An essay on probabilities and their applications to life contingencies and insurance offices

You are reading a page from An Essay on Probabilities and their Application to Life Contingencies and Insurance Offices, Augustus de Morgan (1838)
Part of the American Term Life Insurance History Project
Term Life Insurance

ON DIRECT PROBABILITIES. 31

for the thing itself : thus we talk of a temperature of 60°, when, in fact, each 1° only means a certain length on the tube of the thermometer. It will illustrate this to take a case in which we do not confound the thing and its measure; in the barometer, for instance, we never say that the air's weight is 30 inches, but that the height of the barometer is 30 inches.
The technical words, probability and improbability, must now be considered as meaning the same thing in different degrees. If there be only one white ball out of a thousand, we usually say that to draw the white ball is possible, but not probable. We now speak of it
as having a small probability, namely ; we might
say it has a great improbability, namely , but this phrase is not customary. The moment we obtain either numerical measures, or distinctions which are not verbal, the distinctions which are verbal frequently become suand inconvenient. Thus in the art of book-keeping, profit and loss never appear as separate words, but only as part of a complex term profit-and-loss, meaning, one or the other, according to the side of the account on which the item is found. To a mercantile reader, we should say that probability means probabilitythe first or second, according as its measure is greater or less than !,. When the number of favourable and unfavourable cases is the same, say 50 of each, the probability of the event is --^{5 '„ ^{or ; and in this case we say in common life that there is a balance of probabilities, or that the event has an even chance.
By the word chance with the article (a chance) we mean one single way in which an event may happen, as when we say that every white ball adds a chance to the prospect of drawing a white ball. In the first instance above, the chances of white and black are as 20 to 27. It is also usual to say that the odds are here 27 to 20 in favour of black against white.
Questions on probability are twofold in character : 1. Where we know the previous circumstances and re-quire the probability of an event. 2. Where we know}}

32 ESSAY ON PROBABILITIES.

the event which has happened, and require the probawhich results therefrom to any particular set of circumstances under which it might have happened. The first I call direct, and the second inverse, questions.
We must begin with direct questions, though the in-verse precede the direct in practice. For, as we know the whole range of possible cases in hardly one instance, we cannot proceed with points which have reference to matters of life until we know what presumptions arise with respect to the whole, from observation of a part.
In direct questions of probabilities, the event may either be simple, that is, depending on one indivisible event ; or compound, consisting of several events which may happen together. Thus, suppose four events, either of which may happen, and call them A, B, C, D. Knowing the circumstances of each, I may ask the folquestions, every one of which states an event simple or compound.
1. What is the chance of A. 2. What is the chance that one shall happen and only one. 3. What is the chance that one or more will happen. 4. What is the chance that one at least will happen, and one at least will not. 5. What is the chance that a given pair, and no others, will arrive. 6. What is the chance that a given pair at least will arrive. 7. What is the chance that some pair or other will arrive, but only a pair. 8. What is the chance that a pair at least will happen, &c. &c. All these are most evidently distinct questions when they are clearly proposed ; but it is almost as evident that they are very liable to be confounded.
The mathematical definitions and theorems which will be necessary for our purpose, are the following : —
1. A permutation means a number of cases selected out of all possible cases, in some particular order ; so that different arrangements of the same things make difpermutations. Thus if all the possible cases be A, B, C, and D, we have the following

Permutations of one out of four, A, B, C, and D.
ON DIRECT PROBABILITIES. 33

Permutations of two out of four, AB, BA, AC, CA, AD, DA, BC, CB, BD, D13, CD, DC.
Permutations of three out of four, ABC, A B D,B A C,B AD, ACB, ACD, CAB, CAD, ADB, ADC, DAB, DAC, BCA, BCD, CBA, CBD, BDA, BDC, DBA, DBC, CDA, CDB, DCA, DCB:
Permutations of four out of four (different arrangements of four), ABCD, ABDC, BACD, BADC, ACBD, ACDB, CABD, CADB, ADBC, ADCB, DABC, DACB, BCAD, BCDA, CBAD, CBDA, BDAC, BDCA, DBAC, DBCA, CDAB, CDBA, DCAB, DCBA.
To find the number of permutations of one number

(m)
out of another (n), begin with the whole number
(n)
and write down as many numbers, reckoning down-wards as there are units in the number (m)which are to be in each permutation : then multiply all together. For instance : How many permutations are there of 4, out of 12. The answer is.

12 x 11 x 10 x 9 or 11880

The following table will furnish examples, or serve for reference.

