


What a Coincidence! MATHEMATICAL RECREATIONS by Ian Stewart Scientific American, 1998 Some years ago a friend of mine was on honeymoon, camping in a remote part of Ireland. He and his wife were walking along a deserted beach, when in the distance they saw two people slowly walking towards them. The other couple turned out to be my friend's Head of Department and his new wife. Neither couple knew of the other's plans: it was a coincidence. Such a striking coincidence, in fact, that it always makes a good party story. People seem fascinated by coincidences — nearly everyone can relate several that have happened to them. Should we be impressed by coincidences, however? Especially when they seem to happen to everyone! Robert Matthews, an English journalistmathematician whose work has featured in this column before (The anthropomurphic principle, December 1995; The interrogator's fallacy, September 1996) thinks not. In Teaching Statistics volume 20 Number 1, Spring 1998, he and Fiona Stones of Granada TV examine one of the commonest types of coincidence: people who have the same birthday. Their conclusion is that we are unduly impressed by coincident birthdays because we have a very poor intuition for how likely such events are. The effect is exaggerated because we wait for coincidences to happen, instead of going looking for them. If we went around asking everybody we met for their birthday, for instance, we'd soon find out just how common coincident birthdays are. In fact, these two considerations debunk a great many apparently impressive coincidences.
By 'birthday' I mean day and month, but not year. To keep things simple, I'll ignore 29th February, so there are 365 different birthdays. I'll also assume that each birthday is equally likely to occur, which is not entirely true: more children are born at some times of the year than at others. Taking these extra factors into account would complicate the analysis without greatly changing the general conclusions. OK, how many? A hundred? Two hundred? A survey of university students carried out by Matthews and Susan Blackmore (Why are coincidences so impressive? Perceptual and Motor Skills volume 80 (1995) 11211122) led to estimates centred around 385. This is fascinating, since as soon as the room contains 366 people (or 367 if we include 29 February) a coincidence is guaranteed. As Matthews remarks, this result suggests that 'the probabilistic aspects of the paradox evade many people.' The question is paradoxical in the following sense: the correct answer seems much too small, and people tend not to believe it. In fact, the answer is 23 people. This seems surprisingly small, but it's right nonetheless. Before I explain why, here's another deceptively similar question:
This time the answer is 253 — not counting you — which seems much too big! Why do two such similar questions have such different answers? One of the most useful tricks in probability calculations is to consider the probability that the event under consideration does not happen. If we know this number, then all we have to do to get the probability that the event does happen is to subtract the number from 1. And it's amazing how often it's easier to calculate the probability that an event does not happen. For instance, when does the event 'at least two people have the same birthday' not happen? When all of their birthdays are different. Suppose we start with just one person in the room, and bring extra people in, one at a time. We calculate the probability that the new person has a different birthday from all of the previous ones. Now, an event that is equally likely to happen or not has probability \frac{1}{2} — if the probability exceeds \frac{1}{2}, then the event is more likely to happen than not, but if it is less than \frac{1}{2}, the event is more likely not to happen than it is to happen. So \frac{1}{2} is the breakeven point, and we focus on that value. As people enter the room, the probability of a coincidence increases, so the probability that all birthdays differ steadily decreases. Therefore, as soon as that probability of no coincidences drops below \frac{1}{2}, we know that the probability of the original event — at least two people have the same birthday — has risen above \frac{1}{2}. OK, here goes. With one person only — Alfred, say — there is no problem in making all birthdays different, and the probability is 1. Enter Betty. There are 365 possible birthdays, but Alfred has used up one of them. That leaves 364 possiblilities for Betty, if their birthdays are to be different. So the probability that their two birthdays differ is 364/365. Next, Carla. Now there are only 363 choices that keep all birthdays different, so the probability that Carla has a birthday that differs from the other two is 363/365. The combined probability that Betty's birthday differs from that of Alfred and Carla's birthday differs from those of Alfred and Betty is therefore \frac{364}{365} \cdot \frac{363}{365}.
Now we're starting to see a pattern. When Diogenes comes into the room, this probability must be multiplied by 362/365, to give \frac{364}{365} \cdot \frac{363}{365} \cdot
\frac{362}{365},
and so on. In general, after person n has
entered the room, the probability that all n
birthdays are different is
\frac{364}{365} \cdot \frac{363}{365} \cdot \quad \dots
\quad \cdot \frac{(365n+1)}{365}.
