1.1 Dice rolls and probability
Here’s Marcus to discuss the idea of probability when rolling dice. Watch the video before calculating some probabilities yourself in Activity 1.
Keeping the example numbers of 5 and 12 in mind, let’s refer to any total number as n. If we denote the event frequency – the number of ways of getting n – by F(n), then in the first case F(5) = 4, and in the second F(12) = 1. Since in each case the number of possible outcomes is 36 but the event frequency is different, the probability of getting a total of 5 cannot equal the probability of getting a total of 12.
Activity 1 Rolling two dice
Complete the following table, where the notation (1, 2) denotes a throw of 1 followed by a throw of 2.
Table 1 Results when rolling two dice
| Desired result (n) | Ways to obtain the result | Event frequency F(n) | Probability of obtaining desired result P(n) |
|---|---|---|---|
| 1 | |||
| 2 | |||
| 3 | |||
| 4 | |||
| 5 | (2, 3), (3, 2), (1, 4), (4, 1) | 4 | 4/36 = 1/9 |
| 6 | |||
| 7 | |||
| 8 | |||
| 9 | |||
| 10 | |||
| 11 | |||
| 12 | (6, 6) | 1 | 1/36 |
Answer
Here’s a completed version of the table. As a check, the sum of the 12 probabilities in the final column should add up to 1.
Table 1 Results when rolling two dice
| Desired result (n) | Ways to obtain the result | Event frequency F(n) | Probability of obtaining desired result P(n) |
|---|---|---|---|
| 1 | None | 0 | 0 |
| 2 | (1, 1) | 1 | 1/36 |
| 3 | (1, 2), (2, 1) | 2 | 2/36 = 1/18 |
| 4 | (1, 3), (3, 1), (2, 2) | 3 | 3/36 = 1/12 |
| 5 | (2, 3), (3, 2), (4, 1), (1, 4) | 4 | 4/36 = 1/9 |
| 6 | (1, 5), (5, 1), (2, 4), (4, 2), (3, 3) | 5 | 5/36 |
| 7 | (1,6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3) | 6 | 6/36 = 1/6 |
| 8 | (2, 6), (6, 2), (3, 5), (5, 3), (4, 4) | 5 | 5/36 |
| 9 | (3, 6), (6, 3), (4, 5), (5, 4) | 4 | 4/36 = 1/9 |
| 10 | (4, 6), (6, 4), (5, 5) | 3 | 3/36 = 1/12 |
| 11 | (5, 6), (6, 5) | 2 | 2/36 = 1/18 |
| 12 | (6, 6) | 1 | 1/36 |
As you can see from the completed table, if you roll two dice, the probability of achieving a total of 7 is greater than the probability of achieving any other total. How much more likely are you to achieve a total of 7 than a total of 10?
Answer
Since the probability of achieving a total of 7 is P(7) = 6/36 = 1/6, and the probability of achieving a total of 10 is P(10) = 3/36 = 1/12, and 1/6 equals twice 1/12, you are twice as likely to achieve a total of 7 than you are to achieve a total of 10.
Next, let’s see what happens when three dice are rolled.
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