Hannah’s Sweets—Anatomy of a Math Problem

The following problem has the internet all abuzz.

hannahs sweets

The problem comes from the Edexcel Maths GCSE paper—which, as I understand it is an examination for UK 16-year-olds with aspirations of studying at a university.

The buzz is fascinating for a number of reasons: some positive; some not so positive. But all are worth thinking about. First off, many students found the problem to be very difficult. I’m not sure why this is, but I’ll offer some speculations as we move through this post.

(I’ve had to convert my Word document into graphics because the equation objects wouldn’t paste into the WordPress format.)

Why I Like the Problem

There are a few very nice things about this problem. First, it asks students to demonstrate something, rather than to find a single solution. As a math teacher, I am always looking for demonstrations of understanding, rather than the consequence of a search that may or may not simply be a matter of “getting lucky”. On that note, I am pleased that this question requires a human marker. No multiple choice or numerical response item could replicate this. To demonstrate understanding, students must use natural language (i.e. English) and structured symbolic representation (i.e. numbers and letters with mathematical connectives) to convince another human mind that they understand. This is good.

Second, there are at least four straightforward methods of solution available to an average post-secondary-bound high school student. No one is bound to a single method. If I were to use this question in class, I would have choices for how to approach it, depending on the goals of the lesson.

Solution #1—Elementary Probability

This is probably the most straightforward way to approach the problem. We use two pieces of basic knowledge.


This is not hard, and it involves very easy manipulation. I’ll go over the next two solutions quickly, as they are very similar in form, but not in conception.

Solution #2—The Fundamental Counting Principle

This is a close cousin to #1. The idea is that if you have n ways to do one thing, and m ways to do another distinct thing, then you have n×m ways to do both. So if you have 3 pairs of pants and 4 shirts, you have 12 different outfits you can wear. We now see that Hannah has n(n-1) ways to pick two sweets out of n, (She ate the first sweet, so she only has n-1 left for her second choice.) By similar reasoning she has 6×5=30 ways to pick and eat two orange sweets. From here, the arithmetic and algebra are the same as #1.

Solution #3—Combinatorics


Solution #4—All Guts no Math


The Downside

One of the standard complaints about this problem (and others like it) is that it is so artificial. Sure it gives us a protagonist, a storyline and some talk of yummy sweets, but this is a million miles from real-world mathematics.

I hear this kind of objection all the time, and frankly I don’t get it. One of the other complaints about mathematics is its coldness and distance from human creativity. Well this is a silly story. It’s a flight of fancy; it’s a goofy narrative that paints a picture. Yes, very few people will do math when eating candy; we all know that. But no one ever complains that fiction is unlike real life, because it reminds us of real life and gives us something to think about. As do silly problems like this one.

Now it is often easy to create a problem that can be stated with or without the narrative. Indeed, it is often the case that the narrative obscures the mathematics. That isn’t the case here. A scenario free statement of this question would be quite difficult to write without absolutely obscuring the clear and simple question being asked.

So why did students find it hard? My guess is because the question is surprising. Students are often lulled into calculating numbers, solving equations, concretely answering problems. In this case, they were given a concrete problem and were asked to show that the problem leads to an unexpected equation. The work isn’t difficult, but the interpretation and strategy-selection were surprising. At least that’s my supposition at this time.

Whew. This is more concrete mathematics than I aim for in this blog. Hope it brought some thought and (dare I say it?) some fun to your day.

If any students who wrote the paper happen to stumble onto this blog, I’d love to hear your reflections on the problem.


3 thoughts on “Hannah’s Sweets—Anatomy of a Math Problem

  1. John, I was puzzled for about 10 seconds, because for some reason, I didn’t relate the n in the sweets problem to the quadratic equation I saw below. For a moment it seemed like a non-sequitur.

    That said, it’s an okay problem. But not very rich (as the current buzzword goes). I think a good problem allows you to start fiddling with the conditions and gets you asking further questions. For example, our old friends the Fox, geese, and the bag of corn. What happens if there are two foxes, or two geese or two bags of corn. What if m, n, and p of the above are involved? What if only x number of animals/bags of corn can fit in the boat? What if you have two boats, three boats, z boats?

    And so on. There’s a great book about problem posing called The Art of Problem Posing that I think you would enjoy reading if you haven’t come across it.


    Liked by 1 person

  2. I must admit I’m confused about Hannah’s sweets. I have a different solution which I believe is correct but can’t understand that everyone else should have it so wrong. For what it’s worth the following is my solution.

    The probability of any sweet in the bag being orange is 6/n. That’s true of any sweet you pick out of the bag, first one, second one and any subsequent ones. Each of them has an equal probability, 6/n, of being orange. Removing one doesn’t change the probability of the ones remaining being orange. Of course if you know the first one was orange then that does change the probability for the subsequent ones. But in the examination question you don’t know whether the first one was orange or not. The probability that it was is 6/n.

    After the first sweet is removed from the bag there is n sweets either in the bag or out of it, we don’t know which are orange. There is 6/n chance that the outside one is orange and there is a 6/n chance for each of the inside sweets to be orange. When you take out a second sweet then you have two sweets that might or might or might not be orange. The probability of both of them being orange is 36/(n x n). That probability can never equal 1/3 since n must be an integer. Consequently the quadratic equation is wrong since it applies to the probability being 1/3, which is impossible.


    • Hi Belinda. Thanks for your comment. If I understand your comment, you have misinterpreted the important point that Belinda eats the sweets after choosing them, making them unavailable for later selections. Because of the sweet-eating, there are n sweets for the first choice, but only (n-1) for the second choice.

      Now, if we look at the possibility of choosing two orange sweets in a row, there are 6 orange sweets available for the first choice, but once Hannah eats it, there are only 5 orange sweets available for the second choice.

      I think you are aware of this, but miss its importance. The two events are separate events and the first choosing has an effect on the second choice. To clarify this, consider two scenarios.

      1. Hannah, who is pregnant, has a daughter. What is the probability that her second child will also be a daughter? Here, the probability is clearly 1/2; the first daughter has no bearing on the probability of the second child being a daughter.

      2. There are two boys and two girls in a room. One girl leaves the room, and a second child is chosen at random. What is the probability that the randomly chosen child is a girl? Here, even though there are equal numbers of boys and girls, the correct probability is 1/3.

      In the first example, the two events are completely independent. In the second example, the probability of the second event is dependent on the first event. (If a boy had left the room, then there would have been a 2/3 probability of a girl being chosen in the second event.)

      I believe that you were not convinced that the second sweet-choosing event is dependent on the first. But it is.

      Does this help? Or am I misunderstanding your question?



Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s