expectation of $K$ dice game Suppose that k dice are rolled simultaneously by each player on his turn, and now 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.). What is the expect number of turns needed to complete this game? 
$\textbf{My attempt:}$ I assume that this would follow the negative binomial distribution and $E[X] = \sum xP_r[X\geq x]$. But I believe I'm supposed to get a pure probability and with this method I'm going to have some variables. Any Hints at to what I may be doing wrong?
 A: We give a sketch of an unsatisfactory answer. For concreteness we imagine that $3$ dice are being used, but modification for more dice is straightforward. 
For $i=1,2,3,4,\dots$ define random variable $X_i$ as follows. (Well, the definition doesn't quite work for $i=1$ and $i=2$, but it can be easily modified.)
If $i$ is odd, let $X_i=1$ if there is a total of $\le k+2$ successes, that is, $6$'s, in the odd-numbered tosses $\le i$, and there is a total of fewer than $k$ successes both in the odd-numbered tosses $\lt i$ and the even-numbered tosses $\lt i$. Let $X_i=0$ otherwise. 
If $i$ is even, define $X_i$ analogously, but reversing the roles of odd and even.
Then the total number of $3$-dice tosses is $\sum_1^\infty X_i$, and therefore by the linearity of expectation the expected number of $3$-dice tosses is $\sum_1^\infty E(X_i)$.
It is not hard to write down an expression for $\Pr(X_i)=1$ for odd $i$, and also for even $i$. The expression is obtained by combining some negative binomial probabilities. So we have expressions for $E(X_i)$. 
The quite messy sum may simplify.  
