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
USE OF TABLES.    69
CHAPTER IV.

USE OF THE TABLES AT THE END OF THIS WORK.

I HAVE endeavoured to accumulate in this chapter a considerable part of the uninteresting details of comwhich accompany the solution of complicated problems. It is at the readers' pleasure to omit the whole of it, referring to it afterwards in cases where its assistance may be necessary.
In Table I. we see — (I.) a column headed t, conthe series 00, .01     99, P00,     2, or
every hundredth of a unit from 0 to 2 —(II.) a column headed H, deduced in the manner pointed out in page 17, — (III.) columns headed . and A"-, which are only the differences of the numbers in column H (marked A), and the differences of those differences, (marked mil'=). The following is an extract from the table : —
L    R
47 49374 52 900 46 8 61
48 *50274 98 891 85     
49 51166 83     
The columns A and A2 must be made up to 7 places of decimals'' by means of ciphers : thus, 90046 means 0090046 ; and 861 means 0000861. The formation
of ..    and ,,    is then as follows : —    0090046
    5027498    5116683    
    4937152    5027498    0089185
Subt.    0090046    0089185    0000861
* A little practice will show how to dispense with the decimal points altogether till the end of the process.
F
70    ESSAY ON PROBABILITIES.
Since it is very seldom necessary to use more than five places of the table marked t, the sixth and seventh places, and those which arise from them in the differences, are separated from the rest by a blank space. The sixth and seventh places are allowed to remain, on account of the use which will hereafter be made of the differences derived from them.
In Table II. we find to five places only a column marked Ii, and another marked containing the difof the former column. This is a modification of the former table, the reason of which will hereafter appear. In the meanwhile, however, observe that we can directly find the value of H and Ii by these tables only, when t is 00 or 01, &c. ; that is, when t is a given decimal of two places. But supposing it required to find H when t lies between two of the values in the table; suppose, for instance, we ask what is H when t = 47694 ? The method is as follows :
QUESTION. What is the value of H (Table I.) when t = -47694, correct to five places of decimals ?
RULE.
Take out of Table I. the value of I-I answering to the two first decimal places and the whole number preceding them, if there be one. Retain only five places of decimals.
Take the figures of the first difference (as far as the blank space), and multiply them by the remaining figures in the value of t, and cut away as many places from the result as there were remaining figures.
Add the figures in the last result to the right hand of the first, and the sum is the anrequired.
EXEMPLIFICATION.
Opposite to 47 we find
.49375 *
Three figures remaining, 694.
900 x 694 =624600.
Cut away three figures, 625.
.49375 C'.i5

50000
When t = .47694, H = 50000.
+ Whenever decimal figures are rejected, if the first rejected be five or upwards, the last retained is increased by a unit
USE OF TABLES.    71
As another example, suppose the value of t to be 1.31209
When t=1.51,    H=.96728    1    A=114
    24        209
    When t=1.51209,    H= 96752        1026
23,826
We shall now take the inverse problem, and sup-posing H to be intermediate between two values in the table, require the value of t. For instance, let H = 92972.
2`28
RULE.
Find in the table the value of H next below the given value; note the corresponding value of t, and subtract the nearest value of H from the given value.

Annex three ciphers to the difference just found, and di-vide by the figures of the difin the table which come before the blank space, rejecting fractions, and taking the nearest whole number.
The quotient cannot have more than three places : if three, annex to the value of t already found; if less than three, place ciphers at the be-ginning to make up the defiand annex.
EXEMPLIFICATION.
H 93972, next below    1
in the table 93807t = 32
165 Tabular cliff. 195.

195)165000(846 1560

900 780

1200 1170

30 t=1.32 846
F 4
72    USE OF TABLES AT END OF WORK.
Let H = 97169; required the value of t?
H= 97169
Nearest below = 97162     t =1.55
101)7000(69    t=1.55069 Answer. 606
940
909
31
The Table II. is used in exactly the same way, except towards the end, from t = 340 upwards ; in which case the cipher at the end oft must be neglected, and only one decimal place taken out of the table in the value of t. For instance, to find the value of t answering to K =.98222.
K = 98222
Next below '98176     t=3.5
306 )46000(150    t = 3.5150
306
1540
1530
100
In the first table, there is another result which will frequently be wanted, and which I shall call H'. It arises from adding half the second difference to the first difference *, if the value of t be in the tables, and making five decimal places. But if the value of t be not in the tables, then H' must be formed for the values of t immediately above and below ; and by means of the first and the difference of the two, H' must be found in exactly the same manner as H is found in the first of the preceding rules, page 70., remembering to subtract at the last step instead of adding, if the second H' thus previously determined be less than the first.


