# Equally Likely Results¶

We will think now of more theoretical examples, to learn how to calculate some theoretical probabilities.

We have said in a previous lesson that the probability of getting heads when we toss a fair coin is $\dfrac{1}{2}$. And this was because getting heads is **one** of the **two** possible outcomes (*heads* or *tails*).

But the key here is that the coin is *fair*, meaning that the two outcomes have the same chances of coming up.

This kind of experiments are going to be called *experiments with equally likely results*, which only means that every outcome (or result) in the sample space will have the same probability of coming up.

More examples of this kind of experiments are:

- Throwing a fair dice: in this case, the outcomes are $6$, but they all have the same probability of being rolled.
- Picking a card from a deck: for example if the deck has $52$ cards, there will be $52$ outcomes, all equally likely.

Let's dig deeper into the dice example:

The sample space will be $\Omega = \{ 1, 2, 3, 4, 5, 6 \}$

We know from a previous lesson that the probabilities of all of them will add up to $1$, since there are six outcomes and all of them are equally likely, to get each probability individually we can divide $1$ into $6$.

This means: $P(1)=1/6$, $P(2)=1/6$, $P(3)=1/6$, $P(4)=1/6$, $P(5)=1/6$, $P(6)=1/6$.

We can clearly see that $P(1)+P(2)+P(3)+P(4)+P(5)+P(6)=\dfrac{6}{6}=1$.

Now let's think of the general case:

### If $\Omega$ has equally likely results and $|\Omega|=n$, then $P(\omega)=\dfrac{1}{n}$ $\forall\omega\in\Omega$.¶

This just means that the probability of each outcome individually is $1$ over the total amount of outcomes in our sample space.

And this is because if there are $n$ equally likely options, and all add up to $1$, each one must be $\dfrac{1}{n}$.