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 INVERSE PROBABILITIES.    53
CHAPTER III.

ON INVERSE PROBABILITIES.
IN the preceding chapter, we have calculated the chances of an event, knowing the circumstances under which it is to happen or fail. We are now to place ourselves in an inverted position : we know the event, and ask what is the probability which results from the event in favour of any set of circumstances under which the same might have happened. This problem is freenunciated as follows : - An event has happened, such as might have arisen from different causes : what is the probability that any one specified cause did produce the event, to the exclusion of the other causes ? By a cause, is to be understood simply a state of things antecedent to the happening of an event, without the introduction of any notion of agency, physical or moral.
In order that we may secure a problem of sufficient simplicity, we must limit the number of possible antecedent states. Let us suppose that there is an urn, of which we know that it contains balls, three in number, and either white or black, all cases being equally pro: that is, before any drawing takes place, all we can say is, that we are going to draw out of one of the following, having no reason for supposing one in preto another  —
A    B C    DEF
I. ( . . . )    (o..) III. (oo .) IV. (000)
A drawing takes place, and a white ball is produced,
consequently I. is immediately excluded; for from it
the observed event could not have been produced. This
much is certain; but we are also tempted to say that
E 3
54    ESSAY ON PROBABILITIES.
II. is rendered unlikely, because, from such an antecedent state of things, a black ball would have been more likely than a white one. On the same prinIII. is more likely than II., and IV. the most likely of all. We have then to decide the relative proof II., III., and IV.
Before the drawing took place, the probability of each set of circumstances was y ; and, the lottery being given, the probability of any one ball in it was '-. Thus the chance of III. being the lottery, and the second white ball being drawn from it, was    x    or
1?, The same of other balls : so that, in fact, our primitive position was that of having to draw from 12 balls, 6 white and 6 black, all equally probable. But the observed event changed that position ; a white ball was drawn : was it a given ball (namely, the white ball in II.), or was it one of two given balls (those in III.), or was it one of three (those in IV.) ? There are six cases in question, namely, A, B, C, D, E, F, and one of them happened,— we do not know which. We have used all the knowledge we have (namely, that a white ball was drawn,) in excluding the black balls.
Hence the chance that A was drawn, or that
II. was the lottery    -    -    -    is    s :
    That either B or C was drawn, or that III.    is    n :
was the lottery    -    -    -        
    That either D, B, or F was drawn, or that    is    6.
IV. was the lottery    -    -    -        
In the preceding instance, owing to the number of balls being the same in every lottery, the antecedent probaof each ball was the same. Previous to deducing a rule, I take an instance in which this is not the case.
PROBLEM. A white ball has been drawn, and from one or other of the two following urns :
(2 white, 5 black)    (3 white, I black).
What are the probabilities in favour of each urn ?
The case is not now that of a lottery of 5 white and 6 black balls; for the chance of our going to the first urn (which is a), and thence drawing a given white
ON INVERSE PROBABILITIES.    55
ball (chance ), is x -' or 1'; ; while our chance of going to the second urn (which is -I2), and thence drawing a given white ball (chance 4), is s x 4 or $. But since we do not alter the chance of producing a white ball from either urn, if we double, or treble, &c. the number of white balls, provided we at the same time double, or treble, &c. the number of black balls, let us put four times as many balls into the first, and seven times as many into the second, as there are already. Thus we have :
(8 white, 20 black)    (21 white, 7 black).
There are now 28 balls in each : every individual ball has the antecedent probability x zlf; ; and since our knowledge of the event (a white ball was drawn) excludes the black balls, the question is simply this : — Out of 29 possible, and equally probable cases, was the event which did happen one out of a certain 8, or one out of the remaining 21 ? The chances of these
are and =-; consequently it is 21 to 8 that the second lottery was that which was drawn, and not the first.
On looking at the resulting chance for the first urn, namely, A, or the (21 + 8)th part of 8, we see that 8 and 21 are in proportion to the two chances for a white ball being drawn, when we know that we are drawing from the first urn, or from the second. For these chances are * and -, which, reduced to a common denominator, are A and £S, which are in the proporof 8 to 21. The same reasoning may be applied to any other cases, and the result is as follows : —
Principle V. — When an event has happened, and the state of things under which it happened must have been one out of the set A, B, C, D, &c., take the different states for granted, one after the other, and ascertain the probability that, such state existing, the event which did happen would have happened. Divide the probability thus deduced from A by the sum of the probabilities deduced from all, and the result is the
E 4
56    ESSAY ON PROBABILITIES.
probability that A was the state which produced the event : and similarly for the rest. [Or, reduce the reof the first part of the rule to a common denominator, and use the numerators only in the second part of the rule.]
