# 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?

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One can get an expression for the mean length of the game, as a very unattractive but explicit infinite sum. The expression is not simple. Perhaps it simplifies. – André Nicolas Mar 17 '14 at 23:32
Can you tell me how you got this sum perhaps? – Millardo Peacecraft Mar 18 '14 at 3:40
Let $X_i=1$ if the sum up to the $i$-th set of $3$ tosses is \le k$, and the two previous sums are each less than$k$. Let$X_i=0$otherwise. [Comment: for$i=1$,$2$the definition of$X_i$is marginally different. And$3$is not imprtant.] Then the length of the game is$\sum_1^\infty X_i$, so by linearity of expectation we want$\sum_1^\infty E(X_i)$. It is not hard to find an expression for the probability that$X_i=1$. – André Nicolas Mar 18 '14 at 3:48 Maybe you should post it as an answer. It cuts off after a point. – Millardo Peacecraft Mar 18 '14 at 3:53 I messed up, don't like comments. Here goes again, and I can fix an imprecision of the bad TeX. Define random variable$X_i$as follows. The definition is slightly different for$i$odd and$i$even. For$i$odd, let$X_i=1$if the number of successes in odd trials up to and including the$i$-th is$\le k+2$, and the number of odd successes by the$i-2$-th trial is$\lt k$, and the number of even successes up to the$i-1$-th is$\lt $k$. efine $X_i$ for even $i$ similarly. The the length of the game is $\sum X_i$.(more) – André Nicolas Mar 18 '14 at 3:59

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.

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I've been thinking about it, and what if I use the CDF of the negative binomial distribution which is the regularized incomplete gamma function and calculate $E[x]=\int k\cdot (1-P_r[X<k])$? – Millardo Peacecraft Mar 18 '14 at 19:06