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.
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 $4 \leq k \leq 6$: $$\sum_{k=4}^{6} \binom{6}{k} \left(\frac{2}{3}\right)^k \left(\frac{1}{3}\right)^{6-k} = \frac{31\cdot 2^4}{3^6} $$