109

765

₁

109

765

112

9072

4230

126

720504

336

2101201'

50403024

1680

840360 120

5 30240

15120

6720

2520720 120

151200

60480

20160

5040720

7i 604800

181440 40320

5040

18144001

362880

403201

9 3628800 362880

36288001

:31ESSAY ON PROBABILITIES.Thus the number of permutations of 6 out of 9 is opposite to 6 under 9, or 60480.

2. By a combination is meant the same thing as a permutation, except only that arrangement is no part of the idea. Thus, of all the permutations of three out of four in A, B, C, D, the following, A B C, B A C, A C B, C A B, B C A, and C B A, are all the same combination. Thus we have

a permutation : a selection in a certain order.a combination : a selection without reference to order.

To find the number of combinations, divide the number of permutations by the product of all the numbers up to the number in each combination. Thus, how many combinations can be made of 4 out of 12 ? The answer is found by

dividing 12 x l l x 10 x 9} which gives 495. by lx 2x 3x4

The process may always be shortened by striking common factors from the dividend and divisor. Thus, if we wish to know the number of different hands which can be held at whist (combinations of 13 out of 52) we must

divide 52. 51. 50. 49. 48. 47. 46. 45. 44. 43. 42. 41. 40.
by 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.

The shortest way is to decompose every number in both into its factors, when it will appear that all the factors of the divisor are found among those of the dias follows:

13. 2.2. 17.3. 5.5.2.7.7. 2.2.3.2.2. 47. 2.23. 5.3.3. 11.2.2. 43. 2.7.3. 41. 2.2.2.5.
1. 2. 3. 2.2. 5. 3.2. 7. 2.2.2. 3.3. 5.2. 11. 2.2.3. 13. ;

and the factors which must be multiplied to give the result are

17. 7. 47. 23. 5. 43. 2.7. 41. 2.2.2.5.
or 17. 47. 23. 43. 41. 35. 560 giving 635013559600.
When the number in each combination is more than half of the whole number, the rule may be shortened.
ON DIRECT PROBABILITIES. 35

Thus, if I ask how many combinations of 21 can be taken out of 25, I do in effect ask how many combinations of 4 may be taken. For there are just as many ways of taking 21 as there are of leaving 4.

The following table corresponds to the one preced—
    10    9 8     7    6    5        4    3    2    1    I
    1    10    9!    8    7    6    5    4    3    2    1
    2    45    36    28    21    15    10    F    6    1    3    1
    3    120    84 56    35    20    10    4    1
    4    210    126    70    35    15    5    1
    5        126    6    1        -
    252    .56    21
    6    210    84 1 28 !    7    1
    7    120    36    8    1
    8    45    9    l

    10    ¹1                I^TThus the combinations of 5 out of 9 are 126 in number.

3. When the same event may be repeated in a per-mutation, the number of permutations is the product of as many numbers, all equal to the whole number of possible events, as there are of events in each permutation. Thus, of four events, A, B, C, D, the number of per-mutations of two, with repetition, is 4 x 4 or 16 ; with-cut repetition 4 x 3 or 12. The additional 4 in the first case, are A A, B B, C C, and D D. The number of permutations of three, with repetition, is 4 x 4 x 4, or 64; without repetition, 4 x 3 x 2 or 24. Of the addi40, 10 are where A only is repeated; namely, A A B, AAC,AAD, ABA, ACA, ADA, BA A, CA A,

D 2

36 ESSAY ON PROBABILITIES.

D A A, that is nine where A is repeated twice, and A A A, one, where A is repeated three times. Ten more are those in which B is repeated twice, &c.

4. The number of combinations in which there are repetitions must be determined without rule, in every particular case. Suppose I wish to know how many comincluding repetitions, can be made of four out of seven. Let the seven be A, B, C, D, E, F, G.

I.
The number, without repetition, is 35.
II.
Those in which A is repeated twice, and no other, are evidently as many as the number of pairs in B, C, D, E, F, and G, that is, 15. Thus we have A A B C, A A C D, &c. There are as many in which B is retwice, &c., so that 7 x 15 or 105, is the numin which one only is repeated twice.
III.The number in which A is repeated three times is evidently 6, and the number in which one or other is repeated three times only, is 6 X 7 or 42.
IV.The number in which one is repeated four times, is 7.
V.The number in which two are repeated twice, is as many as the number of pairs in 7, or 21.

