Core Probability Concepts in 5 Minutes with Dice

Ever wondered how we can predict chances of events happening? Let's learn the core concepts of probability using something we're all familiar with - dice! Whether you're playing board games, analyzing data, or making decisions under uncertainty, these concepts will help you understand the mathematics behind chance.

What is Probability?

Probability is the branch of mathematics that deals with the likelihood of events occurring. It's a measure of the certainty that an event will happen, expressed as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty. For example:

What is the probability of raining tomorrow based on today's weather forecast?
What are the chances of flipping a coin and getting heads?
If you roll a die, how likely is it that you'll roll a 6?

We use probability every day, often without even realizing it. Whether you're playing a game of cards, deciding if you need an umbrella, or calculating the odds of winning a lottery, you're dealing with probability.

Sample Space

The sample space is the set of all possible outcomes of a probability experiment. For example, in the experiment of rolling a six-sided die, the sample space is:

Sample Space

A event is a specific outcome or a group of outcomes from the sample space. In the experiment of rolling a six-sided die, the event of rolling an even number includes the outcomes:

Probability of an Event

If a sample space has $n$ possible outcomes and an event $A$ has $k$ favorable outcomes, the probability of event $A$ , which is denoted as $P(A)$ , is calculated as: $P(A) = \frac{\text{Number of outcomes in A}}{\text{Total number of possible outcomes}} = \frac{k}{n}$

For example, let

A

be the event of rolling a 4 on a six-sided die. The probability of rolling a 4 is:

P(A) = \frac{1}{6} = 0.167

This is because rolling a 4 () is just one one favorable outcome out of six possible outcomes on the die.

Permutation

A permutation is an arrangement of objects where order matters. For example, (1, 2) is different from (2, 1). The number of permutations of $n$ distinct objects taken $r$ at a time is denoted as $P(n,r)$ or $^nP_r$ and is calculated as: $P(n,r) = \frac{n!}{(n-r)!}$

For example, let's consider rolling three dice and getting different even numbers (2, 4, 6) on each die. The number of ways to do this is:

P(3,3) = \frac{3!}{(3-3)!} = 3! = 3 \times 2 \times 1 = 6

This means there are 6 different ways to arrange the even numbers 2, 4, and 6 on three dice:

which are:

The order matters because (2, 4, 6), (4, 6, 2), and (6, 2, 4) are all considered different arrangements.

Combination

A combination is a selection of objects where order doesn't matter. The number of combinations of $n$ distinct objects taken $r$ at a time is denoted as $C(n,r)$ or $^nC_r$ or $\binom{n}{r}$ and is calculated as: $C(n,r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}$

For example, let's consider selecting 2 different even numbers from the set 6. The number of ways to do this is:

C(3,2) = \binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3 \times 2}{2 \times 1} = 3

This means there are 3 different ways to select 2 even numbers from 2, 4, and 6:

which are:

The order doesn't matter because () is considered the same as ().

Mutually Exclusive Events

We say that two events are mutually exclusive if they cannot occur simultaneously. In a dice roll, for example, the event of rolling a 2 () and the event of rolling a 5 () are mutually exclusive because they cannot happen at the same time. If we let $A$ and $B$ be two mutually exclusive events, then: $P(A \cap B) = 0$ where $P(A \cap B)$ is the joint probability, or the probability of both events occurring together, which is zero for mutually exclusive events.

Click on the dice to highlight either 2 or 5. They can't be active together!

Independent Events

Two events are independent if the occurrence of one event does not affect the probability of the other. Rolling a die and flipping a coin are two independent events because the outcome of one does not influence the other. If we let $A$ and $B$ be two independent events, then: $P(A \cap B) = P(A) \times P(B)$

For example, let

A

be the event of rolling a

4

on the first die, and

B

be the event of rolling a

6

on the second die.

First Die: Event A

Second Die: Event B

The probability of rolling a

4

and a

6

is:

P(A \cap B) = P(A) \times P(B) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} \approx 0.028

Conditional Probability

The probability of an event occurring given that another event has already occurred. It is denoted as $P(A|B)$ , where $A$ is the event and $B$ is the condition. For example, the probability of rolling a 6 () given that the die roll is even () is:

Even numbers highlighted, 6 emphasized

P(6|\text{Even}) = \frac{1}{3} = 0.333

The formal definition of conditional probability is given by the formula:

P(A|B) = \frac{P(A \cap B)}{P(B)}

By rearranging the formula, we can obtain the joint probability of

A

and

B

, which is also known as the multiplication rule:

P(A \cap B) = P(A|B) \times P(B)

If events $A$ and $B$ are independent, then the conditional probability of $A$ given $B$ is equal to the probability of $A$ : $P(A|B) = P(A)$ . Moreover, if $A$ and $B$ are mutually exclusive, then the conditional probability of $A$ given $B$ is zero: $P(A|B) = 0$ .

