It gets more interesting when you have two dice. One thing that you can do is work out what the total of the dice is. The dice experiment allows you to simulate throwing pairs of dice and see what the result is. This is a good introduction to probability, since you can see which combinations are more likely. But the real world, or even a simulated real world, never matches completely with calculated probability. So how do we calculate it? The first thing is to work out what the range is. You can't have a total less than 2 (both dice being 1) and you can't have a total more than 12 (both dice being 6). The easiest way to see what the probabilities is to write out the possible totals. There are 36 of them in all (6 x 6).

Total on dice |
Pairs of dice |
Probability |

2 |
1+1 |
1/36 = 3% |

3 |
1+2, 2+1 |
2/36 = 6% |

4 |
1+3, 2+2, 3+1 |
3/36 = 8% |

5 |
1+4, 2+3, 3+2, 4+1 |
4/36 = 11% |

6 |
1+5, 2+4, 3+3, 4+2,
5+1 |
5/36 = 14% |

7 |
1+6, 2+5, 3+4, 4+3,
5+2, 6+1 |
6/36 = 17% |

8 |
2+6, 3+5, 4+4, 5+3,
6+2 |
5/36 = 14% |

9 |
3+6, 4+5, 5+4, 6+3 |
4/36 = 11% |

10 |
4+6, 5+5, 6+4 |
3/36 = 8% |

11 |
5+6, 6+5 |
2/36 = 6% |

12 |
6+6 |
1/36 = 3% |

There are plenty of other conditions that you can have with two dice. Often, the easiest way to get the probability of a condition is to list all possible throws, and count the ones that fit the condition. The best way to make sure you have remembered all throws is to list them in a 6 x 6 table, like below.

Imagine two dice, red and green. This table shows all possible throws of these dice, with red numbers for the red dice, and green for the green dice. The background is black if the condition is satisfied. You can change the numbers and the condition to see what happens. You find the probability of this condition by counting the black squares, and dividing by 36 (all possible throws).

Probability of this condition =
See below for an explanation |

Return to interactive exercise for conditions

The probability of one dice being a particular number is **1/6**. You would assume that it would be twice as likely that either of two dice being a particular number, or **1/3**, but this would be wrong. If you select the first condition above, you will see why. While there are six conditions where one dice is a number, and six conditions where the other dice is that number, there is one condition in both. So the probability is **(12-1)/36** or **11/36**.

There is a better way to calculate this. If you click on the second condition, then back on the first, you will find that they are opposites of each other. That means that their probabilities add up to one. The second condition is easier to calculate directly (see below), as **25/36**. So this probability is **(36-25)/36** = **11/36**.

Return to interactive exercise for conditions

The probability of one dice not being a particular number is **5/6**. The probability of two dice not being a that number is **5/6** x **5/6** = **25/36**. You can do this as they don't overlap as all (the correct term is that they are independent events).

Return to interactive exercise for conditions

The probability of one dice being a particular number is **1/6**. The probability of two dice being the same particular number is **1/6** x **1/6** = **1/36**. This is not the same as saying that both dice are the same number. There are six different possible numbers, so that would be **6/36** or **1/6**.

Return to interactive exercise for conditions

This is the opposite of both dice being the same particular number, so the probabilities will add up to one. So the probability of at least one dice isn't a particular number = **1 - (1/36)** = **35/36**.

Return to interactive exercise for conditions

This is a fun conditions, but a little strange to work out! Try experimenting with different numbers, and you will get different collections of throws fitting the conditions, and so different probabilities. To take an example, considering one dice, if the particular number is **5** then there are **5** throws which fit the condition. There are another **5** for the other dice, and there is one overlap, so the result is **(5 + 5 - 1)/36** = **9/36** or **1/4**. But this varies for other numbers.

Return to interactive exercise for conditions

The probabilities vary considering what the number is. But the probability of the lowest dice being a particular number is the same as the probability of the highest dice being six minus that particular number.

Return to interactive exercise for conditions

Like the first condition, there are two ways to work this out. The first is doing it directly. Each dice has **6** throws for each number, but there are four overlaps. So the probability is **(4 x 6 - 4)/36** = **20/36**.

An easier way is to realise that this condition is the opposite of the next one, dice are not either number. That probability is **16/36** so this one is **1 - 16/36** = **20/36** or **5/9**.

Return to interactive exercise for conditions

The probability of one dice not being either of two numbers is **4/6**, so the probability of both dice fitting this condition is **4/6** x **4/6** = **16/36**. We can do this as the conditions don't overlap, so they are independent events.

Return to interactive exercise for conditions

There are only two ways that two dice can have two values (try the exercise above to see why), so the probability is **2/36** or **1/18**.