100 Essential Things You Didn't Know You Didn't Know Page 4
Suppose that Ali, Bob and Carla decide to invest together in a second-hand car and go out to look at three different possibilities: an Audi, a BMW and a Reliant Robin. They don’t all agree on what to buy, so they decide that the outcome must be decided democratically. They must vote on it. So, each of them writes down their order of preference for the three makes:
The voting looks promising at first: Audi beats BMW by 2 preferences to 1, and BMW beats the Reliant Robin by 2 preferences to 1. But, strangely, the Reliant Robin beats the Audi by 2 preferences to 1. Preferring, like ‘admiring’, is an intransitive relationship that can create awkward paradoxes if not used with care. Small elections to decide who you prefer among candidates for a job, who captains a sports team or even what car to buy are fraught with paradox. Let the voter beware.
Faced with this trilemma, Ali, Bob and Carla decided to give up on the car buying and put their resources into renting a house together. Alas, more decisions were soon necessary. Should they decorate the living room? Should they tidy the garden? Should they buy a new TV? There was no consensus so they decided to vote ‘yes’ or ‘no’ on each of the three issues in turn. Here’s what they said:
All seemed clear. There was a majority of two-to-one to do all three things. All three things should be done. But then money seemed to be running short, and the trio realised that they needed two more people to share the house if they were to pay the rent. After just a few phone calls they had found new house-mates Dell and Tracy, who rapidly moved in with all their belongings. Of course, they thought that it was only fair that they be included in the household vote on the decoration, gardening and TV purchase question. They both voted ‘No’ to each of the three proposals while Ali, Bob and Carla stuck to their earlier decisions. A very strange situation has now been created in the household.
Here is the table of decisions after Dell and Tracy have added their ‘No’s:
We see that their negative votes have tipped the scales on each question. Now there is a majority of 3 to 2 not to decorate the house, not to tidy the garden, and not to buy a TV. But more striking is the fact that a majority (each of Ali, Bob, and Carla) are on the losing side of the vote on two of the three issues. These are the issues on which they don’t vote ‘No’. So Ali is on the losing side over the house and garden, Bob is on the losing side over the garden and TV, and Carla is on the losing side over the house and the TV. Hence a majority of the people (three out of five) were on the losing side on a majority of the issues (two out of three)!
15
Racing Certainties
‘There must be constant vigilance to ensure that any legalised gambling activity is not penetrated by criminal interests, who in this connection comprise sophisticated, intelligent, highly organised, well briefed and clever operators with enormous money resources which enable them to hire the best brains in the legal, accountancy, managerial, catering and show business world.’ I am not quite sure about the last two; nevertheless, that is as true now as it was then.
Viscount Falkland quoting from the Report of Rothschild Commission on Gambling (1979)
A while ago I saw a TV crime drama that involved a plan to defraud bookmakers by nobbling the favourite for a race. The drama centred around other goings on, like murder, and the basis for the betting fraud was never explained. What might have been going on?
Suppose that you have a race where there are published odds on the competitors of a1 to 1, a2 to 1, a3 to 1 and so on, for any number, N, of runners in the race. If the odds are 5 to 4, then we express that as an ai of 5/4 to 1. If we lay bets on all of the N runners in proportion to the odds so that we bet a fraction 1/(ai + 1) of the total stake money on the runner with odds of ai to 1, then we will always show a profit as long as the sum of the odds, which we call Q, satisfies the inequality
Q = 1/(a1 + 1) + 1/(a1 + 1) + 1/(a3 + 1) + . . . + 1/(aN + 1) < 1
And if Q is indeed less than 1, then our winnings will be at least equal to
Winnings = (1/Q – 1) × our total stake
Let’s look at some examples. Suppose there are four runners and the odds for each are 6 to 1, 7 to 2, 2 to 1 and 8 to 1. Then we have a1 = 6, a2 = 7/2, a3 = 2 and a4 = 8 and
Q = 1/7 + 2/9 + 1/3 + 1/9 = 51/63 < 1
and so by betting our stake money with 1/7 on runner 1, 2/9 on runner 2, 1/3 on runner 3 and 1/9 on runner 4 we will win at least 21/51 of the money we staked (and of course we get our stake money back as well).
