It may sound like an easy question but the answer will surprise you! Live interview with BBC Radio Oxford.

Image credit: Thurner Hof

Maths, but not as you know it…

It may sound like an easy question but the answer will surprise you! Live interview with BBC Radio Oxford.

Image credit: Thurner Hof

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 28^{th} 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 A | Russia |

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 |

Group G | England |

Group H | Poland |

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 (14^{th} 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 – 14^{th} 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).

**Kai Laddiman **

The 2018 World Cup in Russia kicks off today and so I bring you a special double-edition of Throwback Thursday looking at the science behind the perfect penalty kick… Fingers crossed the England players listen/read my website and we don’t lose to Germany in a penalty shootout (though let’s be honest we probably will).

Live interview with BBC Radio Cambridgeshire looking at the ‘unsaveable zone’ and the best way to mentally prepare for a penalty.

And if that wasn’t enough, here’s a full description of the ‘Penalty Kick Equation’…

For all of the footballers out there who have missed penalties recently, I thought I would explain the idea of the science behind the perfect penalty a little further, and in particular the maths equation that describes the movement of the ball. On the radio of course I couldn’t really describe the equation, so here it is:

If you’re not a mathematician it might look a little scary, but it’s really not too bad. The term on the left-hand side, *D*, gives the movement of the ball in the direction perpendicular to the direction in which the ball is kicked. In other words, how much the ball curves either left or right. This is what we want to know when a player is lining up to take a penalty, because knowing how much the ball will curl will tell us where it will end up. To work this out we need to input the variables of the system – basically use the information that we have about the kick and input it into the equation to get the result. It’s like one of those ‘function machines’ that teachers used to talk about at school: I input 4 into the ‘machine’ and it gives me 8, then I put in 5 and I get 10, what will happen if I input 6? The equation above works on the same idea, except we input a few different things and the result tells us how much the ball will curl.

So, what are the inputs on the right-hand side? The symbol *p* just represents the number 3.141… and it appears in the equation because footballs are round. Anytime we are using circles or spheres in maths, you can bet that *p* will pop up in the equations – it’s sort of its job. The ball itself is represented by *R* which gives the ball’s radius, i.e. how big it is, and the ball’s mass is given by *m*. We might expect that for a smaller ball or a lighter ball the amount it will curl will be different, so it is good to see these things are represented in the equation – sort of a sanity test if you will. The air that the ball is moving through is also important and this is represented by *r,* which is the density of the air. It will be pretty constant unless it’s a particularly humid or dry day.

Now, what else do you think might have an effect on how much the ball will curl? Well, surely it will depend on how hard the ball is kicked… correct. The velocity of the ball is given by *v*. The distance the ball has moved in the direction it is kicked is given by *x*, which is important as the ball will curl more over a long distance than it will if kicked only 1 metre from the goal. For a penalty this distance will be fixed at 12 yards or about 11m. The final variable is *w* – the angular velocity of the ball. This represents how fast the ball is spinning and you can think of it as how much ‘whip’ has been put on the ball by the player. Cristiano Ronaldo loves to hit them straight so *w* will be small, but for Beckham – aka the king of curl- *w* will be much larger. He did of course smash that one straight down the middle versus Argentina in 2002 though…

So there you have it. The maths equation that tells you how much a football will curl based on how hard you hit it and how much ‘whip’ you give it. Footballers often get a bad reputation for perhaps not being the brightest bunch, but every time they step up to take a free kick or a penalty they are pretty much doing this calculation in their head. Maybe they’re not quite so bad after all…