Grasshopper sits in a square of a 5 X 4 rectangular board. Each minute the grasshopper jumps at random to a neighboring square, with any of the 2, 3 or 4 (2 at corners) (3 at edge) (4 at center). All equally likely to be chosen.
Grasshopper always remains on the board. After many jumps, calculate what is the long run average probability that they will end up on a corner square of board?
On
A distribution which had the probability of a particular edge being crossed in a particular direction being constant would clearly be stable. This would have the probability of being in a particular square being proportional to the number of crossable edges it has, and so the stable probability of being in the four corners would be $\frac{4 \times 2}{4\times 2 + 10\times 3+ 6\times 4} = \frac{4}{31}\approx 0.129$. The general solution for an $m\times n$ board (both at least two, to ensure four corners) would be $\frac{4}{2nm-n-m}$
The underlying Markov Chain is in fact periodic with period $2$, so we have to be slightly careful, but a chequerboard argument with a $5\times4$ board shows that two pairs of corners would have opposite colours, and so this means that the limiting combined probability of being in the four corners is also $\frac{4}{31}\approx 0.129$
If instead we had a board with an odd number of squares (e.g. $5 \times 3$), we might instead have to talk about the limiting average of two successive positions
Approach: there are six types of square after we consider the symmetry. The probability of landing in each square can be computed from the probability of being in each neighboring square. For example, a corner square will get visited $\frac 13$ of the time the grasshopper leaves one of the neighboring squares. This will give you six equations in six unknowns, but you will have one redundant combination. Add in the requirement that the sum of all the probabilities on the board is $1$ and you have enough information to solve the system.