EXAMPLE I. There is a lottery which is one or other of the two following :
(3 white, 7 black)    (all white).
A ball is drawn, and restored ; this takes place five times, and the result is always a white ball. What are the chances for each lottery ?
Upon the supposition that the first lottery was that in question, the chance of the observed event is the product of y v, 7SO, i30 and Tab, or Tiff )air When the second is the lottery, the observed event is certain, and its probability is 1 or UM. Consequently, the probability for the second lottery is o ° Q °3, or the second has the odds 100000 to 243, or more than 411 to 1 in its favour.
EXAMPLE II. Two witnesses, on each of whom it is 3 to 1 that he speaks truth, unite in affirming that an event did happen, which of itself is equally likely to have happened or not to have happened. 'What is the probability that the event did happen ?
The fact observed is the agreement of the two witin asserting the event : the two possible antecedents (equally likely) are,—1. The event did happen. 2. It did not happen. If it did happen, the probability that both witnesses should state its happening is that of their both telling the truth, which is a x 4, or ,Q's If it did not happen, then the probability that both witnesses should assert its happening is that of their both speaking falsely, which is a x 4, or 116. Consethe probability that the event did happen is the (9 + 1)th part of 9, or 19, ; that is, it is 9 to 1 in favour of the event having happened.
EXAMPLE III. There are two urns, having certainly 3 and 2 white balls; and in one or other, but which
ON INVERSE PROBABILITIES.    57
is not known, is a black ball. A ball is drawn and replaced; and this process is repeated, but whether out of the same urn as before is not known. Both drawings give a white ball: what is the probability of the several cases from which this result might have happened?
Since the black ball is as likely to be in one as in the other, the antecedent state of things is (so far as a single drawing is concerned,) the same as if there were four urns, as follows :
I. (3 white)    II. (3 white, 1 black)    III. (2 white)
IV. (2 white, 1 black).
There are 16 possible cases, PP [2, 4,] numbered in the first columns following, described in the second, and having the probability which each would give to the observed event (both drawings white) registered in the third, together with the numerator, when all the fractions are reduced to a common denominator 144.
    
1    I.    I.    1, 144    5    II.    I.    4, 108
    2    I. II.    3, 108    6    II.    II.    A, 81
1                    
    31    I.    III.    1, 144    7    II.    III.    3, 108
    4 ! I. IV.    4,    96    8    H.    IV.    6
                    -    ,    72
    9    III. I.    1, 144        13    IV. I.    4, 96
    10    III. II.    4, 108        14    IV. II.    TT' 72
    11    III. III.    1, 144        15    IV. III.    4, 96
    12    III. IV.    14,    96        16    IV. IV.    64
That is to say, if (case 8), II. and IV. were the urns of the first and second drawings, the chance of the observed event is ? or r'';. But, it must be rememthat we do not suppose the black ball may have been removed from one urn into the other before the second drawing takes place. Most of the preceding cases are, therefore, to be rejected ; in fact, I. can combine with nothing but I. or IV., and II. with nothing but II. or III. Reject, therefore, cases 2, 3, 5, 8, 9, 12, 14, 15, and the sum of the numerators in the rest is 841. Hence the probability (for instance,)
58    ESSAY ON PROBABILITIES.
I. That the black ball is with the three white ones. 2. That the first drawing is from the lottery which has the black ball, and the second from the other, is (case 7)
To find the total probability that the black ball is with the three white ones, we must add the probaof all the cases (as to drawings) which can take place under this arrangement, namely F }7, in al, AV, giving -t,4. Consequently, from the observed event, it is slightly more probable that the black ball is with the three white ones, than with the two.
The principle which we have illustrated, though a mathematical consequence of those which precede, is nevertheless received in common life upon its own evidence. `When an event happens, we immediately look to that cause or antecedent which such event most often follows. When it rains, we suspect the barometer must have fallen ; because, when the barometer falls, it usually rains.
