Charlie and Alison's version, Aug 2011
OLD VERSION??
a. 1-4, 1-4, 1-5 and 1-5
b. 1-9 and 1-9 (sum)
c. 1-13 and 1-4 (difference)
d. 1-13 and 1-13 (difference)
e. 1-4, 1-4 and 1-4 (sum)
f. 1-2, 1-4 and 1-6 (sum)
Edgar from Dr Challoners Grammar School was the first to send in a detailed answer.
He writes "For A, the first step I took was to note that the lowest value recorded was $2$. This told me that the sum of two spinners had been recorded; the lowest possible value each spinner can have is $1$, and $1+1=2$. The highest value recorded was $18$, and this told me that because there was a 'triangle' shape on the graph, the numbers were both $9$ because $9+9=18$. The 'triangle' shape I
will now proceed to explain.
"If there are two spinners of nine sides spun at once, there are $9\times 9=81$ possibilities for combinations they can produce. For $2$ or $18$ respectively, there is only one possibility in which they are produced - $1+1$ and $9+9$ respectively, giving them a probability of $\frac{1}{81}$. For $3$ and $17$, however, there are two possibilities - $1+2$, $2+1$ and $8+9$, $9+8$ respectively,
giving them each a probability of $\frac{2}{81}$. This pattern continues all the way to $10$ which has $9$ possibilities - $1+9$, $2+8$, $3+7$, $4+6$, $5+5$, $6+4$, $7+3$, $8+2$ and $9+1$, and then the probability declines again down to eighteen. Providing each spin of the spinner is totally random, two spinners of the same number should always come up with this 'triangle'.
"For B, the solution I came up with was the difference of two spinners of thirteen. The 'triangle' for differences is a right angled one, but it has an anomalous probability for zero, which can only be achieved when both numbers are the same. The reason for this is similar to the corresponding one for the sum, only reversed. Similarly, this 'triangle' shape can be achieved only when both the
spinners must be the same.
"For C, the spinners are a four and a thirteen, and the difference is taken. The graph is a sort of plateaued 'triangle' shape. This shape is created because certain numbers only be achieved in four ways (e.g. $8=12-4=11-3=10-2=9-1$, $7=11-4=10-3=9-2=8-1$) so their probabilities are not different, so the shape created is basically a 'triangle' with the middle stretched out in a sort of plateau
shape. This 'plateau' can also be created when adding the spinners.
"For D, the plateau shape appears again, only this time in the form of a symmetric sum graph. The two spinners involved are a fourteen and a four. I could tell that there were two spinners being added together because the lowest value is two ($1+1=2$), one of them was a four (because there are four steps up to the plateau) and then the other one must be the highest value eighteen minus four which
gives fourteen.
"For E, I first concluded that there must be three spinners because the lowest outcome was three ($1+1+1=3$). Because there was a bit of a jump up to the middle four on the graph, I concluded that the spinners might all be the same (when there are more than two spinners, intermediate results are made more likely). The highest outcome was twelve, and $\frac{12}{3}=4$, so therefore all the spinners
must be four.
"For F, there was no 'jump up' in the middle of the graph, and there were three spinners involved, and because of the absence of the 'jump up', I surmised that the numbers ought to be consecutive (this makes the 'triangle' shape) and the only three consecutive numbers I know which sum to $12$ are $3$, $4$ and $5$, so I concluded that these were the numbers of the spinners."
An outstanding solution, Edgar, well done! Thomas from A Y Jackson also sent in a full, detailed solution to this problem. Interestingly, he had a different answer for F, suggesting that the spinners used were three, three and six.