Probability or expected value of hirings the company will make? This is the full question:

A company is hiring for an open position and has n interviews set up, one
per day. Each day, if the candidate is better than the current employee, the employee
is fired and the candidate is hired. Otherwise, the current employee keeps the job.
What is the expected number of hirings the company will make?

I got the answer of $\frac{1+(n-1)}{2}$. Is this correct? I used the probability of no new hires and all new hires.
 A: The key question is whether the chance on Day-3 of a new hire is $(1/3)$ or $(1/2)$.  $(1/3)$ makes the most sense, in the absence of a specific premise to the contrary, since the Day-3 hire has to be better than both the Day-1 and Day-2 interviewees to get the job.
Based on this, and based on the premise that the Day-1 hire is automatically hired, given that the position starts out open, the enumeration is
$$1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}.$$
The alternative interpretation, which seems to be the assumption made by the OP (i.e. original poster) is that the expected number of hires is
$$1 + \frac{1}{2} + \frac{1}{2} + \cdots + \frac{1}{2},$$
where the above enumeration has exactly $n$ terms.
A: Take k potential employees. If one of them comes to the interview on day k, they will be hired if they are the best of the employees, otherwise the best of the j employees has already been hired and keeps their job. So the probability that someone is hired on day k is 1/k.
The expected number of people hired within n days is the sum of 1/k over 1<=k<=n.
In practice, someone who does the job will improve doing it. Hire the tenth best candidate, and after some time they will become better than the nine candidates that were initially better. On the other hand, whoever you hire will eventually move to another job or retire, and you start all over.
And in civilised countries your plan doesn’t work at all because you can’t just fire people like that.
