Here's a good reference for exactly your problem: http://www.sosmath.com/matrix/markov/markov.html
Since landing on 4 or 8 moves the player immediately (i.e. without requiring another turn, Your transition matrix $P$ should actually be the black numbers below:
$$
\begin{matrix}
&
\color{red}0&
\color{red}1&
\color{red}2&
\color{red}3&
\color{red}5&
\color{red}6&
\color{red}7&
\color{red}9\\
\color{red}0&0 & .5 & .5 & 0 & 0 &0&0&0\\
\color{red}1&0&0&.5&.5&0&0&0&0 \\
\color{red}2&0&0&0&.5&0&0&.5&0\\
\color{red}3&0&0&0&0&.5&0&.5&0\\
\color{red}5&0&0&0&0&0&.5&.5&0\\
\color{red}6&0&0&.5&0&0&0&.5&0\\
\color{red}7&0&0&.5&0&0&0&0&.5\\
\color{red}9&0&0&0&0&0&0&0&1
\end{matrix}
$$
Spaces 4 and 8 are removed since it's impossible to move to and stay on them in the same turn. Then the initial state of the player on space 3 is defined as $X$:
$$
\begin{bmatrix}
0&0&0&1&0&0&0&0
\end{bmatrix}
$$
The formula $XP^n$ yields the probabilities of the player being at each state after $n$ moves. In your case, starting at space 3 and tossing the coin 4 times:
$$
\begin{bmatrix}
0&0&{1\over 8}&{1\over 8}&
{1 \over 16}& 0&{3 \over 16}&
{1 \over 2}
\end{bmatrix}
$$
The probability of you reaching space 9 within 4 turns is ${1 \over 2}$.
Let $\ R_{ni}\ $ be the number of times player $\ i\ $ gets chosen and tosses tails, and $\ R_{n\,i+3}\ $ the number of times he or she gets chosen and tosses heads. The random variables $\ R_{nj}\ $ are multinomially distributed with parameters $\ n\ $ and $\ \frac{1}{6}, \frac{1}{6},\dots, \frac{1}{6}\ $:
\begin{align}
P\big(R_{n1}=n_1,R_{n2}=n_2,\dots,&R_{n6}=n_6\big)\\
=&\cases{\displaystyle
\frac{n!}{\big(n_1!n_2!\dots n_6!\big)6^n}&if $\ \displaystyle n=\sum_{i=1}^6n_i$\\
\hspace{3em}0&otherwise, }
\end{align}
and
\begin{align}
y_n&=\sum_{j=1}^3\big(R_{nj}-R_{n\,j+3}\big)\\
x_{in}&= 4\big(R_{n\,i+3}-R_{ni}\big)+ \sum_{j=1}^3\big(R_{nj}-R_{n\,j+3}\big)\ .
\end{align}
Therefore,
\begin{align}
P\big(y_n>\max_{i\in\{1,2,3\}}x_{in}\big)&=P\big(R_{n1}>R_{n4}, R_{n2}>R_ {n5},\ R_{n3}>R_{n6}\big)\\
&=\frac{1}{8}P \big(R_{n1}\ne R_{n4}, R_{n2}
\ne R_ {n5},\ R_{n3}\ne R_{n6}\big)
\end{align}
by symmetry. Now
\begin{align}
P\big(R_{n1}\ne R_{n4},\ R_{n2}
&\ne R_ {n5},\ R_{n3}\ne R_{n6}\big)\\
=1-&3P\big(R_{n1}=R_{n4}, R_{n2}\ne R_ {n5},\ R_{n3}\ne R_{n6}\big)\\
-&3 P\big(R_{n1}=R_{n4}, R_{n2}= R_ {n5},\ R_{n3}\ne R_{n6}\big)\\
