# Troublesome Dice

When dice land edge-up, we usually roll again. But what if we
didn't...?

## Problem

The numbers on the dice talked about in this problem are arranged in the traditional manner where the sum of the numbers on opposite faces is always seven.

Jenny and her family often roll a die to decide who gets to eat the last slice of pie.

There are three people in the family so it's easy to ensure it's fair: each person is allocated two of the six numbers.

However, when they roll it, the die always seems to land "edge-up" on a napkin, and they have to roll again.

One day, Jenny suggests that instead of rolling again, they simply add the two faces that meet at the top edge.

Image

Edge-up score: $6 + 2 = 8$

What different scores could Jenny and her family get?

Before they throw, everyone chooses some of the scores so that if one of their numbers comes up, they get the last slice of pie.

Can the scores be allocated fairly?

What if there are more, or fewer, people at dinner?

The die might also land corner-up like this:

Image

Corner-up score: $6 + 5 + 3 =
14$

How could the scores be allocated fairly this time?

This problem is based on an
idea shared by Paul Stephenson.

## Getting Started

It might help to look at a die (or make one) to refer to when you
are working out the edge totals.

## Student Solutions

Mustafa from Wilson's School sent us this excellent and comprehensive solution.

Michael, also from Wilson's, came to the same conclusion about the possible results when the dice lands on an edge:

If two sides were showing, the following scores could be possible:

+ | 1 | 2 | 3 | 4 | 5 | 6 |

1 | - | 3 | 4 | 5 | 6 | - |

2 | - | - | 5 | 6 | - | 8 |

3 | - | - | - | - | 8 | 9 |

4 | - | - | - | - | 9 | 10 |

5 | - | - | - | - | - | 11 |

6 | - | - | - | - | - | - |

The reason there are no numbers down the left is because 1+2 is the same as 2+1, etc.

There is one 3, one 4, two 5's, two 6's, NO 7's (since the numbers adding up to 7 are on opposite sides of the dice), two 8's, two 9's, one 10 and one 11.

Altogether there are 12 possibilities; 12 is divisible by the 3 family members and can therefore be split equally and fairly between them.

Matthew from The King's School in Grantham followed this up by showing how the numbers could be shared out:

They could be allocated like the following:

'Person A': 3, 4, 10, 11. Total of 4 chances.

'Person B': 5, 6. Total of 4 chances.

'Person C': 8, 9. Total of 4 chances.

If you have 2 people at the table, they could be split like the following:

'Person A': 3, 4, 5, 6. Total of 6 chances.

'Person B': 8, 9, 10, 11. Total of 6 chances.

If you have 4 people at the table, they could be allocated like the following: 'Person A': 3, 6. Total of 3 chances.

'Person B': 4, 5. Total of 3 chances.

'Person C': 8, 10. Total of 3 chances.

'Person D' : 9, 11. Total of 3 chances.

IT IS IMPOSSIBLE TO HAVE 5 PEOPLE AT THE TABLE, AND SHARE THE TOTALS OUT FAIRLY.

If you have 6 people at the table, the totals could be shared out fairly.

Rian and Ben from Waverley School agreed:

Sum of scores on the edge:

Chances of getting

3 = $\frac{1}{12}$

4 = $\frac{1}{12}$

5 = $\frac{2}{12}$

6 = $\frac{2}{12}$

8 = $\frac{2}{12}$

9 = $\frac{2}{12}$

10 = $\frac{1}{12}$

11 = $\frac{1}{12}$

You can split them fairly:

if person 1 gets a score of 3, 4, 10 or 11 the chances of winning are $\frac{4}{12}$

and then person 2 gets a score of 5 or 8 the chances of winning is $\frac{4}{12}$

and then person 3 gets 6 or 9 and the chances of winning are $\frac{4}{12}$.

Sum of scores at the corner:

For this situation, you cannot split them fairly because each score is unique so the probability of getting each number is $\frac{1}{8}$.

There are 8 numbers and 3 people so you can't split it fairly but you can split it between 2, 4 and 8 people.

Irene and Remminbi from Dulwich College in Beijing and Callum from Cholsey Primary School also sent us partial solutions to this problem.

Well done to you all.

## Teachers' Resources

### Why do this problem?

This problem offers students the chance to visualise in three dimensions (using the context of dice), list possibilities systematically, and work on probabilities in a less familiar context.### Possible approach

Begin with some visualising of dice:"Imagine a six-sided die, where the opposite faces add up to 7. Rotate your die so that the top face shows 1. What does your bottom face show?"

"Now rotate it round so that the front face shows 5. What will the back face show?"

"What numbers are on the other two faces? Think about what's on the left and right faces of your die. Are you all picturing the same die?"

These questions tease out an understanding that all 6-sided dice whose opposite faces sum to seven are essentially the same, the only possible difference being that you can make a "right-handed" and "left-handed" version that are mirror images of each other (read more here.)

The problem Right or Left? explores the differences between right- and left-handed dice.

Now introduce the context of the problem and set the class to working out all the possible edge scores and how these could be distributed so that everyone has an equal chance of getting the pie, when there are two, three, four... people sharing the meal.

After everyone has shared their work on this task, the same analysis can be done with the die that always lands "corner-up".

### Key questions

Numbers on opposite faces sum to 7. Does this imply that numbers NOT on opposite faces DO NOT sum to 7?The smallest edge total is 3, and the largest is 11. How do I know that at least one of the numbers between 3 and 11 can be made in more than one way?

The smallest corner total is 6, and the largest is 15. How do I know that at least two of the numbers in between must be impossible to make?