We have not so much to do with combinations allowrepetition, as with permutations of the same kind.To avoid the perpetual occurrence of long phrases for sim^{ple ideas, I shall use the following abbrevia: — By P € 4,20 1 is meant the number of per-mutations of 4 out of 20, without repetition : by P P € 4,201 the same with repetition. By C € 4,20 1 is meant the number of combinations of 4 out of 20, without repetition : by C C € 4,201 the same with repetition. Again, by [4,20], as in page 15., is meant the product of all numbers from 4 to 20, both inclu; and, by 7^{10, ^{as in algebra, is meant ten sevens multiplied together. Thus, a reference to the preceding rules will show the following : —}}}

ON DIRECT PROBABILITIES.

P {5,121 is [12, 8],PP-(5,12} is 12^{5 C(4,25} is [25, 22] divided by [I, 4]}

5. If there be an event composed of several others in succession, of which the first may happen in, say 10 different ways, the second in 14, and the third (let there be but three) in 6, then the compound event must be one out of 10 x 14 x 6, which is the number of difways in which it may happen. When all the component events may happen in the same number of ways, this reduces itself, both in principle and rule, to the case of permutation with repetition.
I now proceed to some problems requiring nothing but a knowledge of the measure of probability and the preceding rules. Any event of known circumstances may be familiarly represented by a lottery of balls of different colours. Thus, suppose ten Russian ships, twelve French, and fourteen English, are expected in port, or may have arrived. Let one ^{.be as likely to aras another, and suppose it known that two have arrived, but not of which nation they are. Let a ceradvantage accrue to A, if they should happen to be Russian and French, of which he is desirous of selling his chance immediately. The question is precisely the following : — Let there be a lottery of green, white, and red balls, 10, 12, and 14 in number. Let two have been drawn, and suppose a certain advantage to arise to A, if they be green and white. What should be given to A, certain, in consideration of his chance ?
Here we must first consider the number of ways in which two can be drawn. All the balls are 36 in numand the whole number of pairs is 36 x 35 - 2 or 630. Of these a green ball (1 out of 10) may be paired with a white ball (1 out of 12) in 120 different ways. Consequently, the probability of his gaining the advantage is or A. The probability against it is therefore ',?- : or out of every 21 possible events, 4 are in favour of, and 17 against, the advantage. It is there
D 3
38    ESSAY ON PROBABILITIES.

17 to 4, or 41 to 1, against the advantage. If the congain were 211. the value of the chance would be 41., as will hereafter be more fully shown.
Of all the pairs which can be drawn, there are
10 x 12 =120 green and white. l0 x 9=2=45 both green.
12 x 14= 16S white and red.    12 x 11=2=66 both white.
14 x 10=140 red and green.    14 x 13=2=91 both red.
The sum of these is 630, as it should be. The reader should explain to himself the reason of the difference of process. The least probable case is both green, the most probable white and red. Nothing is more common than the idea that the event most likely to happen, which is compounded of two or more events, is a repetition of the event which is individually the most probable. This is true of repetitions : red being most probable, it is more probable that red should be repeatthan that white should be repeated, or that green should be repeated ; but that two white ones should be drawn is not so probable as that a white and red should be drawn. The drawing, whatever it may be, is conindependently of succession —and this makes an important difference. If the balls were drawn succesboth red is more probable than white followed by red, or than red followed by white, but not more probable than the chance of one or other, which is the preceding case.
I here also take occasion to notice the common error, that because an event is more probable than any other, it is the one to be looked for. The question ought to be, Is that event more probable than some one or other out of all the other events which may happen ? If ten persons engage in a competition, with an equal chance of success, and if two of them, A and B, enter into partnership, it is now more probable that the firm of A and B will win than that C will win, or that D will win, &c. But the chances against the firm are still 8 to 2 or 4 to 1. If a hundred halfpence be tossed up into the air, the result which is more probable than any other, is 50 heads and 50 tails. But comsense will tell us that the chances of this result are
ON DIRECT PROBABILITIES.    39

very small, out of the whole, and calculation would con-firm it.
Before laying down any more specific rules, I shall take an instance of a somewhat complicated deduction, in order to show that we do not really require any other principle than is contained in the measure of probability. Suppose A and B to play at whist against
C and D. A begins, holding ace, king, and queen of trumps, and these trumps only : what right has he to exthat, by playing them out in succession, he will force all the trumps of the other party ? That is to say, ten trumps being distributed among B, C, and D, and any single card having the same chance of belonging to either of the three, what is the probability ,that neither C nor
D holds more than three trumps ?
Firstly, as to the number of tenures, which are perdistinct. To find the number of ways in which any number of distinct objects can be divided among any number of persons, use the following RULE : —
Multiply together numbers equal to the number of persons as often as there are things to be divided among them. Thus, to find in how many different ways ten distinct cards can be divided among three persons, find

