From the Exploring Data website - http://curriculum.qed.qld.gov.au/kla/eda/
© Education Queensland, 1997

Solution to Prisoners!

Teacher Notes

1. If you have never used this activity before, you might be surprised at what appears to be the best answer.

2. This activity could be used with year 8 students in an introduction to probability, or with year 12 Maths C students in the study of expected value and probability distributions, or with any group in between.

3. The theoretically best answer still leaves room for some discussion as the theoretical number of prisoners in each cell is not an integer.

Solution

The grid showing all possible outcomes is given below:

 

1

2

3

4

5

6

1

0

1

2

3

4

5

2

1

0

1

2

3

4

3

2

1

0

1

2

3

4

3

2

1

0

1

2

5

4

3

2

1

0

1

6

5

4

3

2

1

0

 

The proportion of each value is:

Value

Fraction

Frac / 6

0

6/36

1 / 6

1

10/36

1.7 / 6

2

8/36

1.3 / 6

3

6/36

1 / 6

4

4/36

.7 / 6

5

2/36

.3 / 6

Based on the above fractions, one good solution would be to put one prisoner in each of 0, 2, 3 and 4, and two prisoners in 1. Since 5 occurs so rarely , put no prisoners in cell 5.