Markov Chain Exercise Help In working with a particular gene for fruit flies, geneticists classify an individual fruit fly as $\small \text{dominant, hybrid or recessive}$. In running an experiment, an individual fruit fly is crossed with a hybrid, then the offspring is crossed with a hybrid and so forth. The offspring in each generation are recorded as dominant,hybrid or recessive
(a) What is the probability the third generation offspring is dominant given the first generation offspring is recessive?
Transition Table is :-
$$\begin{bmatrix} .5 & .5 & 0    \\
                  .25 & .5 & .25 \\
                  0   & .5 &.5   \\
 \end{bmatrix}$$
where states 1, 2, and 3 are “dominant”, “hybrid” and “recessive” respectively. 
According to book , solution for (a) is to find $p(2)13$, means starting from state "dominant" and finishing at state "recessive", but I think it should be $p(2)31$  means starting from state "recessive" and finishing at state "dominant" ! 
In short the question is , what should be the initial state for this problem ? and why ? 
thanks!
here are the snap shots from the book 


 A: Answer edited to fix the multiplication order error noted by Michael.
If I understand your notation correctly, the answers you are considering are $P^2_{(1,3)}$ and $P^2_{(3,1)}$.  You are correct.  As we begin with the first generation offspring, a recessive fly, the initial state vector is 
$$v_1=\begin{bmatrix}  0,0,1
 \end{bmatrix}$$
Then we compute the state vector for two generations in the future as
$$ v_3=v_1P^2=\begin{bmatrix} P^2_{(3,1)},
                  P^2_{(3,2)},
                  P^2_{(3,3)} 
 \end{bmatrix}$$
Finally, as we are interested in the probability the third generation is dominant, we are interested in the first entry of $v_3$, which is $P^2_{(3,1)}$.
A: I hope you don't mind it if I dumb it down (mainly to make sure I get it)...
So we have 
$$P=\overset{\begin{matrix}\color{blue}{\text{Dom}}&\color{blue}{\text{Hyb}}&\color{blue}{\text{Rec}}\end{matrix}}{\begin{bmatrix}
&.5  & .5 & 0  & \\
&.25 & .5 & .25 & \\
&0   & .5 & .5 &
\end{bmatrix}}
$$
with every row spelling a PMF of the distribution among the three possible states after one step. For example, for the first row, if the distribution at zero time only included $\small\text{Dominant}$, after one step $.5$ of the population would remain $\small\text{Dominant}$, while $.5$ would transition to $\small\text{Hybrid}$, and none to $\small\text{Recessive}$.
After two steps,
$$\small P^2=\begin{bmatrix}
.5  & .5 & 0   \\
.25 & .5 & .25 \\
0   & .5 & .5
\end{bmatrix}
\begin{bmatrix}
.5  & .5 & 0   \\
.25 & .5 & .25 \\
0   & .5 & .5
\end{bmatrix}
=\begin{bmatrix}
.375  & .5 & .125   \\
.25 & .5 & .25 \\
.125   & .5 & .375
\end{bmatrix}$$
so that with an initial state of only $\small\text{Recessive}$, i.e. $v1=\begin{bmatrix}0&0&1\end{bmatrix}$, the distribution of interest would be found in the third row of $P^2$, and the probability of $\small\text{Dominant}$ in $P^2_{(\color{red}{3,1})}$:
$$v3 = v1\,P^2=v1=\begin{bmatrix}0&0&1\end{bmatrix}\begin{bmatrix}
.375  & .5 & .125   \\
.25 & .5 & .25 \\
\color{red}{.125}   & .5 & .375
\end{bmatrix}=\begin{bmatrix}\color{red}{.125}&.5&.375\end{bmatrix}$$
