TITLE: Probability/Combinatorics
QUESTION [6 upvotes]: The Question is:

An urn contains 6 black and an unknown number ($\le6$) of white balls. Three balls are drawn one by one without replacement and all of them are found to be white. What is the probability that the next ball drawn is black?

I propose the following solution:
We calculate the probability that there are r white balls in the urn using Bayes' Theorem. So this will be:
$$ f(r) = \frac{\frac{{r}\choose{3}}{{6+r}\choose{3}}}{\sum_{r=3}^6\frac{{r}\choose{3}}{{6+r}\choose{3}} }$$
where f(r) denotes the probability that their are r white balls.
Now, if their were r white balls, the probability that the next draw will be black is $g(r) = \frac{6}{6 + (r-3)} = \frac{6}{3+r}$, because 3 white balls have already been picked. Now to get the total probability that the next ball drawn is black can be given by: $$ \sum_{r=3}^6 f(r)\cdot g(r)$$
Is this approach correct? The answer obtained (I calculated using a python script) does match in decimal value with one of the options ($\frac{677}{909})$ which is equal to 0.744. If it is correct, I would like to know if there is a more elegant/less calculative way of approaching this question.
Thanks!

REPLY [2 votes]: Your answer of ${\large{\frac{677}{909}}}$ is correct, assuming as a prior distribution that $r$ is equally likely to be $0,1,2,3,4,5,6$.

Thus, if we assume, before drawing the $3$ balls, that for each $r\in\{0,1,2,3,4,5,6\}$, the probability that the urn contains $r$ white balls is ${\large{\frac{1}{7}}}$, then for $r\in\{3,4,5,6\}$, the function
$$
f(r)=\frac
{
{\displaystyle{\frac{\binom{r}{3}}{\binom{6+r}{3}}}}
}
{
\;\;
{\displaystyle{
\sum_{r=3}^6
\frac{\binom{r}{3}}{\binom{6+r}{3}}
}}
\;\;
}
$$
is the conditional probability that the urn initially contained $r$ white balls, given that the first $3$ balls drawn were white.

The rest of your analysis is correct.