A very challenging probability question In a certain 2-player game, the winner is determined by rolling a single 6-sided die in turn, until a
6 is shown, at which point the game ends immediately. Now, suppose that k dice are now rolled simultaneously by each player on his turn, and the
ﬁrst player to obtain a total of k (or more) 6’s, accumulated over all his throws, wins the
game. (For example, if k = 3, then player 1 will throw 3 dice, and keep track of any 6’s that
show up. If player 1 did not get all 6’s then player 2 will do the same. Assuming that player
1 gets another turn, he will again throw 3 dice, and any 6’s that show up will be added to
his previous total). Compute the expected number of turns that will be needed to complete the game.
I've analysed this problem as follows: The problem can be modeled by a negative binomial distribution with probability $p=\frac{1}{6}$. Now, X is a random variable representing the  number of dice being thrown. I need to find the cdf $Pr[X\geq k]$, and then find the expectation as follows $E[X] = \int_0^\infty kPr[X\geq k]$. The problem here is that the cdf of a negative binomal distribution is a regularized beta function and this is quite messy to deal with. I'm wondering  is there another way to approach this problem that wouldn't involve that? 
 A: It might be easier to approach this problem with generating functions.
When you roll a die, you have 1/6 chance to roll a 6, and a 5/6 chance to roll any other face. Let a represent rolling a 6, and let b represent rolling any other face. 
Rolling a single die, the outcomes may be represented as $({a\over 6} + {5b\over 6})$ 
For k dice, we have $p(a)=({a\over 6} + {5b\over 6})^k$
Each monomial of the expansion will be of the form $a^{r_1}b^{r_2}$, such that $r_1+r_2=k$. The coefficient gives the probability of obtaining exactly $r_1$ 6's and $r_2$ not-6's. Since we don't earn any points when we don't roll a 6, we can ignore these outcomes by setting $b=1$.
$p(a)=({a\over 6} + {5\over 6})^k$
We can obtain the expected number of 6's per turn by first differentiating with respect to a, and then evaluating it at a=1. If we want to score w points to win, we divide by the expected points per turn to obtain the expected number of personal turns a player should expect before winning.
If you meant how many cumulative turns we should expect to complete the game: if each player expects $t$ turns to win, and one player goes first, presumably the game should take $2t-1$ turns total to complete. Alternately, manipulate the generating function to account for expected cumulative totals.