3 x 3 x 3, &c. (ten threes) or 3^{i0 which is 59049.}
The question now is, how many of these 59049 ways favour the supposition that neither C nor D holds more than three out of ten. Both together they cannot hold more than six : if, then, we pick out any given six of the ten trumps, that set may be divided among C and D in 2^{6 or 64 ways. But of these, there are two ways in which 0 and 6 may be held, twelve ways for 1 and 5, and thirty ways for 2 and 4, none of which must be included. It must be observed, that in the last sentence, the six distinct ways into which 6 may be divided into parcels of 1 and 5 must be doubled, because * each gives two of our cases, ac}
* It is important to observe that no duplication must take place on this account, if the two numbers be the same; for instance, in dividing 6 into parcels old.

D 4

40    ESSAY ON PROBABILITIES.

cording as C holds 1 and D holds 5, or C holds 5 and D holds 1 : and so of the other cases. Consequently, out of a given set of six, there are 64 all but 44, that is, 20 ways, in which, when they happen, A could force all the adversary^{'s trumps. But every set of six yields 20 such cases : hence the ways in which C and D can}
together hold six trumps, are 20 times C € 6,101 or 20 x 210, or 4200. Similarly, a given set of 5 trumps may be held by C and D in 2^{3 or 32 ways, of which 2 and 10 must be excluded. Hence 20 x C _{t 5,10 1 or 20 x 252, or 5040 sets are the number favourable to the event in this case. Four trumps can be held in 2^{4 or 16 ways, of which 2 must be excluded, or 14 x C € 4,10 } that is, 2940, is the number of favour-able cases. On the supposition that C and D together hold only 3, or 2, or 1 trump, no exclusions are necesand the number of cases are 2^{3 x C _{i 3,10 1 or 960, and 2^{= x C € 2,10 1 or 180, and 2 x C _{1 1,101 or 20 ; and there is one case in which C and D hold no trumps. All the favourable cases are, therefore, in number}}}}}}}
4200 + 5040 + 2940 + 960 + 180 + 20 + 1 or 13341.

The chance of A being able to force all the adversary's trumps is Si;?^{Jg, or nearly 3i to 1 against it.}
Given a fraction less than unity, and which has high numbers in its terms, required a set of fractions which shall be very nearly equal to it, and each of which shall be nearer than any other fraction of the same order of simplicity. Required, also, a near estimate of the error committed in each case.

RULE. First perform the process for finding the greatest common measure of the numerator and deno
ON DIRECT PROBABILITIES.    41

13341),59049(4=4
53364

5685)13341(2 X 4+1=9 11370

2 x2+1=5    1971)5685(4 x 9+4=22 3942

5x1+2=7    1743)1971(1x22+9=31 1743

7x7+5=54    228)^-1743(7x31+22=239 1596

54x1+7=61    147^-)228(1x239+31=270 147

81)147(1 81

66 &c.

Opposite to the first quotient write 1, and proceed to form columns out of the several quotients, in the folmanner : —
1 =1st Numerator.    1st Quotient = 1st Denomina
2nd Qu.=2d Num.    2d Qu. x 1st Qu. + 1 = 2d Den.
2d Num. x 3d Qu. + 1st Num. 3d Qu. x 2d Den. + 1st Den. =3d Num.    =3d Den.
3d Num. x 4th Qu. + 2d Num. 4th Qu. x 3d Den. + 2d Den. =4th Num.    =4th Den.
4th Num. x 5thQu. + 3d Num. 5th Qu. x 4th Den. + 3d Den. =5th Num.    =5th Den.

Any numerator multiplied by Any denominator multiplied
the next quotient, and pro-    by the next quotient, and
duct increased by preceding    product increased by pre
numerator, gives succeeding    ceding denominator, gives
numerator.    succeeding denominator.

Thus, —'_{n, _{-, _{h, 3 1, 154g, s i o, &c. &c., is a set of fractions which approach nearer and nearer to ally^.The first is always too great, the second too small, the third too great, the fourth too small : every odd one too great, every even one too small.
Test of correctness. Take any two successive numerand denominators, multiplied crosswise ; they give products which differ by unity.}}}
42    ESSAY ON PROBABILITIES:
54    61    54 x 270=14580
    239 270    239 x 61 =14579
1

Estimation of the error. The error of _{a is less than _{3^{1_{6- _{(4 x 9 = 36) ; that of is less than --- (9 X 22 = 198) ; that of _{_sz is less than _{61-_{T_7. (22 x 31 = 682), &c.}}}}}}}}
The error of any Numerator is less than    ^{1
its Denominator    That Den. x the next.

