim count has negative binomial mean $\mu=5$ and size $r=5$, using the failures-before-successes convention. Find the probability of zero claims. (Answer: 0.0313)
Independent periods have claim probability $0.3$ and claim-free probability $0.7$. Let $M$ count claim periods before the third claim-free period. Find $P(M=2)$. (Answer: 0.1852)
A claim count has mean $\mu=8$ and size $r=2$, with failures-before-successes parameterization. Find $P(N=3)$. (Answer: 0.0819)

What You'll Learn