Read 100 Essential Things You Didn't Know You Didn't Know Online
Authors: John D. Barrow
100 Essential Things You Didn't Know You Didn't Know (6 page)
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 a
1
to 1, a
2
to 1, a
3
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 a
i
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/(a
i
+ 1) of the total stake money on the runner with odds of a
i
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/(a
1
+ 1) + 1/(a
1
+ 1) + 1/(a
3
+ 1) + . . . + 1/(a
N
+ 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 a
1
= 6, a
2
= 7/2, a
3
= 2 and a
4
= 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 a
i
= 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 U
2
= 2gH, where g is the acceleration due to gravity. The energy of motion of the jumper at take-off is 1/2 MU
2
and this will be transformed into the potential energy MgH gained by the jumper at the maximum height H. Equating the two gives U
2
= 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 increase 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 MU
2
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 MU
2
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 = U
2
/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 = πr
2
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 p
2
≥ 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.