Little's formula seems to contradict the PASTA property

Question

$\newcommand{\E} {\mathbb E}$I'm working on the following queueing problem. Group of customers arrive according to Poisson Process with rate $3$ per hour. With probability $2/3$ only one customer needs to be served and with probability $1/3$ two customers need to be served. Customers within the group get served in a random order and they leave individually. The service time is exponentially distributed with parameter $10$.

Let $L$ be the number of customers in the system in the steady state. One can easily find $\E [L] =\frac 5 6$. Now I want to find the mean sojourn time of an arbitrary customer $\mathbb E [S] $. To that end I need to find the mean number of customers per hour, that is: $$\hat \lambda = \frac 2 3 \cdot 3 + 2\cdot \frac 1 3 \cdot 3 = 4$$ So by Little, one has $$\E [S] =\frac 1{\hat\lambda}\E[L]= \frac 5{24}$$

Okay that is nice. But I wanted to also try it using the PASTA property. I say, $$\E[S] =\E[L] \frac 1{10} + \frac 1{10}+ \mathbb {P} (\text{ someone else is with me who is served first}) \frac 1{10}$$ It is clear, I need to wait for the ones who are already in the system, that is the first term. The second term is my own service time. The last term is also clear because I may be within a group. I have said $$\mathbb {P} (\text{ someone else is with me who is served first})=\mathbb {P} (\text{ I am with someone else }) \frac 1 2 = \frac 1 3 \cdot \frac 1 2 = \frac 1 6$$ The factor $1/2$ because the other one or I get served in a random order. If I do this, then I get $\E[S] = \frac 1{5}$ which is contradicting Little's Law.

It seems that my second approach using PASTA would work if I set $$\mathbb {P} (\text{ someone else is with me who is served first}) \stackrel{?} {=} \frac 1 4$$

Question. How is my second approach contradicting Little's Law. If my approach using PASTA is good, how is the last probability mentioned equal to $1/4$ instead of $1/6$ which is more intuitive?

Answer 1

For illustration, consider a typical ensemble of three groups of customers. This is two groups of individual customers and a group of two customers. If you now sample from these four customers, the probability that your sampled customer is from the group of $2$ is $1/2$, not $1/3$. This corrects the discrepancy in exactly the manner that you noted (the probability someone else is with you and is served first becomes $1/4$).

(Of course the argument does not really depend on the probabilities of the associated groups being rational, it just makes the picture more concrete.)