Given a 5% employee turnover per year: What is the average tenure time? Is it possible, based on a given turnover rate per year (in %), to calculate the average time (in years) an employee stays in an organization?
I have looked at the Coupon Collector's problem:
https://en.wikipedia.org/wiki/Coupon_collector's_problem
but so far haven't been able to figure this out.
Turnover could basically be calculated like this:
Turnover Calculation
 A: I don't think this has anything to do with the coupon collector problem.
Say the company has $100$ employees. After $1$ year they have employed $100$ people for a total of $100$ person-years. After $2$ years they have employed $105$ people for a total of $200$ person years. After $n$ years they have employed $100+5n$ people for a total of $100n$ person-years. So what is the average number of years worked, if $n$ is very large?
A: This strikes me as application of Little's Law
$$L = \lambda W$$
where $L$ is the long-term average of the number of customers in a queuing system, $\lambda$ is the arrival rate, and $W$ is the average time a customer spends in the system.  Assuming the company size is stable, the turnover is the arrival rate of employees divided by the number of employees, i.e. turnover is
$$\tau = \lambda / L$$
so by Little's Law,
$$W = L / \lambda =1/ \tau$$
Put in words, the average time in the company is the reciprocal of the turnover rate.
I don't see any connection with the Coupon Collector's Problem.
A: Trying to apply basics of probability here, not sure whether I got everything right.
When


*

*The events are $\Omega = \{L_1, L_2, L_3, ..\}$, where $L_1$ is the employee leaving after the first year, $L_2$ is the employee leaving after the second year, and so on;

*$f = 0.05$ for a turnover rate of 5%.


Then the probability for leaving after the $i$th year is the probability for staying $i-1$ years and leaving in the $i$th year:
$P(X=L_i) = (1-f)^{i-1}*f$
The expected value is then 20 years for 5%, which would be inline with the answer for Little's law above.
Interestingly, when I draw a sample based on the probability distribution above, and make a boxplot, I get a median of ~13.
