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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.