However, suppose that in the next race the odds on the four runners are 3 to 1, 7 to 1, 3 to 2 and 1 to 1 (i.e. ‘evens’). Now we see that we have
Q = 1/4 + 1/8 + 2/5 + 1/2 = 51/40 > 1
and there is no way that we can guarantee a positive return. Generally, we can see that if there is a large field of runners (so the number N is large) there is a better chance of Q being greater than 1. But large N doesn’t necessarily guarantee that we have Q > 1. Pick each of the odds by the formula ai = i(i+2)–1 and you can get Q = ¾ and a healthy 30 per cent return, even when N is infinite!
But let’s return to the TV programme. How is the situation changed if we know ahead of the race that the favourite in our Q > 1 example will not be a contender because he has been doped?
If we use this inside doping information we will discount the favourite (with odds of 1 to 1) and place none of our stake money on him. So, we are really betting on a three-horse race where Q is equal to
Q = 1/4 + 1/8 + 2/5 = 31/40 < 1
and by betting 1/4 of our stake money on runner 1, 1/8 on runner 2 and 2/5 on runner 3 we are guaranteed a minimum return of (40/31) – 1 = 9/31 of our total stake in addition to our original stake money! So we are quids in.fn1
fn1 It has been suggested to me that some criminal money-laundering is performed by spreading on-course bets over all runners, even when Q>1. There is a loss but on average you can predict what it will be and it is just a ‘tax’ on the money-laundering exchange.
16
High Jumping
Work is of two kinds: first, altering the position of matter at or near the Earth’s surface relatively to other such matter; second, telling other people to do so. The first kind is unpleasant and ill paid; the second is pleasant and highly paid.
Bertrand Russell
If you are training to be good at any sport, then you are in the business of optimisation – doing all you can (legally) to enhance anything that will make you do better and minimise any faults that hinder your performance. This is one of the areas of sports science that relies on the insights that are possible by applying a little bit of mathematics. There are two athletics events where you try to launch the body over the greatest possible height above the ground: high jumping and pole vaulting. This type of event is not as simple as it sounds. Athletes must first use their strength and energy to launch their body weight into the air in a gravity-defying manner. If we think of a high jumper as a projectile of mass M launched vertically upwards at speed U, then the height H that can be reached is given by the formula U2 = 2gH, where g is the acceleration due to gravity. The energy of motion of the jumper at take-off is 1/2 MU2 and this will be transformed into the potential energy MgH gained by the jumper at the maximum height H. Equating the two gives U2 = 2gH.
The tricky point is the quantity H – what exactly is it? It is not the height that is cleared by the jumper. Rather, it is the height that the jumper’s centre of gravity is raised, and that is a rather subtle thing because it makes it possible for a high jumper’s body to pass over the bar even though its centre of gravity passes under the bar.
When an object has a curved shape, like an L, it is possible for its centre of gravity to lie outside the body.fn1 It is this possibility that allows a high jumper to control where his centre of gravity lies and what trajectory it follows when he jumps. The high jumper’s aim is to get his body to pass cleanly over the bar while making his centre of gravity pass as far underneath the bar as possible. In this way he will make optimal use of his explosive take-off energy to incr
ease the height cleared.
The simple high-jumping style that you first learn at school, called the ‘scissors’ technique, is far from optimal. In order to clear the bar your centre of gravity, as well as your whole body, must pass over the bar. In fact your centre of gravity probably goes about 30 centimetres above the height of the bar. This is a very inefficient way to clear a high-jump bar.
