1.2 Adding more dice to the rolls
Since there are thirty-six possible outcomes for two dice, for each outcome of these two dice there will be six possible outcomes for the third die. This makes a total of 6 x 6 x 6 = 216 outcomes.
There will be three different types of outcome. Those in which:
all three dice are the same
two dice are the same and one is different
all three dice are different.
These three types can be labelled as [a, a, a], [a, a, b] and [a, b, c] respectively, where a, b, c each represent a different number on the dice.
Activity 2 Three types of outcome
In the first case [a, a, a], all three dice produce the same number, so there is only one way of rolling this combination. What about the other two cases? Can you work out how many ways you can roll the dice in each of these cases?
Discussion
In the second case [a, a, b], the dice can be rolled in three ways: (a, a, b), (a, b, a) and (b, a, a).
In the third case [a, b, c], the dice can be rolled in six ways: (a, b, c), (a, c, b), (b, a, c), (b, c, a), (c, a, b), (c, b, a).
Table 2 shows the probabilities of obtaining certain results when rolling three dice. For example, there are ten ways of achieving a total of six: three ways of rolling [1, 1, 4], six ways of rolling [1, 2, 3], and one way of rolling [2, 2, 2]. Thus P(6) = 10/216 = 5/108.
Can you calculate P(9) and P(13)?
| Desired result (n) | Ways to obtain the result using three dice | Event frequency F(n) | Probability of obtaining desired result P(n) |
|---|---|---|---|
| 1 | None | 0 | 0 |
| 2 | None | 0 | 0 |
| 3 | (1, 1, 1) | 1 | 1/216 |
| 4 | (1, 1, 2) | 3 | 3/216 = 1/72 |
| 5 | (1, 1, 3), (1, 1, 2) | 3 + 3 = 6 | 6/216 = 1/36 |
| 6 | (1, 1, 4), (1, 2, 3), (2, 2, 2) | 3 + 6 + 1 = 10 | 10/216 = 5/108 |
| 7 | (1, 1, 5), (1, 2, 4), (1, 3, 3), (2, 2, 3) | 3 + 6 + 3 + 3 = 15 | 15/216 = 5/72 |
| 8 | (1, 1, 6), (1, 2, 5), (1, 3, 4), (2, 2, 4), (2, 3, 3) | 3 + 6 + 6 + 3 + 3 = 21 | 21/216 = 7/72 |
| 9 | |||
| 10 | (1, 3, 6), (1, 4, 5), (2, 3, 5), (2, 2, 6), (3, 3, 4), (4, 4, 2) | 6 + 6 + 6 + 3 + 3 + 3 = 27 | 27/216 = 1/8 |
| 11 | (1, 4, 6), (2, 3, 6), (2, 4, 5), (1, 5, 5), (3, 4, 4), (5, 3, 3) | 6 + 6 + 6 + 3 + 3 + 3 = 27 | 27/216 = 1/8 |
| 12 | (1, 5, 6), (2, 4, 6), (2, 5, 5), (3, 3, 6), (3, 4, 5), (4, 4, 4) | 6 + 6 + 3 + 3 + 6 + 1 = 25 | 25/216 |
| 13 | |||
| 14 | (2, 6, 6), (3, 5, 6), (4, 4, 6), (4, 5, 5) | 3 + 6 + 3 + 3 = 15 | 15/216 = 5/72 |
| 15 | (3, 6, 6), (4, 5, 6), (5, 5, 5) | 3 + 6 + 1 = 10 | 10/216 = 5/108 |
| 16 | (5, 5, 6), (6, 6, 4) | 3 + 3 = 6 | 6/216 = 1/36 |
| 17 | (5, 6, 6) | 3 | 3/216 = 1/72 |
| 18 | (6, 6, 6) | 1 | 1/216 |
Answer
| Desired result (n) | Ways to obtain the result using three dice | Event frequency F(n) | Probability of obtaining desired result P(n) |
|---|---|---|---|
| 9 | (1, 2, 6), (1, 3, 5), (1, 4, 4), (2, 2, 5), (2, 3, 4), (3, 3, 3) | 6 + 6 + 3 + 3 + 6 + 1 = 25 | 25/216 |
| 13 | (1, 6, 6), (2, 5, 6), (3, 4, 6), (3, 5, 5), (4, 4, 5) | 3 + 6 + 6 + 3 + 3 = 21 | 21/216 = 7/22 |
Probability calculations on the roll of three dice date back centuries, with the earliest known appearance being a Latin poem titled ‘De Vetula’, written during the 13th century. It is thought that the reason many copies of this poem survived is because it provided medieval gamblers with a certain way to make money, even though they almost certainly did not understand it!