The naive definition of probability

We've seen that two identical dice, once labelled, give $36$ outcomes. Grouping them by the numbers that appear gives $21$ events instead: "a $5$ and a $6$" holds $(5, 6)$ and $(6, 5)$, while "double $6$" holds only $(6, 6)$.

Leibniz, one of the two founders of calculus, claimed that "it is equally likely to throw twelve points, than to throw eleven; because one or the other can be done in only one manner". He was wrong. A sum of $11$ comes from die $1$ showing $5$ and die $2$ showing $6$, or the other way round, while a sum of $12$ comes only from two $6$s. Identical dice are still two separate dice; labelling them lets us count both ways.

This is where the difference between an outcome and an event matters. An outcome is one result the experiment can produce. An event is a set of outcomes, picked out by a description we choose: "a sum of $11$" is the event $\{(5, 6), (6, 5)\}$. Leibniz treated events as if they were outcomes.

The words fair, chosen at random, shuffled and dealt state that the labelled outcomes are equally likely. A stated bias, such as a die weighted towards $6$, states that they are not.

For two fair dice, the $36$ outcomes are equally likely. An event's chance depends on how many of them it holds, so "a $5$ and a $6$" is twice as likely as "double $6$", and the $21$ events are not equally likely.

Two fair coins are tossed. Are $0$ heads, $1$ head and $2$ heads equally likely?

Solution

We label the coins, coin $1$ and coin $2$, so that an outcome records what each coin shows. The coins are fair, so these labelled outcomes are equally likely:

Each number of heads is an event, so we count the outcomes it holds:

• $0$ heads is $\{TT\}$, $1$ outcome.
• $1$ head is $\{HT, TH\}$, $2$ outcomes.
• $2$ heads is $\{HH\}$, $1$ outcome.

The three events hold different numbers of outcomes, so they are not equally likely: $1$ head has a greater chance than $0$ heads. The equally likely outcomes are the four labelled ones, $HH$, $HT$, $TH$ and $TT$.

Once the outcomes of an experiment are equally likely, finding the chance of an event is a matter of counting. We picture this with pebble world, a framing from Blitzstein and Hwang's Introduction to Probability.

We draw the sample space $\Omega$ as a pile of pebbles, one per outcome. Every pebble has the same mass, and the whole pile has mass $1$. An event is a handful of those pebbles, and its probability is the share of the mass that the handful holds.

On a fair die each of the six pebbles has mass $\dfrac{1}{6}$. The even faces $2$, $4$ and $6$ are three of them, so the chance of an even roll is $\dfrac{3}{6} = \dfrac{1}{2}$.

So nothing here is new: a probability is one count divided by another, the number of outcomes in $A$ divided by the number in $\Omega$, and we already know how to count.

We read $P(A)$ as "the probability of $A$" and give it as a fraction in lowest terms.

Three fair coins are flipped and the sequence of results is recorded. Find the probability of getting exactly two heads.

Solution

The coins are fair, so every sequence of three results is equally likely, and each sequence is one outcome. We list them:

The event $A$ is "exactly two heads", so we keep the sequences with two $H$s and one $T$. Reading down the list:

These are the favourable outcomes, so