Probabilities, CHANCES, or the THEORY OF AVERAGES.

Probabilities, CHANCES, or the THEORY OF AVERAGES. To assign a number which measures the probability of a future event may at first seem impossible; and yet the whole business of many large companies instituted in every civilised country for the 'insurance' or 'assurance' of lives, &c. is mainly based upon the methods of assigning such a number. When it is certain that a future event will take place, or will not take place, a fixed number is selected for each case to indicate that then the probability amounts to certainty: and these two measures are the limits of our scale. Will the sun rise to-morrow morning in the east? Probability = 1, certainty in favour. Will full moon be seen to-morrow morning in the east? Probability = 0, certainty against. Between these two limiting numbers, 0 and 1, lies the number (a proper fraction) which measures the probability of any undecided event. The number, then, by which we mark the chance, or expectation, or probability of anything occurring in the future, must be a fraction like $\frac{3}{4}$ , $\frac{5}{8}$ , $\frac{14}{27}$ , or '273, and can never be so large as 1, which was fixed as the higher limit, certainty: and by the fractional number assigned to any event we can readily compare its probability with those of other future occurrences.

To assign the proper fraction to any future event will, in general, imply knowledge of a large number of similar events. Thus, in January, what is the probability that on next 12th April the sun will rise bright and unclouded? Relying on the constancy of nature and the doctrine of averages, we consult the calendars and weather-notices of the last 50 years, say, and find that in 17 of these the result was favourable and in 33 unfavourable. On these data the probability required is $\frac{17}{50}$ , rather over $\frac{1}{3}$ . In other words, the odds are nearly 2 to 1 against the event. The fraction $\frac{33}{50}$ measures or shows the probability that the event will not happen. More generally, if any future event may occur in 12 ways and fail in 15 ways, then the probability of its occurring is $\frac{12}{12+15} = \frac{4}{9}$ ; and the probability of failure, $\frac{15}{12+15} = \frac{5}{9}$ . In such a case the 27 ways are supposed to have each the same chance of occurrence: and, since the event must either happen or fail, the sum of the two probabilities = certainty—i.e. $\frac{4}{9} + \frac{5}{9} = 1$ . Thus, if $\frac{4}{9}$ is the chance of an event, $1 - \frac{4}{9} =$ chance that it will not occur. In a certain town only 4 days of May—taking the average of many years—are rainless: what will be our chance of finding next 15th May rainless? Chance = $\frac{4}{31}$ ; and $1 - \frac{4}{31} =$ chance of having rain. The principle involved in such simple solutions is the foundation of the mathematical treatment of chance or probability. Of all the occurrences, all equally possible, which relate to a future event, if $a$ are favourable and $x$ unfavourable, then $p = \frac{a}{a+x}$ , where $p$ stands for probability of the event occurring. Sometimes it is easier to find the probability of the event failing, and subtract that result from 1 as in the examples just given.

Out of 100 sailors who mutinied there were 10 ringleaders. If 2 are selected by lot for capital punishment, find the chance that both will be ringleaders. The total number of pairs is $\frac{100 \cdot 99}{1 \cdot 2}$ , and the number of pairs among the ringleaders is $\frac{10 \cdot 9}{1 \cdot 2}$ .

Hence chance required = $\frac{10 \cdot 9}{1 \cdot 2} \div \frac{100 \cdot 99}{1 \cdot 2} = \frac{1}{110}$ ; i.e. the odds are 109 to 1 against the event. A bag contains 5 sovereigns and 4 shillings: if a child is asked to draw three coins at random, what is the probability that 2 will be sovereigns and 1 a shilling? Here the total number of groups of 3 which can be formed out of all the 9 coins is $\frac{9 \cdot 8 \cdot 7}{1 \cdot 2 \cdot 3}$ , or 84, which forms our denominator. Of the sovereigns there are $\frac{5 \cdot 4}{1 \cdot 2} = 10$ pairs, each of which may be drawn with each of the 4 shillings, giving 40 groups of 3, which forms our numerator. Hence chance required is $\frac{40}{84} = \frac{10}{21}$ ; i.e. the odds are 11 to 10 against the event.

Sometimes actual trial seems to throw discredit on the mathematical measure of a chance. Thus, if a die be thrown, the chance of a 5 or any other number turning up must be $\frac{1}{6}$ by our definition: whereas a person may cast a die, say 20 times in succession, with the result: ace, 4 times; 6 and 4, each 3 times; 2 and 3, each 5 times; 5 not at all. How then explain the mathematical estimate? Simply that 20 is much too small a number to take an average from, and the result 'chance = $\frac{1}{6}$ for each side of the die' refers to the most general case possible—i.e. a very large number or even an infinite number of throws. Register for 10,000 throws, then for 100,000 or 1,000,000, and the results would more and more approximate to the mathematical result, and prove that each side has chance = $\frac{1}{6}$ —the die being of course a perfect cube.

