Does statistician Piggy wait longer when she is in a hurry?

Question

The statistician Piggy has to wait an amount of time $T_0$ at the post office on an occasion when she is in a great hurry. In order to investigate whether or not chance makes her wait particularly long when she is in a hurry, she checks how many visits she makes to the post office until she has to wait longer than the first time. Formally, let $T_1, T_2, . . . $be the successive waiting times and $N$ be the number of times until some $T_k > T_0$, that is, $${N = k} = {T_j ≤ T_0, 1 ≤ j < k, T_k > T_0}$$ What is the distribution of $N$ under the assumption that $(T_n, n ≥ 0)$ are i.i.d. continuous random variables? What can be said about $E(N)$ ?

This question confuses me. Does it have all information needed to solve it? Nothing is said about the distribution of $T_i$. The answer in my book is : $$P(N=n)=\frac{1}{n(n+1)}$$ which does not even depend on $T_0$, which as i understand it, is a given time, not a random variable. Where is the trick?

Answer 1

We are told the $T_i$ come from a continuous uniform distribution though in fact all that matters is that the distribution is continuous so there are no ties. To have $n$ be the first time the waiting time exceeds $T_0$ we must have $T_n$ be the greatest waiting time and $T_0$ be the second greatest. If $T_0$ is not the second greatest we have exceeded it before. If we draw $n+1$ random numbers, the chance the last one is greatest is $\frac 1{n+1}$. Given that it is, the chance the first is second largest is $\frac 1n$, so the chance that $n$ is the first time we exceed $T_0$ is $\frac 1{n(n+1)}$

Answer 2

Ross has already given a nice answer; here's another way to look at it:

The probability that $T_0$ is the longest of the $n$ times $T_0$ through $T_{n-1}$ is $\frac1n$. Thus

$$ \mathsf P(N\ge n)=\frac1n\;, $$

so

\begin{eqnarray*} \mathsf P(N=n) &=& \mathsf P(N\ge n)-\mathsf P(N\ge n+1) \\ &=& \frac1n-\frac1{n+1} \\ &=& \frac1{n(n+1)}\;. \end{eqnarray*}