The Lady or the Lions

The King showed the Princess a map of the maze and the Princess was allowed to decide which room she would wait in. She was not allowed to send a copy to her lover who would have to guess which path to follow. Which room should she wait in to give her lover the greatest chance of finding her?

Four fair dice are marked differently on their six faces. Choose first ANY one of them. I can always choose another that will give me a better chance of winning. Investigate.

Squaring the Circle

Bluey-green, white and transparent squares with a few odd bits of shapes around the perimeter. But, how many squares are there of each type in the complete circle? Study the picture and make an estimate.

Nines and Tens

Stage: 3 Challenge Level:

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 $\frac{27}{216}\geq\frac{25}{216}$