5. Nathaniel and Obediah play a game in which they take turns rolling a fair six-sided die and keep a running tally of the sum of the results of all rolls made. A player wins if, after he rolls, the number on the running tally is a multiple of 7 . Play continues until either player wins, or else indefinitely. If Nathaniel goes first, determine the probability that he ends up winning.
the number on the tally at the beginning of his turn is $k$ modulo 7 . The probability that the total is $l$ modulo 7 after his roll is $\frac{1}{6}$ for each $l \not \equiv k(\bmod 7)$; in particular, there is a $\frac{1}{6}$ chance he wins immediately. The chance that $A$ will win if he leaves $l$ on the board after his turn is $1-x_{l}$. Hence for $1 \leq k \leq 6$
$$
x_{k}=\frac{1}{6} \sum_{1 \leq l \leq 6, l \neq k}\left(1-x_{l}\right)+\frac{1}{6} .
$$

Letting $s=\sum_{l=1}^{6} x_{l}$, this becomes $x_{k}=\frac{x_{k}-s}{6}+1$ or $\frac{5 x_{k}}{6}=-\frac{s}{6}+1$. Hence $x_{1}=\cdots=x_{6}$, and $6 x_{k}=s$ for every $k$. Plugging this in gives $\frac{11 x_{k}}{6}=1$, or $x_{k}=\frac{6}{11}$.
Since Nathaniel cannot win on his first turn, he leaves Obediah with a number not divisible by 7 . Hence Obediah's chance of winning is $\frac{6}{11}$ and Nathaniel's chance of winning is $\frac{5}{11}$.