* Meaning the whole differences; not the parts which precede the blank space, as in the preceding rules.
USE OF TAPLES.    73
EXAMPLE 1. When t is 1.56, what is H' ? '= 9745 4A2= 151
H'= 09896
EXAMPLE 2. When t = F23412, what is II' ?
    t = 1.23    H' _ 24852
    t=1.24    Difi. 607
    H'= 24245
    412    24852
    607    250
    2884    Subt. 24602=H' when t=1.23412
2472 250,084
We have already seen that when two events, A and B, one of which must happen at every trial, have severally happened m times and n times in m + n trials, it is m + 1 to n + 1 that A shall happen at the next trial. But m + 1 to n + 1 is very nearly m to n, when m and n are considerable numfor instance, 248 to 117 is very nearly 247 to 116. That is, when a great many trials have been made, the numbers of times which A and B have hapexpress very nearly the odds (relative probafor A against B ; or, inverted, for B against A. Let us convert the problem, and supposing that we know beforehand the chances of A and B, are we to suppose that in a great many trials A and B will happen in proportion to their respective probabilities ? Common sense tells us that such will always be nearly the case, but that the odds are great against an exact result. Suppose 3000 drawings to be made from a lottery contwo As and one B. We must then, it seems clear, draw A twice as often as B, in the long run. Our reason convinces us thus. Let one of the As be distinguished from the other by an accent, so that we have A, A', and B. If the urn be well shaken before each drawing, it is impossible to believe that, in the
74    ESSAY ON PROBABILITIES.
whole result of 3000 trials, we shall have drawn the three in very unequal numbers ; so that, destroying the distinction between A and A', we feel secure of drawing A twice as often as B ; and it is obviously two to one in favour of A at each trial. The following phrases seem to common sense to mean the same thing.
It is two to one that A shall happen, and not B.
It is an even chance for head or tail.
It is more than a hundred to one, that a ship at sea will not be lost.
In the long run, A will happen twice as often as B. In a large number of tosses, the heads and tails will occur in nearly equal numbers.
Of all the ships which sail, the number which is not lost exceeds that which is lost more than a hundred times.
I now proceed to some problems, which will exhibit the method of applying the tables, and will illustrate and confirm the preceding notions.
PROBLEM I. The odds for A against B being a to b, to find the chance that in n times a + b trials, A shall happen exactly n x a times, and B n x b times.
RULE. Divide the H' belonging tot = 0 (page 72) by the square root of the following : 8 times the product of n, a, and b, divided by a + b.
Suppose, for instance, a die is thrown 6000 times ; what is the chance that exactly 1000 of the throws shall give an ace? Here it is 1 to .5 that an ace shall be thrown in any one trial, and 6000 is 1000 times I + 5. Hence a = 1, b = 5, n = 1000: 8 times the product of n, a, and b is 40,000, the sixth part of which is 6667 (sufficiently near), and the square root of this is 8 P65. Again, when t = 0, we have in Table I.

