The problem of Hannah's sweets
A few weeks ago, teenagers who had just sat a GCSE maths exam took to Twitter to talk with each other about their test. Most specifically, a question regarding Hannah and her sweets. Adults across the internet decried the teenager's lack of skills but was the question really as straightforward as they made it out to be? Jordan Ellenberg, author of How Not to be Wrong, our Non-Fiction Book of the Month, explains the problem with Hannah and her sweets.
Teenagers around the UK were flummoxed recently by a mystifying question on the math portion of the GCSE. The notorious “Hannah’s Sweets” problem goes like this:
Hannah has n sweets. 6 of them are orange, the rest yellow. Hannah chooses two sweets at random from the bag. If the probability that these two sweets are both orange is 1/3, show that n^2 – n – 90 = 0.
The quadratic equation seems to barrel in out of nowhere, imposing itself on an innocent question about a girl and her candy. What is it doing there? How can you “get this”?
First, I’m going to explain what the writers of this question had in mind. There are n ways to choose the first sweet, all equally likely, and n-1 ways to choose the second (one sweet having already been removed.) So there are n(n-1), or n^2 – n ways to choose the two sweets.
How many ways are there to choose two orange sweets? When you pick the first sweet, there are 6 orange ones to choose from. Having pulled one of the oranges out of the bag, 5 of your choices for the second sweet yield an orange one. So there are 6 x 5 or 30 ways of choosing two orange sweets.
So the chance of drawing two orange sweets is
30/(n ^2 – n).
If it is to be equal to 1/3, we need
30/(n^2 – n) = 1/3
90 = n^2 – n
n^2 – n – 90 = 0
Simple, right? So am I saying this is actually an easy question and kids shouldn’t have complained?
No! This is a terrible question. There are two big reasons why.
First of all – why ask the student to show that n^2 – n – 90 =0? Is that a question you would ever ask about a bag of sweets? Obviously not. You would ask – and the GCSE question should have asked – how many sweets are there?
The answer is 10. A student can figure this out in various ways. They can, for instance, do the formal computation above, and then try to figure out a value of n for which n^2 – n = 90. They can do this using the quadratic formula, or (better, in my opinion!) they can just think about what number n satisfies that equation. A student can also approach the problem by trial and error, trying various values of n, seeing if they result in a chance to two oranges higher or lower than the 1/3 demanded by the problem, and thereby zeroing in on the right value of n.
But even if the problem had simply asked “How many sweets are there?” it would still be a terrible problem. Because suppose, in practice, you pulled two sweets out of the bag. Would you know what the probability had been that the two sweets were both orange? No! You would only know whether the sweets were orange or not. Math is supposed to help you turn observations into knowledge; but the setup of this problem refers to an observation it’s hard to imagine actually happening!
It’s a fun challenge to try to write a version of this problem that makes any sense. Here’s one go:
Hannah has 100 dispensers of sweets to distribute. She knows that all the dispensers have the same number of sweets, and that each dispenser has 6 orange sweets, and the rest are yellow. But the number of yellow sweets in each dispenser is written in illegible handwriting and Hannah doesn’t know what it is. Hannah wants to know how many yellow sweets are in each dispenser, but she doesn’t want to just empty one and find out, because then one of the 100 children in her care won’t get a dispenser and will throw a gigantic tantrum. Instead, she dispenses two sweets from each of the dispensers. She finds that 33 of the dispensers give two orange sweets, 13 give two yellow sweets, and the other 54 give one orange and one yellow. How many yellow sweets should she estimate are in each bag?
Do you think the GCSE students would find this version of the problem harder, easier, or about the same?