-& P\big(R_{n1}=R_{n4}, R_{n2}= R_ {n5},\ R_{n3}= R_{n6}\big)\\
=1-&3\big(P\big(R_{n1}=R_{n4}, R_{n3}\ne R_{n6}\big)\\
&\hspace{2em}-P\big(R_{n1}=R_{n4}, R_{n2}= R_ {n5},\ R_{n3}\ne R_{n6}\big)\big)\\
-& 3 P\big(R_{n1}=R_{n4}, R_{n2}= R_ {n5},\ R_{n3}\ne R_{n6}\big)\\
-& P\big(R_{n1}=R_{n4}, R_{n2}= R_ {n5},\ R_{n3}= R_{n6}\big)\\
=1-&3P\big(R_{n1}=R_{n4}\big)+3P \big(R_{n1}=R_{n4}, R_{n3}= R_{n6}\big)\\
-& P\big(R_{n1}=R_{n4}, R_{n2}= R_ {n5},\ R_{n3}= R_{n6}\big)\ ,\\
P\big(R_{n1}=R_{n4}\big)=&\sum_{j=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\frac{n!}{(j!)^2(n-2j)!}\left(\frac{1}{6}\right)^{2j}\left(\frac{2}{3}\right)^{n-2j}\ ,\\
P \big(R_{n1}=R_{n4},\ R_{n3}&= R_{n6}\big)\\
=\sum_{j=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor-j} &\frac{n!}{(j!)^2(k!)^2(n-2(j+k))!} \left(\frac{1}{6}\right)^{2(j+k)} \left(\frac{1}{3}\right)^{n-2(j+k)}\ ,\\
P\big(R_{n1}=R_{n4}, R_{n2}&= R_ {n5},\ R_{n3}= R_{n6}\big)\\
=&\cases{0&if $\ n\ $ is odd\\
\displaystyle\frac{1}{6^{2r}}\sum_{j=0}^r\sum_{k=0}^{r-j}\frac{(2r)!}{(j!)^2(k!)^2((r-j-k)!)^2}&if $ n=2r$}\ .
\end{align}
Putting all this together, we have
\begin{align}
g(n)= \frac{1}{8} &\Bigg(1-3 \sum_{j=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\frac{n!}{(j!)^2(n-2j)!}\left(\frac{1}{6}\right)^{2j}\left(\frac{2}{3}\right)^{n-2j}\\
+&3 \sum_{j=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor-j} \frac{n!}{(j!)^2(k!)^2(n-2(j+k))!} \left(\frac{1}{6}\right)^{2(j+k)} \left(\frac{1}{3}\right)^{n-2(j+k)}\Bigg)
\end{align}
if $\ n\ $ is odd, or
\begin{align}
g(2r)= \frac{1}{8} &\Bigg(1-3 \sum_{j=0}^r\frac{(2r)!}{(j!)^2(2(r-j))!}\left(\frac{1}{6}\right)^{2j}\left(\frac{2}{3}\right)^{2(r-j)}\\
+&3\sum_{j=0}^r\sum_{k=0}^{r-j} \frac{(2r)!}{(j!)^2(k!)^2(2(r-j-k))!} \left(\frac{1}{6}\right)^{2(j+k)} \left(\frac{1}{3}\right)^{2(r-j-k)}\\
-&\frac{1}{6^{2r}}\sum_{j=0}^r\sum_{k=0}^{r-j}\frac{(2r)!}{(j!)^2(k!)^2((r-j-k)!)^2}\Bigg)\ ,
\end{align}
if $\ n=2r\ $ is even.
The values $\ g(n)\ $ for $\ n=1,2,\dots,10\ $ are given in the following table
$$
\begin{array}{c|c|c|c|c|c|c|c|}
&1,2&3,4&5,6&7&8&9&10\\
\hline
\text{exact}&0&\frac{1}{36}&\frac{55}{1\,296}&\frac{2\,401}{46\,656}&\frac{7\,217}{139\,968}&\frac{97\,147}{1\,679\,616}&\frac{16\ 235}{279\,936}\\
\hline
\text{approx.}&&0.028&0.042&0.05146&0.05156&0.0578&0.05800\\
\hline
\end{array}
$$
Best Answer
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.