Negative Binomial with $4$ white faces before $3$ black faces Suppose that a fair $6$-sided die having $2$ black faces and $4$ white faces will be rolled repeatedly. What is the probability that $4$ rolls resulting in a white face occur before $3$ rolls resulting in a black face?
Attemped Solution:
I'm trying to make use of the following negative binomial formula:
$n$ trials, given $k$ success: ${n-1}\choose{k-1}$$p^k$$(1-p)^{n-k}$
In our case, $n$ can be $4,5$, or $6$ and $k$ is fixed at $4$.
$3\choose{3}$$(\frac{2}{3})$$^4$(${1}\over{3}$)$^0$+$4\choose3$(${2}\over{3}$)$^4$(${1}\over{3}$)+$5\choose3$(${2}\over{3}$)$^4$(${1}\over{3}$)$^2$ = $.680$
Is this a valid solution? I would also be interested in alternative solutions.
 A: It is correct.
What your solution is doing is adding up the probabilities that the 4th white roll will be exactly the 4th, 5th, or 6th roll, respectively. For the $k$-th white roll to be the $n$-th overall, in the previous $n-1$ you must have $k-1$ whites exactly, that's why you are doing binomial with $n-1$ and $k-1$, but then you have to multiply by an extra $2/3$ for the actual $n$-th roll being white, and you get your formula. The possibilities for $n$ are 4, 5, and 6, because the 7th roll would be too late, already failed.
You can also proceed like this: to have 4 whites before 3 blacks is the same as to have at least 4 whites among the first 6 rolls: if it happens, you have 4 whites before (at most) 2 blacks, and if it doesn't, then you have at least 3 blacks and no more than 3 whites in these first 6 rolls, so you have failed.
So it's just standard binomial with $n=6$ and $4 \leq k \leq 6$:
$$\sum_{k=4}^{6} \binom{6}{k} \left(\frac{2}{3}\right)^k \left(\frac{1}{3}\right)^{6-k} = \frac{31\cdot 2^4}{3^6} $$