Our next step is to inquire, what is the probability which an event gives to its several possible anteceupon the supposition that they are not all equally likely beforehand ; as in the following instance.
PROBLEM. A white ball is drawn, and from one or other of the following urns :
(3 white, 4 black)    (2 white, 7 black);
but before the drawing was made, it was three to one that the drawer should go to the first urn, and not to the second. What is the chance that it was the first urn from which the drawing was made ?
We may immediately reduce the preceding to the case where all the antecedent circumstances are equally probable, by introducing urns enough of the first kind to make it 3 to 1 that the drawing is made from one or other of them. Let us suppose the urns to be as follows :
(3 white, 4 black)    (3 white, 4 black)    (3 white, 4 black)
(2 white, 7 black):
these urns being equally probable, the hypothesis of
ON INVERSE PROBABILITIES.    59
the problem exists. If we number the urns AI, A0, A3, B, the chances which they severally give to the observed event are „ 7, 4, and -, the numerators of which, reduced to a common denominator, are 27, 27, 27, and 14. Consequently, the probability that A , was chosen, is; and the same for A: and A3. There-fore, the chance that one or other of the three, At, A0, and A1, was chosen, is ; which is the probability of the ball having been drawn from the urn (3 white, 4 black,) in the first statement of the problem.
The rule to which the preceding reasoning conducts us is as follows : 'When the different states under which an event may have happened are not equally likely to have existed, then having found the probability which each state would give to the observed event, multiply each by the probability of the state itself before using the rule in page 55. The following is another example.
An event has happened, the possible preceding states of which are represented by A, B, and C. The chances of the existence of these different states (independently of all knowledge of the observed event) are, say, -, -, and : the probabilities that the observed event would have happened are 3, and 1='1, if A, B, or C were certainly existing. Form the three products
Probability that the event would
Probability of A x { happen if A were known to exist,
there are
x 7T' 3 x T, and ,l x T3;
the numerators of which (the denominators being comare 20, 12, and 4- Then the probability that A was the state under which the event happened, is 20 divided by 20 + 12 + 4, or R, ; those of B and C are
3' and 346
Let us now suppose, that having only a first event by which to judge of the preceding state of things, we ask what is the probability of a second event yet to come. For instance, an urn contains two balls, but whether white or black is not known; the first draw-
60    ESSAY ON PROBABILITIES.
ing gives a white ball, and the ball is replaced. What is the chance that a second drawing shall give a black ball ?
The preceding states under which the first event may have happened, are —
    (2 white)    (I white, 1 black);
and 1 and 4. are the chances of a white ball, if one or other state were absolutely known to exist. Hence, by the last principle, - and -'r are the chances which the observed event gives to the two states ; that is, it is two to one that both balls were white. Now, the black ball can only appear at the second drawing, upon the supposition of the second state existing ; and this supposition being made, the chance of a black ball at the second drawing is 1. Hence, page 43., the second event depending upon two contingencies, of which the chances are -1 and 1, its chance is i, or it is five to one against the second drawing being black. But let us now ask what is the chance of a white ball at the second drawing ? Either of the preceding states admit of such an event, and, in fact, the event proposed — a white ball at the second drawing — means
One or other of these (2 white) and white drawn.
two combinations 1 (1 white, 1 black) and white drawn.
In the first combination, the first contingency (the chance of which is ) ensures the second : so that ?- x 1 is the chance of a white ball being drawn, and of (2 white) being the lottery from which it is drawn. In the second combination, the chances of the two contingencies are and whence } is the chance of a white ball being drawn, and being drawn from (1 white, 1 black). But the event proposed happens if either of these cases occur ; therefore, + , or , is the chance of a white ball at the second drawing, as might have been inferred from the probability already obtained for a black ball. By such reasoning as the preceding, the following principle is established :
Principle VI. Having given an observed event A,
ON INVERSE PROBABILITIES. 61 to find the probability which it affords to the suppothat a coming event shall be B, find the proba
bility which A gives to every possible preceding state ; multiply each probability thus obtained by the chance which B would have from that state, and add the results together.