A=112833, A"=11 ; whence H' is 1.12844

and P12814 divided by 8P65 gives 014 very nearly. This is very near the real probability that 6000 throws with a the shall give exactly 1000 aces : for such an event there are only 14 chances out of a thousand ; and
USE OF TABLES.    75
it is 1000 — 14 to 14, or about 701- to 1, against the event. This result is rather above what we should have expected ; we might have imagined it to be more than 71 to 1 against 6000 throws giving exactly 1000 aces.
As another example, let us find the probability that, out of 200 tosses with a halfpenny, there shall be ex100 heads and 100 tails. Here a = 1, b = 1, n = 100, and 8n a b'' is 800, which divided by a + b, or 2, gives 400, the square root of which is 20. And H', when t = 0 (or 1.12844) divided by 20, gives 056. It is therefore about 944 to 56, or 17 to 1, against the proposed event; and (page 42.) we must repeat 200 throws 12 times to have an even chance of the equality of heads and tails happening once.
Generally speaking, the rules in this chapter are very accurate only when the number of trials is considerable. Suppose only 12 tosses ; required the chances of 6 heads and 6 tails. Here a = 1, b = 1, n = 6, 8 nab = 48, which divided by 2 gives 24, whose square root is 4.9 very nearly. And 1.12844 divided by 4.9 gives 23, or 77 to 23, that is 318; to 1 against the event. That is (page 47), this rule is not very inaccueven when the number of trials is as low as 12.
We shall call the event whose chance is sought in the preceding problem, the probable mean; understanding by that term the event which is more likely to happen than any other. Thus, when 12 halfpence are thrown up, 6 heads and 6 tails is the probable mean, being the event which is more likely than any other, though not in itself more likely than not. When 6000 throws are made with a die, the probable mean is 1000 aces, 1000 deuces, &c.
PROBLEM II. The odds for A against B being a to b, required the chance that in n times a + b trials, the As shall fall short of the probable mean by a given
* Juxtaposition of numbers, in algebra, stands for their product.
76    ESSAY ON PROBABILITIES.
number 1 : 1 being small, compared with the whole number of As in the probable mean.
RULE. Divide twice 1 by the square root obtained in the last example ; and find the value of H' from Table I., taking the preceding quotient for t. Divide H', so found, by the square root just used, and the quotient is the answer required.
N. B. This rule also applies when the number of A s is to exceed the probable mean by 1.
EXAMPLE I. In 6000 throws with a die, what is the chance that the aces shall fall short of (or exceed) 1000 by exactly 50 ? Here a = 1, b = 5, n = 1000, and the square root is 81.65, as before. And twice 1, or 100, divided by 81.65, gives 1.22: to which the value of H' is 25162 + 1- of 611, with five decimal places, or 25468. This last, divided by 81.65, gives 0031 ; so that it is about 997 to 3, or 332 to 1, against the proposed event : and the 6000 throws must be re232 times to give an even chance of succeeding once.
EXAMPLE II. What is the chance that in 200 tosses, there shall be exactly 95 heads ? Here a = 1, b = 1, n = 100, 1 = 5, and the square root, as before, is 20. And twice 1, or 10, divided by 20, gives 50, which being t, the value of H' is 87882, which divided by 20 gives 044 very nearly. It is, then, 956 to 44, or about 22 to 1, against the proposed event.
EXAMPLE III. In 12 tosses, what is the chance of exactly 7 heads ? Here a= 1, b= 1, n = 6, 1= 1, the square root, as before, is 4.9, and 2 divided by 4.9 is 41 nearly ; which being t, H' is 95384, which divided by 4.9 gives 194. It is therefore 806 to 194, or 4-i" to 1, against the proposed event. In page 47. it is 3304 to 792 against this event, or 44 nearly. Hence the incorrectness of our rule is very small.
PROBLEM III. The odds for A against B being a to b, required the chance that in n times a + b throws, the number of A s shall not differ from the probable mean by more than 1.
USE OF TABLES.    77
RULE. Divide one more than twice 1 by the square root already mentioned, and the quotient being made t, the value of H in Table I. is the probability re
EXAMPLE I. In 6000 throws with a die, what is the chance that the number of aces shall not differ from 1000 by more than 50 ; that is, shall lie between 950 and 1050, both inclusive. Here a = 1, b = 5, n =1000,1= 50, and the square root as before is 81.65. Divide 2 1+ 1, or 101, by 81.65, which gives P237, which being t, H is 91977. Hence it is 920 to 80 in favour of the proposed event, or about 112 to 1.
EXAMPLE II. In 200 tosses, what is the chance that the number of heads shall lie between 97 and 103, both inclusive ? Here a = 1, b = 1, n = 100, 1= 3, and the square root, as before, is 20. And 2 1 + 1, or 7, divided by 20, gives 35, which being t, H is 379. Hence it is 621 to 379, or about 31 to 19, against the proposed event.
EXAMPLE III. In 12 tosses, what is the chance of the heads being either 5, 6, or 7 in number ? Here a = 1, b = 1, n = 6, 1 = 1, and the square root, as before, is 4.9. And 2 1 T 1, or 3, divided by 4.9, gives 61, which being t, H is 6117. Hence it is about 612 to 388, or 153 to 97, in favour of the proposed event. In page 47 the chance of this event is
792+924+792    2508
or - or 612 4096    4096
PROBLEM IV. The odds for A against B being a to b, and n times a + b trials being to be made, for what number is there a given probability H that the As shall not differ from the probable mean by more than that number ?
RULE. Find in Table I. the value of t answering to that of H (page 71), multiply it by the square root already described, subtract 1, and divide by 2 : the
78    ESSAY ON PROBABILITIES.
quotient, or its nearest whole number, is the answer required.
EXAMPLE. In 6000 throws with a die, within what limits is it two to one that the aces shall be con? The square root is 81.65, and H is -° or 66667, to which the value of t is 68409, found as follows (page 71) : —
H = 66667
Nearest below 66378 t=68
Tab. Diff. 706) 289000(409 t=68409 say 6841 2824    81.65
6600    34`-05
6354    41046
6841
246    54728
55.856765 1
2)54.86 27.43=1.
Answer. It is a little more than 2 to 1, that the aces shall lie between 1000 — 28 and 1000 + 28, and a little less than 2 to 1 that they shall lie between 1000 — 27 and 1000 + 27.
