The birthday problem asks a simple question: if you have a room full of people, how many people do you need so that there’s a 50% probability that at least two of them share the same birthday? In the simplest setting, we make a few assumptions: for example, we usually ignore the presence of leap years, and we assume that each day of the year is equally likely to be someone’s birthday.
With these assumptions, solving the problem is not difficult. The probability that at least 2 people in a group of n share a birthday is 1 minus the probability that all n people have different birthdays. In other words, the probability is
If instead you want to know how many people you need in a room so that there’s a 50% chance one of them will have the same birthday as you, this probability is given by the new expression
As I said, this is a well-known problem in probability theory. A few months ago, however, I was asked about a variant of the birthday problem by my friend Gabe over at Motivated Grammar. Rather than looking for identical birthdays, what if you look for different birthdays? More specifically, how many people do you need in a room to guarantee with, say, a 90% certainty, that every day of the year is someone’s birthday?
Of course, you could always pick poorly so that everyone in the room has the same birthday, but the odds of this happening are quite low. In fact, this problem is more commonly known by another name: the coupon collector’s problem.
For this problem, suppose you are clipping coupons from a newspaper (any newspaper except for USA Today). There are n different coupons you can collect, but each newspaper only has 1 coupon, and you can’t see which coupon the newspaper has until you’ve bought the newspaper. In this context, the question becomes: what’s the probability that you’ll need to buy at least newspapers to collect n coupons?
If you prefer, you can think about this problem in terms of trading cards as well. Each pack of cards is akin to buying a set of newspapers, and you want to know the probability that you’ll need to buy at least a certain number of packs in order to collect all the cards. From baseball players to Pokemon, this same problem governs the distribution (assuming that all cards in the back are equally likely to be in the pack, which may not always be the case).
What’s known about this problem? Well, as I said above, even if you buy hundreds of thousands of cards, or stuff millions of people in a room, there’s no guarantee that you’ll collect every card or every date. However, on average, the number of cards you’ll need to go through to complete a set of size n is about n*logn.
In terms of birthdays, this says that if you want to collect every date, on average you’ll need to pool together around 2,153 people. Why such a large number? It’s not unreasonable to expect something like this – when you first begin collecting people, it won’t be hard to get people with different birthdays. However, as your numbers increase, you’ll get a new birthday less and less frequently. Finding that last birthday could prove to be quite elusive.
The same analysis works for trading cards. Trying to complete your collection of series 2 Teenage Mutant Ninja Turtles Animated Series trading cards? Well, my friend, with 88 cards total and 5 cards per pack, you can expect to buy around 79 packs of cards. Perhaps this box set would be a better investment.
While the expected values are easy to calculate, it may be that you need to greatly exceed the expected value in order to complete your collection. However, one can use Markov inequalities to get bounds on the probabilities. For example, there’s a 90% chance that you’ll be able to hit every birthday if you take no more than about 21,535 people. To bump those odds up to 95%, take 43,069 people.
So, for parents whose children who have gotta catch ‘em all, you can use these methods to get a rough estimate for how much that completion will cost you. And if you’re trying to get a room full of people together so that every day of the year is someone’s birthday, I’d strongly suggest not picking people at random. What an awkward party that would be.