Cat and mouse question 
I got the following recurrence relations, letting $t_i$ be the expected number of mouse moves to get to room 6 from room $i$:
$$
t_6 = 0\\
t_5 = \frac{1}{2} + \frac{1}{2}(1+t_2)\\
t_4 = 1 + t_1\\
t_3 = 1 + t_2\\
t_2 = \frac{1}{3}(1 + t_5) + \frac{1}{3}(1 + t_1) + \frac{1}{3}(1 + t_3)\\
t_1 = \frac{1}{2}(1 + t_2) + \frac{1}{2}(1 + t_4)
$$
which I solved to find $t_1 = 19$. Does this seem right and does anyone have a quicker way to solve these kinds of questions?
 A: There is indeed another way to solve this – whether it’s quicker than yours depends on how quickly you can solve systems of linear equations. This one doesn’t require any equations, only symmetry considerations.
The only fact we need is that when the mouse is in a room, before going to some particular adjacent room, by symmetry it will on average go to all other adjacent rooms once. (Another perspective onto this is that the mean number of trials to reach one out of $n$ equiprobable adjacent rooms is $n$.)
If the mouse is in room $1$, it will on average make one visit to $4$ before going to $2$. That visit takes $2$ steps, so on average it will take $3$ steps to go from $1$ to $2$.
At $2$, it will on average make one visit each to $1$ and $3$ before going to $5$. The visit to $3$ costs $2$ steps, and as we found, the visit to $1$ costs $4$ steps ($1$ step there and $3$ to get back). So it will on average take $2+4+1=7$ steps to go from $2$ to $5$.
In $5$, the mouse will on average make one visit to $2$ before going to $6$. The visit to $2$ costs $8$ steps ($1$ there and $7$ to get back), so on average it takes $9$ steps to go from $5$ to $6$.
Adding up $3$ steps from $1$ to $2$, $7$ steps from $2$ to $5$ and $9$ steps from $5$ to $6$, we find that on average the mouse takes $3+7+9=19$ steps to go from $1$ to $6$, in agreement with your result.
