I found this question while going through a book:
In a casino game Ruben rolls a die and whenever a one or a six is rolled he receives a token. The game ends when Ruben has received y tokens; he then receives $x, where x is the number of rolls he has made.
(a) The probability of the game ending on the sixth roll is 40/729. Find the value of y.
According to the negative binomial distribution representation, we need to calculate $y$ in $X$ follows $NB(y, 1/3)$. Here is the formula for negative binomial distribution for this quesiton:
$$ \begin{pmatrix} 6-1 \\ y-1 \\ \end{pmatrix} (\frac{1}{3})^y(1-\frac{1}{3})^{6-y} = \frac{40}{729} $$ The first paranthesis above is for the combinations equation. How would one go about solving from this point, or is this the right path to approach this question. How should I find the value of $y$?