But the most convenient way of solving this problem is by first finding for what degree of departure from the probable mean there is an even chance. In this case, since H = .5 (page 70), t is = 476936, which the method in page 41, will show to be very nearly a-. It will be worth while to re-state the whole process.
The odds for A against B being a to b, and the pro-posed number of trials being n times a + b, required the limits of departure from the probable mean na, within which it is an even chance that the number of As shall be contained.
RULE. Multiply together 8,n,a, and b, and divide by
USE OF TABLES.    79
a + b : extract the square root of the quotient, and multiply it by 31 : subtract 65, and divide by 130: the nearest whole number is the answer required. Thus in the preceding instance, where the square root is 81.65, multiply this by 31, which gives 2531.15 ; and 65 less is 2466.15, which divided by 130 gives 18.97. Hence it is very little less than an even chance that the aces in 6000 throws shall be between 1000 + 19 and 1000 — 19, or 1019 and 981.
Having found the limits of departure for which there is an even chance, we can now use Table II. as follows. The values of t in Table II. are the proportions of various departures (each increased by 5) to that deparwhich has an even chance, as just ascertained, and also increased by 5 : the values of K are the probaof the departures answering to those of t. Having then ascertained 18.97 to be the departure for which there is an even chance, suppose I ask what is that limit of departure within which it is two to one that the aces shall be contained. Two to one gives for the chance, or 66667: I look into Table II., and find that when K is 66667, t is 1.43433, found as follows : —
K= 66667 t=1 43 Next below    66521
Tab. Diff. 337) 146000(433 t=1.43433 1348
1120 1011
1090 1011
79
This is the proportion which the departure in quesincreased by 5, bears to 18.97 increased by 5 or 19.47. Multiply 1.43433 by 19.47, giving 27.93; from which subtract 5, giving 27.43 for the limit of departure, the same as in page 78.
80    ESSAY ON PROBABILITIES.
Suppose the question to be that of page 77, namely, what is the probability that the number of aces in 6000 throws shall lie within 50, one way or the other, of the probable mean 1000 ? Now, 18.97 + 5 is 19.47, and 50 + 5 is 50.5, and 50.5 divided by 19.47 gives 2.594, which being t (in Table II.), K is 91981, exnear to the result in page 77.
There are, therefore, two distinct methods of treating these problems, connected with the two tables : and this is a great advantage, since it is a very strong presumpof a correct answer, when the results of the tables agree. The problems III. and IV. being of great importI shall now recapitulate their details, with the adof some new phraseology. Let the instance be 6000 throws of a die, and the event A the arrival of an ace, and B the arrival of some other face. The most probable number of aces is 1000, though the arrival of that exact number is not probable in the common sense of the word. There will then most likely be a defrom the number 1000 in the number of aces thrown ; of which departure we are now entitled to say, that it is very improbable it should be considerable. Let the term neutral departure mean that degree of defor which it is just an even chance that the actual event shall be contained within its limits : in the present instance it is 18.97. We may explain the fraction as follows : suppose a person to receive 1001. for every unit by which the number of aces falls short of or exceeds 1000. Then, supposing him to try this stake a great many times, he will in the long run reless than 18971. at a trial, as often as he receives more. But his receipts will oftener exceed than fall short of 18001. ; while they will oftener fall short of than exceed 19001. Roughly speaking, there is here the same probability that the aces shall not lie between 1000 — 19 and 1000 + 19 (both inclusive), and that they shall lie between these numbers.
In all these problems there is a square root to be found, which we call the square root, as there is no other
USE OF TABLES.    81
The odds are a to b, for A against B, and n (a + b) trials are contemplated. Though we have only instanced whole values of n, yet it may be a fraction : thus, if the odds are 3 to 2, and 96 trials are contemplated, n (3 + 2) must be 96, or n must be 19. In this case, the promean is that A shall happen 57, and B 38; ; by which it must be understood, that a person who should repeat 96 throws a great many times, receiving 11. for every A, would, in the long run, gain on the average 5741. per trial of 96 throws.
The square root in question, represented algebraiis
Snab !üa+b
or the square root of the product of 8, n, a, and b divided by a + b. I now subjoin the two principal prowith the two rules in parallel columns.
PROBLEM. What is the chance that the number of As in n(a +b) trials shall lie between na + l and na -1, both inclusive ? or what is the chance that the departure from the probable mean shall not exceed 1 ?
BY TABLE I.
Find the square root, and divide one more than twice by it; call the result t, and find H answering tot in the table. (Use the rule in p. 70. if necessary.) This H is the probability required.
By TABLE II.
Find the square root, and multiply it by 31 ; then divide by 130. To 1 add 5, and divide by the preceding quotient; call the result t and find the value of I< answering to t : this is the probability required.
N. B. The neutral departure is 5 less than the quotient first found.
PROBLEM. What is that degree of departure within which it is p to q that the number of As in n (a + b) trials shall lie ?
a
82    ESSAY ON PROBABILITIES.
BY TABLE I.
Divide p by p + o, and callthe result II, find the corvalue of t in the table. Multiply it by the square root, subtract 1 and divide by 2; the quotient being called 1, it is then p to q that the A s in n (a +b) trials shall be contained between na—1 and na+1, both inclu
By TABLE II.
Find the square root ; mulby 31, and divide by 130. Divide p byp+q, and calling the quotient K, find the corresponding value of t in the table ; multiply t by the preceding quotient, subtract 5 from the product, and 1 being the remainder, it is then p to q that the As in n (a+b) trials shall be contained between na —land na + 4 both inclusive.
I shall conclude this problem with an example of each case, worked by both methods, without explanation. There is a lottery containing 3 white and 2 black balls : what is the chance that in 50,000 drawings the numof white balls shall be between 30,000 + 100 and 30,000- 100 ?
a=3 6=2,n=- 10,000, 1=-IOU

