6. 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.
Answer: $\frac{5}{11}$ For $1 \leq k \leq 6$, let $x_{k}$ be the probability that the current player, say $A$, will win when 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 $\frac{5}{11}$.