The high-jumping techniques used by top athletes are much more elaborate. The old ‘straddle’ technique involved the jumper rolling around the bar with their chest always facing the bar. This was the favoured technique of world-class jumpers up until 1968 when the American Dick Fosbury amazed everyone by introducing a completely new technique – the ‘Fosbury Flop’ – which involved a backwards flop over the bar. It won him the Gold Medal at the 1968 Olympics in Mexico City. This method was only safe when inflatable landing areas became available. Fosbury’s technique was much easier for high jumpers to learn than the straddle and it is now used by every good high jumper. It enables a high jumper to send his centre of gravity well below the bar even though his body curls over and around it. The more flexible you are the more you can curve your body around the bar and the lower will be your centre of gravity. The 2004 Olympic men’s high-jump champion Stefan Holm, from Sweden, is rather small (1.81 m) by the standards of high jumpers, but he is able to curl his body to a remarkable extent. His body is very U-shaped at his highest point. He is able to sail over a bar set at 2 m 37 cm, but his centre of gravity goes well below the bar.
When a high jumper runs in to launch himself upwards, he will be able to transfer only a small fraction of his best possible horizontal sprinting speed into his upward launch. He has only a small space for his approach run and must turn around in order to take off with his back facing the bar. The pole vaulter is able to do much better. He has a long, straight run down the runway and, despite carrying a long pole, the world’s best vaulters can achieve speeds of close to 10 metres per second at launch. The elastic fibreglass pole enables them to turn the energy of their horizontal motion 1/2 MU2 into vertical motion much more efficiently than the high jumper can. Vaulters launch themselves vertically upwards and perform all the impressive gymnastics necessary to curl themselves in an inverted U-shape over the bar while sending their centre of gravity as far below it as possible. Let’s see if we can get a rough estimate of how well we might expect them to do. Suppose they manage to transfer all their horizontal running kinetic energy of 1/2 MU2 first into elastic energy by bending the pole, and then into vertical potential energy of MgH. They will raise their centre of mass a height H = U2/2g.
If the Olympic champion can achieve a 9 m/s launch speed then, since the acceleration due to gravity is g = 10 ms-2, we expect him to be able to raise his centre of gravity by H = 4 metres. If he started standing with his centre of gravity about 1.5 metres above the ground and made it pass 0.5 metres below the bar then he would be expected to clear a bar height of about 1.5 + 4 + 0.5 = 6 metres. In fact, the American champion Tim Mack won the Athens Olympic Gold Medal with a vault of 5.95 metres (19’6” in feet and inches) and had three very close failures at 6 metres, knowing he had already won the Gold Medal, so our very simple estimates turn out to be surprisingly accurate.
fn1 One way to locate the centre of gravity of an object is to hang it up from one point and drop a weighted string from any point on the object, marking where the string drops. Then repeat this by hanging the object up from another point. Draw a second line where the hanging string now falls. The centre of gravity is where the lines of the two strings cross. If the object is a square then the centre of gravity will lie at the geometrical centre but if it is L-shaped or U-shaped the centre of gravity will not generally lie inside the boundary of the body.
17
Superficiality
The periphery is where the future reveals itself.
J.G. Ballard
Boundaries are important. And not only because they keep the wolves out and the sheep in. They determine how much interaction there can be between one thing and another, and how much exposure something local might have to the outside world.
Take a closed loop of string of length p and lay it flat on the table. How much area can it enclose? If you change its shape you notice that by making the string loop longer and narrower you can make the enclosed area smaller and smaller. The largest enclosed area occurs when the string is circular. In that case we know that p = 2πr is its perimeter and A = πr2 is its area, where r is the radius of the circle of string. So, eliminating r, for any closed loop with a perimeter p which encloses an area A, we expect that p2 ≥ 4πA, with the equality arising only for the case of a circle. Turning things around, this tells us that for a given enclosed area we can make its perimeter as long as we like. The way to do it is to make it increasingly wiggly.
If we move up from lines enclosing areas to surfaces enclosing volumes then we face a similar problem. What shape can maximise the volume contained inside a given surface area? Again, the biggest enclosed volume is achieved by the sphere, with a volume V = 4πr3/3 inside a spherical surface of area A = 4πr2. So for any closed surface of area A we expect its enclosed volume to obey A3 ≥ 36πV2, with equality for the case of the sphere. As before, we see that by making the surface highly crenellated and wiggly we can make the area enclosing a given volume larger and larger. This is a winning strategy that living systems have adapted to exploit.
