text: # A $M/M/\infty$ queue with fixed probability going back to the end of the queue
  
  Suppose for a $$M/M/\infty$$ queue with arrival rate $$\lambda$$ and service rate $$\mu$$, with probability $$p$$ the service is successful and the person will leave the queue, and otherwise she will go back to the end of the queue. I am wondering if it is equivalent to another $$M/M/\infty$$ queue? If this is the case, what is this queue's arrival and service rates and how to prove it?
  
  Let $$S$$ be the (total) service time of a customer and $$N$$ the number of times a customer is served. Then $$S\mid N=n\sim\mathsf{Erlang}(n,\mu)$$ and $$N\sim\mathsf{Geo}(p)$$, so by the law of total probability \begin{align} f_S(t) &= \sum_{n=1}^\infty f_{S\mid N=n} (t)\mathbb P(N=n)\\ &= \sum_{n=1}^\infty \frac{\mu(\mu t)^{n-1}}{(n-1)!} e^{-\mu t}(1-p)^{n-1}p\\ &= pe^{-\mu t}\sum_{n=0}^\infty \frac{((1-p)(\mu t))^n}{n!}\\ &= p\mu e^{-p\mu t}, \end{align} and hence $$S\sim\mathsf{Expo}(p\mu)$$.
  The resulting detailed balance equations $$\pi_n = \frac{\lambda}{np\mu}\pi_{n-1}$$, $$n\geqslant 1$$ differ from those of a regular $$M/M/\infty$$ queue only by the factor of $$\frac 1p$$ on the right-hand side and accordingly the stationary distribution has Poisson distribution with mean $$\frac\lambda{p\mu}$$.