Disclaimer: I have been thinking about this problem for 3 days and my previous thoughts, as well as the details of the problem, are in this question, so please take a look there to understand the context. Thanks.
Long story short, I had thought of a state assignment that turned out to be wrong because there is some dependency that is not being accounted for.
Now, I was solving the MIT OCW problem set on Markov Chains, and I noticed that the first problem is so similar to the umbrella problem. However, the state assignment used in the solution is different.
Applying the same logic to the umbrella problem, I will draw the Markov chain for a single place (assume it's home, there is no loss of generality since the other place will have the remaining umbrellas anyway). State $i$ means that this place has exactly $i$ umbrellas.
This should be correct since (1) half of the time I'm going from home to office, and the other half I'm going from office to home, (2) For state $0$, the probability that home stays at $0$ is that [it doesn't rain, or it rains but I'm going from home to office], so $p_{00} = 1-p + \frac{1}{2}p = 1-\frac{1}{2}p$, and so on.
Therefore, the steady state probabilities are each $\frac{1}{5}$, and the probability that I get wet is that [home is at state $0$ and I am going from home to office and it rains] or [home is at state 4 and I am going from office to home and it rains]. That is, the probability that I get wet should be $\frac{1}{5}\times \frac{1}{2} \times p + \frac{1}{5} \times \frac{1}{2} \times p = \frac{1}{5}p$.
This does not agree with the famous solution for the umbrella problem, which is $\frac{p(1-p)}{5-p}$.
Could somebody please explain to me whether the two problems are actually essentially different, or are there multiple solutions to this problem depending on the initial set of assumptions? And in the latter case, what assumptions have been made here to make the solutions so drastically different?
I think the problems are different. In the umbrella problem, you always switch between home and office. So you cannot go from the state of being at home and having $0$ umbrellas to being at home and having $1$ umbrella. By defining the states instead to be the number of umbrellas at your current location however, you can go from having $0$ umbrellas to having $4$ umbrellas (and this is one of the state transitions in the solution to the umbrella problem).