Average Waiting Time in a Single Server Queue Consider a single server queue with exponential arrival rate $2$ when the queue is empty and $1$ otherwise, and exponential departure rate $3$. 
To find the invariant distribution, we solve $2\pi_0=3\pi_1$,$\pi_n=3\pi_{n+1} (n\geq 1)$ to get $\pi_0=\frac{1}{2}$, $\pi_n=\frac{1}{3^n} (n\geq 1)$. The average queue length is $L=\sum_{n=0}^\infty n\pi_n=\frac{3}{4}$.
The average time between arrivals is $\pi_0 /2+(1-\pi_0)/1=3/4$, so the average arrival rate is $\lambda:=\lim_{t\rightarrow \infty}\frac{\text{No. of Arrivals up to time } t}{t}=4/3$. By Little's Law, the average waiting time is $W=L/\lambda=9/16$.
On the other hand, a customer who arrives when there are $n$ other customers already in the queue (with probability $\pi_n$) has an expected waiting time of $(n+1)/3$. Thus, we have $W=\sum_{n=0}^\infty (n+1)\pi_n /3=(L+1)/3=7/12$, disagreeing with the above.
Where is the mistake?
 A: Good question.  There are two mistakes here. The first is an incorrect derivation of the rate $\lambda$. The second is because $\overline{W} \neq \sum_{n=0}^{\infty}(n+1)\pi_n/3$. See below for details. 
First, I agree with your computation that:
\begin{align}
&\pi_0=1/2 \\
&\pi_n=(1/3)^n \quad \forall n \in \{1, 2, 3, ...\}\\
&\overline{L} = \sum_{n=1}^{\infty}n\pi_n = 3/4
\end{align}
1) The total arrival rate $\lambda$ is actually: 
$$ \lambda = 2\pi_0 + 1(1-\pi_0) = 1 + \pi_0 = 3/2 $$
Hence, by Little's formula:  $\overline{W} = \frac{\overline{L}}{\lambda} = 1/2$. 
2) Your alternative derivation for $\overline{W}$ implicitly assumes that the fraction of jobs that "see" $n$ jobs in the queueing system when they arrive is equal to $\pi_n$.  If arrivals were Poisson, we could claim this by PASTA (Poisson Arrivals See Time Averages).  However, arrivals are not Poisson. 
Here is a correct alternative derivation:  Let's make the arrivals Poisson of rate $\lambda_{new}=2$ by adding "virtual arrivals" of rate 1 when the system is busy.  To make the new system equivalent to the old, suppose it works this way:   If an arrival occurs when the system is empty, then it is a "true" arrival and it begins its service in the queueing system.  If it occurs when the system is busy, with probability  1/2 it is a "true" arrival and it joins the queue.  With prob 1/2 it is a "virtual" arrival and immediately exits with total delay 0.  Now, PASTA holds for this system, and $\pi_n$ is the same as before for this system. So: 
$$ \overline{W}_{new} = \pi_0 (1/3) + \sum_{n=1}^{\infty} \pi_n\frac{n+1}{6} = \frac{3}{8} $$
where we have used the fact that the expected delay, given we arrive when there are $n\geq 1$ jobs, is $\frac{1}{2}0 + \frac{1}{2}\frac{n+1}{3}$.
However, $\overline{W}_{new}$ can be decomposed into a weighted sum of the delay of true arrivals and virtual arrivals.  The total rate due to true arrivals is $2-(1-\pi_0)= 3/2$ and the total rate due to virtual arrivals is $1-\pi_0=1/2$.  Thus: 
$$ \overline{W}_{new} = \frac{3/2}{2}\overline{W}_{true} + \frac{1/2}{2}0 $$
Since we already know $\overline{W}_{new}=3/8$, this gives $\overline{W}_{true} = 1/2$ which is consistent with part 1 above.
3) Why is your way of computing average time between arrivals incorrect? 
You give an expression $\pi_0/2 + (1-\pi_0)/1$.  This of course assumes a PASTA-like property that is not necessarily true.  But even with PASTA, how do you get this expression?  If you consider a job arriving to an empty system, the next inter-arrival time is not exponentially distributed with rate $1$, nor is it exponentially distributed with rate $2$.  
