# Probability: Equally Likely Outcomes

Starting a new term in university comes with 4 brand new topics to learn. One of these is Probability. I’d like to share a little bit of (simple) maths that I learnt in my lecture last week on calculating probabilities when there are equally likely outcomes.

I will not define what a probability measure or space is rigorously, as it is quite irrelevant for the simple examples I’m considering today. But, I will introduce the following notation:

Let Ω be a set and F be a set of all subsets of Ω. The elements of Ω are called outcomes and the elements of F are called events. As we are using this as a probability model, Ω will be an abstraction of a real set of outcomes, and F will model observable events.

Then let us write, P(A) as the probability of the event A occurring. Note that:

P(A) = |A|/|Ω|.

Now, let us consider a few examples of how we would use this. In the following examples we use symmetry and randomness to argue that there are equally likely outcomes.

### Throwing a Die

The throw of a die has six possible outcomes, and so

Ω = {1, 2, 3, 4, 5, 6} and P(A) = |A|/6 for A ⊆ Ω.

Therefore, for example P({2, 4, 6}) = 1/2.

### Balls from a Bag

A bag contains r balls. Say we draw k balls from this bag without looking. If we consider that the balls are labelled by {1, . . . , n}, we have selected a subset of {1, . . . , n} of size k.

Therefore, Ω is the set of subsets of {1, . . . , n} of size k, so

with the probability of a individual outcome being 1/|Ω|.

### Pack of Cards

In an idealised well-shuffled pack of cards, every possible order is equally likely, so it is modelled by the set Ω of permutations of {1, . . . , 52}.

Because of this, we can see that |Ω| = 52!

What is the probability of the event A that the first two cards are aces?

There are 4 choices for the first ace, 3 choices for the second, and the rest can come in any order. So,

|A| = 4 × 3 × 50!

and

P(A) = |A|/|Ω| = 12/(52 × 51) = 1/221.

### Distribution of Largest Digit

Suppose that r digits are chosen from a table of random numbers. What is the probability that k is the greatest digit drawn?

Firstly, model the set of possible outcomes by

Ω = {0, 1, . . . , 9}n

Consider the event Ak = [no digit exceeds k], and the event Bk  = [the largest digit is k] So:

|Ω| = 10n ;

|Ak| = (k + 1)n ;

|Bk| = |Ak \ Ak−1| = (k + 1)n − kn.

Therefore, P(Bk) = [(k + 1)n − kn]/10n.

Hope you enjoyed today’s post. M x