Likelihood of sum of dice roll is exactly 1 million A board game is set up such that there is a number line with squares numbered 0..1 million. You roll a standard 6 sided die and move forward the number of spaces that you roll. Eventually you will either land on, or pass the 1 millionth spot. What is the probability you land exactly on the 1 millionth spot?
In other words, what is the probability that after a number of rolls the sum is exactly 1 million

Let $S(\omega) \subset \mathbb{N}$ be the spaces that are landed on during trial $\omega$. That is, if $d_k(\omega) \in \{1,...,6\}$ are the die rolls, then $S(\omega) = \{ \sum_{k=1}^n d_k(\omega) \}_{n=1}^\infty $.
For $k \ge 1$, let $r_k(\omega) = \min \{ j-k | j \ge k, \ j \in S(\omega) \}$. Note that $r_k(\omega) = 0$ iff space $k$ was landed on.
Also, note that $r_k(\omega) \in \{0,...,5\}$, and that if $r_k(\omega) >0$, then $r_{k+1}(\omega) = r_k(\omega) +1$.
To simplify notation, for  $n \in \{0,...,5\}$ let $\pi_k(n) = p\{ \omega | r_k(\omega) = n \}$.
We have $\pi_1(n) = \frac{1}{6}$ (initial roll must land on one of $\{1,...6\}$ with probability $\frac{1}{6}$).
Then for $k >1$, we can compute $\pi_k(n) = \sum_{m=0}^{5} p \left(\{ \omega | r_k(\omega) = n \} | \{ \omega | r_{k-1}(\omega) = m \} \right) \pi_{k-1}(m)$.

If $m \in \{1,...,5\}$, then $p \left(\{ \omega | r_k(\omega) = n \} | \{ \omega | r_{k-1}(\omega) = m \} \right) = \begin{cases} 1, & n=m-1 \\ 0, & \text{otherwise}\end{cases}$.
If $m = 0$, then $p \left(\{ \omega | r_k(\omega) = n \} | \{ \omega | r_{k-1}(\omega) = 0 \} \right) = \frac{1}{6}$.
Letting $\pi_k = (\pi_k(0),...,\pi_k(5))^T$, we can write the above as $\pi_k = M \pi_{k-1}$, where
$M= \begin{bmatrix}
\frac{1}{6} & 1 & 0 & 0 & 0 & 0 \\
\frac{1}{6} & 0 & 1 & 0 & 0 & 0  \\
\frac{1}{6} & 0 & 0 & 1 & 0 & 0 \\
\frac{1}{6} & 0 & 0 & 0 & 1 & 0  \\
\frac{1}{6} & 0 & 0 & 0 & 0 & 1  \\
\frac{1}{6} & 0 & 0 & 0 & 0 & 0 \\
\end{bmatrix}$.
The solution is then given by $\pi_{1000000}(0) = e_1^T M^{999999} \pi_1$.
Solving $M \hat{\pi} = \hat{\pi}$, gives the solution $\hat{\pi} = c(1, \frac{5}{6}, \frac{4}{6}, \frac{3}{6}, 
\frac{2}{6}, \frac{1}{6})^T$. Choosing $c= \frac{7}{2}$ gives the stationary distribution $\hat{\pi} = \frac{2}{7}(1, \frac{5}{6}, \frac{4}{6}, \frac{3}{6}, 
\frac{2}{6}, \frac{1}{6})^T$, hence we get the limiting $\hat{\pi}(0) = \frac{2}{7}$, as Ross pointed out with much less work.
(Thanks to my friend Takis Konstantopolos for his 'random set' approach ($S(\omega)$ in this instance) and introducing the $r_k$ to simplify my life.) 