8 x 3 x 2 x 10,000
5    = 96,000

96,006= 309.84 309.84)201( -6487S = t H = 64109
309.84 31

130)9605.04(73.885 73.885)100.5(1.3602=t K = 64109
This question shows how nearly a great many trials may be expected to agree with the probable mean : in 50,000 trials, it is nearly two to one against the number of white balls differing from 30,000 by more than a hundred.
In 100,000 tosses, between what limits is it 99 to 1 that the heads shall be contained ?
arc 1, b=I, n=50,000,p=99, q=1 100)99(99=H t-1.8215
8x1x1x50,000
2    — 200'000
447°21 31

130)13863.51(106.64 100)99(99—K
USE OF TABLES.    83
J 200,000 = 447'21    t=3'821
    106.64
    1'8215    3.821
    447.21    
    814.6    407'47
    2)813'6    406'97 =
    406'8 =1    
Answer. Between 50,000—407 and 50,000 + 407.
Now for the inverse method attached to the preceding problem. If I be totally unacquainted with the naof the events A and B, except only that one or other, and not both, must happen every time, it is then clear that, as the matter stands, it is to me 1 to 1 for A against B, with a very great chance that, if I were better informed, I should form a different opinion. At the same time (page 10), I choose 1 to I as my rule of action, because, though coming events may not justify my pre-diction, I know of nothing to warrant my assuming that the odds are in favour of A, rather than in favour of B. A trial takes place, and A happens; it becomes immost safe to assume that the odds for A against B are 2 to 1, but still that safety is not very decided. But if 1000 trials be made, and if A have happened .520 and B 460 times, I can then confidently say, that the odds for A against B are very nearly, if not exactly 520 + 1 to 480 + I, which is nearly 520 to 480. The notion then formed has a strong presumption that it is nearly correct.
PROBLEM. In a + b trials A has happened a times and B b times : from which, if a and b be considerable numbers, it is safe to infer that it is a to b nearly for A against B. 'What is the presumption that the odds for A against B really lie between a—k to b + k and a+k tob —k?
G 2
84    LSSAy ON PROBABILITIES.
RULE. (TABLE I.)    RULE. (TABLE II.)
Divide twice the product of a and ,b by their sum, and ex-tract the square root of the quotient, by which divide k. Then the last quotient being t, the H of the table is the probability required.
Having found the square root, and divided k by it, as opposite, from seven times the quotient, take the hundredth part of the quotient, and take three tenths of the remainder. Make the result t, and K in the table is the probability re
Suppose that in a thousand trials, A has happened exactly 600 times, and B 400 times ; what is the prethat the odds for A against B lie between 570 to 430 and 630 to 370 ?
a= 600, b = 400, k = 3O.
    2 x 600 x 400=480,000    1.369
    7
    480,000    
    1000 =480    9.583
    014
    x/480=21.91    
    9.569
    30    3
    
