The Maths of Red Noses
Karate Konk and the Coupon Collector's Problem
06 March 2015
[This blog is about the 2015 Red Nose collection. In 2021 there are once again nine noses plus a rare tenth, but this time the theme is 'the great outdoors']
Are you collecting the set of Comic Relief noses for Red Nose day? No, nor am I. But my kids are.
Altogether there are nine noses to collect, each with its own nasal-punning name, for example 'Snout Dracula' and 'The Snotty Professor'. There is also a tenth nose that is painted gold and which, if you are lucky enough to get one, qualifies you for a fantastic prize, but since the chance of getting one of those is 12 in several million I'm going to ignore it.
So far between them my offspring have bought a total of 14 noses, of which eight are different. They are still waiting to get the ninth, Karate Konk.
It made me wonder how many more they are going to have to buy before they finally get that last nose. With a bit of luck, it'll appear in the next purchase, but of course there is no guarantee that even if they bought a thousand packets there would be a Karate Konk inside, though the chances of failing would be ( 8/9)^1000, an incredibly small number.
The challenge of collecting a full set of something is known as The Coupon Collector's Problem, and I was alerted to it by John Haigh, who explains the maths behind it in detail in his excellent book Taking Chances.
The Coupon Collector's Problem has been around for a long time. Before and after the second World War, cigarette card collection was a national pastime, and in the 1970s I still remember the excitement of collecting the badges of all the league football clubs from Esso Service Stations - and the joy when Dad came home with the final badge (Chester) after weeks of hunting.
But here is the key question. How many coupons, badges or noses should you expect to have to buy before you have the full set? John Haigh reveals that the answer is a rather elegant formula. The average (mean) number N that you have to buy depends on how many items, K, there are in the complete set, as follows: N = K x ( 1 + ½ + 1/3 + ¼ + 1/5 + ... + 1/K).
I've set out the results below:
No. of items in set Average no. you need to buy
1 1
2 3
3 5.5
4 8.3
5 11.4
6 14.7
7 18.2
8 21.7
9 25.5
Notice that if you double the number of items in the full set (from four to eight, say), you more than double the average number you should expect to buy, which I find rather surprising.
The table shows that on average to get the full set of nine Red noses you need to buy between 25 and 26 of them, which is great as a fund-raiser for Comic Relief, but frustrating for collectors.
The calculation makes one huge assumption - that there are equal numbers of each nose, randomly distributed around shops. You can imagine why this might not be true. For example, the organisers might want to hold back the supply of one of the noses for a couple of weeks to reduce your chance of the full set and get you buying more.
Anyway, as it stands, my kids can expect to have to buy nine more nose packets before they get Karate Konk. Either that or - my recommendation - they start asking around their friends and arrange a swap.