I now take the following problem : A die is thrown time after time ; in how many times have we an even chance of throwing an ace. The common error attached to this problem is, that since there are six faces, it is most likely all will have come up in six throws.
In the first throw there are six events, five of which are unfavourable. In the first two throws, considered as giving one event, there are 6 x 6 or 36 possible cases, for every possible case of the first throw may comwith any case of the second. But of these 36 throws, any one of the five unfavourables of the first throw may combine with any one of the second throw, and there are therefore 5 x 5, or 25 unfavourable comevents. Hence the following table : -
One throw gives    6 cases ;    5 unfavourable
Two throws give    36 cases ;    25 unfavourable
Three         216     ; 125     (odds still
against.)
Four         1296     ; 625     (odds
turned in favour.)
Five     7776     ; 3125 &c.

Answer : There is not quite an even chance of doing it in three throws, but more than an even chance of doing it in four.
RULE. When the odds against success in one trial are n to 1, then _{o of n. (or the nearest whole numto it) is about the number of trials in which there is an even chance of one success: more correctly ^{ioo of n. Thus, if it be 144 to 1 against a single attempt,}}
ON DIRECT PROBABILITIES.    43
there is about an even chance of one success in 100 throws.
A table of the number of trials (odds being n. to 1) in which there are various odds of succeeding once : —}}Ito 1    1 J    Li to 1    4264    50 to 1    243    1000 to 1    691
    2to1    110    14 tot    271    40 to 1    371    21100 to 1    760
    3 to 1    139    15 to 1    277    50 to 1    393    3000 to 1    801
    4 to 1    161    16 to 1    283    60 to 1    411    4000 to 1    829
    5 to 1    179    17 to 1    299    70 to 1    426    5000 to 1    852
    6 to 1    195    18 to 1    29-1    811 to 1    439    60110 to 1    870
    7 to 1    208    19 to 1    304)    90 to 1    451    7000 to 1    885
    8 to 1    220    20 to 1    934    100 to 1    462    8000 to 1    899
    9 to 1    230    21 to 1    309    110 tot    471    9000 tot    911
    lo to 1    240    22 to 1    314    120 to 1    4110    10,000 to 1    921
    ll tol    248    23 to1    318    130 tot    4)03
    12 tot    256    24 tot    322    140 tot    495

The method of using this table is as follows : -- Sup-pose the odds to be 20 to 1 against success in a single trial : required in how many throws it is 100 to 1 there shall be one success (or more). Look in the table opposite to "100 to I,^{" and We see 462 ; multiply 20 by 462 and divide by 100 (which is always to be done) this gives 92_{14,,, or 92 is about the number of throws required.
I new proceed to the method of compounding the probabilities of single events, so as to find those of comevents : that is, the way of methodising the reof actual inspection.
Let there be two events, one of which may happen in 7 ways, and may not happen in 5; and the other of which may happen in 4 ways, and may not happen in 9 : whence the probabilities of the two events (A and B) are and - ^{4 . ^{Now, the compound event may hap-pen in 12 13 different ways, for any case of the first}}}}

(12 in all) may come up with any case of the second

(13 in all). By similar reason, the compound case in which both A and B happen may come up in 7 x 4 different ways ; hence the probability of A and B both happening, is

7 x 1 7 4

-- which is the product of and

12 x 12 13
RULE. When events are perfectly independent, so
4I. ESSAY ON PROBABILITIES.

that the happening of one has nothing to do with tha of the other, the probability that both will happen i the product of the probabilities that each will happen.
This rule applies to any number of independen events.
It is indifferent whether the events are to happen to gether, or one after the other: thus the chances of compound drawing out of two lotteries, one drawing out of each, are the same whether two persons drat simultaneously, or one person first draws out of one, ani then out of the other.

