Negative Binomial with $4$ white faces before $3$ black faces

Question

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.

No, 4 white rolls total occurs before getting 3 black rolls total. — Remy, Oct 03 '17 at 22:33
Actually, the idea is that out of the first 6 rolls, at least 4 should be white (because otherwise you have at least 3 black, and less than 4 white). So think about the meaning of your $n$ and $k$ — Nick Pavlov, Oct 03 '17 at 22:37
But don't you stop after you get 4 whites before 3 black? For example, WBWWWW isn't in the event space? — Remy, Oct 03 '17 at 22:40
it doesn't hurt to continue, what happens after will not affect your odds, because WBWWWB is also in the accepted winning combos — Nick Pavlov, Oct 03 '17 at 22:41
They give the same result, so I guess yours is also good; basically you are adding up the probabilities that the 4th white will be exactly on the 4th, 5th,or 6th roll — Nick Pavlov, Oct 03 '17 at 22:48
Gotcha so on the 5th it would be $4\choose{3}$$(2/3)^3$$1/3$? — Remy, Oct 03 '17 at 22:58
Because you'd have to get 3 whites in the first 4 rolls and then a white on the 5th. — Remy, Oct 03 '17 at 22:58

Nick Pavlov · Accepted Answer · 2017-10-03T23:04:33.320

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

Standard binomial makes much more sense to me than negative binomial. — Remy, Oct 03 '17 at 23:02

Negative Binomial with $4$ white faces before $3$ black faces

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