PROBLEM. There is a lottery of 10 balls, each one white or black, but which is not known : drawings are made, after each of which the ball is replaced. The first five drawings are white ; what chance is there that the next two drawings shall be white ?
Let S (20) denote the sum of all numbers up to 20 ; S (202) the sum of the squares of all numbers up to 202 or 400; and so on. The possible preceding states are —
(1W,9B)    (2W,8B.)....(10W,0B);
and the probabilities of IV five times running from each, are
Tb ~'o TT'To. 1115, T1't'T5.r2 T ,&e.
up to - 1 R.- 1 -i . ° -1'O . °, or 1, the event being certain, if the last state existed. The numerators of these products (the common denominator being 105,) are 15, 25, 105 : whence, page 55., the probaof the several states are —
15    25    105
S105'    S105'    5105
By the same reasoning, the probabilities of the proposed events (two more white balls,) are —
12    22    102
102    102     102 ;
the different preceding states being successively sup-posed to exist; whence the actual chance which the observed event gives to the proposed is —
1    2    25    22    105
x +—X—+    }    xl
S 105 102 S 105 102    S 105 102 ;
S 107 which is 102 s 105.
62    ESSAY ON PROBABILITIES.
By precisely the same reasoning, if there had been 1000 balls in the lottery, and if 157 had been drawn white, the probability that 27 more drawings would have given white balls, would have been
S 1000184 (157+27=184)    100027 S 1000157
The difficulty of calculating S 1000184 is insuperable : but a mathematical theorem which we shall proceed to explain, makes it very easy to find a near approximato the preceding result, and the nearer the greater the numbers in question.
Take the sums of the powers of the different numas follows :
    First powers 1+ 2+ 3+ 4+ 5+ 6+    
Squares    1+ 4+ 9+ 16+ 25+ 36+    }
Cubes    1+ 8+ 27+ 64+ 125+ 216+    
    Fourth powers 1 + 16 + 81 + 256 + 625 + 1296 +     
    Fifth powers 1 +32+243+1024+3125+7776+    
and examine the sum of any number of terms in any line, as compared with the term immediately below the last in the sum ; thus : —
1+2+3+4    1+16+81+256+625
16    3125
Form fractions with such sums as numerators, and their compared terms as denominators, and observe how much each fraction, so formed, differs from the fraction written in the last column, as follows : —
1+2    3    1    1    1+2+3    6    1    1
    4    4    2+4    9        9    26
    S 4    10    1    1    s5    15    1    and so on ,
    16    16    2+8    25    25    2}10
whence    S (any number) _ 1    1
square of that number    `2 +twice that number
Hence it follows, that when the number is large, the preceding fraction is very nearly one half, or 1 + 2 + S +     up to a large number, is very nearly one half the square of that number.
ON INVERSE PROBABILITIES.    63
    Again :— 1+4_5I+    7    1+4+9_1+    5
8    8    3    24    27    3    27
    S42    30=1    13    S 52    1    8    
    64 or 43    64    3    + 96    125 or 5';    3 + 75    
    S 102    385    1    155    > and so or,.    
        + —        
    103    1000    3    3000        
In this way it appears that the sum of all the squares of numbers is nearly one third of the cube of the last number, and that the greater the number of squares taken, the greater the proximity in question. This proposition is general, namely, that the sum of the nth powers of numbers is nearly the (n + 1)th part of the (n + 1)th power of the last of the numbers : thus, the sum of all the 13th powers 113 + 213 + ---up to 100013, is very nearly the 14th part of 100014. This proposition, never absolutely true, may be made as near the truth as we please, by taking the number of terms sufficiently great ; and the error made by the substitution, is nearly such a fraction of the whole as has one more than the index of the power for its nuand twice the number stopped at for its denominator. Thus, if the tenth powers of all numwere summed up to 10,00010, the substitute for this sum given by the theorem, namely, _11 of 10,0001 1, would be too small by about
10+
]      or    11     of the whole.
2 x 10,000 20,000
PROBLEM. A lottery contains 10,000 balls, each of which may be white or black. A ball is drawn and then replaced, and 100 such drawings give nothing but white balls : what is the chance that the five next drawings shall all be white ?