EXAMPLE. Let there be three lotteries, as follows : — 6 white l white l white

1 5 black } { 2 black } 10 black }

What is the chance of drawing from the three, white black, and white ? The probabilities of these events are _{1^{6_{1, _{- _{, _{and the product of which is _{1y^{fi , or about 1' to 1 against the compound event.}}}}}}}}

The probability that A will happen, and B will not is the product of the chance for A, and that against B The probability that one will happen, and one only, i the sum of the probabilities—1. that A will happen ant B will not ; 2. That A will not happen and B wil happen. The probability that one or both will hap. pen is the remainder when the probability that neithe will happen is subtracted from unity. The following are evident results of the measure of probability : —

When either P, Q, or R must happen, the sum o their probabilities must be unity, which is always tin representative of certainty. Thus, if a lottery contain white, 5 black, and 3 red, either white, black, or re( must be drawn, and _{i^{7_{5, _{- _{, _{and _{i^{3„ _{together make 1 and the same if the events be rnore or less than three it number.}}}}}}}}}

When of two events, each excludes the other the probability that one or other of them will happen i the sum of their probabilities. For to say that eacl excludes the other, is to say that they are conuectet events, or different possible cases of the same set. Thus

ON DIRECT PROBABILITIES.45in the preceding lottery, if I draw white I cannot draw black, and vice versa, there being one drawing only supposed. Hence the probability of drawing black orwhite is+_{T⁵-_{5 or E. But suppose there had been _{17_{5 two lotteries, as follows : —
(7 white, 8 red) (5 black, 10 red),and I draw in one without knowing in which I am drawing. The individual probabilities of white in the two, if the lottery be known, are ; and _{T^{5_{5 as before ; but the question, what is the chance of drawing white or black, and not red, is a different one from the pre-ceding. If I draw without knowing in which lottery I am drawing, niy position is the same as if one lottery had been thrown into the other, and I had drawn from both in one, giving(7 white, 5 black, 18 red).This union, which will readily be admitted to make no alteration in the probabilities, is not admissible unless the total number of balls in both be the same. For, in the mixture, the 8 red balls of the one and the 10 of the other, are considered as severally of equal probaBut this they are not; unless the number of possible cases in the two were the same. Suppose, for instance, I had the following lotteries : —(10 white, 10 red)(4 black, 5 red);the probability of each ball of the first isbut of each of the second -. Before I can draw a white ball, two events must happen. 1. I must happen to select the first lottery. 2. I must happen to draw a white ball rather than a red. The chances of these are ; andor ^{Jr ; consequently, —, x or ? is my chance of a white ball. But if I mix the two lotteries, that chance will be y, which is more than a. The reason is, that I have mixed together balls which had unequal chances of being drawn, and treated them as if they were to have equal chances. But it is evident, from the measure of probability, that I do not alter the chances for an eventESSAY ON PROBABILITIES.or the general chances which any set of circumstances affords, if I multiply the chances for an event any numof times, provided I multiply the chances against it as often. Reduce the preceding lotteries to common numof balls, by putting 20 balls for one into the second, and 9 for one into the first. We then have —(90 white, 90 red)(80 black, 100 red).Now mix the two, and we have —(90 white, 80 black, 190 red) ;and the chance of a white is 1_6^9₀or '-₁, that is, the prois not altered by the mixture. And the same may be shown in any other case.Let us now suppose that the chance of A isandthat of BThe chances against A and B are there
fore ; andThe possible cases areThe probability of which isThat A and B shall both happen4 x 5 or z{That A shall happen, and not BIs _{T ^{or tThat B shall happen, and not Ax q or niThat neither shall happen.;x4 or zTOne of these cases must happen, and the sum of the chances is =-i or I, each compound event being exof the others. Again, we find(Both or neither isI One One or other, but only one s7 One or other, or both4The probability of One or neiltera _{6j A or both _TA or neither _lTII_{I B or both _{aiB or neither 7_I A or neither, or both ^'siLB or neither, or bothThe latter set of events does not consist of those which are mutually exclusive. Any thing which happens falls under six of them, that is, six of them must happen. Thus, A, and not B, actually arriving, would secure any gain which depended upon either of the following : —One or other, but only oneA, or bothOne or other, or bothA, or neitherOne or neitherA, or neither, or both,ON DIRECT PROBABILITIES.4.7If we add together all the probabilities of the preceding, we find or 6. This is the indication that every possible event enters in six different ways into the contingencies whose probabilities united amount to six.ExAMPLE. Twelve halfpence, A_{I A. ... A_{10, _{are thrown up, required the probability of all the cases which can happen, and which we shall symbolise thus : (Ha T₉) means that there are three heads and nine tails. The chance of H or 'I' in any one piece it i; consequently, the chance of the several pieces A _{i, _{A,, &c., yielding each any particular letter is ,^{1- ^{x ^{-4 xx (twelve factors), or a,_1^1g₆. Now, the 4096 pos
