Transition matrix of "Three machines, one repairman"

Question

This is an example given in Durrett's textbook. I can only understand half of it.

Suppose that an office has three machines that each break with probability $.1$ each day, but when there is at least one broken, then with probability $.5$ the repairman can fix one of them for use the next day. If we ignore the possibility of two machines breaking on the same day, then the number of working machines can be modeled as a birth and death chain with the following transition matrix:

\begin{matrix} i/(i+1) & 0 & 1 & 2 & 3\\ 0 & .5 & .5 & 0 & 0\\ 1 & .05 & .5 & .45 & 0\\ 2 & 0 & .1 & .5 & .4\\ 3 & 0 & 0 & .3 & .7 \end{matrix}

The number in the first column represents the number of working machines today; the number in the first row represents the number of working machines on the next day.

My problem is to understand the last two rows. In my calculation, $p(2,2)=p(\text{nothing is newly broken today and nothing is repaired today})+p(\text{one working machine is newly broken today and a broken machine is repaired})=(1-0.1)\times (1-0.1)\times (1-0.5)+(1-0.1)\times 0.1\times 0.5=0.45\neq 0.5$.

The same problem with other two nonzero entries in the third row. Also, for the entry $p(3,2)$, I am wondering why we do not need to times $.5$, which ensures this broken machine is not repaired immediately?

Thanks in advance!

Answer 1

First off, you flipped the matrix. In fact, each column represents the number of working machines $\textbf{tomorrow}$ and each row represents the number of working machines $\textbf{today}$

Let's start with $p(2,2)$. There are two ways we can end up at two working machines:

1. No new machine breaks, repairman fails
2. A machine breaks, repairman succeeds.

The first case, we see that the $p(\textrm{no machine break})=0.8$ and $p(\textrm{repairman fails})=0.5$, so $p(\textrm{case 1})=0.4$. The second case, we see that $p(\textrm{one machine break})=0.2$ and $p(\textrm{repairman succeeds})=0.5$, so $p(\textrm{case 2})=0.1$. So, the total probability of $p(2,2)=\color{red}{0.5}$

For $p(1,2)$, we can only reach this state if the sole working machine doesn't break, and the repairman succeeds. So, $p(1,2)=0.9\cdot0.5=\color{red}{0.45}$.

I think you can figure out the rest from here.