A two-level Markov chain The diagram below illustrates a chain with two levels: Level One that has states $1,2,...$ and Level Two that includes $1',2',...$ and State $0$ that connects these two levels. Note that once the system is in Level Two, it cannot visit any of Level One states, unless it visits $0$ first. 
If the system is currently in State $1$, I am interested in the expected number of visits to State $1$ before visiting $0$. In other words, I am interested in the expected number of visits to $1$ when it starts from $1$ and stays within red states. 

I can think of two solution methods, but I cannot understand why they lead to different answers:
I will use the following notations: Event $R$ is returning to State $1$ again before visiting $0$; event $A$ is an arrival; event $D$ is a departure, and $U$ is moving up from $i$ to $i'$. Also, define the random variable $Z$ to be the number of visits to State $1$ before visiting $0$.
Method 1:
the system is currently in State $1$, so I put $1$ and then I condition on the next step:
$E[Z] = 1 + E[Z|A] P(A) + E[Z|D] P(D) + E[Z|U] P(U) = 1+E[Z|A] P(A)$ (note that $E[Z|D] = E[Z|U] = 0$ because once $U$ or $D$ occurs the system leaves red states).
When $D$ occurs, the system goes to State $2$. The expected number of visits to $2$ before visiting $1$ would be equal to $E[Z]$, and each time that the system visits $2$, it goes to $1$ with probability $P(D)$, so $E[Z|A] = E[Z] P(D)$. Therefore:
$E[Z] = 1 + P(A)P(D)E[Z]$, which gives me $E[Z] = \frac{1}{1-P(A)P(D)}$. 
Method 2:
The system is currently in State $1$, so it visits $1$ before visiting $0$ exactly once with probability $1-P(R)$. The system visits State $1$ exactly twice if it comes back to $1$, with probability $P(R)$, and then never comes back to $1$ before visiting $0$, with probability $1-P(R)$, and so on. That is: 
$E[Z] = 1 (1-P(R))+ 2 P(R)(1-P(R)) + 3 P(R^2)(1-P(R)) + \cdots \\
   \  \ = (1-P(R)) \sum_{i=1}^\infty i P(R)^{i-1}=\frac{1}{1-P(R)}$.
To calculate $P(R)$, I use first-step analysis:
$P(R) = P(R|A) P(A) + P(R|D) P(D) + P(R|U) P(U) = P(R|A) P(A)$. (Note that $P(R|D) + P(R|U)=0$ because once $U$ or $D$ occurs the system leaves red states). When $A$ occurs and the system goes to $2$, the probability of visiting $2$ again before visiting $1$ would be the same as $P(R)$. And each time that the system visits $2$, it goes down to $1$ with $P(D)$. So:
$P(R) = P(R|A) P(A) = P(A)(P(D) + P(D)P(R)^1 + P(D)P(R)^2 + \dots)
\ \ = P(A) P(D) (1+P(R)^1 +P(R)^2+ \dots) = \frac{ P(A) P(D)}{1-P(R)}$. I then obtain $P(R)$ by solving the quadratic equation.
If I replace $P(R)$ in the $E[Z] = \frac{1}{1-P(R)}$ equation, it is obvious that the result is different from that of Method 1. Can anyone help me understand why?
 A: Your method 2 is correct, method 1 is incorrect. 
Another way to solve it
Define
\begin{align}
P[R] &= \mbox{probability we return to 1 before going to a black state (given we start in 1)} \\
q &= \mbox{probability that we eventually get to 1 before going to black (given we start in 2)}\\
P[GoRight] &= \frac{\lambda}{\lambda + \mu + \gamma} \\
P[GoLeft] &= \frac{\mu}{\lambda + \mu + \gamma}
\end{align}
Then we get
\begin{align}
P[R] &= P[GoRight] q \quad (Eq. 1) \\
q &= P[GoLeft] + P[GoRight]q^2 \quad (Eq. 2) 
\end{align}
Multiplying (Eq. 2) by $P[GoRight]$ and using (Eq. 1) reduces to: 
$$ \boxed{P[R] = P[GoLeft]P[GoRight] + P[R]^2}$$
which is the same as your method 2 equation if we interpret $P[A]=P[GoRight]$ and $P[D] = P[GoLeft]$. 
Fixing method 1
You have a typo/mistake when you write "1+" but later ignore the 1.  There is also no reason to multiply $E[Z]$ and $P[D]$.  A correct approach of this type is: 
$$ Z = 1 + Remaining$$
$$ E[Z] = 1 + E[Remaining] = 1 + \underbrace{E[Remaining|GoRight]}_{qE[Z]}P[GoRight]$$
where $q$ is defined as above, and so 
$$ E[Z] = \frac{1}{1-qP[GoRight]} $$
Of course here we still need to compute $q$ somehow, such as from (Eq. 2). Of course, from (Eq. 1) we also know $qP[GoRight] = P[R]$. 