All we have to do now is compute successive values of this expression,
and see when it drops below \frac{1}{2}. Here's what happens:
With 22 people the probability of all birthdays being different is 0.524, slightly bigger than \frac{1}{2}, but with 23 people it is 0.493, slightly less than \frac{1}{2}. So when the 23rd person (Wilma) enters, the probability that at least two of the people present have the same birthday becomes 1  0.493 = 0.507, slightly bigger than \frac{1}{2}. Try it at parties with over 23 people. Take bets. In the long run, you'll win. At big parties you'll win easily. Matthews and Stones tested the probability prediction in another way. Simon Singh — author of the bestselling Fermat's Last Theorem [I think it's called The Fermat Enigma in the USA?] pointed out to Matthews that in a soccer match there are 23 people on the field — two teams of eleven players each, plus the referee. (Two other officials occupy the sidelines, just off the field, but let's ignore them.) So the probability prediction is that among such a group, more often than not, two birthdays coincide. They looked at soccer matches in the UK's Premier League, played on 19 April 1997. Out of ten matches, there were six with birthday coincidences, four without. In fact, for one match there were two coincidences! This was Liverpool versus Manchester United. Two players had birthdays on August 1st, and two had birthdays on January 21st. Using probability theory we can estimate the chances of more than one coincidence with 23 people. The chance of exactly two birthday coincidences is 0.111, that of three is 0.018. The chance of three people out of 23 having the same birthday, by the way, is 0.007. Notice that one match out of the ten featured a double coincidence, and \frac{1}{10} = 0.1, close to the theoretical 0.111. These calculations show that roughandready guesses of the probability of a coincidence often go sadly astray. Why? In this case it is probably because we focus on a misleading aspect of the problem: the number of people. Although 23 is small, there are 253 different pairings among 23 people (253 = 23 \cdot 22/2). That's a lot larger — and a lot more relevant to the question. Now for my second question: you are in a room and people start to enter. How many of them must there be in order for the probability that one of them has the same birthday as yours to be more than \frac{1}{2}? Is the answer 364/2, or 182 people? After all, there are 364 birthdays that differ from yours, and once half of those have been used up... No, it is 253. To see why, we use the same trick: find the probability that the birthdays remain different from yours, and then subtract from 1. Suppose that you're already in the room and everyone else comes in one by one — Alfred, Betty, Carla, Diogenes, and so on. The probability that Alfred has a different birthday from yours is 364/365. The probability that Betty has a different birthday from yours is also 364/365. And the same goes for Carla, Diogenes... everybody. We're not interested in coincidences between the birthdays of the people that come in — say Fred and Gina both have birthdays on 19 May. Doesn't matter: all that counts is whether their birthday is the same as yours. So the probability that after n people have entered they all have different birthdays from yours is (364/365)^n. The first value of n for which this number is less than \frac{1}{2} is n = 253; in fact, \left( \frac{364}{365} \right)^{253} = 0.499,
but
\left( \frac{364}{365} \right)^{252}
is slightly bigger than 0.500.
Incidentally, the fact that this answer to the second problem is the same as the number of pairings in the first problem (253 pairings for 23 people) seems not to have any mathematical significance. In fact, it seems to be a coincidence. The number 253 is much bigger than the 23 that answers the first problem. I suspect that our intuition gets the two problems mixed up. When thinking about the first problem, we subconsciously simplify it by thinking of just one person's birthday — perhaps our own, or perhaps one of the two (supposed) people who have the same birthday — and wonder how many people are needed before there is a better than evens probability that this particular birthday is duplicated. But that's a very different question. The room can be full of people whose birthdays differ from ours, but if any two of them have the same birthday as each other then the game is up. By focusing on just one date, we forget how difficult it is to keep missing all the others as the number of people grows. Why do we underestimate when it comes to our own birthday being duplicated? We make the opposite mistake, and think of those 364 slots that are not our birthday as filling up, one by one. However, as new people come in, many slots get filled more than once as the new arrivals' birthdays duplicate previous ones. So it is harder to duplicate that one birthday out of 365 than we expect. What do such calculations teach us about other kinds of coincidence? First, not to be unduly impressed by things that seem unlikely: maybe they're not. Those soccer players would have been amazed to discover their coincident birthdays, even though everything conformed entirely to the laws of probability. But there's another lesson: a coincidence seems striking to you if it involves you. The two players with the same birthday would have remembered the coincidence and talked about it at dinner parties; the 252 other pairs of players, whose birthdays did not differ, would not have recounted such a boring fact. So when we notice coincidences, but ignore noncoincidences, we make the coincidences seem more significant than they are. My friend's honeymoon encounter seems much less striking when you think of just how many other people he must have encountered during his life who were not his boss and spouse. Nonetheless, every so often a coincidence happens that seems truly striking, even when you take these effects into account. Coincidences that cannot easily be debunked are definitely more interesting than those that can. Whether they have some deep significance must be a matter of personal opinion. But before you interpret your holeinone as a sign of special favour from the Gods of Golf, do bear in mind that there are an awful lot of golfers, and according to the laws of probability such an event will, every so often, happen to one of them. Who will then be enormously impressed to have been 'chosen' for this honour. 