    21'91 =1'369 = t    
    28.707 t=2. 871
H =94713    K = 94719 Answer. About 95 to 5, or 19 to 1 in favour of the odds being between the limits specified.
In the preceding problem, A and B have happened a and b times; whence the most likely of all individual cases is, that the odds for A against B are a to b; or, in other words, the result which has the strongest presumption in its favour is, that
Probability of A was 72 + b. Probability of B was— b Now we have found, in the preceding problem, the presumption that
Probability of A lies between a + bb and a a
+ b '
11SE OF TABLES.    85
or, which is the same thing, that
Probability of B lies between a +b and a+b
For since it is our hypothesis, that either A or B must happen at every trial, whatever presumption there is that the chance of A is x, there is the same presumption that the chance of B is 1 x.
But we might ask the following questions : A and B having happened a and b times in a + b trials, what are the values of the following presumptions ?


    
a+    a 1. That the probability of A lies between + b and a+b;


or its equivalent, that the probability of B lies between
b— k and b
a+b    a+b.
a+ k
?. That the probability of A lies between a and —
    a+b    a+b'
b or or that the probability of B lies between a+b and a+b
To solve these by the help of the following rule, remember that, if a be greater than b, it is more likely that the chance of A falls short of a=(a-f b) than exit : and if a be less than b, then it is more likely that the chance of A exceeds a=(a+b) than falls short of it.
RULE. First find the result of the preceding proand find from Table I. the H' (p. 72) belongto the value of t. Subtract this from the H' derived from 0 in the table (which is F12844) ; mulby the difference between b and a, and divide by the product of the square root used in the preceding problem and three times the whole number of trials : call the result V. To one half of the result of the preceding problem add V ; and from it subtract V : and call these results zH - V and 2H — V.
Then, if B have happened most times, z H + V is the
G
86    ESSAY ON PROBABILITIES.
a
presumption that the chance of A lies between a and
a+ b
+b or that the odds for A against B lie between a to b
a
and a+k to b—k. But in this case, ill—V is the presumption that the chance of A lies between a—k    a
¢+b and —f— or that the odds for A against B lie
between a— k to b +k and a to b.
But if A have happened most times, make I H + V and 1, H -- V change places in the preceding paragraph, every thing else remaining the same.
EXAMPLE. In a thousand trials, A has happened 600 times, and B 400 times. What is the pre1. that the odds for A against B lie between 600 to 400 and 630 to 370 ; 2. that the odds for A against B lie between 570 to 430 and 600 to 400 ?
From the preceding problem t—1.369, say—1.37; the square root is 21.91, and H is 9471.
t=1.37    17036    1.12844    b—a=200    
X02    232    17268        
        3000 x 21.91 =    65730
    H'= 17268    95576        
    200        
65730)191.152(0029 =V '4736=1H
4765 =H+ V 4707 =1H—V
Hence (since A has happened most times), it is 4707 to 5293, or about 47 to 53, that the odds for A against B lie between 600 to 400 and 630 to 370. And it is 4765 to 5235, or about 48 to 52, that the odds for A against B lie between 570 to 430 and 600 to 400.
The problems which the preceding part of this chapter has enabled us to solve, are the determination of the chance which exists (under known circumstances) for the happening of an event a number of times which
USE OF TABLES.    87
lies between certain limits, and its converse. The latter problem contains a consideration of some diffinamely, the probability of a probability, or, as we have called it, the presumption of a probability. To make this idea more clear, remember that any state of probability may be immediately made the expression of the result of a set of circumstances, which being ininto the question, the difficulty disappears. Thus, suppose a large number of urns containing various proportions of black and white balls. Let there be 100 urns, and let one of them only contain equal numbers of black and white balls. If then I lay my hand upon one of these urns with the intention of drawing, it is, before the drawing, 99 to 1 against my having placed my hand upon an urn from which, in the long run, equal numbers of both sorts of balls will be produced : the presumption that black and white balls have an even chance is only ; the prethat the probability of a white ball is , is
3'00 t

