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# Nine or Ten?

With 2 dice, there are actually 4 ways of getting a score of 9, because the pairs can be reversed:

9 = 3 + 6

9 = 6 + 3

9 = 4 + 5

9 = 5 + 4

(so getting a 3 on the red die and a 6 on the blue die is not the same as getting a 6 on the red die and a 3 on the blue die).

But there are only 3 ways of getting a score of 10, because the pair 5 + 5 cannot be reversed.

With 3 dice, you can use a systematic approach to find all of the ways of making 9.

If we start with 1 on the first die, and then with the smallest possible number on the second die, we can list them all - and we can avoid counting any twice if we make sure the numbers never get smaller as we list them.

1, 1, is an impossible start because next we would need a 7

1, 2, 6

1, 3, 5

1, 4, 4

1, 5 won't work because next we would need 3, but then the numbers are getting smaller as we list them (3 is smaller than 5)

2, 2, 5

2, 3, 4

3, 3, 3

Possible trios of numbers that add up to 9 and to 10 are shown in the table.

**Using logic**

So there are 3 ways to get 9 or 10 using 3 different numbers, and at least two ways to get 9 or 10 using two different numbers.

Additionally, there is 1 way to get 9 with all 3 numbers the same, and 1 more way to get 10 with two numbers the same.

There will be more different orders of this last way to get 10 than of this last way to get 9 - so you are more likely to get a 10 than to get a 9 with 3 dice.

**Counting**

If all the numbers are different, use 1 + 2 + 6 = 9 as an example.

Options with the 1 first: 1 + 2 + 6, 1 + 6 + 2

Options with the 2 first: 2 + 1 + 6, 2 + 6 + 1

Options with the 6 first: 6 + 1 + 2, 6 + 2 + 1

That is a total of 6 options.

If two of the numbers are the same, use 1 + 4 + 4 = 9 as an example.

The 1 could go first, middle or last: 1 + 4 + 4, 4 + 1 + 4, 4 + 4 + 1

That is a total of 3 options.

If all of the numbers are the same, then there is no way to change the order.

So to get 9, there are a total of 3$\times$6 + 2$\times$3 + 1$\times$1 = 18 + 6 + 1 = 25 ways.

To get 10, there are a total of 3$\times$6 + 3$\times$3 = 3$\times$9 = 27 ways.

So you are more likely to get a score of 10 than a score of 9 with 3 dice.

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With 2 dice, there are actually 4 ways of getting a score of 9, because the pairs can be reversed:

9 = 3 + 6

9 = 6 + 3

9 = 4 + 5

9 = 5 + 4

(so getting a 3 on the red die and a 6 on the blue die is not the same as getting a 6 on the red die and a 3 on the blue die).

But there are only 3 ways of getting a score of 10, because the pair 5 + 5 cannot be reversed.

With 3 dice, you can use a systematic approach to find all of the ways of making 9.

If we start with 1 on the first die, and then with the smallest possible number on the second die, we can list them all - and we can avoid counting any twice if we make sure the numbers never get smaller as we list them.

1, 1, is an impossible start because next we would need a 7

1, 2, 6

1, 3, 5

1, 4, 4

1, 5 won't work because next we would need 3, but then the numbers are getting smaller as we list them (3 is smaller than 5)

2, 2, 5

2, 3, 4

3, 3, 3

Possible trios of numbers that add up to 9 and to 10 are shown in the table.

Sum of 9 | Sum of 10 | |
---|---|---|

3 different numbers | 1, 2, 6 1, 3, 5 2, 3, 4 |
1, 3, 6 1, 4, 5 2, 3, 5 |

2 numbers the same | 1, 4, 4 2, 2, 5 |
2, 2, 6 2, 4, 4 3, 3, 4 |

All numbers the same | 3, 3, 3 |

So there are 3 ways to get 9 or 10 using 3 different numbers, and at least two ways to get 9 or 10 using two different numbers.

Additionally, there is 1 way to get 9 with all 3 numbers the same, and 1 more way to get 10 with two numbers the same.

There will be more different orders of this last way to get 10 than of this last way to get 9 - so you are more likely to get a 10 than to get a 9 with 3 dice.

If all the numbers are different, use 1 + 2 + 6 = 9 as an example.

Options with the 1 first: 1 + 2 + 6, 1 + 6 + 2

Options with the 2 first: 2 + 1 + 6, 2 + 6 + 1

Options with the 6 first: 6 + 1 + 2, 6 + 2 + 1

That is a total of 6 options.

If two of the numbers are the same, use 1 + 4 + 4 = 9 as an example.

The 1 could go first, middle or last: 1 + 4 + 4, 4 + 1 + 4, 4 + 4 + 1

That is a total of 3 options.

If all of the numbers are the same, then there is no way to change the order.

So to get 9, there are a total of 3$\times$6 + 2$\times$3 + 1$\times$1 = 18 + 6 + 1 = 25 ways.

To get 10, there are a total of 3$\times$6 + 3$\times$3 = 3$\times$9 = 27 ways.

So you are more likely to get a score of 10 than a score of 9 with 3 dice.

You can find more short problems, arranged by curriculum topic, in our short problems collection.