The Geometric & Negative Binomial Distributions

Run independent Bernoulli($p$) trials — each a success with probability $p$ and a failure with probability $1-p$ — and let $X$ be the trial on which the first success occurs. For $X=k$, the first $k-1$ trials must all fail and the $k$-th must succeed; by independence these multiply: $\mathbb{P}\{X=k\}=\underbrace{(1-p)(1-p)\cdots(1-p)}_{k-1\text{ failures}}\;p=(1-p)^{k-1}p,\qquad k=1,2,3,\dots$ This is Definition 12.9, written $X\sim\mathrm{Geom}(p)$ with $0<p\le 1$. The possible values start at $k=1$ — at least one trial is needed — and run through all positive integers, so unlike the binomial the geometric is a genuinely infinite discrete distribution. The masses decay by the constant factor $1-p$ at each step, $\mathbb{P}\{X=k+1\}=(1-p)\,\mathbb{P}\{X=k\}$, yet still account for all the probability, because the total is a geometric series: $\sum_{k=1}^{\infty}(1-p)^{k-1}p=p\sum_{j=0}^{\infty}(1-p)^{j}=p\cdot\frac{1}{1-(1-p)}=\frac{p}{p}=1.$

The geometric has an unusually clean tail probability. The event $\{X>k\}$ says the first success has not happened by trial $k$ — equivalently, the first $k$ trials are all failures. By independence that is just $\mathbb{P}\{X>k\}=(1-p)^{k}.$ (The same value drops out of the series $\sum_{j=k+1}^{\infty}(1-p)^{j-1}p=(1-p)^{k}$, but the 'all $k$ fail' reading is faster and more memorable.) The complementary event $\{X\le k\}$ — the first success arrives on or before trial $k$ — therefore has cumulative distribution function $\mathbb{P}\{X\le k\}=1-\mathbb{P}\{X>k\}=1-(1-p)^{k}.$ These two formulas settle most geometric questions without summing anything: 'more than $k$', 'at least $k$', and 'within the first $k$' all collapse to a single power of $1-p$. For instance $\mathbb{P}\{X\ge k\}=\mathbb{P}\{X>k-1\}=(1-p)^{k-1}$.

Now wait longer: let $X$ be the trial on which the $r$-th success occurs in the same stream of independent Bernoulli($p$) trials. For $X=k$, two things must hold. The $k$-th trial is a success (the $r$-th and last one we count), and among the first $k-1$ trials there are exactly $r-1$ successes (hence $k-r$ failures). One specific such pattern has probability $p^{r-1}(1-p)^{k-r}$ over the first $k-1$ trials times $p$ on the last, i.e. $p^{r}(1-p)^{k-r}$; and there are $\binom{k-1}{r-1}$ ways to choose which of the first $k-1$ trials carry the $r-1$ early successes. Multiplying, $\mathbb{P}\{X=k\}=\binom{k-1}{r-1}p^{r}(1-p)^{k-r},\qquad k=r,r+1,r+2,\dots$ This is Definition 12.10, written $X\sim\mathrm{NegBin}(r,p)$ with $0<p\le 1$ and $r\in\mathbb{N}$. The support starts at $k=r$: you cannot gather $r$ successes in fewer than $r$ trials. Note the coefficient is $\binom{k-1}{r-1}$, not $\binom{k}{r}$ — the final trial is pinned as a success, so only the first $k-1$ trials are free to be arranged.

Setting $r=1$ recovers the geometric exactly. With a single success to wait for, $\binom{k-1}{0}=1$ and the negative binomial mass becomes $\mathbb{P}\{X=k\}=1\cdot p^{1}(1-p)^{k-1}=(1-p)^{k-1}p$ — the geometric p.m.f. This is Remark 12.4: $\mathrm{Geom}(p)=\mathrm{NegBin}(1,p)$. Conceptually the negative binomial is a sum of waits: the trial of the $r$-th success is the wait to success $1$, plus the wait from there to success $2$, and so on — $r$ independent geometric waits laid end to end. Two cautions keep the formulas straight. First, mind the support: geometric values begin at $1$, negative binomial values at $r$. Second, mind the convention: here $X$ is the trial number of the $r$-th success, so the exponent on $1-p$ is $k-r$ (the number of failures). Some books and software instead let the variable count only the failures before the $r$-th success (values $0,1,2,\dots$); the two differ by the shift $k\mapsto k-r$, and mixing them is the most common source of error.