by Rob Eastaway
There are 128 players at the start of a Wimbledon singles tournament, so the number of matches is 128 – 1 = 127.
In the men’s and women’s doubles championships there are 64 pairs in the first round, and so 63 matches in total, and there are 48 in the mixed doubles, so 47 matches.
That makes the total number of main-event matches at Wimbledon:
127 + 127 + 63 + 63 + 47 … let’s just call it 400.
Men’s Wimbledon matches can last anything between three and five sets (call it four on average), while matches involving women will be two or three sets (call it two and a half on average). Combining the two, let’s say an average Wimbledon match has three sets.
The number of games in a set can range from a minimum of six (when the set is 6–0) to a maximum of 13 when it goes to tie-break (7–6). That’s an average of 9½ games, but let’s round it up to 10, because that’s easier, and also because the final set doesn’t have a tie-break at 6–6, which pushes up the average number of games.
So an average Wimbledon match has about:
10 games per set × 3 sets per match
= 30 games per match.
One bit of technical knowledge that the apocryphal Accenture interviewers surely couldn’t expect anything but a tennis nut to know is that the umpire asks for a set of six new balls (‘New balls please!’) every nine games. In other words, our tennis match lasting 30 games is going to require three or four sets of six balls – let’s say 20 balls per match. Four hundred matches.
20 balls per match × 400 matches
= 8,000 balls.
That’s a lot of balls, but this estimate is way below the 50,000 that Wimbledon claim to use each year. How come? For a start, there’s the wheelchair, junior and celebrity tournaments that also go on during the fortnight – so you can probably double our 8,000 just with that. Then there’s balls that players use for warming up, plus the balls that get hit into the crowd and get kept as souvenirs … and no doubt Wimbledon keep a few spares in the shop as well.
Put it this way: if you are in a job interview and you are asked how many balls are needed for Wimbledon, and if you use the reasoning above to get you to somewhere in the 5,000 to 50,000 range, you’ll probably get a big tick in the assessment sheet.
HOW MANY ATTENDED DONALD TRUMP’S INAUGURATION?
Crowd counting can be very political. When there is an organised protest and march (be that for more pay, lower taxes or better rights for animals) the organisers want to be able to claim as big a number as possible, while the thing or people being protested against – often the government – would much prefer that number to be as small as possible. The police, acting on behalf of government, have a tendency to downplay the numbers too. As a result, you can almost guarantee that the size of crowds claimed by the protestors and protestees will always be very different, sometimes by a factor of two, depending on who you ask.
For example, there was a huge protest in London when Donald Trump made his first visit to the UK. The protestors were quick to claim that 250,000 were there. The police, according to the Independent, were reluctant to give a figure, but were prepared to agree that ‘more than 100,000’ had turned out.
Which raises the question: did more people turn out to protest against Trump in London than turned out to celebrate him at his inauguration in 2017?
One person whose answer would be no is, of course, Donald Trump, who was adamant that there had never been a bigger crowd. As far as (most of) the rest of us could tell, however, Trump’s inauguration crowd was a fraction of that of his predecessor, Barack Obama.
But how big was it?
One unreliable way to estimate a crowd size is to view it from a low angle, for example when speaking to it from a raised podium. Looked at this way, even partially filled space can look full because the floor is not visible. Given that this is the view that Trump had of his crowd, you can understand his over-estimate. Well, maybe.
Professional crowd estimators prefer to view the crowd from above, from a photo taken by a drone, for example. This way it’s possible to see how packed the crowd is. Common practice is to divide the photo into squares of (say) five metres, look at the crowd density within each square, then group the squares into categories from ‘packed’ to almost empty.
There are commonly used guideline figures for a crowd’s density, based on the number of people per square metre:
No. per square metre Nature of outdoor crowd
1 Blanket/picnic crowd (e.g. outdoor classical concert).
2 Crowd on Henman Hill at Wimbledon.
3 Political rally.
4 By the barriers as the Queen drives past.
5–6 Fans at the front of a One Direction concert.
>6 An uncomfortable crush, individuals likely to be pressed up against each other.
But for Trump’s crowd we don’t have the luxury of aerial photos, so our estimate is going to be very crude. Here’s a sketch of the National Mall:
Since Trump’s crowd by no means filled the National Mall in Washington DC, we can assume that people had no need to pack themselves to maximum density. It was very full in the big semi-circular section in front of the Washington Capitol, and the crowd also stretched back into the straight section between the Reflecting Pool and the Washington Monument.
The semi-circular section is about 100 metres radius, so its area is roughly:
π × 1002÷ 2 ≈ 15,000 m2.
That section was very full, and therefore probably comfortably packed, so let’s assume its density was three people per square metre. That means a crowd of around 50,000 (a decent Premier League crowd) immediately in front of Trump.
The straight Mall beyond the lake was not packed – the event planners had put down white flooring, so it was easy to pick out the empty sections. The straight is around 1 km (1,000 metres) long, and the covered segment perhaps 100 metres wide. So the total area of the straight is around:
100 × 1,000 = 100,000 m2.
