Consider a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ where $\Omega$ is finite, $\mathcal{F}$ is the power set of $\Omega$ and $\mathbb{P}(F)\equiv |F|/|\Omega|$ $\forall F \in \mathcal{F}$.
In what follows:
$$ \Omega\equiv \{1,2,3\} $$ $$ \mathcal{F}\equiv \Big\{\emptyset, \Omega, \{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{2,3\}\Big\} $$ $$ \mathbb{P}(\emptyset)=0, \mathbb{P}(\Omega)=1, \mathbb{P}(\{1\})=\mathbb{P}(\{2\})=\mathbb{P}(\{3\})=1/3, \mathbb{P}(\{1,2\})=\mathbb{P}(\{1,3\})=\mathbb{P}(\{2,3\})=2/3 $$
Suppose we draw from $\Omega$ a simple random sample of size $N=2$.
1) It is intuitive to say that "The probability of drawing $\{1,2\}$ is $\frac{1}{\binom{|\Omega|}{N}}=\frac{1}{3}$". However, what is the formal link between this probability (which seems to me somehow related to the classical definition of probability) and the measure $\mathbb{P}$ defined above?
2) I found in some books the following: let $W_i$ be a random variable equal to $1$ if the $i$th element of $\Omega$ is drawn and zero otherwise. Then, the probability that $W_i=1$ is $\frac{\binom{|\Omega|-1}{N-1}}{\binom{|\Omega|}{N}}$. But, again, what is the exact measure-theoretic definition of $W_i$ as random variable with respect to the probability space $(\Omega, \mathcal{F}, \mathbb{P})$?
I am puzzled to find a relation. I was wondering whether the only solution would be to define ANOTHER probability space $(\Omega_1,\mathcal{F}_1,\mathbb{P}_1)$ where $$ \Omega_1\equiv (A,B,C) \hspace{1cm} \text{[all possible outcomes of the drawing experiment]} $$ $$ \mathcal{F}_1\equiv \Big\{\emptyset, \Omega_1, \{A\}, \{B\}, \{C\}, \{A,B\}, \{A,C\}, \{B,C\}\Big\} $$ $$ \mathbb{P}_1(\emptyset)=0, \mathbb{P}_1(\Omega)=1, \mathbb{P}_1(\{A\})=\mathbb{P}_1(\{B\})=\mathbb{P}_1(\{C\})=1/3, \mathbb{P}_1(\{A,B\})=\mathbb{P}_1(\{A,C\})=\mathbb{P}_1(\{B,C\})=2/3 $$ where $\mathbb{P}_1$ is somehow related to $\mathbb{P}$.
We then define the random variable $$ W:\Omega_1\rightarrow \{\{(1,2)\}, \{(2,3)\}, \{(1,3)\}\} \hspace{1cm}\text{[outcome of drawing experiment]} $$ with $$ W(A)=\{(1,2)\}, W(B)=\{ (2,3)\}, W(C)=\{(1,3)\} $$ Hence, $\mathbb{P}_1(W=\{(1,2\})=\frac{1}{3}$ as expected.
We can also defined another random variable $$ W_1: \Omega_1\rightarrow \{1,0\} $$ $$ W_1(A)=1, W_1(B)=0, W_1(C)=1 \hspace{1cm}\text{[$1$ if first element of $\Omega$ is drawn]} $$ Hence, $\mathbb{P}_1(W_1=1)=\frac{2}{3}$ as expected.