Mike Hosken's

"Hinges and Loops"

CHAPTER TWO

NUMBER PUZZLES

  1. Funny Numbers
  2. Car Costs
  3. Dice
  4. Computer-simulated Dice
What I want to do in this Chapter is to share with you an exploration of some cases in which it is sensible to question what simple arithmetic proves. Now I know some forms of figuring have a very bad press - that comment about "Lies, damned lies, and statistics" just for one. "You can prove anything with figures" may be true in the hands of an expert, but that doesn't include me.

We shall not need to go into any complicated or advanced mathematics: no calculus or trigonometry or even Cartesian co-ordinates. If you should want to check absolutely everything in the Chapter all you will need is a pocket calculator and three dice. Briefly, we shall have a look at some of the funny ways of numbers themselves, consider the profits of a possible taxi business, and then go gambling.

§2.1 Funny numbers

Simple arithmetic using every-day sorts of numbers produces straightforward answers. That is such a commonplace observation that it needs no examples. But the advent of the electronic calculator has made it easy to explore simple arithmetic with not-quite-so-every-day decimals. Surprises soon appear.

We can take any decimal between zero and one (other than exactly 0.5): we are going to multiply it by itself and by two and take one away. Then we're going to repeat the squaring and multiplying and taking one away - repeatedly. So for an example run the keys to press on an ordinary simple calculator might be . (decimal point) 6 5 4 3 2 1, then × (multiply) and = does the squaring, × 2 doubles, and − (minus) 1 = does the taking away. My calculator produces -0.1437282: nothing remarkable. The first repeat (or 'iteration') is done without clearing that number so it's just × = × 2 − 1 =. The second answer? -0.9586846. The next × = × 2 − 1 = takes it to 0.8381522. And so on, to 0.4049982 then -0.671953 and -0.0969584 and -0.9811982 and 0.9254998 and so on - and on - and on with no pattern whatsoever. Indeed, the series goes on for ever without ever having values of 1 or 0 or 0.5. There is no magic in the .654321 starting point: any other decimal, positive or negative, produces a similarly chaotic sequence.

A computer can speed things up, of course. So if a spreadsheet is preferred the starting figure can be put into cell A1. Then the formula for A2 is =A1 × A1 × 2 − 1. It is then a very simple matter to get the computer itself to copy that into every cell as far down column A as you like. (So cell A50, for example, will get =A49 × A49 × 2 − 1.) Whenever you change the start value in A1 the sequence zips down the column faster than thought. Better still perhaps, if the spreadsheet can display the results as a graph it is even easier to appreciate the chaotic irregularity of the results. It may even be possible to watch the (absence of) pattern as the start figure in A1 is changed.

At a very much more sophisticated level "chaos" is apparently a branch of mathematics nowadays. None of this has any great philosophical significance: I am not suggesting that the world is chaotic. Rather am I hoping to show that, just as seeing and hearing, tasting and feeling are not as simple as they seem to be, so 'simple' mathematics needs only a very modest extension to reach strange lands - to turn counter-intuitive.

But we can turn now to perhaps even simpler sums, certainly in a more down-to-earth context.

§2.2 Car costs

Let's suppose you want to know how much it costs to run your car. You'd know how to set about it, probably along these lines:
  • You would have a good idea as to how many miles the car does per litre of fuel, and the cost of each litre. So divide the one by the other and you have the fuel cost per mile.
  • Longer-term, you could probably record or estimate the average cost of a routine service which would include oil and perhaps the occasional new tyre. Divide that figure by the 6,000 or whatever the service mileage is and you have the maintenance cost per mile.
  • But there are also a number of nasty lumps -
    • There's the annual licence. No problem: you know what that cost.
    • The case is similar with the insurance premium - annual, fixed cost regardless of mileage covered. There might be the membership charge for a breakdown organisation too.
    • But then there's that horrible thing called depreciation. Even if we don't call it that, we do know that sooner or later we'll need to replace the car with a new (or newer) one: we ought to be setting aside some savings so that the cash is there when the time comes. It's really a bit of future-predicting so it can't be exact: a guess must do.
      Put those three together and you get a figure to add to the day-by-day, month-by-month costs.
  • Purists might want to add some other odds and ends - perhaps quite significant odds and ends in some cases. Like the cost of garaging. It can be expensive to rent a garage. Or if you have your own as part of your home then a proportion of any mortgage ought really to be charged to the car. And mentioning the mortgage raises the subject of interest payments. Was the car bought on hire purchase or with a loan perhaps? If so there will be interest to pay, but if you bought it outright then you could, otherwise, have had that money invested to produce interest on the credit side for you. So either way the car ought to bear some form of interest charge, even if it has to be pretty much of a guess rather than anything very exact.
If you've been able to put figures in under all these headings you will have finished up with a total cost. Divide that by the total number of miles and there you are - total cost per mile! Let's suppose, for the sake of argument, that you have finished up with 60 pence per mile. So if you were thinking of offering a taxi service you'd need to charge a minimum of 61 pence per mile to make a profit. Obviously. But wrong!

