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Nines and Tens

Stage: 3 Challenge Level: Challenge Level:2 Challenge Level:2

Congratulations to Josh Tattersall , age 10 from Amptill, Bedfordshire and to D. Crowe from the K.S.3 Maths Club, Strabane Grammar School for the solutions which follow:

Joshua writes: With two dice you are more likely to throw a 9 than a 10 because in order to throw a 9 you can get a 3, 4, 5 or 6 with the first die. That is a 2 in 3 chance. But to throw a 10, you can only throw a 4, 5 or 6. That is only a 1 in 2 chance.

Whichever number you throw on the first die, the odds of getting the required number on the second die is 1 in 6.

So the chance of throwing a 9 with two dice is 1 in 9 and the chance of throwing a 10 is 1 in 12.

With three dice it is different and then there would be a better chance of throwing a 10 than a 9.

This is D. Crowe's solution. When 2 dice are thrown you are more likely to get a total of 9 than of 10 because there are 4 ways of getting a total of 9

(3,6), (4,5), (5,4), (6,3)

but only 3 ways of getting a total of 10

(6,4), (5,5), (4,6).

With 3 dice a total of 10 is more likely than a total of 9 because there are 27 ways of getting a total of 10 but only 25 ways of getting a total of 9 as described below:

Total of 9

6 arrangements of 1, 2, 6 - {(1, 2, 6), (1, 6, 2), (2, 6, 1), (2, 1, 6), (6, 2, 1), (6, 1, 2)}

6arrangements of 1, 3, 5

6arrangements of 2, 3, 4

3arrangements of 1, 4, 4 - {(1, 4, 4), (4, 1, 4), (4, 4, 1)}

3arrangements of 2, 2, 5

1 arrangement of 3, 3, 3

25 possible arrangements

Total of 10

6arrangements of 1, 5, 4

6arrangements of 1, 3, 6

6arrangements of 2, 3, 5

3arrangements of 2, 4, 4

3arrangements of 2, 2, 6

3arrangements of 3, 3, 4

27 possible arrangements

James Page, Hethersett High School, Norwich and Nisha Doshi, Year 9, the Mount School, York gave similar solutions but included a mention of the probabilities:

2 dice

The probability of scoring 9 is $\frac{4}{36}=\frac{1}{9}$

The probability of scoring 10 is $\frac{3}{36}=\frac{1}{12}$

and so you are more likely to score 9 because $\frac{1}{9}\geq\frac{1}{12}$

3 dice

The probability of scoring 9 is $\frac{25}{216}$

The probability of scoring 10 is $\frac{27}{216}$

and so you are more likely to score 10 because $\frac{27}{216}\geq\frac{25}{216}$