Copyright © University of Cambridge. All rights reserved.

William from Churchers College Junior School sent in his good explanations.

$1$a You can roll numbers between $2$ and $12$ with two six-sided dice

b $7$: Because you can roll combinations which use all the numbers on the dice.

$1-6$, $6-1$, $5-2$, $2-5$, $4-3$, $3-4$

$6$ combinations give $7$.

There are $36$ possible combinations when you threw two six-sided dice SO $6$ /$36 \times 100 = 16.67$ % chance of rolling $7$ with two six-sided dice.

$2$a With two ten-sided dice you can roll numbers between $2$ and $20$.

b $11$; because there are ten combinations which give $11$ which use all the numbers on the dice

There are are $100$ possibilities when you roll two ten-sided dice.

So $10$/$100 \times 100 = 10$ % chance of rolling $11$ with $2$, $10$ sided dice.

From Oscar at CCJS (which I think is Cheltenham College Junior School - but forgive me if I'm wrong!) also worked on this and sent in the following;

I wrote down every possible throw I could get using a black and white six-faced dice - it came to $36$ possibilites.

I counted how many of each total number I could get and found that the commonest total number to get was $7$

Two six-faced dice gave me $36$ possibilities. That is $6$ squared or $6 \times 6$.

The two six-faced dice had $7$ as the commonest number that we could get - that is, one greater than the number of sides on each dice.

Finally Tom from Fen Ditton C.P. School wrote to tell us what he already knew or what he found out by doing the activity:

$6$ numbered dice: the most likely number is $7$ because is has got the most combinations.

$10$ numbered dice: $11$ is the most likely because it has the most combinations.

The formula for this problem is:n $+1$ n= no. of sides on the die.

Well done all of you, three excellent solutions sent in, keep up the good work!

$1$a You can roll numbers between $2$ and $12$ with two six-sided dice

b $7$: Because you can roll combinations which use all the numbers on the dice.

$1-6$, $6-1$, $5-2$, $2-5$, $4-3$, $3-4$

$6$ combinations give $7$.

There are $36$ possible combinations when you threw two six-sided dice SO $6$ /$36 \times 100 = 16.67$ % chance of rolling $7$ with two six-sided dice.

$2$a With two ten-sided dice you can roll numbers between $2$ and $20$.

b $11$; because there are ten combinations which give $11$ which use all the numbers on the dice

There are are $100$ possibilities when you roll two ten-sided dice.

So $10$/$100 \times 100 = 10$ % chance of rolling $11$ with $2$, $10$ sided dice.

From Oscar at CCJS (which I think is Cheltenham College Junior School - but forgive me if I'm wrong!) also worked on this and sent in the following;

I wrote down every possible throw I could get using a black and white six-faced dice - it came to $36$ possibilites.

I counted how many of each total number I could get and found that the commonest total number to get was $7$

Two six-faced dice gave me $36$ possibilities. That is $6$ squared or $6 \times 6$.

The two six-faced dice had $7$ as the commonest number that we could get - that is, one greater than the number of sides on each dice.

Finally Tom from Fen Ditton C.P. School wrote to tell us what he already knew or what he found out by doing the activity:

$6$ numbered dice: the most likely number is $7$ because is has got the most combinations.

$10$ numbered dice: $11$ is the most likely because it has the most combinations.

The formula for this problem is:n $+1$ n= no. of sides on the die.

Well done all of you, three excellent solutions sent in, keep up the good work!