Surprisingly perhaps, there's no one single figure that can sensibly be taken as a cost per mile: the truest figure depends on what you want the information for. Let's take some examples.
  • If a friend wanted to come on a journey with you, what would be the extra cost to you? Virtually nothing: the reduction in miles per litre would increase the fuel used by an almost negligible amount which would be more than adequately covered at 1 penny per mile.
  • Sharing the cost of the petrol is another basis on which journeys are sometimes shared. Just a few pence per mile covers it.
  • But it's much less clear what it would mean to "share the costs equally". Would the friend be expected to contribute towards the interest charges and the garaging costs? The car depreciates anyway, regardless of how many miles it goes let alone how many people are in it for each mile. And the same goes for tax and insurance of course. It's more likely that fuel and maintenance costs were intended. Certainly it's more a problem of defining terms than of simple arithmetic.
  • All those journeys were shared. But what about that possible taxi service? There the meaningful questions are not about cost per mile but of cost per EXTRA mile. Clearly all the annual charges can be left out of account completely since none of them increase with the extra mileage. What we need to do is to charge enough, over and above the fuel and maintenance costs, to contribute something towards those fixed annual costs. If charges are low enough sufficient customers should be attracted to produce an income which covers all the costs and leaves something over, at the end of the year, as profit.
The point of the discussion is not to learn about taxi finances: it is simply to realise that obvious answers are not necessarily the meaningful answers, even in matters of simple arithmetic. Business management experts embroider these ideas with more precise definitions, involving fixed costs and variable costs, gross margins and perhaps gross profits. First-blush intuition is not likely to be useful.

§2.3 Dice

When we go gambling with dice, as promised in the Introduction, we shall be entering a blind alley. There are logical (even if counter-intuitive) reasons for asking particular questions in management accounting: what we are going to look at next is just simply a mystery. It is most certainly counter-intuitive, but if any sensible questions can be found which have reasonable answers then I have yet to find them or hear of anyone who has.

I suppose by this time you are on your guard against my making some very questionable statements. But if I said that I can show you that something called "A" is better than another similar one called "B" which is in turn better than a third, "C", there would be nothing at all remarkable about it. There would just be a perfectly ordinary series A, B, C in decreasing order of quality. You know from the heading that we are going to be considering dice so the "better" must refer to being a better bet - to yielding higher scores when thrown at random. And they are perfectly ordinary dice with no bias nor weighting, and the throwing is indeed purely at random since we cannot put on any twists or prevent any spinning.

So OK then, we have got three dice and -

A consistently beats B
B consistently beats C

- and you don't need to have a degree in logic to realise that -

A must consistently beat C.

Now what is actually going to happen, in accordance with purely mathematical rules, is that -

C consistently beats A.

If you find that remarkable, read on: if you don't you may as well skip to Chapter Three. It is exactly on a par with the second floor ("A") being above the first floor ("B") and the first floor ("B") being above the ground floor ("C") AND the ground floor ("C") being above the second floor ("A")! It is illogical, unreasonable, incoherent - but not impossible, as we are about to see.

Please note that it is NOT on a par with the children's game in which stone blunts scissors, scissors cut paper, and paper wraps stone: in that case a different criterion in used for each pair, whereas in the dice paradox we are simply to use numerical scores in every case.

Dice always have six faces with six (normally different) numbers on them. But consider three special dice with these following numbers on their six faces:

Die A 4 4 4 4 1 1
Die B 3 3 3 3 3 3
Die C 5 5 2 2 2 2

(The reason for choosing B to have 3 on every face is simply to keep our later sums as simple as possible. I know you wouldn't go gambling with such a die. And I know we are supposed to be dealing with a game of chance when there is no chance of B showing anything but 3. But please bear with me for the time being - as I say, for the sake of simplicity: we shall have a look at the exactly similar case of dice with different numbers on their faces towards the end of §2.4.)

Each die has a dot-total of 18, such as 4+4+4+4+1+1. And since every face is equally likely to appear it is clear that no die can be any "better" or higher-scoring than any other. False! And what is worse, the impossible but true situation is that over any reasonably long series of throws:

A consistently beats B
B consistently beats C
C consistently beats A

There are two ways in which we can investigate this situation - theory and practice. In either case it is not necessary actually to manufacture the dice, though you can if you like. First of all, then, the from-first-principles pure argument....

Since the dice are not weighted in any way, each face of any die is equally likely (in the long run) to appear. So if we were to throw A 3600 times the chances are that the outcome would be something like -
600 of the first face with score 4
600 of the second face with score 4
600 of the third face with score 4
600 of the fourth face with score 4
600 of the fifth face with score 1
600 of the sixth face with score 1
So from the 3600 throws there will about -
2400 scores of 4
1200 scores of 1
We already know that B is going to score 3 every time it is thrown.

So when A (4 4 4 4 1 1) plays B (3 3 3 3 3 3) we logically and mathematically expect that:

A would win about 2400 times
B would win about 1200 times
IN ANY LONG RUN, A BEATS B.

