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? 

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.)
