A simple game of snakes and ladders is played on a board of nine squares. At each turn a player tosses a fair coin and advances one or two places according to whether the coin lands heads or tails. If you land at the foot of a ladder you climb to the top, but if you land at the head of a snake you slide down the tail. how many turns on averages does it take to complete the game? What is the probability that a player who reaches the middle square will complete the game without slipping back to square 1?
This is a very famous Markov chain question. The original problem and the its image is here. it's the problem 1.3.3 in Markov Chains.
We can represent the chain with the following table $$\begin{array}{ccc} \text{Start Square} & \text{If Head go to} & \text{If Tail Go to} \\ 1&2&5\\ 2&5&4 \\ 3&4&5\\ 4&5&1\\ 5&1&7\\ 6&7&4\\ 7&4&9\\ 8&9&9\\ 9&9&9 \end{array}$$
There are some general methods, but at that stage in the book I think you're supposed to work it out manually.
Let $p_i$ be the probability of reaching square nine before square one given that I'm on square $i$.
I need to come up with an equation involving $p_5$ that I can solve. From square five I can reach only squares four five and seven before I reach square one or nine.
Looking at the table we may write $$\begin{array}{rcl} p_4 &=& \tfrac 12 p_5 \\ p_5 &=& \tfrac 12 p_7 \\ p_7 &=& \tfrac 12 p_4 + \tfrac 12 \end{array}$$
It remains to solve this system of equations, which you shouldn't have much trouble with.