In speaking of compound probabilities, writers have employed six words synonymously—probability, chance, presumption, possibility, facility, and expectation. Reonly the word possibility, as indicating a thing of which there cannot be different degrees, the five remaining terms have their advantages, each one pointing out a peculiar and useful view of the main idea. Thus the word presumption refers distinctly to an act of the mind, or a state of the mind, while in the word probability we feel disposed rather to think of the external arrangements on the knowledge of which the strength of our presumption ought to depend, than of the presumption itself. When, therefore, having observed an event, we want to know how strongly we are to suppose that the observed event was preceded by a given arrangement of circumstances, the term preof probability is very appropriate. The word facility applies particularly to the notion which we form when we see one event happen more often than
0 4
88    ESSAY ON PROBABILITIES.
another, namely, that it is easier to produce the firs' than the second. In our problems, however, the facility is not that arising from art, but from previous (it may be accidental) distribution of means. The word expectation will be applied throughout this work to that state of things for the production of which there is an even chance. If (p. 79), 6000 throws be made with a die, it is an even chance that the number of aces lies between 981 and 1019: the odds are against any smaller amount of departure on both sides of the probable mean, and against any greater amount ; this is then our expectation of the number of aces.
When one of two possible events happens oftener than the other, it being understood that one, and only one, can happen each time, we are led to suppose that the excess of one event is the consequence of some arrangewhich would, had we known it, have made us count that event more probable than the other. If A or B must happen, and if in a thousand trials the As outnumber the Bs very much, we feel perfectly certhat such must have been the case. The theory of probabilities confirms this impression, as will appear by the solution of the following
PROBLEM. In a + b trials, the number of As was a, and that of Bs was b. If a exceed b considerably `, required the presumption that there was at the outset a greater probability of drawing A than of drawing B, in any one single trial ?
RULE. Divide the difference of a and b by the square root of twice their sum, and let the result be t. Find (page 72) the H' corresponding to t : multiply the result by the sum of a and b, and divide by the product of 8, t, and the square root of the product of a and b. The result subtracted from unity gives the answer required. Suppose, for instance, that out of 50 trials A occurs 32 times, and B 18 times. Then,
* In order that the result may be very correct, a must exceed b so much that the excess of a above b, multiplied by itself, may considerably exceed the sum of a and b.
USE OF TABLES.    89
50x2=100, 41100=10    82—18 =1.40
10    =
H'= 15891 15891 x 50=7.9455
ü32 x 18=24, 8 x 24 x 1.40=268.8
7.9 divided by 268.8 is sh nearly: 1—45 is bg Hence it is about 261 to 8 that A was more probable than B.
ADDITIONAL RULE. When a and b are nearly equal, find t, as in the last rule ; find H (not H') correspondto t, add 1, and divide by 2 : the result is the prorequired.
The additional rule belongs to the more important case of the two, namely, that in which A has not hapso much oftener than B as to justify an immeconclusion that it was the more probable event of the two. Suppose, for instance, that A has occurred 10,100 times out of 20,000 trials, and B 9,900 times: then t = 200 divided by 200, or 1 ; to which H is 843, and this increased by 1, and the result divided by 2, gives 922. It is, therefore, about 112 to 1 that A was the more probable.
The preceding solution can be applied to various species of observations ; of which we shall see more hereafter. The following may be considered as closely connected with it. If we make two different sets of trials, in circumstances which we suppose to be the same, it will generally happen that the A s will not bear the same proportion to the Bs in both sets. If, for instance, we find 1000 As arrive in 2000 trials, the odds are very much against the arrival of exactly 5000 A s in a new set of 10,000 trials, though the expectation is that something near that number of As will arrive. Suppose that the first and second sets of trials give—1st, 50As,30Bs; 2nd, 112 As, 61 Bs.
In the second set the As bear a larger proportion to the whole than in the first : and our present question is what presumption thence arises that there is some difof circumstances between the two sets, which
9d    ESSAY ON PROBABILITIES.
gives A a greater facility in the second than in the first, or a greater probability of being drawn at any one trial ? Or, if in a first set of a + b trials, A happen a times and B happen b times ; and if in a second set of a' + b' trials, A happen a' times, and B happen b' times ; and if a' be a larger proportion of a' + b' than a is of a -}- b ; required the presumption that there was a greater chance of drawing A at a single trial in the second set than in the first ?
RULE. Divide the cube of the sum of a and b by twice their product : do the same with a' and b' : multhe two results together, and add them together : divide the product by the sum, and extract the square root of the quotient.
Divide a' by a' + b', and a by a + b, and subtract the less result from the greater. Multiply the difference by the square root previously found, and let the product be t. Then the H corresponding to t, increased by I, and divided by 2, is the presumption required.
In the example a = 50, b = 30, a' = 112, b' = 61.
80 x 80 x 80 = 512000    2 x 50 x 30 = 3000, 512000 = 170'7 3000
173 x 173 x 173 = 5177717 2 x 112 x 61 = 13664 5177717 = 3759 13664 1707 x 378.9 = 64678.23, 170'7+ 3789 = 549'6