This chance, by what precedes, is


But S 10,000100+5 = 1 x 10,000106 very nearly. 106
S 10,000100+5
10,0005 S 10,000100
61    ESSAY ON PROBABILITIES.

S 10,000100 =— x 10,000101    
101
Consequently, the chance is To 6     of 10,000106 or 101 nearly.
    T6T of 10,000106    106
If the number of balls had been a million, instead of 10,000, the preceding odds, namely, 101 to 5, would still more nearly have represented the chance that after 100 drawings, all white, the next five should be white also. if the number of balls had been absounlimited, the preceding odds will correctly ex-press that same chance. But a lottery with an unlimited number of balls, each of which may be either white or black, is a lottery which may be anything whatever. For instance: [S IV, 7 B] is equivalent to an unlimited lottery, in which for every 7 black balls there are 3 white ones. Again, an unlimited lottery in which any number of balls may be black, and the rest white, is one in which the chance of drawing a white ball may be any whatever, and is absolutely unknown. The drawing of a white ball from such a lottery may be likened to the occurrence of an event, about the preceding chances of which we are in total ignorance. The pre-ceding process furnishes us with the following theorem. When an event which may, for any thing we can see to the contrary beforehand, happen in either of two different ways, happens one way m times in succession, it is nt + 1 to n that it shall happen It times more in the same way, if it happen n times more at all. Thus, suppose a person on the bank of a river, not knowing in what country he is, and not having the smallest reason to know whether the vessels which come up the river carry flags or not : the first ten ships which come up all carry flags (m = 10) ; then it is 10 + 1 to 3, or 11 to 3 that the next three ships shall carry them, and 10 + 1 to 1, or 11 to 1 that the next ship shall carry a flag. And it is always to + 1 to 1, that an event which has occurred in one out of two possible ways 11t times in succession, shall happen the same way on the next occasion.
    
ON INVERSE PROBABILITIES.    65
The preceding affords some view of the way in which chances are obtained, in cases where the antecedent probability of the events stated may be any whatever. The following are conclusions upon the same subject, obtained by a more complicated reasoning of the same kind.
If an event, each repetition of which may be either A or B, have happened m+n times, and if A have occurred m times, and B n times ; then it is m + 1 to n +1 that the next event shall produce A, and not B. And in the same case the chance that out of p - q events to come, p shall produce A, and q shall pro-duce B, is (see pages 15 and 16, for explanation of [ ]).
[p+q] x[m+1,m+p] x [n+1,n+q] [p] x [q] x [m+n+2,m+n+p+q+1]
and     [m+l,m+p]      and    [n+1,n+q]
[m+n+2,m+n+p+1]' [m+n+2,m+n+q+l] are the chances that in p new events, all shall give A and that in q new events all shall give B.
EXAMPLE. In a lottery containing an unlimited number of balls, in which the proportion of black and white is absolutely unknown, six drawings give four white and two black ; what are the chances that four drawings more shall give all white, or one only black, or two only black, &c.
Let us first take the case of three white and one black : here m=4,n=2,p=3,q=1.
[p+q] = 1.2.3.4, [m+1, m+p] = 5.6.7, [n+1,n+q] = 3[p] = 1.2.3. [q] = 1,
[m+n+2, m+n+p+q+1] = 8. 9. 10.11.
Chance required is     1.2.3.4. x 5.6.7. x 4. _ 7
1.2.3. x 1. x 8.9.10.11.    22
That two shall be white and two black (m = 4, n = 2 p = 2, q = 2), the chance is
1 2.3.4. x 5.6. x 3.4.    3
1.2.x1.2.x8.9.10.11.'or 11
That one shall be white and three black (p = 1; = 3), the chance is
F
66    ESSAY ON PROBABILITIES.
1.2.3.4 x 5 x 3.4.5 or 5
1 x 1.3.3 x 8.9.10.11    33
That all shall be white, or all black (p = 4, q = 4, second and third formulae, the chances are
    5.6.7.8    7 and     3.4.5.6
or
or_
    8.9.10.11    33    8.9.10.11    22
and the verification of the whole is
7    3    5    7    1
22 + 11 + 33 + 33 + 22 1'
which must be, since one or other of the cases considered must happen.
