A Trick With Money

Everyone loves money! Everyone loves magic! Everyone loves mathematics!

Sadly not all of this is true. Actually most of it is not true. Imagine how difficult it must be for David and me to earn a living. Here is a trick we use in bars when people are very drunk. And we are desperate for cash.

We say to a drunk person, “Here are 20 coins, all showing heads. While I look away you turn over eight of them to show tails. I will keep looking away, you can even blindfold me, and I will divide the coins into two piles, each having the same number of tails face up. If I can’t do it you keep the 20 coins. If I can then you give me the same amount. Deal or no deal?”

How is this possible? How can we get two piles with same number of tails if we can’t see anything?

I will give the solution and then prove it is possible by algebra. Algebra is when you use letters instead of numbers. We’ll talk about symbols mathematics and numbers mathematics later.

But would you mind taking just a minute or two to have a go yourself? I know, I know. I also hate it when I’m asked in a book to spend a minute breathing deeply, or making a list of all the things I’d like to change about myself. But we aren’t going to do this often. Honest.

Ok, the solution. All you have to do to win the money from the drunk person is to move any eight of the coins to another pile, and turn all of them over. You will see that both piles have the same number of tails. Ok, it won’t be four in each pile, maybe that is what you expected. But I did not say that.

Here is the mathematical proof in a table. We don’t have to use 20 coins and turn over eight. Here we start with m coins and turn over n. When we move n into the new pile (on the right) x is the number of heads that get moved across, we don’t know what this number is but it doesn’t matter. See how the number of tails, coloured red, is the same on the left and the right. And it doesn’t depend on m or n.

Usual deal, if you do this in a bar and win then I get my 10% commission. If you mess it up and lose, I don’t pay you. It’s a bit like the performance-related pay in a hedge fund.

Mathemagical Thinking Lesson

This lesson is about thinking in different domains.

Did you figure out how to do trick? If you hadn’t then I wonder if it would have been easier if we’d used the numbers five and two instead? Or even two and one! What about 137 and 59?

With two and one, I think you would have got it. But then it wouldn’t have looked special, just dumb. With five and two you might have got the solution if you could be bothered. With 20 and eight we reckon you might have given up. That they are both even numbers and there are two piles might seem to be a clue, but is misleading. But then with 137 and 59 you might have thought there’s something special about those two numbers, maybe it’s all about prime numbers. Now there’s a garden path you wouldn’t want to be led along.

The mathematician has two ways to approach this.

The first method: Given the 20/eight version the mathematician might wonder if the numbers are special. If not, then they’d reduce the problem to the two/one version. Then move on to three/one, etc. Then 20/eight and finally the general solution, using letters instead of numbers.

The second method isn’t just for mathematicians, and is the way I approached it initially. We have four numbers to play with: Four (half of eight…there’ll be two piles); Six (half of 20-8); Eight; And 10 (half of twenty). (Twelve isn’t special, moving 12 is the same as moving eight.) And there’s one optional thing to do, turn coins over. There aren’t many experiments to go through. Moving four and turning them over is quickly shown to not work. And moving and turning eight works. Solution found.

We’ll talk about numbers people and symbols people later. There is a great advantage to being able to think in both domains. The mathemagician knows this, and knows how to distract either with the tedium of dealing with numbers or the fear of dealing with symbols.