a random variable, the law of total probability tells us that the expected probability of success in the next experiment is just the expected value of p. Since p is conditional on the observed data Xi for i = 1, ..., n, we have The same calculation can be performed with the (improper) prior that expresses total ignorance of p, including ignorance with regards to the question whether the experiment can succeed, or can fail. This improper prior is 1/(p(1 − p)) for 0 ≤ p ≤ 1 and 0 otherwise. If the calculation above is repeated with this prior, we get Thus, with the prior specifying total ignorance, the probability of success is governed by the observed frequency of success. However, the posterior distribution that led to this result is the Beta(s,n − s) distribution, which is not proper when s = n or s = 0 (i.e. the normalisation constant is infinite when s = 0 or s = n). This means that we cannot use this form of the posterior distribution to calculate the probability of the next observation succeeding when s = 0 or s = n. This puts the information contained in the rule of succession in greater light: it can be thought of as expressing the prior assumption that if sampling was continued indefinitely, we would eventually observe at least one success, and at least one failure in the sample. The prior expressing total ignorance does not assume this knowledge. To evaluate the "complete ignorance" case when s = 0 or s = n can be dealt with, we first go back to the hypergeometric distribution, denoted by . This is the approach taken in Jaynes (2003). The binomial can be derived as a limiting form, where in such a way that their ratio remains fixed. One can think of as the number of successes in the total population, of size . The equivalent prior to is , with a domain of . Working conditional to means that estimating is equivalent to estimating , and then dividing this estimate by . The posterior for can be given as: And it can be seen that, if s = n or s = 0, then one of the factorials in the numerator cancels exactly with one in the denominator. Taking the s = 0 case, we have: Adding in the normalising constant, which is always finite