Cat / mouse probability question 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?

I'll take a shot at this........
1) The eventual fateful outcome occurs when the total number of moves, lower versus higher, differs by 3.
So, a way to look at this as an $E(x) = n\cdot p$ type problem is to sum the $(n\cdot p)$s for all possible outcomes.
This will be:
$3(\frac{1}{3})+5(\frac{2}{9})+7(\frac{4}{27})+9(\frac{8}{81})+ .....\text{etc}$ which is an infinite arethmetico-geometric series whose infinite sum is: $$S = \frac{dg_2}{(1-r)^2} + \frac{a}{1-r} = \frac{2\cdot (\frac{1}{3}\cdot \frac{2}{3})}{\frac{1}{3}^2} + \frac{3\cdot \frac{1}{3}}{\frac{1}{3}} = 2\cdot 2 + 3 = 7$$
2) $$P(n\ge 100) = 1 - P(n<100)$$
$$P(n<100) = S_n = \frac{1}{3}+\frac{2}{9}+\frac{4}{27}+ .......+\frac{2^{n-1}}{3^n}$$ 
This turns out to be a geometric series, where $a_1 = \frac{1}{3}, r = \frac{2}{3}$ and $n = 49$ (odd from $3$ to $99$) a different n from the n moves.
Example calculation for $3$rd term is: $9(\frac{2}{3})^5(\frac{1}{3})^2 + 9(\frac{1}{3})^5(\frac{2}{3})^2 = \frac{4}{27}$
$$P(n\ge 100) = 1 - \frac{\frac{1}{3}(1-(\frac{2}{3})^{49})}{1-(\frac{2}{3})}$$
$$P(n\ge 100) = 1 - .9999999976 = 2.4\cdot 10^{-9}$$
