Coupon collector’s problem
Since 2010, LEGO has been releasing series of minifigs for people to collect. The current one has The Simpsons characters.
I thought it would be fun to have at least Bart and Homer, but of course you quickly end up wanting all 16 of them. The minifigs come individually in sealed bags, making it impossible to guess what’s inside. So the problem obviously is that the more characters you collected so far, the more likely it will be that you’ll buy one you already have. This is known as the coupon collector’s problem. What a boring name. But go ahead and try to read the wikipedia article. They lost me quite easily. I understand the probabilities: the chance to get your very first minifig is obviously 100%, or 16/16. For the second one, it’s 15/16. The third, 14/16. And so on. My head starts to hurt when the article mentions expectations. I give up here, and go straight to the result.
How many minifigs do I have to buy to get all 16?
55, on average. At CHF 3.50 a piece, that’s CHF 192.50 for 16 minifigs. Crazy! The formula supposes that all minifigs have an equal probability of being selected by you in the store, which is most likely not the case. Lego might produce the same amount of each, but then every store basically gets a random sample in which there might be more of some particular minifig. What I’m trying to say is that 55 is probably not realistic, even as an average.
My experience so far – and I will stop buying them I think – is as follows: I bought 4 bags the first time, and got 4 different minifigs. The second time, I got 3 new minifigs out of 4. The third time, I got 4 duplicates out of 6 bags. So, in total, I had to “draw” 14 times to get 9 minifigs.
A similar problem, linked to from the wikipedia article, is the birthday problem:
find the probability that, in a group of N people, there is at least one pair of people who have the same birthday.
I had looked this one up one night while drunk in a bar and hearing people “singing” Happy Birthday for the second time in the evening. Of course, the group of N people is less random in a bar or a restaurant, because people tend to go out on their birthday. But anyway, if you’re curious, unlike what Fat Mike says (at the end of the song) in their live album the answer is not that you need 700 people (roughly 2x365) to have 2 people with the same birthay, but 367 (=366+1). Which seems obvious. The fun part however is that with 70 people, you’re already at 99.9%. With 50 people at 97%. With 23 people, you have a 50-50 chance.