I was solving problems on Markov Chains found here. Problem 5, which talks about umbrellas, seems to be pretty famous as I found many variations on it in google and even multiple Stack Exchange questions including this.
However, everybody seems to take this assignment of states for granted. When I first thought of the problem, I thought as follows:
For any single umbrella, I could define a 2×2 matrix
$$\begin{matrix}
\begin{matrix}\;\;\;\;\;\;\;&&\textrm{home}&&\textrm{office}\end{matrix}\\
\begin{matrix}\textrm{home}\\\textrm{office}\end{matrix}\begin{pmatrix}1-p&&p\\p&&1-p\end{pmatrix}
\end{matrix}$$
since any umbrella could be transported from home to office (or the opposite way round) if it rains, and stays in the same place if it doesn't.
I computed the stationary distribution, which turns out to be $$\pi = \begin{pmatrix}0.5&&0.5\end{pmatrix}$$
which I think means that any umbrella is equally likely to be at home or in the office at any given time, regardless the starting place.
Okay, so returning to the question of the probability of getting wet, which I call $P_{req}$. Let $h$ be the probability that I am at home and going to office. The probability of getting wet is the probability of (going from home to office and it rains and all umbrellas are in the office) or (going home from office and it rains and all umbrellas are at home). Mathematically:
$$
P_{\textrm{req}} = h\times p \times 0.5^4 + (1-h)\times p \times 0.5^4 \\
= p \times 0.5^4
$$
Clearly, this is not true. And I feel that the problem is with the $h$ that I have assumed, but I cannot pinpoint the problem exactly.
Best Answer
Ok, so it suddenly hit me that an umbrella, by itself, is not independent of the others.
Meaning, I cannot so simply say that the probability of a transition from home to office is $p$, because the probability of the transition is the probability that [it rains and this particular umbrella is chosen out of all the umbrellas that are in the same place]. This is the reason why my transition matrix is wrong, and making it correct would require too much work compared to the known solution (I would have to account for the case an umbrella is by itself, or with one other, etc.).