Prickles, Lizards, Ogres, Collectible Cards and Calculus

[This is a back-issue of one of this site’s newsletters]

The other day, after a walk, I found a double-gee seed stuck in my shoe. A nasty little weed. Lucky I was wearing shoes!

It reminded me of a story of how kids used to sell these seeds to visiting sailors, claiming they were the eggs of a lizard called the Thorny Dragon. This lizard is quite amazing. One of its tricks is that water condensing on its body makes its way via capillary action to the mouth – a useful trick for a desert reptile!

This in turn reminded me of a promotion by a big national supermarket chain (Woolworths) – collecting Aussie Animals cards. There were 100 or so cards to collect, and you’d get one card for every – oh, I don’t know – let’s say $10 you spent.

Of course, kids wanted to collect the whole set, and the supermarket chain even sold albums to keep the cards in. The promotion was so successful that they launched another one – a collection of 42 Dreamworks Heroes cards, with characters from movies like Shrek, Turbo, The Croods and others. It’s one card per $20 spent now.

If you never swapped cards with friends, how much would you have to spend to complete the collection? It’s actually quite easy to work out how much you’d need to spend on average.

Before I show you how, let me tell you about a new video series I’ve started uploading to Youtube.

Remember Ruler-and-Compass constructions?

I’ve been making a videos showing how to draw various shapes (regular polygons) using only a ruler and a straightedge. If you’ve been following my blog, facebook page or Youtube Channel you might have already seen some. If not, check out the first one: how to construct an equilateral triangle inside a given circle.

Anyway, back to Dreamworks Hero collectible cards.

The way to work out how much you’ll need to spend is to ask this question: if you already have some cards in your collection, how many cards do you have to get to find a new one you don’t already have?

If you don’t have any, that’s easy. You just need one. Suppose, however, you had 21 cards in your collection. There’s a 50% chance you get a new card in one go. There’s a 25% chance that you’ll only get your new card on your second try. There’s a 12.5% chance it will take three tries, and so on. On average, you’ll need two cards. You’re missing precisely half the cards, the number you need to add ojne to your collection is one over a half.

If you had 41 cards, most of the time you went shopping, you’d get repeats. On average, you’d need to buy 42 cards to get that last one to complete your collection.

The total you’d need from beginning to end would be would be 42 times (1/42 + 1/41 + 1/40 + … + 1/2 + 1). If there were N cards, the formula would be N times (1/N + … + 1/2 + 1).

The sum in the brackets works out to be approximately ln(N), the natural logarithm of N. Yes, calculus even appears in questions about movie characters. More accurately, it’s ln(N) plus about 0.5772 plus about 0.5/N.

You can use this kind of calculation whenever you’re trying to collect the complete set of something, and the item you get is random. Lego Minifigures, anyone? For the whole collection of Dreamworks heroes, people need to collect 181.72 cards on average, spending $3634 on groceries.

It’s no wonder the promotion is so successful, and that parents desperately try to swap cards with each other.

I predicted to my wife that Woolworths’ major competitor will follow up soon enough with a collectible set of their own, maybe of Disney Heroes.

We shall see.