sible cases are thus classified : —Ho T12_{and H_{i. T_{o happen in1 case each.H, T_{11 H,}t T, 12 cases each.H2T,oH10T,C{2,12 or 66 H, T_{o _H,T, C 3,12 or 220 TH_{6 T_{4 _{C 4, 12 or495
..H₅ T₇ 117 T C 5, 12 or 792
H₆ 'I'₆ happens in 924 cases.It appears, then, that the most likely individual reis He T₆, against which, however, it is about 3 to 1. But if we ask for the probability either of this or of a single variation on one side or the other, that is, of the following event —Either H_{5 T,, or H₆ T₆, or H₇ T, —we find 792 x 2 + 924 or 2508^{'(more than half of 4096) cases in which the event arrives. That is, the chances are in favour of one or other of these arriving. As the number of events increases, a given degree of nearness to the most probable event becomes more and more likely. To illustrate this : suppose 24 halfpence thrown up, the notation remaining as before. The total number of cases is now 2 " or 16777216: calculating the number48ESSAY ON PROBABILITIES.of combinations of 1,2,3, &c. out of 24, we have the following summary :H_{O ^{'1^'R4 and I1_{23 T_{o happen in1 case each.H _{1 1 23 11 _{R3 T, 24 cases each.112 _{T22H_{R2 T_{2 276 H_{3 T_{21 H_{21 T_{3 2024 1-1_{4 T_{ROH_{2O T_{4 10626 H_{5 T_{19H_{19 T_{S 42504 H_{6 T_{181I_{18 Ts 134596 II, T,_{7 _{H,, 1'_{7 _{346104 119 T1O H_{15 T_{9 735471 H _{9 1_{15 11,_{5 T_{9 1307504 ^{H10 T1^{H 14 '1'10 1961256 ^{H11 ^{T1a^{H13 Tu 2496144
11_{12 T_{12 happens in 2704156 cases.The odds are now about 6 to 1 against the even di-vision of the pieces into heads and tails. But let us consider the same degree of departure from the most probable case as we took before. An alteration in one piece out of twelve answers to that of 2 pieces out of 24. Now the number of cases in which either H 1 oT_1t,orH_{11T_{1,₁,orH_{1,_T₁,,,orH_{11T_{1„orH_{14T_{1 arrives, and the odds of one or other of them, is computed as follows :H10 T_{14 or _{H14 T_{10392251216777216 whole No. of cases11n _{T13 ^{or H13 ^{Tn499228811618956 No. favourableIlT_{627041565158260 No. unfavourable11618956or it is now more than 2 to 1 in favour of the heads lying between 10 and 14, both inclusive.I now put together the principles on which we have hitherto gone, adding two more, the first of which (Principle III.) is obviously a consequence of the pre-ceding, and the second of which will be presently exPrinciple I. When all the ways in which an event may happen are equally probable, the chance of its happening is the number of ways in which it may happen, divided by all the number of ways in which it may happen and fail.ON DIRECT PROBABILITIES.49Principle II. The probability of any number of in-dependent events all happening together, is the product of their several probabilities.Principle III. The probability of two events arriving together being known ; and also, that of one of them : the probability of the other is found by dividing the first mentioned probability by the second.Principle IV. When an event may happen in several ways, 'whether equally probable or not, the probability of the event is the sum of the probabilities of its hapin the several different ways.The best way of illustrating the last principle is by beginning with one of the numerous errors into which we may fall, either in proceeding towards it, or applying it. Let there be an event which may happen in two different ways, for each of which there is an even chance. Then, according to the principle, it is certain (',- ± =1) that the event must happen. In this there is nothing inconsistent. Let there be a lottery containing ten white balls, and let them be sub-divided into two sets of five by a mark. Then there are two ways of drawing a white ball, for each of which there is an even chance ; namely, I may choose a ball of one mark, or of the other. But it is evidently certain that in this case a white ball must be drawn. Now, suppose an event can happen in three different ways, for each of which there is an evenchance. Then the event is more than certain G + + ? — I ) ; which is absurd. But the absurdity is inthe supposition : an event can only have an even chance of happening in one particular way, when that way involves half of the total number of individual cases of the event; and it is impossible that three different ways of arriving can each contain half of the whole number of possible cases. Consequently, when we have made a calculation of the probabilities which different ways of arriving give to an event, there has certainly been an error if the sum of the probabilities exceed unity.But when we throw three half-pence into the air, are there not three different