TITLE: Cat / mouse probability question
QUESTION [9 upvotes]: There exist 7 doors numbered in order from 1 to 7 (going from left to right).  A mouse is initially placed at center door 4.  The mouse can only move 1 door at a time to either adjacent door and does so, but is twice as likely to move to a lower numbered door than to a higher numbered door each time it moves 1 door.  There are cats waiting at doors 1 and 7 that will eat the mouse immediately after the mouse moves to either of those 2 doors.
So for example, the mouse starts at door 4.  He could then move to door 3, then to door 2, then back to 3, then back to 2, then to door 1 where he gets eaten.  That counts as 5 moves total.  Skipping doors is not allowed. 
So there are 2 questions I have regarding this:
1) What is the expected average number of moves before the mouse gets eaten?  (do not count the initial start at door 4 as a move but count any final move to doors 1 or 7 and any "intermediate" moves between those 2 states).
2) What is the probability that the mouse will survive for 100 or more moves?

REPLY [10 votes]: I'll renumber the doors $0$ to $6$ to simplify the calculations.
For the first question, we can set up a recurrence for the expected number $a_n$ of moves the mouse will make starting at door $n$:
$$
a_n=1+(1-p)a_{n-1}+pa_{n+1}\;,
$$
with $p=\frac13$. A particular solution of the inhomogeneous equation is $a_n=3n$, and the homogeneous equation can be solved using the ansatz $a_n=\lambda^n$, leading to the charateristic equation
$$
p\lambda^2-\lambda+1-p=0
$$
with solutions $\lambda=1$ and $\lambda=\frac1p-1=2$. Altogether,
$$
a_n=c_1+c_22^n+3n\;.
$$
The boundary values are $a_0=a_6=0$, which yields
\begin{eqnarray*}
c_1+c_2&=&0\;,\\
c_1+64c_2+18&=&0\;,
\end{eqnarray*}
with solution $c_1=-c_2=\frac27$. Thus the life expectancy in the middle is
$$
a_3=\frac27-\frac27\cdot2^3+3\cdot3=7\;.
$$
For the second question, you can group the $100$ steps into $50$ pairs, reducing the process to doors $1$, $3$ and $5$. The transition matrix for each pair of steps is
$$
\frac19\pmatrix{2&4&0\\1&4&4\\0&1&2}\;,
$$
which conveniently happens to have a nice eigensystem. The initial state decomposes as
$$
\pmatrix{0\\1\\0}=\frac16\left(\pmatrix{4\\4\\1}-\pmatrix{4\\-2\\1}\right)\;,
$$
where the first vector is an eigenvector with eigenvalue $\frac23$ and the second vector is an eigenvector with eigenvalue $0$. Thus after $50$ pairs of steps the distribution on doors $1$, $3$ and $5$ is
$$
\frac16\left(\frac23\right)^{50}\pmatrix{4\\4\\1}\;,
$$
and the sum of these probabilities is
$$
\left(\frac23\right)^{49}\approx2.4\cdot10^{-9}\;,
$$
so your guess had the right order of magnitude.
We can also use this eigenanalysis to derive the life expectancy another way. After the first pair of steps, the distribution is
$$
\frac19\pmatrix{4\\4\\1}\;.
$$
From then on, the mouse gets eaten with probability $\frac13$ in each pair of steps, so the expected number of pairs after the first one is $3$. Since death occurs on the first step of a pair, that translates to $7$ steps.