About the turning point in a sequence of random variables. Let $X_1,X_2,..$ be a sequence of random variables drawn i.i.d. Let $N$ be the random variable which given a sequence outputs the smallest $n$ such that $X_{n} \geq X_{n-1}$. What is $\mathbb{E}[N]$? 

On solution I was reading says that we can calculate $P[N \geq k] = P[ X_1 \geq X_2 \geq..\geq X_{k-1} ] =  \frac{1}{(k-1)!}$ because apparently one can equivalently think of being given $k-1$ numbers and this is the probability of them being arranged in decreasing order. 
I don't understand this argument at all. Even the first equality above is not clear to me! Also $N$'s domain is the space of all (infinite) sequences of numbers. This above calculation seems to totally miss the fact that the sequence itself was chosen randomly. 
One closeby case that I can think of is $P[N = k] = \frac{1}{(k-1)!}$ being true for a given set of numbers when the randomization is only over their arrangement. But here there is a randomization over the sequences it self. 

Because the $X_i$ are continuous random variables, the sequence $(X_1,\dots,X_{k-1})$ consists of distinct real numbers almost surely, and all $(k-1)!$ orderings of these real numbers are equally likely because the $X_i$ are i.i.d (so the joint law is symmetric). Since only one of the orderings gives us the outcome $N\geq k$, it follows that $\mathbb{P}(N\geq k)=\frac{1}{(k-1)!}$.
Here is an example when $k=3$ in the situation when the $X_i$ have pdf $f$: We have
$$ \mathbb{P}(X_1\geq X_2)=\int_{-\infty}^{\infty}\int_{-\infty}^xf_{X_1}(x)f_{X_2}(y)\;dydx=\int_{-\infty}^{\infty}\int_{-\infty}^xf(x)f(y)\;dydx $$
and on the other hand
$$\mathbb{P}(X_2\geq X_1)=\int_{-\infty}^{\infty}\int_{-\infty}^yf_{X_1}(x)f_{X_2}(y)\;dxdy=\int_{-\infty}^{\infty}\int_{-\infty}^yf(x)f(y)\;dxdy $$
These two integrals have the same value (since the only thing that's different is the variables of integration) so the two probabilities are the same. The argument for larger $k$ is essentially the same.
