Consider the situation where we have observed *s* successes and know the probability of success *p*, but would like to know how many trials were actually done to have observed those successes. We wish to estimate a value that is fixed, so we require a distribution that represents our *uncertainty* about what the true value is. There are two possible situations: we either know that the trials stopped on the *s*^{th} success or we do not.

If we know that the trials stopped on the *s*^{th} success, we can model our uncertainty about the true value of *n* as:

*n* = NegBinomial(p,*s*)

If, on the other hand, we do not know that the last trial was a success (though it could have been), then our uncertainty about *n* is modeled as:

*n* = NegBinomial(p, *s*+1) - 1

Both of these formulae result from a Bayesian analysis with Uniform priors for *n*.

Let *x* be the number of trials that were needed to obtain the *s*^{th} success. We will use a uniform prior for *x*, i.e. *p(x)* = *c*, and, from the binomial distribution, the likelihood function is the probability that at the (*x*-1)^{th} trial there had been (*s*-1) successes *and* then the *x*^{th} trial was a success, which is just the Negative Binomial probability mass function:

\( l(X \mid x)=\left( \begin{array}{c} x-1 \\ x-s \end{array} \right) p^{s}(1-p)^{x-s} \) |

Since we are using a uniform prior (assuming no prior knowledge), and the equation for *l(X|x*) comes directly from a distribution (so it must sum to unity) we can dispense with the formality of normalizing the posterior distribution to one, and observe:

\( p(x)=\left( \begin{array}{c} x-1 \\ x-s \end{array} \right) p^{s}(1-p)^{x-s} \) |

i.e. that *x* = NegBinomial(p,*s*).

In the second case, we do *not* know that the last trial was a success, only that in however many trials were completed, there were just *s* successes. We have the same Uniform prior for the number of trials, but our likelihood function is just the binomial probability mass function, i.e.:

\( l(X \mid x)=\left( \begin{array}{c} x \\ s \end{array} \right) p^{s}(1-p)^{x-s} \) |

Since this does not have the form of a probability mass function of a known distribution, we need to complete the Bayesian analysis, so:

f \big(x \mid X \big)= \frac{\left( \begin{array}{c} x \\ s \end{array} \right) p^{s}\big(1-p\big)^{x-s}}{\displaystyle\sum_{i=s}^{\infty}\left( \begin{array}{c} i \\ s \end{array} \right) p^{s}(1-p)^{i-s}} |

Look at the denominator, and substituting *j = i+1* gives:

\displaystyle\sum_{i=s}^{\infty}\left( \begin{array}{c} i \\ s \end{array} \right) p^{s}(1-p)^{i-s} = \displaystyle\sum_{j=s+1}^{\infty}\left( \begin{array}{c} j-1 \\ s \end{array} \right) p^{s}(1-p)^{j-s-1}=\bigg( \frac{1}{p}\bigg).\displaystyle\sum_{j=s+1}^{\infty}\left( \begin{array}{c} j-1 \\ s \end{array} \right) p^{s+1}(1-p)^{j-s-1}=\bigg(\frac{1}{p}\bigg) |

since

\left( \begin{array}{c} j-1 \\ s \end{array} \right) p^{s+1}(1-p)^{j-s-1} |

is the probability mass function for the NegBinomial(s+1,p) distribution for j and therefore sums to 1. The posterior distribution then reduces to:

f(x \mid X)=\left( \begin{array}{c} x \\ s \end{array} \right) p^{s+1}(1-p)^{x-s} |

For x = y - 1:

f(y \mid X)=\left( \begin{array}{c} y-1 \\ s \end{array} \right) p^{s+1}(1-p)^{y-s-1} |

i.e. y = NegBinomial(p,s+1), and therefore x = NegBinomial(p,*s*+1) -1 distribution.