A gambler bets half the money in his pocket on the toss of a coin,
winning an equal amount for a head and losing his money if the
result is a tail. After 2n plays he has won exactly n times. Has he
more money than he started with?
You have two bags, four red balls and four white balls. You must
put all the balls in the bags although you are allowed to have one
bag empty. How should you distribute the balls between the two bags
so as to make the probability of choosing a red ball as small as
possible and what will the probability be in that case?
To win on a scratch card you have to uncover three numbers that add
up to more than fifteen. What is the probability of winning a
Shaun from Wilson's School observed:
Every graph of the sum of two spinners would bulge in the middle because there is more chance of making the middle number. There is little chance of getting the highest numbers so that would mean the higher and, equally, the lower numbers would come up less often.
Thomas and Holly from Hymers College explained their thinking clearly:
After figuring out the first few graphs and their corresponding spinners, we soon found out that the maximum sum of the spinners equalled the highest bar value for the sum.
We also found that a sample space diagram could be applied to the graphs involving sums.
For A and B, we saw a graph going up and down with equal steps, peaking at 8 and 12 for A and B respectively.
Therefore, we deduced that the two spinner numbers had to be of equal value. As we had already figured out, the sum of A had to be 14, so we could see that the sum was 7 + 7. For B, the sum had to be 20, so the sum had to be 10 + 10.
For C, we knew the sum was 14, and we counted 3 steps up to the peak, so we thought it was 3 + 11, still measuring the sum. It is very hard to get a graph exactly like shown in C, but 3 + 11 is the way to get it.
For D, we followed the same method, but knew the sum was twenty with 6 steps, so after inputting 6 + 14; we got a graph similar to the one shown in D.
For the remaining four, it is obvious from the shape of the graph that it must be on the difference setting, as it would be impossible to get 0 on sum setting. Since E has a maximum difference of 6, it has to be 7 + 7 (because the difference of 6 would be between 1 and 7 (as it is impossible to get 0 and 6)).
We worked on a similar basis for the remaining 3. F was 11 + 11, G was 3 + 7 and H was 4 + 11.
Jennifer from Taipei European School had a go at the final challenge:
The difference between two 1-20 spinners.
There will be 20 bars because the smallest bar is 1-1=0 and the biggest bar is 20-1=19 then 19+1=20 because the smallest number is 0 isn't 1. The second bar will be the longest because 1, the answer, appears most often when finding the difference between two numbers and the last bar will be the shortest because the only way to make 19 is 20-1.
The sum of a 1-20 and a 1-30 spinner
There will be 49 bars because the smallest is 1+1=2 and the biggest is 20+30=50
The first and the last bar will be the same and the two bars will be the shortest because only 1+1=2 and only 20+30=50.
The difference between a 1-20 and a 1-30 spinner.
There will be 30 bars because 1-1=0 and 30-1=29.
The last bar (29) will be the shortest because the only way to make 29 is 30-1. The second bar will be the longest because 1, the answer, appears most often when finding the difference between two numbers.