You all know the Markov umbrella puzzle about professor having umbrellas at home and at office and taking one whenever it rains. But how about if we have more than two places? Of course then the question arises how do they choose to which place to go. Let's assume they choose uniformly at random among the places they aren't currently in.
More concretely, let there be $n$ places and $u$ umbrellas distributed among these places (It doesn't matters how they start out). Let the probability of rain be $p$. A person moves from one place to another (one step is one move) by choosing the next place uniformly at random from the other $n-1$ places. If it rains and there is an umbrella available at their current place, they take it with them to the next place. If it doesn't rain, they don't take an umbrella with them (i.e., they leave the umbrella they possibly brought from the previous place). If there isn't an umbrella at their current place, they have no option but to go without and in case of rain they get wet. On average in the long run, how often do they get wet?
By brute force calculating small cases, I conjecture the formula
$$\frac{p(1-p)}{\frac{n+u-1}{n-1}-p} = \frac{p(1-p)}{\frac{u}{n-1}+ 1-p} $$
This seems to indicate that the problem is equivalent to the two place problem with $\frac{u}{n-1}$ umbrellas (if we could have fractional umbrellas). Obviously all the places are symmetric in the limit distribution but can we say that when the place we're leaving from is empty, the umbrellas are divided equally among the remaining places and it doesn't matter to which place we go, so we can group the other places into just one place with $\frac{u}{n-1}$ umbrellas. This is of course very vague speaking.