Every pair
Roll two fair dice, a red one and a blue one. Each outcome is a pair: what the red dice shows and what the blue dice shows. A possibility diagram lays out every pair on a grid, with the red dice down the side and the blue dice across the top, so each cell is one pair.
The red dice has 6 outcomes, and for each of them the blue dice has 6, so there are 6 × 6 = 36 cells. Each dice is fair, so all 36 pairs are equally likely, and each has probability . A probability is then a count of cells over 36.
The red dice down the side and the blue dice across the top: 6 rows of 6 cells, one cell for each of the 36 pairs.
Counting a total
Which pairs total 6? Whatever the red dice shows, the blue dice must make up the rest of 6: red 1 needs blue 5, red 2 needs blue 4, and so on to red 5 with blue 1. Red 6 would need blue 0, which is impossible. That gives 5 pairs: (1, 5), (2, 4), (3, 3), (4, 2) and (5, 1).
On the diagram these 5 cells lie on one diagonal, where the two numbers always add to 6. So the probability of a total of 6 is .
The 5 pairs that total 6, on a diagonal from red 1 with blue 5 to red 5 with blue 1.
Every total
Write the total in each cell of the grid, and each total fills one diagonal. The total 2 has a single cell, (1, 1), and so does 12. The longest diagonal is the total 7, with 6 cells: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1).
From 2 up to 12 the numbers of cells are 1, 2, 3, 4, 5, 6, 5, 4, 3, 2 and 1, and they add to 36. So 7 is the most likely total, with probability , and 2 and 12 are the least likely, with each.
Each cell holds the total of its row and column. The 6 cells that total 7 make the longest diagonal.
Events that are not totals
The diagram answers any question about the pair, not only totals. What is the probability of at least one six? The cells with a six on the red dice fill one row, 6 cells, and the cells with a six on the blue dice fill one column, 6 more. The double six is in both, so count it once: 6 + 6 − 1 = 11 cells, and the probability is .
The last row and the last column are the pairs with at least one six. They share one cell, the double six, so 11 cells are colored.
Other two-stage experiments
The same grid works for any two stages. Spin a fair spinner numbered 1, 2 and 3, roll a fair dice, and multiply the two numbers. The grid has 3 rows and 6 columns, so 3 × 6 = 18 equally likely outcomes.
A product is odd only when both numbers are odd: 2 odd numbers on the spinner and 3 on the dice give 2 × 3 = 6 odd cells. The other 12 cells are even, so the probability of an even product is .
The spinner down the side and the dice across the top, with the product in each cell. The 12 even products are colored.
The usual mistakes
Treating the 11 totals from 2 to 12 as equally likely. A total of 7 comes from 6 pairs and a total of 12 from only 1; the 36 pairs are the equally likely outcomes.
Merging (3, 5) and (5, 3) into one outcome. They are different rolls, because the red dice shows 3 in one and 5 in the other. Merging every such pair leaves 21 outcomes that are not equally likely.
Going one cell too far along a diagonal. A total of 6 has 5 pairs, not 6: a sixth pair would need one dice to show 0.
Counting the double six twice when counting "at least one six". It is in the six row and the six column, but it is one outcome.
Leaving jail
In the application below, a player in a board game leaves jail on a total of 8 or more with two dice. The diagram of totals gives the probability, and a changed rule, a total of exactly 7 or exactly 11, is compared with it.
Worked example: Leaving the Jail Square in a Board Game with Two Dice, Read from a Possibility Diagram
Question In a board game, a player in the jail square throws two fair dice and leaves only if the total is 8 or more. (a) Draw a possibility diagram of the totals and find the probability that the player leaves on one throw. (b) The rule is changed so that the player leaves on a total of exactly 7 or exactly 11. Find the new probability, and say which rule makes leaving more likely.
1.Draw a 6 by 6 grid with the first dice across the top and the second dice down the side, and write the total in each of the 36 cells.
The first dice goes across and the second down. Each of the 36 cells is one equally likely throw, with its total written in it. 2.Shade the cells with a total of 8 or more. They lie on diagonals: 5 cells total 8, 4 total 9, 3 total 10, 2 total 11 and 1 totals 12, which is 5 + 4 + 3 + 2 + 1 = 15 cells.
A total of 8 or more is shaded. The shaded cells lie on diagonals of 5, 4, 3, 2 and 1 cells, which is 15. 3.(a) The probability of leaving is 1536 = 512.
(a) The probability of leaving is 1536 = 512. 4.Under the new rule, 6 cells total 7 and 2 cells total 11. No cell has both totals, so the two events are mutually exclusive and their probabilities add: 636 + 236 = 836.
A total of 7 fills 6 cells and a total of 11 fills 2. No cell has both, so the events are mutually exclusive. 5.(b) The new probability is 836 = 29. Since 836 is less than 1536, the first rule makes leaving more likely.
(b) 636 + 236 = 836 = 29, which is less likely than 1536.
Answer: (a) 1536 = 512; (b) 29, so the first rule makes leaving more likely
Common mistakes
- Counting the totals 2 to 12 as 11 equally likely outcomes, so that a total of 8 or more has probability 511. The totals are not equally likely: a total of 7 comes from 6 pairs and a total of 12 from only one. The 36 pairs are the equally likely outcomes.
- Counting (3, 5) and (5, 3) as one outcome. They are different throws, because the first dice shows 3 in one and 5 in the other. Merging such pairs leaves 21 outcomes that are not equally likely, and gives the wrong answer 921 for part (a).
More probability with several events problems, worked step by step →