I have a job where the service time follows an exponential distribution with mean $1/\mu$. At the same time, there are new customers coming in with Poisson rate $\lambda$. As soon as they arrive, they start a job whose service time follows an exponential distribution with mean $1/\mu$. I would like to know the probability that I finish the job first (i.e., there are no incoming customers, if any, finishes their job before me).
Things I tried:
(1) Conditional on my own service time, then integrate over my service time. This looks very complicated.
(2) Assume $x_k$ is the probability that when $k$ people (other than me) are currently doing their job and I am still the first to finish. The original question asks for $x_0$.
I can write the system of equations according to the exponential race:
$$ x_n=\frac{\mu}{\mu+\lambda+n\mu}+\frac{n\mu}{\mu+\lambda+n\mu}x_{n-1}+\frac{\lambda}{\mu+\lambda+n\mu}x_{n+1} $$ where the first term stands for the case where I finish the first, the second term stands for the probability that a customer other than me finishes the first, and the third term stands for a new arrival happens first. Then I am not sure how to solve this.
Any hints will be extremely helpful. Thank you!
Edit: $$ x_n=\frac{\mu}{\mu+\lambda+n\mu}+\frac{\lambda}{\mu+\lambda+n\mu}x_{n+1} $$