Is this an Expectation Problem? Or is it more complex? Following question is off a probability/statistics review sheet:
Every day a professor leaves their home in the morning and walks to their
office. Every evening they walk home. They take their umbrella with them only if it is raining. If it is raining and they do not have their umbrella with them (at their home or office), then they must walk in the rain.
Suppose that it rains with probability $\frac{1}{3}$ at the beginning of any given trip independently of all other trips. Show that $\frac{63}{16}$ $\approx$ 4 is the expected number of days until the professor must walk in the rain without their umbrella (either that morning or evening), supposing that initially they have their umbrella with them at home.
Here is a hint I was given: Let $μ$ be the expected number of days supposing they initially have their umbrella with them at home, and let $v$ be the expected number of days supposing that they do not. Explain why $$μ = \frac{2}{9} + \frac{5}{9}(1-μ) + \frac{2}{9}$$ and then, similarly, find an equation for $v$ in terms of $μ$. Use these equations to solve for $μ$.
My thoughts:
At first glance to me this looks like it could be done with the expectation formula, but given the details, I'm not sure how to structure a daily $\frac{1}{3}$ probability of it raining up until the professor doesn't have an umbrella on hand. Would you need to keep track of where the umbrella would be based on probability of it raining on trips on different days?
I'm guessing that since $v$ and $μ$ are each a calculation of amount of expected days, one with and one without the umbrella, maybe the sum of these expectations would total 1, since these are the only two states the professor could start in? I'm also guessing this relationship would be how we calculate $v$ in terms of $μ$.

Follow the hint.  In the starting case, with probability $1/3$ it rains, the professor takes the umbrella, and with probability $2/3$, it does not rain when it is time for the professor to return home.  So with probability $2/9$ the professor has not walked in the rain but the umbrella is at the office.
Similarly, with probability $1/9$, it has rained both to and from work and the umbrella has made a round trip.
With probability $2/9$, it did not rain on the way to work but it rained on the way back from work, meaning the professor got wet.
With probability $4/9$, it did not rain either way, and the professor is back home.
We can summarize this in a table for the round trip:
$$\begin{array}{ccccc}
\text{Umbrella} & \text{Rain} & \text{Got wet} & \text{Probability} \\
\hline
\text{Office} & \text{Yes, No} & \text{No} & 2/9 \\
\text{Home} & \text{Yes, Yes} & \text{No} & 1/9 \\
\text{Home} & \text{No, Yes} & \text{Yes} & 2/9 \\
\text{Home} & \text{No, No} & \text{No} & 4/9 \\
\end{array}$$
Therefore, with probability $5/9$, we have returned to the initial state (not wet, umbrella home), except a day has passed.  Thus the expected number of additional days until getting wet is still $\mu$.  With probability $2/9$, the professor got wet that day.  With probability $2/9$, the professor has survived a day but now the umbrella is at the office.  Since $v$ represents the expected number of days until getting wet when the professor is home but the umbrella is not, we summarize the expected number of days until getting wet is $$\mu = \frac{5}{9}(1 + \mu) + \frac{2}{9}(1) + \frac{2}{9}(1 + v).$$
Now for $v$, we suppose the professor begins the day at home but the umbrella is at the office.  Then with probability $1/3$, the professor must walk in the rain to work.  With probability $2/9$, the professor makes it to the office, and takes the umbrella home because it rains when it is time to leave.  With probability $4/9$, it does not rain at all and the professor survives a day but returns to the state where the umbrella is not at home.  So the expected number of days until getting wet in this case is...?  I have not given the formula so that you have a chance to do the rest.
