A Flea moves around the vertices of a triangle in the following manner: Whenever it is at vertex i it moves to its clockwise neighbor vertex with probability $p_i$ and to the counterclockwise neighbor with probability $q_i=1-p_i$, $i = 1,2,3.$ Let $p_1=p_2=\frac{1}{2}, p_3=\frac{2}{3}$
$(a)$ Find the proportion of time that the flea is at each of the vertices.
$(b)$ How often does the flea make a counterclockwise move that is then followed by five consecutive clockwise moves?
My solution:
For $(a)$:
Let the states $1,2,3$ be the three vertices of a triangle. We then have the following transition probability matrix:
\begin{bmatrix}
0 & \frac{1}{2} & \frac{1}{2} \\[0.3em]
\frac{1}{2} & 0 & \frac{1}{2} \\[0.3em]
\frac{2}{3} & \frac{1}{3} & 0
\end{bmatrix}
Then by solving the linear equations(omitted) for the limiting probabilities, I got $\pi_1=\frac{10}{27},\pi_2=\frac{8}{27},\pi_3=\frac{1}{3}.$
For $(b)$:
Here's the place I am not so sure of. And I did:
Let $cc$ be the event that the flea moves counterclockwise and $c$ be the event that the flea moves clockwise.
Then I computed the long-term probabilities for moving clockwise and counterclockwise as follows:
\begin{align*}
P(cc)
&=P(cc|\text{at vertex 1})P(\text{at 1})+P(cc|\text{at vertex 2})P(\text{at 2})+P(cc|\text{at vertex 3})P(\text{at 3})\\
&=\frac{1}{2}\frac{10}{27}+\frac{1}{2}\frac{8}{27}+\frac{1}{3}\frac{1}{3}\\
&=\frac{12}{27}
\end{align*}
\begin{align*} P(c) &=P(c|\text{at vertex 1})P(\text{at 1})+P(c|\text{at vertex 2})P(\text{at 2})+P(c|\text{at vertex 3})P(\text{at 3})\\ &=\frac{1}{2}\frac{10}{27}+\frac{1}{2}\frac{8}{27}+\frac{2}{3}\frac{1}{3}\\ &=\frac{15}{27} \end{align*}
And if we want the probability of a counterclockwise move followed by five consecutive clockwise moves, we simply multiply the probabilities: $\frac{12}{27}\left(\frac{15}{27}\right)^5$
Is this the correct approach? Thanks in advance!
You have the probability of a counter-clockwise (widdershins) turn :$$\mathbb P({\rm w})= \pi_1q_1+\pi_2 q_2+\pi_3q_3$$
And the probability of a clockwise turn: $$\mathbb P({\rm c})=\pi_1p_1+\pi_2 p_2+\pi_3 p_3$$
As I read it, the event of a counter-clockwise (widdershins) turn followed by five subsequent clockwise turns would likewise have a probability of:
$$\begin{align} \mathbb P({\rm wccccc}) & = \pi_1 q_1 p_3 p_1 p_2 p_3 p_1 + \pi_2 q_2 p_1 p_2 p_3 p_1 p_2 + \pi_3 q_3 p_2 p_3 p_1 p_2 p_3 \\[1ex] & = (\pi_1 q_1/p_2+\pi_2 q_2/p_3+\pi_3 q_3/p_1)(p_1 p_2 p_3)^2 \\ & = \left(\frac {10}{27}\frac{1}{2}\big/\frac{1}{2}+\frac{8}{27}\frac{1}{2}\big/\frac{2}{3}+\frac{1}{3}\frac{1}{3}\big/\frac{1}{2}\right)\left(\frac{1}{2}\frac{1}{2}\frac{2}{3}\right)^2 \\ & = \frac{11}{486} \end{align}$$
[Addendum]
Let's consider the probability of making two subsequent turns: widdershins then clockwise. By your method (1) and my method (2) that would be:
$$\begin{align} \mathsf P_1(wc) & = (\pi_1 q_1+\pi_2 q_2+\pi_3 q_3)(\pi_1 p_1+\pi_2 p_2+\pi_3 p_3) \\ & = \bigg(\frac {10}{27}\frac{1}{2}+\frac{8}{27}\frac{1}{2}+\frac{9}{27}\frac{1}{3}\bigg)\bigg(\frac {10}{27}\frac{1}{2}+\frac{8}{27}\frac{1}{2}+\frac{9}{27}\frac{2}{3}\bigg) \\ & = \frac{20}{81} \\[2ex] \mathsf P_2(wc) & = (\pi_1 q_1 p_3+\pi_2 q_2 p_1+\pi_3 q_3 p_2) \\ & = \bigg(\frac {10}{27}\frac{1}{2}\frac 2 3+\frac{8}{27}\frac{1}{2}\frac 1 2+\frac{9}{27}\frac{1}{3}\frac 1 2\bigg) \\ & = \frac{41}{162} \end{align}$$
The difference is that your method considers each subsequent step to begin from an unknown start and concatenates them together.
My method observes that each subsequent step depends on where the previous step ended with only the original step being from a completely unknown start.