Is it possible to estimate $e$ based on $N$? 
Consider a sequence of random numbers $u_1,\dots,u_n$ obtained from a continuous distribution $F$. Let $N$ be the first one that is greater than its immediate predecessor. In othe words,
  $$N=\min_n\{n:n \ge 2,u_n >u_{n-1}\}$$
  Is it possible to estimate $e$ based on $N$.

I know some other way to estimate $e$. In this some way are discuss. But how to estimate $e$ based on upper defined $N$.  
 A: What a nice question. The answer is "Yes."
Let me explain:
If you choose two numbers, these can be ordered in two ways: $a, b$ or $b, a$. Because the numbers are from a continuous distribution (as opposed to discrete), they will never be equal to each other. One must be greater than the other. Arbitrarily, suppose $a>b$. Then the probability that $N=2$ is $\frac 1 2$ and $P(N>2)=\frac 1 2$.
If you choose three numbers, these can be ordered in $3!=6$ ways. Rather than listing them as $a, b, c$ I am going to list them as integers (for convenience only).
1, 2, 3 yields $N=2$
1, 3, 2 yields $N=2$
2, 3, 1 yields $N=2$
2, 1, 3 yields $N=3$
3, 1, 2 yields $N=3$
3, 2, 1 yields $N>3$
We have  $P(N=2) = \frac 1 2$, $P(N=3) = \frac 1 3$ and $P(N>3)=\frac 1 6$.
It becomes clear that $P(N>r)=\frac 1 {r!}$
Then $P(N=r)=P(N>r-1)-P(N>(r))=\frac 1 {(r-1)!} - \frac 1 {r!}$
So $P(N=r)=\frac r {r!} - \frac {1} {r!}= \frac{r-1} {r!}$
$E(N)=\sum_{N=2}^{N=\infty} r P(N=r)=\sum_{N=2}^{N=\infty} r\frac{r-1} {r!} = \sum_{N=2}^{N=\infty} \frac {r(r-1)}{r!} = \sum_{N=2}^{N=\infty} \frac {1}{(r-2)!}$
$E(N)=\sum_{N=0}^{N=\infty} \frac {1}{r!} = e$
Thus the strategy is to evaluate N a large number of times and calculate the mean. This will be an estimate of $e$. 
