## 35. Being Late for Work

Maurice goes to work either by his own car (and then due to traffic jams, he is late half of the time), or by subway (and then he is late only one time out of four). If on any given day Maurice arrives at work on time, he always uses the same mode of transport the next day as he did the previous day, but if he is late for work, he definitely changes his mode of transport the next day. Knowing all this, how likely is it that Maurice will be late for work when he goes there for the 467th time?
35. Let $p_{n}$ be the probability that Maurice goes to work by his car on the $n$-th day. Then the probability that he will take the metro on the $n$-th day is $1-p_{n}$.

Maurice will go by car on the $n$-th day if he went by car on the $(n-1)$-th day and did not get late (which happens once in two times) or if he went by metro on the $(n-1)$-th day and got late (which happens once in four times).

From this, we obtain the following recurrence formula:

$$
p_{n}=\frac{1}{2} p_{n-1}+\frac{1}{4}\left(1-p_{n-1}\right)=\frac{1}{4}\left(p_{n-1}+1\right)
$$

Now, substituting $p_{n-1}=1 / 4\left(p_{n-2}+1\right)$ into this formula, we express $p_{n}$ in terms of $p_{n-2}$, then similarly express $p_{n}$ in terms of $p_{n-3}$, and so on, until we express $p_{n}$ in terms of $p_{1}$:

$$
\begin{aligned}
p_{n}= & \frac{1}{4}\left(1+\frac{1}{4}+\frac{1}{4^{2}}+\ldots+\frac{1}{4^{3}}+\ldots+\frac{1}{4^{n-2}}\right)+\frac{p_{1}}{4^{n-1}}= \\
& =\frac{1}{4} \cdot \frac{\left(1-\frac{1}{4^{n-1}}\right)}{1-\frac{1}{4}}+\frac{p_{1}}{4^{n-1}}=\frac{1}{3}\left(1-\frac{1}{4^{n-1}}\right)+\frac{p_{1}}{4^{n-1}}
\end{aligned}
$$[^22]

118

If $n$ is very large, then $p_{n}$ is obviously very close to $1 / 3$. Therefore, it can be said that Maurice has one chance in three to be late for work, going there on the 467-th day.