Addition Rule

The probability that either of two events occurs (

A

B

) is the sum of their individual probabilities, minus the probability of both occurring together (if they can overlap).

P(A \cup B) = P(A) + P(B) - P(A \cap B)

For example, let

A

be the event of rolling an odd number(,,) and

B

be the event of rolling a number greater than 4 ().

First Die: rolling an odd number

Second Die: rolling a number greater than 4

The probability of rolling an odd number or a number greater than 4 is:

P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{3}{6} + \frac{2}{6} - \frac{1}{6} = \frac{2}{3} = 0.667

Complement Rule

The probability that an event does not occur is 1 minus the probability that it does occur.

$P(A^c) = 1 - P(A)$

For instance, if we let

A

be the event of rolling a 1, 2, or 3, the complement of

A

is the event of rolling a 4, 5, or 6. The probability of the complement of

A

is:

Event A: Rolling 1, 2, or 3

Complement of A: Rolling 4, 5, or 6

P(A^c) = 1 - P(A) = 1 - \frac{3}{6} = \frac{1}{2} = 0.5

Law of Total Probability

A fundamental rule in probability theory that allows the calculation of the probability of an event based on its relationship with other events.

For a partition of the sample space into events $B_1, B_2, ..., B_n$ , the probability of an event A is given by:

P(A) = \sum_{i=1}^n P(A|B_i) \cdot P(B_i)

Let's illustrate this with an example using a six-sided die:

Die faces: Blue (1-3), Red (4-6)

Let's calculate the probability of event of $A$ (rolling an even number: $2$ , $4$ , or $6$ ) using the Law of Total Probability. We'll partition the sample space into two events:

$B_1$ : Rolling a number from $1$ to $3$ (blue faces)
$B_2$ : Rolling a number from $4$ to $6$ (red faces)

We know:

$P(B_1) = P(B_2) = \frac12$ (equal probability of blue or red)
$P(A|B_1) = \frac{1}{3}$ (only $2$ is even out of $1, 2, 3$ )
$P(A|B_2) = \frac{2}{3}$ ( $4$ and $6$ are even out of $4, 5, 6$ )

Applying the Law of Total Probability:

\begin{align*} P(A) &= P(A|B_1) \cdot P(B_1) + P(A|B_2) \cdot P(B_2) \\ &= \frac{1}{3} \cdot \frac{1}{2} + \frac{2}{3} \cdot \frac{1}{2} \\ &= \frac{1}{6} + \frac{1}{3} \\ &= \frac{1}{2} = 0.5 \end{align*}

This confirms that the probability of rolling an even number on a fair six-sided die is indeed $0.5$ .

Bayes' Theorem

A fundamental theorem in probability theory that describes the probability of an event based on prior knowledge of conditions that might be related to the event. It is denoted as P(A|B) = \rac{P(B|A) \cdot P(A)}{P(B)}, where:

$P(A|B)$ is the posterior probability of $A$ given $B$
$P(B|A)$ is the likelihood of $B$ given $A$
$P(A)$ is the prior probability of $A$
$P(B)$ is the marginal probability of $B$

Let's illustrate Bayes' Theorem with an example using a single six-sided die:

Even numbers highlighted, 2 emphasized

Scenario:

We roll a fair six-sided die.
We're told that the result is an even number.
What's the probability that the number rolled was specifically a $2$ ?

Let's define our events:

$A$ : The event of rolling a $2$
$B$ : The event of rolling an even number

We know:

$P(A) = \frac16$ (probability of rolling a $2$ )
$P(B) = \frac12$ (probability of rolling an even number)
$P(B|A) = 1$ (probability of even given it's a $2$ is certain)

We want to find $P(A|B)$ . Applying Bayes' Theorem:

P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}

Now we can solve:

P(A|B) = \frac{1 \cdot \frac{1}{6}}{\frac{1}{2}} = \frac{1}{3} = 0.3333

This means that given we rolled an even number, there's a $\frac13$ (or about $33.33\%$ ) chance that the number rolled was specifically a $2$ .

Bayes' Theorem allowed us to update our probability based on new information. Initially, the probability of rolling a $2$ was $\frac16$ (about $16.67\%$ ). However, knowing that the result was even increased this probability to $\frac13$ ( $33.33\%$ ).

This example demonstrates how Bayes' Theorem can be used to calculate conditional probabilities and update our beliefs based on new evidence.

Quick Tips

Always identify your sample space first
Check if events are independent or mutually exclusive
Use fractions to keep calculations simple
Verify that probabilities are between 0 and 1

Practice Problems

Question (1 / 10)

In a standard 52-card deck, what is the sample space?

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