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.
Transcript: Video 2 Probability
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.
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.
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.