Let’s say the whole segment was 50% occupied by spectators (that’s generous), and the space was comfortably full in those sections, i.e. two people per m2. We’re looking at 200,000 in the straight section, and 50,000 near the front. That’s maybe as many as 250,000 in total? (Interestingly, 250,000 is the figure that one of the expert sources gave as their estimate as well.)
Since 250,000 is the upper limit of the London protest crowd, it’s likely that there were more at Trump’s inauguration, but it seems that it was a fairly close-run thing.
WHAT ARE THE CHANCES?
People love coincidences. We are drawn to them instinctively, because the human brain is geared to look out for patterns. When we observe something surprising, we want an explanation, and we assume that something has caused it. If there is no obvious physical cause (such as somebody fiddling the lottery) then we reach out for explanations that are supernatural.
And the hunt for an explanation inevitably includes the question: ‘what are the chances?’ – the more remote the odds, the more excited everyone gets.
But how are these odds worked out? If ever there are situations that call for the back of an envelope, it is these. Let’s look at some coincidences in more detail.
WHAT’S THE CHANCE OF WINNING THE LOTTERY TWICE?
On 10 September 2009, the Bulgarian Lottery hit the headlines. Four days earlier, the six numbers drawn had been 4, 15, 23, 24, 35 and 42. Now, in the very next draw, exactly the same six numbers were pulled out. ‘What are the chances?’ asked almost everybody (to which the answer reported in the media was ‘about one in five million’4).
With some coincidences that we’ll see later, all sorts of hand-waving assumptions are needed before you can come up with an estimate of the probability. Not so the result of a single lottery. These games are carefully engineered so that the outcomes are entirely random. And the conditions of the game are well defined: there are a fixed number of balls (typically somewhere between 40 and 60 in most lotteries), each of which has exactly the sa
me chance of being drawn.
This means that any combination of six numbers is equally likely: next week’s numbers in the UK Lottery could be 7, 12, 14, 23, 41 and 58. Or, just as likely, they could be 1, 2, 3, 4, 5 and 6.
If the numbers 1, 2, 3, 4, 5 and 6 were to be drawn, it would be headline news of course, not because this combination is any less likely than any other, but because it makes a far more interesting pattern.
So the reports of the freak occurrence in the Bulgarian Lottery, where the identical numbers came up in consecutive draws, were right. This was indeed a ‘millions to one against’ occurrence – and we haven’t had to do any estimation.
However, there are other lottery coincidences that do require some back-of-envelope thinking.
In June 2018, an anonymous Frenchman won a million euros in his country’s lottery for the second time in two years. The odds against this were quoted in headlines as being 16 trillion to one, that’s 16,000,000,000,000:1 written long-hand. When numbers get this big, we have little sense of what the numbers mean – millions, billions, trillions, they might as well be called gazillions; they all sound the same. Sixteen trillion is more than two thousand times the number of people on the planet. It sounds unlikely as a figure. In fact, it is so big, it sounds suspicious.
Where did ‘16 trillion’ come from?
We do know that the chance of winning the French My Millions Lotto in any given draw is around 20 million to 1. If you want to work out the chance of two events happening, such as tossing heads on a coin and rolling a six on a dice, then, so long as the events are completely independent, you simply multiply the probabilities together (see here). The heads and six combination has a probability of 1/2 × 1/6 = 1/12. This same maths applies to a lottery. The chance of entering the French Lottery exactly twice and winning both times is:
1/20 million × 1/20 million,
… which works out at 1 in 400 trillion – far longer odds than the 16 trillion we saw above.
But our French winner didn’t just play twice and win both times. He won twice over a period of 18 months, and it’s likely that he played the lottery many times during that period. In fact, like most lottery participants, he probably played every week. This is going to significantly reduce the odds, simply because he now had a lot more opportunities for a double win.
To work out the odds of this, we need to start making some guesses.
Let’s suppose he played Lotto every week. This means he played about 75 times over the 18 months. Looking at all the possible combinations of dates, it works out that there were nearly 3,000 different pairs of dates5 on which the Frenchman could have won the jackpot over a period of 18 months. This means the odds were roughly:
reducing the odds to around 130 billion to one.
That’s now far smaller than the 16 trillion figure. Perhaps the academic who came up with the number 16 trillion was only looking at ways of winning the lottery with a gap of exactly 18 months ‘plus or minus a few weeks’. In other words, he was saying ‘What is the chance that somebody would win the lottery, and then win it again about a year and a half later?’ And if you set the level of vagueness right, the answer to that question might legitimately be ‘16 trillion to one’ … but it’s a pretty meaningless figure.
But there’s another huge flaw in quoting the odds as being in the trillions. Yes, that was the chance that this particular monsieur would win the lottery twice in the stated period. But before he won the first time, we had no interest in him as an individual, because we know with 100% certainty that somebody was going to win the Lotto. Only when somebody is already a winner are we interested in whether they win again. In other words, rather than asking ‘What is the chance of a particular person winning the lottery twice in two years?’, we should be asking: ‘What is the chance that somebody who has won the lottery once will win it again inside two years?’