646,8'23 =117.7, x/117.7 = 10'85 549'6 i-g ='6474—'C 50 = C224


'0224 x 10'85 = 24304 = t, H = 2689
(H + 1) = 635, the probability required; and it is therefore about 16 to 9 in favour of the excess of A s at the second set of trials not being accidental fluctuabut arising from some new circumstance or difarrangement of the old ones.
If, in 1000 trials, A should happen .520 times, and B 480 times, there is strong presumption that in any funumber of trials the whole number will be divided among As and Bs nearly in the proportion of 520 to
USE OF TABLES.    91
480. But this is not the same set of circumstances as that of the problem in page 77. We are there sup-posed to know exactly in what proportion As and Bs are contained in an urn ; and with this positive knowledge we can ascertain the probability of drawing any given number of A s in a given number of trials. In the present instance we do not know the contents of the urn, but only the result of a certain number of drawings, from which we can draw presumptions, as in page 53. about the whole contents. The determination of chances relative to a new set of trials depends upon two risks in the latter case, and upon one only in the former. The latter problem is therefore more complicated in its printhough not so in its results.
Let us suppose two different persons, John and Thomas, thus situated with respect to the contents of an urn: John knows that there are as many As as Bs; Thomas has observed a hundred successive drawings, of which (so let it have happened) fifty have given A, and as many have given B. That which John knows is rendered not improbable to Thomas by the result of the trials, while the same result would have been thought not unlikely beforehand by John. But there is this difference between their degrees of knowledge, that John has the certainty of a fact (the equality of As and Bs), of which Thomas can only say that the fact, or some-thing near it, is extremely probable. No one could argue with John against any particular venture in such a lottery upon the ground of the possibility of the As much exceeding the Bs ; while with Thomas it might be urged as possible, though not probable, that the former might exceed the latter a hundred-fold. Again, suppose John and Thomas, having equal fortunes, are disposed to venture as far as produce woulil warrant, upon the results of a hundred (to Thomas a second hundred) trials. It is obvious to common sense that Thomas must not venture so much as John ; for he runs a larger risk, seeing that he assumes as an average result what possibly may have been a rare occurrence.
92    ESSAY ON PROBABILITIES.
The following is the rule pointed out by the theory of probabilities:— The expectation of fluctuation should be greater to a person who proposes to try q new in-stances, upon the assumption that p preceding instances have fairly represented the long run, than it should be to another person, who knows in what proportions the As and Bs really exist ; and greater in the proportion of the square root of p augmented by q to the square root of p. Thus, if in the preceding case, John and Thomas propose to embark in a matter which depends on 300 more trials, the proportion of the square root of 100 + 300 to that of 100, being that of 2 to 1, it follows that, whatever reason John may have to guard against the possibility of 300 drawings giving x more than 150 As, Thomas has as much reason to guard against 2x more than the same number.
PROBLEM (to be compared with that in page 77.). When a + b trials have happened to give a As and b Bs, required the chance that in n times a + b new throws, the number of As shall not differ from na by more than 1.
RULE. Divide one more than twice 1 by a square root to be immediately mentioned, and the quotient being made t, the value of H in Table I. is the probability re
The square root in the The square root in the former problem was that of the present problem, is that of product of 8, n, a, and b the product of 8, n, n+ 1, a,
divided by a + b.    and b divided by a + b.
The additional rule in page 81. may also be applied verbatim, the square root now meaning the second square root above given ; and the inverse rule (p. 82) may be applied in exactly the same way.
EXAMPLE. In 600 drawings A occurred 100 times, and B 500 times ; what presumption thence arises that in 6000 more drawings A would occur somewhere be1000 — 50, and 1000 + 50, or 950 and 1050 inclusive? (See page 77 for the corresponding pro