Probability Measure on Equiprobable Outcomes

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

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

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


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

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 $$\Omega = \bigcup_{i=1}^n \left\{{\omega_i}\right\}$$ we have that:
 * $$\Pr \left({\Omega}\right) = \sum_{i=1}^n \Pr \left({\left\{{\omega_i}\right\}}\right) = \sum_{i=1}^n p = n p$$.

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

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

Now consider $$A \subseteq \Omega$$.

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

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

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

Hence $$\Pr \left({A}\right) = k p$$, and so $$\Pr \left({A}\right) = \frac {\left|{A}\right|} n$$.