Probability space of "talking 3 balls in a box that contain 5 balls" Let $A$ the event : Taking 3 balls from a urn that contains 5 
different balls. I would like to find $(\Omega ,\sigma (A), \mathbb P)$ 
I know that $\Omega =\{\{i,j,k\}\mid \{i,j,k\}\subset \{B_1,...,B_5\}\}$ and all singleton are measurable with probability $\frac{1}{\binom{5}{3}}$. 
In many experiments (e.g. throwing simultaneous indistinguishable dice), we first distinguish them (which we can), and then work around the indistinction. It's what I would like to do here (because it can be always done), but it looks like there are 2 different ways to do it and I don't see the difference :
1) We take $\Omega =\{(i,j,k)\mid i\neq j\neq k, i,j,k\in \{B_1,...,B_5\}\}$. Then, the elementary event associated to the experience are $A_{ijk}=\{(\sigma (i),\sigma (j),\sigma (k))\mid \sigma \in \mathfrak S_3\}$ with $i\neq j\neq k$, and the mesure associated to the experiment is $\mathbb P(A_{ijk})=\frac{3!}{5\cdot 4\cdot 3}.$ Finally our space is $(\Omega ,\sigma (A_{ijk}\mid i\neq j\neq k), \mathbb P)$.  
2) Now I take $\Omega' =\{(i,j,k)\mid i,j,k\in \{B_1,...,B_5\}\}$. I can proceed as before : The measurable sets are going to be $A_{ijk}$ as previously with $i\neq j\neq k$, and $\mathbb Q(A_{ijk})=\frac{3!}{5\cdot 4\cdot 3}$ ? I don't really see the difference here if I take $(\Omega ',\sigma (A_{ijk}\mid i\neq j\neq k),\mathbb Q)$.
 A: *

*Waooouuu, you are doing very complicated for such an easy problem. Moreover, it's not very rigorous. I'll denote $\sigma (A)$ for $\sigma (A_{ijk}\mid i\neq j\neq k\neq i)$. For 2), we fix the probability space $(\Omega', \mathcal P(\Omega '),P)$ where $P(\{(i,j,k)\})=\frac{1}{6^3}$. So the measure $\mathbb Q$ you want to define on $\sigma (A)$ should be rather $$\mathbb Q(A_{ijk})=\frac{3!}{6^3}$$ and not what you wrote (it's important in discret time to be first in the most general space $(\Omega ,\mathbb P(\Omega ),\mathscr P)$ where all singleton are measurable. Otherwise you make mistake as you just done).

*What you can also do (I guess it's what you had in mind) is to take on $\Omega '$ the measure $\mathbf P$ s.t. $$\mathbf P\{(i,j,k)\}=\begin{cases}
\frac{1}{5\cdot 4\cdot 3}&i\neq j\neq k\neq i\\
0&\text{otherwise}\end{cases}.$$ Then, the measure $\mathbb Q$ that you defined in your orignal post works for $(\Omega ',\sigma (A))$. However, the probability space $(\Omega ', \sigma (A), \mathbb Q)$ will not be complete (not all event of measure $0$ are $\sigma (A)-$measurable... for example, $\{(1,1,1)\}$ is not). After, you can always complete it by adding the sets of measure $0$. But honestly, as far as I know, in discrete time you can always manage to have nice space. 
A: 
I know that $\Omega =\{\{i,j,k\}\mid \{i,j,k\}\subset \{B_1,...,B_5\}\}$ an all singleton are measurable with probability $\frac{1}{\binom{5}{3}}$. 

Technically, $\{B_1,B_1,B_2\}$ is generally considered a legitimate alternative way to write the two-element set $\{B_1,B_2\}.$
When you write $\Omega =\{\{i,j,k\}\mid \{i,j,k\}\subset \{B_1,...,B_5\}\}$
after telling us that you want to take three balls from an urn,
you presumably expect us to assume that $i \neq j \neq k \neq i$ even though you have not explicitly stated this assumption.
In $\Omega =\{(i,j,k)\mid i\neq j\neq k, i,j,k\in \{B_1,...,B_5\}\}$,
you have made your implicit assumption (that each selection is different) explicit. (Though you technically omitted $k \neq i$, I suppose you intended the sequence of inequalities to be read as "pairwise distinct.")
The new sample space also specifies the sequence in which the balls are taken, but by partitioning the space into equivalence classes under permutations and declaring each to be an event, you say the sequence doesn’t matter after all. 
What we have above is not just two different ways of writing the same sample space; they are two different sample spaces between which you can easily make a one-to-one correspondence of events. 
I think this is useful; I have used this correspondence (expressed in words rather than set notation) to explain to someone why it’s OK to say “order doesn’t matter” in some probability problems. 
For $\Omega' =\{(i,j,k)\mid i,j,k\in \{B_1,...,B_5\}\},$ it is unclear to me why you say the only measurable sets are those which contain only 
$(i,j,k)$ with $i\neq j \neq k \neq i.$
I cannot give a specific reason why you may not use this set and define several of its elements to have zero probability.
But note that $\Omega \subset \Omega'$ and that $\Omega$ is a countable
(in fact, finite) union of measurable sets such that in the probability space you are constructing over $\Omega'$, you require that $\Omega$ is measurable and $P(\Omega) = 1.$
Therefore $\Omega'\setminus\Omega$ is measurable and 
$P(\Omega'\setminus\Omega) = 0.$
You are not required to assign a measurement to any non-empty proper subset of $\Omega'\setminus\Omega,$ 
but the measurability of that set itself is not optional.
In conclusion, no, the measurable sets in
$(\Omega',\sigma (A_{ijk}\mid i\neq j\neq k),\mathbb Q)$
cannot be the same as in
$(\Omega,\sigma (A_{ijk}\mid i\neq j\neq k),\mathbb Q).$
For one thing, the measurable sets in the $\Omega'$ case include $\Omega'$ itself, which cannot be one of the measurable sets in the $\Omega$ case since it is not a subset of $\Omega.$
It’s perfectly legitimate to define the probability space you want over $\Omega’$;
It just seems more complicated than necessary to me. And it still does not leave any part of the sample space outside the measurable sets. 
