Tough Probability Question: Visiting 4 Friends, Finding Probability Mass Function Question:
Someone lives at the point 0 in the diagram, and has friends who live at each A, B, C, and D. On a day, they randomly select a number from 1 to 4 to visit a specific friend. Once at this friend's house, they return home with probability $\frac{3}{5}$ or proceeds to one of the adjacent houses with probability $\frac{1}{5}$ (i.e., if they're at A, they have probability $\frac{1}{5}$ to go to B, and $\frac{1}{5}$ to go to D).
(a) If X is the number of times that the person in question visits a friend before returning home, find a formula for the probability mass function: P{X = i}, with i = 1, 2, ... .
(b) If Y is the number of straight line segments that the person in question traverses, find a formula for the pmf P{Y = i}, with i= 2, 3, ... . Note that this includes the line segments that are traversed going to/from 0.

My thoughts:
(a) I'm a little more confused on this part.
My first attempt started by breaking the probability like so: $$P(i)=(P(adjacent)P(i|adjacent)) + (P(non-adj.)P(i|non-adj.))$$ with adjacent meaning the first house is adjacent to the house in question, non-adjacent being the house in question, or the house opposite (i.e., if we're considering house B, those adjacent are A and C; non-adjacent are B and D). 
I know we're not really considering specific houses as such but this is how I'm trying to wrap my head around it.
From this, I get, $$\biggl(\frac 12\biggl(\frac 15\cdot\frac 35 + \frac 35 \cdot\biggl(\frac15\biggl)^3 + \cdots \biggl) + \biggl(\frac 12 \biggl( \frac 14 + \frac 34 \cdot \frac 35 \cdot \biggl( \frac 15 \biggl ) ^2 + \frac 34 \cdot \frac 35 \cdot \biggl( \frac 15 \biggl) ^4 + \cdots \biggl)$$
Which I simplified down to
$$\frac 18 + \sum_{i=1}^\infty \frac {9}{40} \cdot \biggl( \frac 15 \biggl ) ^i$$
This didn't work as we shouldn't be summing to infinity for a finite probability, and I know realize as I'm typing this that I should probably be splitting the non-adjacent into two cases: one where the house in question is the first friend's house, and one where the house in question is opposite the first friend's house (here I only had the case where the one in question is the first friend's house in the $\frac 14$ term).
My next attempt is (I hope), more along the right lines. I considered a geometric variable, where we consider a 'success' to be going home. So $p = \frac 35$, and I had: $$P(i)=(P(first)P(i|first))+(P(adj.)P(i|adj.))+(P(opp.)P(i|opp.))$$ which by definition of geometric variables became 
$$= \frac 14 (p\cdot(1-p)^{i-1}) + \frac 12 (p\cdot(1-p)^{i-1}) + \frac 14 (p\cdot(1-p)^{i-1})$$
My (main) concern with this is that the trials aren't independent, as the probability of the next trial depends on what position we are at in the current trial; the probability of moving to A if we are at C is 0, but the probability of moving to A if we are at B or D is $\frac 15$.
So would we be able to consider X as a geometric variable? If not, what type of variable would we be able to consider it as?
(b) I'm more confident in this part of the question so here's the pmf I found: $$P(i)=\biggl(\frac 35 \biggl) \biggl(\frac 25 \biggl)^{i-2}$$ where $\frac 35$ is the probability of returning home, and $\frac 25$ is the probability of not returning home (and thus traversing another line segment).
Many thanks in advance!! 
 A: I agree with your answer to part (b), and note that this is also the probability that the traveler makes a total of exactly $\ i-1\ $ visits before returning home.
To answer part (a), note that if the traveller makes exactly $\ 2n\ $ visits before returning home then exactly $\ n\ $ of those must be to A or C and exactly $\ n\ $ of them must be to  B or D.  On the other hand, if he makes a total of $\ 2n+1\ $ visits then exactly $\ n+1\ $ of those must be to one of the two pairs A and C or B and D and exactly $\ n\ $ to the other pair.  Thus, if $\ V_A\ $ and $\ V_C\ $ are the numbers of visits made to A and C, respectively, then
$$
P\left(V_A + V_C = n\right)=\frac{3}{5}\left(\frac{2}{5}\right)^{2n-1}\hspace{-0.7em}+\frac{1}{2}\cdot\frac{3}{5}\left(\frac{2}{5}\right)^{2n-2}\hspace{-0.7em}+\frac{1}{2}\cdot\frac{3}{5}\left(\frac{2}{5}\right)^{2n}\\
=\frac{147}{40}\left(\frac{2}{5}\right)^{2n}\ ,
$$
for $\ n\ge 1\ $, or $ P\left(V_A + V_C = 0\right)=\frac{3}{10}\ $.
A key observation now is that each visit to A and C is equally likely to be to A or C, and the house visited is independent of all previous visits. Thus, given $\ V_A + V_C = n\ $, $\ V_A\ $ will follow a binomial distribution with parameters $\ n\ $ and $\ p=\frac{1}{2}\ $:
$$
P\left(V_A=v\,\left\vert\,V_A + V_C = n\right.\right)=\frac{n\choose v}{2^n}\ .
$$
Therefore, for $\ v\ge 1 $,
\begin{align}
P\left(V_A = v\right)&=\sum_\limits{n=v}^\infty P\left(V_A=v\,\left\vert\,V_A + V_C = n\right.\right)P\left(V_A + V_C = n\right)\\
&=\frac{147}{40}\sum_\limits{n=v}^\infty {n\choose v}\left(\frac{2}{25}\right)^n\\
&= 735\cdot \frac{2^{v-3}}{23^{v+1}}\ ,
\end{align}
the final equation following from the identity
$$
\sum_\limits{n=v}^\infty{n\choose v}y^n=\frac{y^v}{(1-y)^{v+1}}
$$
for $\ \vert y\vert < 1\ $, and
$$
P\left(V_A = 0\right)=1-735\sum_{v=1}^\infty\frac{2^{v-3}}{23^{v+1}}=\frac{57}{92}\ .
$$
