# Roll a die

Rolling a die (or multiple dice) is an example of a simple experiment that is often studied in probability and statistics. There are many different ways that dice can be used to construct probability experiments, such as rolling a single die multiple times or rolling multiple dice at the same time. From these simple experiments, it is possible to demonstrate the probabilities of independent events, dependent events, compound events, and more.

When trying to calculate probabilities, it is necessary to determine the sample space, which is the set of all possible outcomes of an experiment. For the experiment of rolling a single die, the sample space consists of the six possible values that the die can land on, which can be represented using set notation:

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

Once the sample space is known, it can be used to calculate probabilities such as the probability of rolling either a 2 or a 6, a number greater than 3, and so on. Generally, the probability of an event, A, is calculated as:

Since there are a total of six possible outcomes, the probability of any one of the outcomes in the set occurring is 1/6. The probability of rolling either a 2 or a 6 is the sum of their individual probabilities (1/6 + 1/6 = 2/6 = 1/3). The only numbers greater than 3 are 4, 5, and 6. Summing their individual probabilities, there is a 3/6, or 1/2 chance of rolling a 4, 5, or 6, on each roll of a single die.

## Compound events

A compound event is an event made up of two or more simple events. Rolling a single die is an example of a simple event; rolling two dice is an example of a compound event. Given that two dice are rolled at the same time, the sample space (shown below) will consist of 36 possible outcomes, or 6^{2}, since each die has 6 faces.

1 | 2 | 3 | 4 | 5 | 6 | |
---|---|---|---|---|---|---|

1 | (1, 1) | (1, 2) | (1, 3) | (1, 4) | (1, 5) | (1, 6) |

2 | (2, 1) | (2, 2) | (2, 3) | (2, 4) | (2, 5) | (2, 6) |

3 | (3, 1) | (3, 2) | (3, 3) | (3, 4) | (3, 5) | (3, 6) |

4 | (4, 1) | (4, 2) | (4, 3) | (4, 4) | (4, 5) | (4, 6) |

5 | (5, 1) | (5, 2) | (5, 3) | (5, 4) | (5, 5) | (5, 6) |

6 | (6, 1) | (6, 2) | (6, 3) | (6, 4) | (6, 5) | (6, 6) |

Using this sample space, it is possible to determine the probabilities of various events involving rolling two dice.

Example

Find the probability of rolling a sum of 7 given that 2 dice are rolled.

The highlighted entries in the table below show the outcomes that have a sum of 7:

1 | 2 | 3 | 4 | 5 | 6 | |
---|---|---|---|---|---|---|

1 | (1, 1) | (1, 2) | (1, 3) | (1, 4) | (1, 5) | (1, 6) |

2 | (2, 1) | (2, 2) | (2, 3) | (2, 4) | (2, 5) | (2, 6) |

3 | (3, 1) | (3, 2) | (3, 3) | (3, 4) | (3, 5) | (3, 6) |

4 | (4, 1) | (4, 2) | (4, 3) | (4, 4) | (4, 5) | (4, 6) |

5 | (5, 1) | (5, 2) | (5, 3) | (5, 4) | (5, 5) | (5, 6) |

6 | (6, 1) | (6, 2) | (6, 3) | (6, 4) | (6, 5) | (6, 6) |

The six outcomes highlighted in green are the favorable outcomes of this experiment. There are a total of 36 possible outcomes, thus the probability of rolling a sum of 7 is:

6/36 = 1/6

## Conditional probability

Dice can also be used to demonstrate conditional probability experiments. Conditional probability is the probability of an event occurring given that another event has already occurred.

Example

What is the probability that the sum is 7 given that two dice are rolled and the second die results in a 2 or 3?

From the previous example of rolling 2 dice, only 6 outcomes result in a sum of 7. With the condition that the second dice is a 2 or 3, the original sample space is reduced to the space highlighted below.

1 | 2 | 3 | 4 | 5 | 6 | |
---|---|---|---|---|---|---|

1 | (1, 1) | (1, 2) | (1, 3) | (1, 4) | (1, 5) | (1, 6) |

2 | (2, 1) | (2, 2) | (2, 3) | (2, 4) | (2, 5) | (2, 6) |

3 | (3, 1) | (3, 2) | (3, 3) | (3, 4) | (3, 5) | (3, 6) |

4 | (4, 1) | (4, 2) | (4, 3) | (4, 4) | (4, 5) | (4, 6) |

5 | (5, 1) | (5, 2) | (5, 3) | (5, 4) | (5, 5) | (5, 6) |

6 | (6, 1) | (6, 2) | (6, 3) | (6, 4) | (6, 5) | (6, 6) |

There are 12 outcomes in which the second dice is a 2 or a 3, only 2 of which have a sum of 7: (5,2) and (4,3). Thus, the probability of two dice summing to 7 given that the second dice is a 2 or a 3 is

P(A|B) = 2/12 = 1/6

where A is the event that the dice sum to 7 and B is the event that the second dice is a 2 or a 3, and P(A|B) is the probability of A given B.