Rolling a Die

A die is a small cube, where each face is marked with a number (1 to 6). Rolling a die refers to tossing this small cube to produce a random outcome between the numbers 1 to 6. Each side of the die cube has equal probability of landing face-up while tossing. This process of rolling a die is widely used in board games, gambling, and statistical experiments to model uncertainty and chance.

For example, the table below shows the number of times or occurrences of outcomes when a die is rolled 100 times:

The sum of relative occurrence should be equal to 1. 

\(=\frac15 + \frac5{100} + \frac1{10} + \frac3{20} + \frac14 + \frac14\)
 

\(= 0.2 + 0.05 + 0.1 + 0.15 + 0.25 + 0.25 = 1\)

The value of probability should be between 0 and 1. 

Sample Space

Whenever a die or dice is rolled, all possible outcomes form a set known as the sample space. For example, when we are rolling a standard six-sided die, the sample space includes the numbers {1, 2, 3, 4, 5, 6}. It corresponds to the numbers displayed on the die. This can be understood better with the help of an image showing a die with all its sides and numbers.

When a fair die is rolled, each face is equally likely to appear. So, no single outcome is favored, and each has a probability of 1/6. With many rolls, each number should appear in roughly one-sixth of the trials. We can then apply the general dice probability formula.

The probability of a die formula is written as,

\(\text{Probability} = \frac{\text{No. of Favorable Outcomes}}{\text{No. of total outcomes}}\)

On the probability scale shown below, an event with probability \(\frac{1}{6}\) falls in the low likelihood range. A probability of 0 means the event is impossible, while a probability of 1 means the event is sure to occur. Since a fair six-sided die has six possible outcomes, the probability of rolling any specific number is \(\frac{1}{6},\) and getting an outcome with probability zero is impossible.

Probability of rolling an even number on a die

To find the probability of rolling an even number on a die, we first identify the favorable even-numbered outcomes. 

The even numbers on a die are {2, 4, 6}.

Therefore, we have three favorable outcomes. The total possible outcomes are 6.

Now, let us apply the probability formula:

\(\text{P(Even number)} = \frac{\text{No. of Favorable Outcomes}}{\text{No. of total outcomes}}\)

\(\text{P(Even number)} = \frac{3}{6} = \frac{1}{2}\)

Therefore, we have a 50% of rolling an even number on a fair die. 

Probability of rolling an odd number on a die

Let us follow the same steps as above. 

The odd numbers on a die are {1, 3, 5}

Here, we have three favorable outcomes, and the total number of possible outcomes is 6. 

Now, let us apply the probability formula:

\(\text{P(Odd number)} = \frac{\text{No. of Favorable Outcomes}}{\text{No. of total outcomes}}\)

\(\text{P(Odd number)} = \frac{3}{6} = \frac{1}{2}\)

Therefore, we have a 50% of rolling an odd number on a fair die.

When rolling a single die, each number from 1 to 6 has an equal chance of appearing. We use the following formula to calculate the probability of rolling a die:

\(\text{Probability} = \frac{\text{No. of Favorable Outcomes}}{\text{No. of total outcomes}}\)

If we roll 2 dice, we get the following outcomes:

{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

The above are all the possible outcomes that have equal probability.

For example, let's consider throwing two dice at the same time. Probability of rolling a 1 on the first die and a 5 on the second die (or vice versa) is given as, 

First, let us calculate the probability of rolling a 1 on the first die and a 5 on the second die

\(P(1, 5) = \frac{1}{6} \times \frac {1}{6} = \frac{1}{36}\)

The probability of rolling a 5 on the first die and a 1 on the second die is, 

\(P(5, 1) = \frac{1}{6} \times \frac {1}{6} = \frac {1}{36}\)

Since these events are mutually exclusive, let us add their probabilities. 

\(\)\(P(1, 5) + P(5, 1) =  \frac{1}{36} + \frac{1}{36} = \frac{2}{36}\)

therefore, the probability is, 

\(P = \frac{1}{18}\)