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For a finite stopping time $n$ with target $k\ge 1$, necessarily $n\ge k$. The last toss must be a head and the preceding $n-1$ tosses must contain $k-1$ heads. The negative binomial stopping argument therefore gives likelihood $\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)p^k(1-p{)}^{n-k}$ for the biased coin, and $\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)2^{-n}$ for the fair coin. Applying Bayes theorem again yieldsThe two experiments have different combinatorial coefficients, but each coefficient is independent of the coin parameter. Thus posterior equality for proportional likelihoods explains their identical posterior formula. If $p=0$, a finite positive-head stopping observation has zero likelihood under the biased coin, consistently giving posterior zero. If the target is $k=0$, the experiment stops at $n=0$ without a toss and the posterior remains $1/2$; a positive stopping time would then be impossible.

For $X=x$, the last flip must be the $r$th head. Among the preceding $x+r-1$ flips, exactly $x$ are tails and $r-1$ are heads. There are $\left(\genfrac{}{}{0ex}{}{x+r-1}{x}\right)$ choices for their positions, and every resulting sequence has probability $(1-p{)}^x{p}^r$. The negative binomial stopping argument therefore givesso $X$ has the failures-before-the-$r$th-success negative binomial distribution.

To put the mass function into exponential family form, setSince $p=1-e^{\theta }$,whereThus the natural parameter of an exponential family is $\theta$, and the Fisher-Neyman factorization theorem shows that $T(X)=X$ is a sufficient statistic. This is the negative binomial exponential family.

The exponential-family derivative identities now giveand