Remember that the chance of winning the French Lottery is about 1 in 20 million. If a person who wins plays once a week, they have roughly 100 chances to win the lottery again in the two-year period, so the odds become 20 million divided by 100, or 200,000 to one. Still a big figure, but tiny compared to the 16 trillion we were first presented with.
And that’s if they only buy a single ticket each week. Lottery winners have so much money sloshing around, they can easily afford to buy 100 tickets a week and not notice the cost. Maybe that’s what our Frenchman did – we have no idea, since he kept himself secret. In that case, we would be looking at odds in the low thousands, not the millions.
This mistake of including the first instance of an event in the calculations helps to make the odds sound more impressive, and therefore newsworthy. That’s why newspapers invariably do it when reporting an interesting coincidence.
It reminds me of the story of the man who was terrified of bombs on planes. On his next flight, he tried to take a bomb onto the plane with him. ‘What the hell do you think you’re doing?’ asked the security officer. ‘Well,’ replied the man, ‘I’ve heard that the chance of there being a bomb on a plane is about one in a million. So I thought I’d take on a bomb myself, because I’ve figured out that the chance that there will be two bombs on the plane will be one in a trillion.’
THE EXACT ODDS OF WINNING A LOTTERY
The odds of winning the jackpot in any lottery can be worked out precisely. In the UK, where six balls are drawn from 59, the odds of any particular combination of six balls being selected is worked out by calculating the total number of ways in which six balls can be drawn from 59. The calculation looks like this:
Here, 59! means the product of all the numbers from 1 to 59, otherwise known as 59 factorial. Written out long-hand (skipping the middle numbers) it looks like this:
The 53! can be cancelled out from top and bottom and the calculation simplifies to:
A quick estimate should convince you that this is going to work out to be a very large number. In fact, the (precise) figure is: 45,057,474.
In other words, if you pick six numbers at random, there’s roughly a 1 in 45 million chance that they will come up in the next draw.
WHAT WERE THE CHANCES OF THE ORKNEY BABIES?
On 13 November 2018, two women – Angela Johnston and Karen Daily – both gave birth in Balfour Hospital in the north of Scotland. What made this more unusual was that both women came from Stronsay in the Orkney islands, an island with a population of 350, where babies are a relative rarity. Even more remarkably, the two women had both taken the same ferry to the mainland hospital a few days earlier. Both had (quite independently) decided that if their baby was a boy, they would call it Alexander. And then – to the amazement of everyone – both babies arrived at 11:36 p.m.
‘What were the chances?’, asked everyone, including BBC Radio Scotland, who then rang me for the answer. I guess they were assuming that this is the sort thing that does have ‘an answer’ one can just rattle off.
But as with all the other estimates in this book, we have to make some assumptions before we can do any meaningful calculations.
First of all, how many babies would we expect to be born in Stronsay each year?
If a population is going to be stable, rather than growing or declining, the number of births and deaths per year has to be roughly the same. For a population with a regular age distribution, with life expectancy of 80 years, we might expect 1/80th of the population to die each year. The population of Stronsay is 350, so this first stab at an estimate suggests we might expect
Number of births per year = 350 × 1/80 h 4.
Of course I may be wrong to assume that Stronsay’s population level is stable, and even if it is, those who die might be replaced by migrants rather than by babies. Still, it seems reasonable to suppose that an island like Stronsay might expect two babies each year. Let’s suppose it’s exactly two babies per year.
What’s the chance that both those babies will arrive in the same minute?
The number of minutes in a year is 365 × 24 × 60.
> We can use Zequals to work this out. The number of minutes in a year:
h 400 × 20 × 60 h 500,000,
which is half a million. So, if my assumptions are correct, then the chance of a pregnant mother giving birth in a particular specified minute in the year is about one in half a million.
However, remember that the first baby was inevitably going to be born at some minute. To work out the chance that both babies will be born in the same minute, we only need to consider the second baby (just as we did with the lottery winner and his second jackpot). The chance that the second baby would arrive in the same minute was about one in half a million, as calculated above. So that’s my first pass at an answer.
But what about those other factors? Catching the same ferry to the same hospital? Actually, there is little surprise here: ferries to the mainland probably don’t run that often, and if the mums’ due dates were similar then it’s not a huge surprise they caught the same ferry to the nearest hospital that has a maternity unit.
Then again, what about those names? The chance they’d both call their child Alexander was much more of a coincidence. According to a recent census, perhaps as few as 1 in 100 boys are given the name Alexander6 in Scotland. So this would have carried the odds into lottery-winning territory.
There was just one catch. Mrs Daily gave birth to … a baby girl. There was no second Alexander after all.
WHAT’S THE CHANCE OF TWO HOLES IN ONE?
In October 2017, Jayne Mattey and Clair Shine were enjoying a round of golf in Berkshire. It was hole 13, a number viewed as unlucky for some, but not for Jayne and Clair. Jayne teed off first and to her amazement the ball hit the pin and dropped into the hole. It was the first hole in one of her life. Then Clair took aim: the ball went straight, and to their astonishment, it too rolled in. The chances? According to the National Hole-in-One Registry, which is based in North Carolina, USA but likes to keep tabs of golfing feats across the world, it’s ‘17 million to one’.
