TITLE: Selecting n real numbers between 0 and 1 at random, what is the probability they are in order, increasing?
QUESTION [1 upvotes]: If I select n distinct real numbers $[0-1]$ at random, there are $n!$ arrangements of those numbers. Only one of the arrangements is in order from least to greatest. Hence:
P(n distinct reals increasing order)=$\frac{1}{n!}$ likewise
P(n distinct reals increasing or decreasing order)=$\frac{2}{n!}$
Is this still true if the numbers are not required to be distinct? I suspect yes, but why?
I want to say that getting the same real numbers more than once has probability of effectively 0, but this feels like hand waving and also potentially incorrect.

REPLY [1 votes]: In order to talk about probabilities in this situation, you need to have a measure; the obvious one here is Lebesgue measure on $[0,1]^n$.
For that measure, the subset of $[0,1]^n$ having any two numbers equal has measure 0. (This is not hard to prove with even a modicum of measure theory.) So your intuition is correct, and your "hand-waving" argument is easily turned into a formal proof with the right tools.
Historical note: before Kolmogorov worked out the connections between probability and measure theory, lots of questions like this caused a lot of puzzlement and sometimes even lead to paradoxes.
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