> 
> 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?