Markov Chain with dice rolling board game Simple board game.
S---1---2---3---4---5---H
You start at S. Each turn, your roll a standard six-sided die, and move forward that number of spaces.  Your goal is to reach H. You can only get to H on an exact roll;  if you roll a number that would take you past H, you do not move.  For example, suppose you are sitting at 4.  If you roll a 1, you move to 5.  If you roll a 2, you move to H. If you roll anything else, you do not move.  Once you reach H, you remain there forever.
What is the transition matrix? 
 A: The transition matrix is straightfoward; enumerate the states as $S,1,2,3,4,5,H$, then we have
$$
P= \begin{pmatrix}
0 & 1/6 & 1/6 & 1/6 & 1/6 & 1/6 & 1/6\\
0 & 1/6 & 1/6 & 1/6 & 1/6 & 1/6 & 1/6\\
0 & 0 & 1/3 & 1/6 & 1/6 & 1/6 & 1/6\\
0 & 0 & 0 & 1/2 & 1/6 & 1/6 & 1/6\\
0 & 0 & 0 & 0 & 2/3 & 1/6 & 1/6\\
0 & 0 & 0 & 0 & 0 & 5/6 & 1/6\\
0 & 0 & 0 & 0 & 0 &0 & 1
\end{pmatrix}.
$$
Note that the states $S$, $1$, $2$, $3$, $4$, and $5$ are transient and $H$ is absorbing, so $P$ is in the canonical form 
$$
P = \begin{pmatrix}Q&R\\0&I\end{pmatrix}
$$
where $Q$ is the substochastic matrix corresponding to transitions between transient states, $R$ is the substochastic matrix corresponding to transitions from a transient state to the absorbing state, and $I$ the identity matrix. The probability of transitioning from a transient state $i$ to a transient state $j$ in exactly $k$ steps is simply $Q^k$; summing this for all $k$ yields the fundamental matrix $N=\sum_{k=0}^\infty Q^k$. Now, since $Q$ has norm strictly less than one, it can be shown that $\sum_{k=0}^\infty Q^k = (I-Q)^{-1}$ (recall the formula for a geometric series), and hence
$$
N = \left(
\begin{array}{cccccc}
 1 & \frac{1}{5} & \frac{3}{10} & \frac{1}{2} & 1 & 3 \\
 0 & \frac{6}{5} & \frac{3}{10} & \frac{1}{2} & 1 & 3 \\
 0 & 0 & \frac{3}{2} & \frac{1}{2} & 1 & 3 \\
 0 & 0 & 0 & 2 & 1 & 3 \\
 0 & 0 & 0 & 0 & 3 & 3 \\
 0 & 0 & 0 & 0 & 0 & 6 \\
\end{array}
\right).
$$
Here the $(i,j)$ entry of $N$ is the expected number of visits to state $j$, given that the chain started in state $i$. Moreover, the expected number of steps before being absorbed is given by
$$
N\cdot\mathbf 1 = \left(
\begin{array}{cccccc}
 1 & \frac{1}{5} & \frac{3}{10} & \frac{1}{2} & 1 & 3 \\
 0 & \frac{6}{5} & \frac{3}{10} & \frac{1}{2} & 1 & 3 \\
 0 & 0 & \frac{3}{2} & \frac{1}{2} & 1 & 3 \\
 0 & 0 & 0 & 2 & 1 & 3 \\
 0 & 0 & 0 & 0 & 3 & 3 \\
 0 & 0 & 0 & 0 & 0 & 6 \\
\end{array}
\right)\begin{pmatrix}1\\1\\1\\1\\1\\1\\1\end{pmatrix} = \begin{pmatrix}6\\6\\6\\6\\6\\6\end{pmatrix},
$$
which is the same $6$ regardless of the starting state. (An interesting result, in my opinion.)