An important extension of the theory is that the probability of two independent events both occurring is measured by the product of their separate probabilities. Thus, if A's chance of passing a certain examination is $\frac{2}{3}$ and B's $\frac{1}{2}$ , then (1) the chance that both will pass is $\frac{2}{3} \times \frac{1}{2} = \frac{1}{3}$ —i.e. the odds are 7 to 5 against; (2) the chance that both will fail is $(1 - \frac{2}{3})(1 - \frac{1}{2}) = \frac{1}{6}$ ; (3) the chance that A passes and B fails is $\frac{2}{3}(1 - \frac{1}{2}) = \frac{1}{3}$ ; and (4) the chance that A fails and B passes is $(1 - \frac{2}{3})\frac{1}{2} = \frac{1}{6}$ . By comparing these four results we see that the last event is the most probable of all, the odds being 25 to 24 in favour of it. Moreover, these results exhaust the possible alternatives of double event, therefore the four probabilities should together amount to certainty: and $\frac{1}{3} + \frac{1}{6} + \frac{1}{3} + \frac{1}{6} = \frac{4}{6} + \frac{2}{6} = 1$ , Q.E.D.

By the same principle we solve many useful and curious problems. A town-council of 20, 12 Liberals and 8 Conservatives, have to choose a deputation of 5 by ballot: find the probability that it will contain 3 Liberals and 2 Conservatives. Total number of groups of 5 is $\frac{20 \cdot 19 \cdot 18 \cdot 17 \cdot 16}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5}$ , or 19.3.17.16, which forms our denominator. Number of groups of 3 from the Liberals is $\frac{12 \cdot 11 \cdot 10}{1 \cdot 2 \cdot 3}$ , or 2.11.10, and number of pairs of the Conservatives is $\frac{8 \cdot 7}{1 \cdot 2}$ , or 4.7; therefore, multiplying 2.11.10.4.7 = total number of groups of 5 which fulfil the conditions; and required probability is $\frac{2 \cdot 11 \cdot 10 \cdot 4 \cdot 7}{19 \cdot 3 \cdot 17 \cdot 16}$ , or $\frac{385}{584}$ . In other words, the odds are 584 to 385, or more than 3 to 2 against the event.

When a person buys lottery tickets his chance of success is found as in our opening paragraphs, and if multiplied by the value of the money attainable the product is called his 'expectation.' In this connection may be noted an important distinction between the moral and mathematical values of 'expectation,' owing to the assumption that in such speculations the loss of money paid for tickets is not to be regarded. If one man of moderate means risks £500 in order to gain £5 when the odds are 100 to 1 in his favour, and another risks £5 to gain £500 when the odds are 100 to 1 against, the speculation in the former case appears much more reckless and immoral than in the latter, although in both cases the stake is exactly equal to the expectation.

We now reach the most important of all the applications of the theory of probability, its use in the calculation of life insurances and annuities. During the early part of the 18th century the celebrated London mathematician De Moivre constructed a formula of great simplicity which is still available, although largely superseded by elaborate 'tables of mortality' which have since been compiled in all commercial countries. By De Moivre's hypothesis, out of 86 children born at the same time 1 dies every year until all are extinct. Thus, for a man 40 years old, $86 - 40 = 46$ , 46 years on an average are still before him and 45 others; and his chance of life is the average number between 0 and 46—i.e. $\frac{1}{2} \times 46 = 23$ . Generally a person's probability of life or expectation is $\frac{1}{2}(86 - n)$ , where $n$ is the present age. Actuarial writers have found that this simple formula agrees with their official tables, except in the case of young children and aged persons. The tables are based upon long-continued observations of the mortality in the class of persons dealt with, and from them the theory of probability is easily applied in calculating annuities, reversionary payments, and other results.

For ascertaining the various life contingencies the Institute of Actuaries employ a table giving all the ages from 10 upwards, and, beginning with 100,000 persons alive at the age of 10, place opposite each succeeding age the number of survivors, till at 98 years none are left. At 40, survivors = 82,284; at 50, survivors = 72,726; therefore the chance that a man of 40 shall live to 50 is $72,726 \div 82,284 = .884$ . The Belgian tables give '.832 for the same event in the case of a married man living in town; and if his wife is 30 years old her chance of surviving for ten years is '.862. These data give the following calculation of the chances of the four double events occurring 10 years hence:

Both being alive	$.832 \times .862$	$= .717$
Both dead	$(1 - .832) \times (1 - .862)$	$= .023$
Husband alive only	$.832 \times (1 - .862)$	$= .115$
Wife alive only	$(1 - .832) \times .862$	$= .145$

As we have seen already the sum of these four probabilities must = 1, which verifies the reckoning. The chance of both these persons being alive is evidently more than $\frac{7}{10}$ —i.e. the odds in favour are better than 7 to 3.

Some of the higher applications of the doctrine of probability require a knowledge of the infinitesimal calculus, and are of interest only to experts. It is proved, for example, by integration and the theory of averages that the mean latitude of all places north of the equator is $32.704^{\circ}$ ; and when four points in the circumference of any circle (radius = $r$ ) are taken at random, the mean area of the quadrilateral so determined is $\frac{3r^2}{\pi} = r^2 \times .955$ .

There are works on the subject by De Morgan (1837), Boole (1854), Todhunter (1865), Venn (1866), Whitworth (1886), and Procter (1887).