What is the probability that a third randomly chosen real number is between two earlier randomly chosen real numbers? Question 1:  Suppose you pick three random real numbers A, B and C all at the same time.  
Are the following outcomes all equally likely or are some more likely than others?
A < B < C
A < C < B
B < A < C
B < C < A
C < A < B
C < B < A
Question 2:  Suppose you pick two random real numbers A and B and it happens that A < B.  Next, you pick a third random real number C.  
What is the probability that A < C < B?
 A: I see your dilemma. On the one hand we have:
Argument 1
It does not matter whether you pick all three numbers at once or one at a time. For example, suppose we pick three numbers A,B,D all at once, but reveal only two of A and B. And, also suppose that after we have revealed A and B, we pick a new number C.
Now, it makes little sense to think that the probability of D being between A and B would be any different than the probability of C being between A and B: the only difference is that one number was picked before the reveal, and the other after the reveal, and of course the picking of a number is not going to be affected by the reveal.
Also, as the first question makes clear, the probability that D is between A and B is $\frac{1}{3}$ .. so therefore the probability that C is between A and B is also $\frac{1}{3}$
But on the other hand we have:
Argument 2
Whatever A and B are, their difference is finite, and hence any third random number has a $0$ probability of being between them.
... so ... maybe this is a reductio ad absurdum against the very assumption that we can randomly pick numbers from all real numbers with equal likelihood?
A: You can not choose a real number at random with uniform probabilities. However, there are various probability distributions you can choose from. Assuming you choose one of these PDFs, then for question one, all the possible orderings are equally probable because you are choosing them independently or all at the same time. For the second question the probability of A < C < B completely depends on the PDF.
A: Let A, B and C be uniformly distributed on $(-N,N)$
Then each has pdf = $\frac 1{2N}$
The probability $A<C<B$ is given by 
$$ \begin{eqnarray} P = & &\frac 1 {(2N)^3}\int_{-N}^N\int_A^N\int_A^B dC \; dB \; dA
\\=&&\frac 1 {(2N)^3}\int_{-N}^N  \int_A^N\  (B-A)\; dB \; dA
\\=&  & \frac 1 {(2N)^3}\int_{-N}^N  ( \frac 12 N^2-AN)-(\frac 12 A^2-A^2     )    \; dA 
\\=& & \frac 1 {(2N)^3}\int_{-N}^N  (\frac 12 A^2-AN+ \frac 12 N^2 )    \; dA 
\\= & &\frac 1 {(2N)^3}  (\frac 13 N^3 +N^3) \\ = & &\frac 16  \end{eqnarray} $$
This result is independent of $N$ so you would expect it to apply in the limit $N \to \infty$
