How can I find the distribution for the number of years until the first year's rainfall is exceeded for the first time? Let $X_1, X_2, \cdots$ be jointly continuous and independently distributed with marginal pdf $f(x)$, where each $X_i$ represents annual rainfall at a given location. How can I find the distribution for the number of years until the first year's rainfall $(X_1)$ is exceeded for the first time?
I assume this is a fairly straight forward problem, but I am failing to see what the most logical first step would be. How would you go about thinking of the first step?
 A: Let us assume that annual rainfall can be considered a continuous variable (which requires the probability of no rainfall in a year to be equal to zero.)  The following steps get us to our goal:


*

*The first year's rainfall has cumulative distribution function
$F(X_1)$, and, as is well-known, $F(X_1) \sim \text{Uniform}(0,1)$ (the Probability Integral Transform) 

*The probability that any given successive year's rainfall exceeds the
first year's rainfall is  $1-F(X_1)$, label it $p$.  (This is because the probability of any given successive year's rainfall being $\leq$ year 1's rainfall is $F(X_1)$, so the probability of exceeding it is just $1-F(X_1)$.) Since $F(X_1)
\sim \text{Uniform}(0,1)$, $p$ is too; $1 - $ a $\text{Uniform}(0,1)$ variate is also a  $\text{Uniform}(0,1)$ variate.

*The number of years ($k$) until the first exceedence of the first
year's rainfall, conditional upon $p$, has a Geometric distribution with probability parameter $p$: $p(k | p) = (1-p)^{k-1}p$.
To remove the conditioning upon $p$, we integrate the Geometric distribution with respect to the Uniform distribution on $p$, which "disappears" in the expression below because it is equal to $1$ everywhere:
$$p(k) =  \int_0^1(1-p)^{k-1}p\text{d}p$$
which is the $\text{Beta}(2,k)$ function.  Expanding this function leads to:
$$p(k) = {1 \over k(k+1)}, \, k \geq 1$$ 
A quick check in R that this (plausibly) does sum to $1$:
> sum(beta(2,1:100000))
[1] 0.99999

A: jbowman's answer is correct, though there is an alternative approach to the same answer
By symmetry and assuming a continuous distribution so no ties, the probability that $X_1$ is the largest of $\{X_1,X_2,\cdots X_n\}$ is $\frac1n$ and the probability that $X_1$ is the largest of $\{X_1,X_2,\cdots X_n, X_{n+1}\}$ is $\frac1{n+1}$ 
This means the probability $X_{n+1}$ is the first value to exceed $X_1$ is $\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n(n+1)}$ 
Slightly counter-intuitively,  this implies that the expected number of years until $X_1$ is exceeded is therefore infinite. 
