Following Manchester City’s penalty shootout victory over Liverpool in the Community Shield, I was asked by BBC Oxford to explain how scientists are trying to find the formula for the perfect penalty…
Cast your mind back to the summer of 2018… we saw the warmest ever weather in the UK, Brexit was not yet a complete and utter disaster, and seemingly against all the odds the England football team reached the semi-finals of the World Cup for the first time since 1990. No doubt the team had a huge celebration together afterwards – but it wouldn’t be the first time that two of them have celebrated an occasion at the same time. As well as playing together at the heart of England’s defence, Manchester City duo Kyle Walker and John Stones also share the same birthday! Stones was born on 28th May 1994, making him 24 years old; Walker was born on the same day in 1990, meaning that he is exactly four years older than his teammate. How strange! Or is it…?
On the face of it, it seems quite surprising that in an England squad of just 23 players, two of them happen to share a birthday. However, as we’re about to see, this isn’t a freakish coincidence – maths says that it’s quite likely! What we’re talking about here is commonly known as the birthday problem: if there are a group of people of a certain size, what is the likelihood that at least two of them have the same birthday?
Let’s start by saying that we have a group of N people, and assume that birthdays are equally likely on every day of the year. (There is some evidence to suggest that this isn’t the case for top athletes; some say that they tend to be born early in the school year, such as around September in England. This is because they are slightly older than the other children in the year, and so they have a slight head-start in their physical development. However, we don’t want to make things too complicated, so we’ll ignore that for now.)
The easiest way to think about the problem is to first try to work out what the probability is that none of the N people share a birthday. Suppose our N people walk into a room, that is empty at first, one at a time. When the first person walks in, it’s obvious that they don’t share a birthday with anyone else in the room, because there isn’t anyone else. Therefore, they have the maximum probability of not sharing a birthday with anyone else in the room, which is 1.
Now think of the second person who walks in. The only way that they could share a birthday with someone in the room is if it happens to be exactly the same day as the first person. That means there is a 1 in 365 chance that they do share a birthday, so there is a 364 in 365 chance that they don’t.
Suppose that the first two birthdays don’t match, and then the third person walks in. They now have 2 days that they can’t share a birthday with, so there are 363 possible choices out of 365. Because we assumed that the first two didn’t match, we multiply the probabilities, so now the chance that none of them share a birthday is (364/365) * (363/365).
We can repeat this process until we get to our final person, number N. For example, the fourth person has 3 birthdays that they cannot share, so we multiply by a chance of 362/365; the fifth person has 4 days to avoid, so we include a probability of 361/365… By the time the Nth person walks in, there are N-1 people already in the room, so there are N-1 days that their birthday cannot fall on. This leaves them with 365-(N-1) possibilities out of 365.
To work out the total probability, we multiply all of these terms together which gives the likelihood that none of the N people share a birthday as
1 * (364/365) * (363/365) * (362/365) * … * ((365-(N-1))/365).
You might be thinking that this still looks like quite a big probability that none of them share a birthday, because all of the terms are very close to 1. But, if we try some values of N in a calculator, then it tells a very different story. (The percentages are calculated by finding the probability from the equation above and multiplying by 100.)
When N = 10, we get an 88% chance that none of them share a birthday. However, this drops down to 59% when there are N = 20 people. When we get to N = 23, the number of players in the England squad, the probability reaches just under 50%. That means that, incredibly, the likelihood that at least two of the 23 people share a birthday is just bigger than 50%!
So, in a random group of 23 people, it’s more likely than not that two of them share a birthday! This seems very strange at first; surely you’d need more than 23 people for a shared birthday to be more likely than not?! This is why the problem is commonly known as the birthday paradox – it might be very hard to get your head around, but the maths doesn’t lie!
Perhaps, in order to convince ourselves, we should look at some real-life examples. This is where the World Cup squads come into play: each team is restricted to bringing 23 players to the tournament. (We’ve seen that number before…) If our calculations above are correct, then if we picked any one of the World Cup squads, there would be roughly a 50:50 chance that at least two of the squad members share a birthday, which means that out of all of the squads that went to Russia, we would expect about half of them to have a birthday match. Well, let’s take a look…
Of the 32 teams, which were divided into 8 groups of 4, the following teams have at least one pair of players who share a birthday:
|Group B||Iran, Morocco, Portugal, Spain|
|Group C||Australia, France, Peru|
|Group D||Croatia, Nigeria|
|Group E||Brazil, Costa Rica|
|Group F||Germany, South Korea|
So, not only is there at least one team in every group with a birthday match, but if we count the total, there are 16 squads with a shared birthday pair – exactly half of the teams! The experimental results have matched up with the mathematical theory to perfection. Hopefully that’s enough to convince you that our calculations were indeed sound!
A slightly different question that you might ask is as follows: if I am in a group with a certain number of people, what are the chances that at least one of them shares my birthday? Is it the same idea? What we have worked out above is the probability that any two people in the room share a birthday (or rather, we worked out the opposite, but we can find the right answer from our working). Note that the pair doesn’t necessarily include you; it’s a lot more likely that it’s some other pair in the group.
In order to work out the answer to this similar sounding question, we work the other way around again, by calculating the probability that none of the N people share my birthday. For each of the N people, there is only one birthday that they cannot have, and that is mine (14th November, in case you were wondering), which means there are 364 out of 365 possibilities for each person. We no longer care whether their birthdays match up; we only care if they match with mine. So each person has a 364/365 chance of not sharing my birthday; and the overall probability is just 364/365 * 364/365 * … * 364/365, N times, which we write as (364/365)N.
Once again, we can plug some values of N into a calculator: N = 10 gives a 97% chance that no-one else has my birthday. For N = 50 the probability is still very high: there is an 87% chance that none of these 50 people have the same birthday as me. N = 100 gives 76%; N = 200 gives 58%; you have to go all the way to N = 253 before the probability dips below 50%, and it becomes more likely than not that at least one person will celebrate their birthday with me.
Applying this idea to all 736 players (32 squads of 23 players) involved in the World Cup, we should expect around 3 of them to have been born on the same day as me – 14th November. And I am very happy to confirm that France’s Samuel Umtiti, Switzerland’s Roman Burki, and Belgium’s Thomas Vermaelen all have what is undoubtedly the best birthday of the year… Two similar problems with two very different solutions!
You can check which footballers share a birthday with you at www.famousbirthdays.com/date/monthDD-soccerplayer.html, where you enter the month in words and the day in numbers (no preceding zero required).