There are many situations where a large surface area is important. If you want to keep cool, then the larger your surface area the better. Conversely, if you want to keep warm it is best to make it small – this is why groups of newborn birds and animals huddle together into a ball so as to minimise exposed surface. Likewise, the members of a herd of cattle or a shoal of fish seeking to minimise the opportunities for predators will gather themselves into a circular or spherical group to minimise the surface that a predator can attack. If you are a tree that draws moisture and nutrients from the air, then it pays to maximise your surface area interfacing with the atmosphere, so it’s good to be branchy with lots of wiggly leafy surfaces. If you are an animal seeking to absorb as much oxygen as possible through your lungs, then this is most effectively done by maximising the amount of tubing that can be fitted into the lung’s volume so as to maximise its surface interface with oxygen molecules. And if you simply want to dry yourself after getting out of the shower, then a towel that has lots of surface is best. So towels tend to have a rough pile on their surfaces: they possess much more surface per unit of volume than if they were smooth. This battle to maximise the surface that is enclosing a volume is something we see all over the natural world. It is the reason that ‘fractals’ so often appear as an evolutionary solution to life’s problems. They provide the simplest systematic way to have more surface than you should for your volume.
Is it better for you to stay in one group or to split up into two or more smaller groups? This was a problem faced by naval convoys trying to avoid being found by enemy submarines during the Second World War.
It is better to stay in a big convoy rather than to divide. Suppose that a big convoy covers an area A and the ships are as close together as they can be, so that if we divide the convoy into two smaller ones of area A/2 the spacings between ships are the same. The single convoy has a perimeter equal to p = 2π√(A/π), but the total perimeter of the two smaller convoys equals p×√2, which is bigger. So, the total perimeter distance that has to be patrolled by destroyers to protect the two smaller convoys from being penetrated by submarines is greater than that to be patrolled if it stays as a single convoy. Also, when the submarine searches for convoys to attack, its chance of seeing them is proportional to their diameter, because this is what you see in the periscope. The diameter of the single circular convoy of area A is just 2√(A/π), whereas the sum of the two diameters of the convoys of area A/2 that don’t overlap in the field of view is bigger by a factor of √2, and so the divided convoy is 41% more likely to be detected by the attacking
submarine than is the single one.
18
VAT in Eternity
In this world nothing is certain but death and taxes.
Benjamin Franklin
If you live in the United Kingdom you will know that the sales tax added on to many purchases is called ‘Value Added Tax’ or VAT. In continental Europe it is often called IVA. In the UK it amounts to an additional 17.5% on the price of a range of goods and services and is the government’s biggest source of tax revenue. If we suppose that the 17.5% rate of VAT was devised to allow it to be easily calculated by mental arithmetic what do you expect the next increase in the rate of VAT to be? And what will the VAT rate be in the infinite future?
The current value of the VAT rate sounds arcane – why pick 17.5%? But if you have to prepare quarterly VAT accounts for a small business you ought soon to recognise the niceness of this funny number. It allows for very simple mental calculation because 17.5% = 10% + 5% + 2.5%, so you know what 10% is immediately (just shift the decimal point one place to the left), then just halve the 10%, and then halve the 5% that remains and add the three numbers together. Hence, for example, the VAT on an £80 purchase is just £8 + £4 + £2 = £14.
If the same convenient ‘halving’ structure for mental arithmetic is maintained, then the next VAT increase will be by an amount equal to half of 2.5%, or 1.25%, giving a new total of 18.75% and the new VAT on £80 would be £8 + £4 + £2 + £1 = £15.
We would then have a tax rate that looked like 10% + 5% + 2.5% + 1.25%. To the mathematician this looks like the beginnings of a never-ending series in which the next term in the sum is always half of its predecessor. The current VAT rate is just