Approximate the expected number of jobs in a year. A person is employed for one day at a time. When he is out of work, he visits the job
agency in the morning to see if there is work for that day. There is a job for her with
probability $\frac{1}{2}$. If there is no work, he comes back the next day. When he has a job,
he will be called back to the same job for the next day with probability $\frac{2}{3}$. When he
is not called back, he goes to the job agency again the next morning to look for a new
job. Approximate the average number of jobs the person works in a year.
My approach :
Let us denote $p_n$ as the probability that he has a job on day $n$. By the law of total probability,
$P[$Job on Day $n] = P[$ Job on Day $n \ \cap$ same job as Day $n-1$$] + P[$ Job on Day $n \ \cap$ different or no job as Day $n-1$$]$.
This implies the recursion :
$p_n = \frac{2}{3} p_{n-1} + \frac{1}{2}(1-p_{n-1})$.
Solving this recursion and I guess introducing the total number of jobs as a sum of indicators will give my expected number of jobs in a year.
But I feel a bit weird about this recursion and I think it is not correct. Can anyone have a different approach to this?
 A: I'd look at this as a Markov Chain with two states $J$ (job) and $N$ (no job).  Draw the transition probabilities.  Then assume at steady state there's a $p_J$ and $p_N$.  Set up your equations $(p_J = \dfrac 23 p_J + \dfrac 12 p_N$ and $p_N = \dfrac 13 p_J + \dfrac12 p_N)$.  Solving those I get $p_J = \dfrac 35$, so 60% of the time she has a job on any given day (in steady state).
A: Note that the problem asks for the number of jobs, not the number of days the person has a job. If the person holds the same job for 5 days, that only counts as one job.
We can model this as a Markov chain with three states: new job, same job, and unemployed. Then the expected number of jobs in a 365-day year is $P[\text{same job on day 1}] + \sum_{i=1}^{365} P[\text{new job on day $i$}]$ (by linearity of expectation).
If the person is not called back, the problem is ambiguous as to whether he goes to the job agency (a) the morning after the last day with the previous job, or (b) the morning after the first day without the previous job. The two interpretations lead to different Markov chains.
Interpretation (a). The transitions are

*

*new job → same job with probability $\frac23$, new job with probability $\frac16$, unemployed with probability $\frac16$;

*same job → same job with probability $\frac23$, new job with probability $\frac16$, unemployed with probability $\frac16$;

*unemployed → new job with probability $\frac12$, unemployed with probability $\frac12$.

The steady-state probabilities satisfy the system
$$n = \frac16 n + \frac16 s + \frac12 u, \quad
s = \frac23 n + \frac23 s, \quad
u = \frac16 n + \frac16 s + \frac12 u, \quad
n + s + u = 1,$$
whose solution is $n = \frac14, s = \frac12, u = \frac14$, leading to an expected $\frac12 + 365 · \frac14 = \mathbf{91\frac34}$ jobs in a year.
Interpretation (b). The transitions are

*

*new job → same job with probability $\frac23$, unemployed with probability $\frac13$,

*same job → same job with probability $\frac23$, unemployed with probability $\frac13$,

*unemployed → new job with probability $\frac12$, unemployed with probability $\frac12$.

The steady-state probabilities satisfy the system
$$n = \frac12u, \quad
s = \frac23n + \frac23s, \quad
u = \frac13n + \frac13s + \frac12u, \quad
n + s + u = 1,$$
whose solution is $n = \frac15, s = \frac25, u = \frac25$, leading to an expected $\frac25 + 365 · \frac15 = \mathbf{73\frac25}$ jobs in a year.