When it is known beforehand that either A or B must happen, and out of m + n times A has happened m times, and B n times, then (page 65.) it is m+ 1 to n+ 1 that A will happen the next time. But suppose we have no reason, except what we gather from the observed event, to know that A or B must happen ; that is, suppose C or 1), or E, &c. might have happened: then the next event may be either A or B, or a new species, of which it can be found that the respective probabilities are proportional to m + 1, n+1, and 1; so that though the odds remain m+l to n -}- 1 for A rather than B, yet it is now m + 1 to n + 2 for A against either B or the new event. Thus, suppose a game at which one party or the other must win, and suppose that out of 20 games A has won 13 and B 7 : and this is all we know of the game or of the players. Then, it is 13+1 to 7+1, or 14 to 8, or 7 to 4, that A shall win the 21st game. But suppose that it is possible to have a drawn game; then there is some chance that the 21st may be a drawn game, though but a small one, as might be inferred from such a thing never happening in 20 trials. The 21st game may be either A's or B's, or drawn: of which the chances are as 13+1, 7+1, and 1 ; or as 14, 8, and 1. Consequently, though in the preceding case it was 14 out of 22 in favour of A's
ON INVERSE PROBABILITIES.    67
winning, it is now 14 out of 23, and 1 chance out of 23 remains for the next game being drawn.
When a number of different events have happened, A, B, C, &c., write down each number increased by 1, and the results will express the several relative probaon the supposition that no events can happen except those which have happened. But if new events may happen, write down 1 for the relative probability of such an occurrence at the next trial. Thus, if out of a box, and in 100 drawings, there have appeared 49 white balls, 37 red, and 14 black ; then if it be known that nothing but white, red, or black can appear, consider the chances of these to be as .50, 38, and 15 ; that is, -,5012s is the chance of drawing a white ball at the 101st trial. But if another sort of ball may appear, then the chances of the four cases being as 50, 38, 13, and 1, it follows that o a is the chance of a white ball at the 101st trial.
In judging of future events by those which have passed, we must be extremely cautious always to pre-serve the same method of considering the event pro-posed. If, for instance, in 1 00 trials, A has appeared 49 times, B 37 times, and C 14 times, we know that there is one chance out of 104 that the 101st drawing shall give neither A, B, nor C, but something else. What the new character may possibly be is left un; it may he another letter, or it may be a numa picture, or a blank. Are we to understand that all these are equally probable ? Common sense tells us the contrary; experience makes us feel it much more likely that the letter D should appear at the 101st trial, than any stated number or picture. But we have now changed the question, and, dropping the distinction between A, B, and C, have considered them merely as letters. Having drawn a letter 100 times running, we are to infer (page 64.) that it is 101 to 1 in favour of our drawing a letter at the 101st trial; or that i ;; ,i is the chance of this. But it is already 103 to 1 that the next drawing shall be, not merely a letter, but one
F 2
68    ESSAY ON PROBABILITIES.
of the letters A, B, and C : that is, to all appearance, we have this strange result ; — the chances of drawing one of the three, A, B or C, are greater than those of drawing one of the set, A or B, or C or D, &c. &c. up to Z. This paradox will afford me a good opportunity of again inculcating the maxim, that the probability of an event is the presumption drawn from certain obvious principles, as to what the state of our minds ought to be with regard to belief in the happening of that event, as influenced by our knowledge of previous events. Consequently, if John know that A, B, and C have been drawn 49, 37, and 14 times, and nothing more, he has reasonable ground, with his knowledge, for assenting to the proposition that the 101st trial shall give either A, B, or C, as to a proposition which has of probability, or 103 to 1 in its favour. But if Thomas only know that 100 drawings have all given letters, then he, with his know-ledge, has no ground of inference with respect to A, B, and C, in particular, but may reasonably assent to the proposition, " the 101st drawing will also give a letter,"
as having a probability of or 101 to 1 in its favour. But the paradox in question requires that John should make believe he knows no more than Thomas, and then be surprised that a discordance should arise from his using that knowledge in recona result, which he has suppressed in the method of attaining it.
It very rarely happens that we meet with a case in which we can so distinctly specify the antecedent cirwhich influence our assent or dissent, as to enable us to apply direct calculation to the determination of the rational probability of an event to come. And in most of the practicable cases, the largeness of the numbers employed would check our progress, if it were not for the approximative methods of the higher mathematics, and the tables at the end of this work. I now proceed to the method of using those tables.