Returning to this, I now think perhaps you were speaking of drawing balls from the jar without placing them back (thus, no risk of re-drawing the same ball). But, if so, this would still be a flawed analysis in pretty much the same way. I think what has happened here is perhaps the common mistake of conflating a conditional probability with the converse conditional probability. It’s true that the probability of the first 99 draws being black balls, given that there is precisely one non-black ball, is 1/100. However, it does not follow that the probability of there being precisely one non-black ball, given that the first 99 draws are black balls, is also 1/100, or in either direction bound by such. P(a|b) and P(b|a) are quite separate quantities, related only through Bayes’ Theorem [P(a|b) = P(b|a) * P(a)/P(b)]. They may diverge wildly in any which way.
As before (though I made a slip-up expressing it before), the ratio between the a priori probability of the jar containing 99 blacks and that of it containing 100 blacks can run the full gamut from 0:1 to 1:0, and correspondingly will allow the conditional probability of “All 100 balls in the jar are black, given that the first 99 drawn were” to run the full gamut from 0 to 1.
To illustrate this with one of the simplest, most natural probability distributions to consider, let us suppose our jar is initially filled by the following process: a fair coin is flipped 100 times, with a black or white ball being placed in the jar each time accordingly. We then give our jar a good vigorous shaking and start drawing balls from it at uniform random without replacement. Suppose we’ve drawn 99 balls from it and they’ve all been black. What is the probability that the sole remaining ball is black as well (i.e., that the jar was filled entirely with black balls)?
Answer: 1/2. Not shabby, but not particularly high either. Far short of 99%.
Proof: The a priori probability of the jar containing all blacks (with us then necessarily first drawing 99 blacks) is 1/2^100. The a priori probability of the jar containing precisely 99 blacks but with us happening to first draw those 99 blacks (that is, drawing its sole white ball last) is 100/2^100 * 1/100 [the first factor is because there are 100 coin flip sequences which would result in precisely 99 black balls]. Since these two are equal, the answer follows.