ways of throwing head, for eachE50ESSAY ON PROBABILITIES.of which there is an even chance? If by H we here mean a single H, the three events by which we propose to attain it are not H, H, H, but H T T, T H T, and TTH, the probability of each of which is _{a and, by the application of the principle, + ii + , or II, is the probability of the event,-,-one single head. If by throwing a head we mean one head or more, the ways under which it may be brought about are—one head only (involving the cases H T T, T H T, T T H)—two heads only (involving H H T, H T H, and T H H)—and three heads (involving H H H.) The probabilities of these are t, fl, and ;, or _{H is the chance required.Another error to which we are liable, is the wrong estimation of the probability of the different cases: For instance, a person is to go on until he throws H, and is to win if he do it in less than five throws. There are then five possible cases ; namely, H, T H, T T H, T T T H, T T T T. In four of these cases he wins ; in the fifth he loses. His chance of winning then appears to beBut this supposes the five cases to be equally likely, which is not true. Their several probabilities are +, ;,and - : not 5,-'_{k-, _{&c., as supposed. Consequentlythe sum of the probabilities which the several winning cases actually have is 1- or it is 15 to 1 that he wins. If we put together all the cases, which four throws pre-sent, thus,1.HHHH5.HTHH9.THIIH13.TTHH2.HHHT6.HTHT10.THHT14.TTHT3.HHTH7.HTTH11.THTH15.TTTH4.HHTT8.HTTT12.THTT16.TTTT;all those which begin with H are 8 in number, those which begin with T H are 4, with T T H, 2, and with T T T H, one. Hence 8 + 4 + 2 + 1, or 15, is the number of winning cases. But the argument against us, is this : most of the preceding cases are impossible, for the oondition is, that the play shall stop as soon as H occurs : so that, in fact, the only possible cases are, H, TH, TTH, TTTH, TTTT. Let it be so; we must then represent the several events as follows:—ON DIRECT PROBABILITIES.511st event; certainly happens, and gives either H or T. Probability of H,2nd event; does not certainly happen, but is conupon the first throw being T ; and gives either H or T. Probability that the throw is made, 1- that if it be made it gives H ; probability that the throw is made, and that, being made, it gives H, z3rd event ; contingent upon the two first throws giving 1' and T, of which the chance is -1. Probability of winning at this throw, - x '- = t_{i.4th event; contingent upon the three first throws giving T, T, T. Probability of winning at this throw,1. x i_ — 1o1lIt is obvious enough, when stated, that every con. tingency must enter into the consideration of a question ; so that if some of the circumstances depend upon the manner in which preceding contingencies arrive, this circumstance itself influences the method of proceeding. If we wish to avoid the necessity of considering a conthe trial of which is itself contingent, that is, if we wish to make a contingency certain, we must inall the new events which such change of coninto certainty brings with it. The whole problem is exactly the same as if we made the four throws certain, and made the gain dependent upon one head or more being thrown : but we revert again to the former state of the question, if we agree to mark the first H as the winner. In this point of, view, the distinction between the two is evidently immaterial.EXAMPLE. There are seven lotteries, as follows (W means white, B black) :(1 . (3 W, 2p) (2 W, , 1 B) B) V. IV.11. (2 W,3B)(1 W, 5 B) VI.III. (1 W,3 B) (4W,1 B) VII.; and the conditions of drawing are the following. I. is drawn, and then II. or III., according as I. gives W or B.If II. be drawn, then IV. or V. is to be drawn, according as II. gives W or B. But if III. be drawn, then VI. orE 252ESSAY ON PROBABILITIES.VII. is to be drawn, according as III. gives %V or B. What are the probabilities of the several possible drawings ?If from I. we draw W (of which the chance is 5), we proceed to II. If we still draw IV (chance, 5), we proceed to IV. And here the chance of MT is 1. Hence,the chance of WW W is x i x i= 23Computing all the other chances in the same way, we get the following:

WWW1 i 1_A5. BWW3 J =

WWB ;= AG. BWB j = 3. WBWi=_{7. BBW is⁼}
4. WBB ; 3_ i^{t;^{8. BBB= lbaThe sum of all these is equal to unity, as it should be, since one or other of these cases must happen. And by reducing all to the common denominator 5. 5.4.6, or 600, we have the following chances:1. WWW3.WBWs 5.BWW _{asa 7.BBW_^{2. ^{WWB ZI 4. WBBI 6. BWB aI 8. BBL' s3±,That all shall be white,7 to I against (nearly).
Two white and one black, 3 to 1 against (very nearly) Two black and one white, an even chance (nearly)All black,10 to 1 against (nearly).From what has been said in this chapter, no great difficulty will be found in ordinary questions. The cirare supposed to be fully known, and the probabilities will be found, of the strength which it follows they must have, to those who admit the axioms on which the measure of probability is founded.}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}