Probability of taking white balls until a black shows up. (No replacement) I have an urn with $n$ white balls, and 1 black ball.
What's the probability of taking $x-1 \in \{0,n\}$ white balls, and the last extraction being a black ball? (No replacement)
My reasoning:
$$P(ww...b)=\frac{n}{n+1}\frac{n-1}{n}\cdots\frac{n-(x-2)}{n-(x-3)}\frac{1}{n-(x-2)}=\frac{1}{n+1}$$
However, when I try to think in terms of combinations:
I reason $$\frac{C^n_{x-1}C^1_{1}}{C^{n+1}_{x}}=\frac{x}{n+1}$$
Where is my flaw in the combinatorics reasoning?
 A: In order to select the black ball on the $k$th draw, we must first select $k - 1$ of the $n$ white balls while selecting $k - 1$ of the $n + 1$ balls, then select the only black ball from the remaining $n + 1 - (k - 1) = n + 2 - k$ balls.  Hence, the probability of selecting the black ball on the $k$th draw is
\begin{align*}
\frac{\dbinom{n}{k - 1}}{\dbinom{n + 1}{k - 1}} \cdot \frac{1}{n + 2 - k} & = \frac{\dfrac{n!}{(k - 1)![n - (k - 1)]!}}{\dfrac{(n + 1)!}{(k - 1)![n + 1 - (k - 1)]!}} \cdot \frac{1}{n + 2 - k}\\
& = \frac{\dfrac{n!}{(k - 1)!(n + 1 - k)!}}{\dfrac{(n + 1)!}{(k - 1)!(n + 2 - k)!}} \cdot \frac{1}{n + 2 - k}\\
& = \frac{n!}{(k - 1)!(n + 1 - k)!} \cdot \frac{(k - 1)!(n + 2 - k)(n +  1 - k)!}{(n + 1)n!} \cdot \frac{1}{n + 2 - k}\\
& = \frac{1}{n + 1}
\end{align*}
which agrees with your first answer.
However, there is a simpler way to see this.  There are $n + 1$ balls.  Since the lone black ball is equally likely to be any position in the sequence, the probability that it is in the $k$th position is
$$\frac{1}{n + 1}$$
for $1 \leq k \leq n + 1$.
