Probability Measure on Equiprobable Outcomes

Theorem
Let $\left({\Omega, \Sigma, \Pr}\right)$ be an equiprobability space.

Let $\left|{\Omega}\right| = n$.

Then:
 * $\forall \omega \in \Omega: \Pr \left({\omega}\right) = \dfrac 1 n$


 * $\forall A \subseteq \Omega: \Pr \left({A}\right) = \dfrac {\left|{A}\right|} n$.

Proof
By definition, $\Pr \left({\omega_i}\right) = \Pr \left({\omega_j}\right)$ for all $\omega_i, \omega_j \in \Omega$.

So let $\Pr \left({\omega_i}\right) = p$.

Also, by definition of probability measure, we have:
 * $\Pr \left({\Omega}\right) = 1$

We have that $\left\{{\omega_i}\right\} \cap \left\{{\omega_j}\right\} = \varnothing$ when $i \ne j$ and so, by definition of definition of probability measure:
 * $\Pr \left({\left\{{\omega_i}\right\} \cup \left\{{\omega_j}\right\}}\right) = \Pr \left({\left\{{\omega_i}\right\}}\right) + \Pr \left({\left\{{\omega_j}\right\}}\right)$

Using the fact that $\displaystyle \Omega = \bigcup_{i \mathop = 1}^n \left\{{\omega_i}\right\}$:
 * $\displaystyle \Pr \left({\Omega}\right) = \sum_{i \mathop = 1}^n \Pr \left({\left\{{\omega_i}\right\}}\right) = \sum_{i \mathop = 1}^n p = n p$

But we have that $\Pr \left({\Omega}\right) = 1$, and so $1 = n p$.

Hence:
 * $\forall \omega \in \Omega: \Pr \left({\omega}\right) = \dfrac 1 n$

Now consider $A \subseteq \Omega$.

Let the cardinality of $A$ be $k$, i.e. $\left|{A}\right| = k$.

Thus:
 * $\displaystyle \Pr \left({A}\right) = \sum_i \Pr \left({\omega_i}\right) \left[{\omega_i \in A}\right]$

where $\left[{\omega_i \in A}\right]$ uses Iverson's convention.

Hence:
 * $\Pr \left({A}\right) = k p$

and so:
 * $\Pr \left({A}\right) = \dfrac {\left|{A}\right|} n$