We can also work out what the balance of scores is likely to be when B (3 3 3 3 3 3) plays C (5 5 2 2 2 2)

Again, all the B scores are bound to be 3s: the expected 3600 scores for C will be around:
600 of the first score 5 so C wins
600 of the second score 5 so C wins
600 of the first score 2 so B wins
600 of the second score 2 so B wins
600 of the third score 2 so B wins
600 of the fourth score 2 so B wins
so

B would win about 2400 times
C would win about 1200 times
IN ANY LONG RUN, B BEATS C.

Now for the crunch!
It takes a bit longer to work out the chances with A (4 4 4 4 1 1) playing C (5 5 2 2 2 2): we no longer have the convenience of a die which always scores 3. But we can carry on at the very basic no-tricks-up-my-sleeve level: the method is still straightforward even if it is a bit more tedious. As before, 3600 throws of A are likely to produce something like -
600 of the first score 4
600 of the second score 4
600 of the third score 4
600 of the fourth score 4
600 of the first score 1
600 of the second score 1
While A scores 600 of its first "4"s C will score something like -
100 of the first score 5 so C wins
100 of the second score 5 so C wins
100 of the first score 2 so A wins
100 of the second score 2 so A wins
100 of the third score 2 so A wins
100 of the fourth score 2 so A wins
- to give A 400 wins, and C 200 wins.

Exactly the same chances occur with A's other three "4"s, to give a total so far of A 1600 wins, and C 800 wins.

Combining the other two options, while A scores 1200 of its "1"s C is bound to win every time with its scores of "2" or "5". So all told -

A would win about 1600 times
C would win about 2000 times
IN ANY LONG RUN, C BEATS A.

So there we are.  A beats B, B beats C, yet C beats A.  I can't explain it.  It is simply an example of logic and mathematics being inconsistent.

§2.4 Computer-simulated Dice

It would be rather a pointless waste of time actually to construct dice to this specification and play hundreds of 'games' to verify or disprove all this. But with a fairly simple spreadsheet a lot can be achieved with each click of a mouse. The dice used so far have been rather too unrealistic, especially the all-threes die. So a modified set might, slightly more realistically, be to this specification -

Red 5 5 5 5 1 1
Blue 4 4 4 4 3 3
Green 6 6 2 2 2 2

Those would be normal six-sided dice. But the relationships would hold true in exactly the same way if it were possible to conceive of three-sided dice. Although impractical in reality, the idea serves to simplify the spreadsheet a great deal. So we'll adopt the simplified specification -

Red 5 5 1
Blue 4 4 3
Green 6 2 2

The dot totals happen to be eleven for Red and Blue but only ten for Green. The 6 could be changed to 7 but it would not make any difference to the relationships: feel free to use 7 in your own spreadsheet if you prefer. The important thing is that there are no equal numbers on different dice, so there can never be a tie. In a similar way, the number of 'throws' in each 'game' must be odd so that, again, no ties are possible. And the actual number of 'throws' must be large enough to give randomness a chance to work itself out but need not be huge: 41 will do.

Many spreadsheets share identical or similar codings so the following are likely to fit your set-up.
Cell A1 RED
Cell C1 BLUE
Cell E1 GREEN
Cell A2 '5-5-1
Cell C2 '4-4-3
Cell E2 '6-2-2
Cell A3 =INT(RAND()*3+1)
Cell B3 =IF(A3=3,1,5)
Cell C3 Copy and Paste Cell A3
Cell D3 =IF(C3=3,3,4)
Cell E3 Paste the copy of Cell A3
Cell F3 =IF(E3=1,6,2)

That has set up the first throws of the dice. Thanks to the magic of spreadsheets all these formulae can then be copied right down to and including row 43. So cell F43, for example, stores =IF(E43=1,6,2).

Now we need to find and record the winners of each throw. There will be three pairs to compare so we can set up the headings and then make the comparisons.

Cell G1 Red/Blue
Cell H1 Blue/Green
Cell I1 Green/Red
Cell G3 =IF(B3>D3,1,0)
Cell H3 =IF(D3>F3,1,0)
Cell I3 =IF(F3>B3,1,0)

As before, G3, H3 and I3 can be copied and pasted right down to G43, H43 and I43. Then all that remains is to find out which member of each pair has 'won'. Since there have been 41 'throws' in each case the winner is whichever has scored more than 20. It will be user-friendly if the outcome can be shown in writing, preferably at the top where is it easily visible, perhaps along these lines -

Cell G2 =IF(SUM(G3:G43)>20,"Red>Blue","Blue>Red")
Cell H2 =IF(SUM(H3:H43)>20,"Blue>Green","Green>Blue")
Cell I2 =IF(SUM(I3:I43)>20,"Green>Red","Red>Green)

The first 'game' will have been revealed as the spreadsheet was being built. After Saving it, any "Recalculate" (possibly F9) produces a whole new 'game'. Experiment!

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Please contact Mike Hosken at
[email protected]
with your observations, comments, criticisms and suggestions, or to request an A5 printed copy of "Hinges and Loops".

The next chapter concerns Paradoxes.

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