DE PROBABILITATE CAUSARUM AB EFFECTIBUS ORIUNDA

Transcription

1 DE PROBABILITATE CAUSARUM AB EFFECTIBUS ORIUNDA A MATHEMATICAL INQUIRY Jean Trembley Commentationes Societatis Regiae Scientiarum Gottingensis Vol. XIII, 795/ Excellent men of Geometry have treated this material, and esecially the Celebrated Lalace in the Memoirs of the Academy at Paris. Since however in Problems to be solved of this kind the higher and difficult analysis will have been alied, I have considered to undertake the rice of the trouble to aroach the same questions by an elementary method and by roer use of the theory of series. By which reason this other art of the calculation of robabilities is reduced to the theory of combinations, and as I have reduced the first art in the case of the discussion transmitted by the Royal Society. I will attemt to undertake these rincial questions here briefly, having been directed chiefly by a method to be illuminated.. Let there be a vessel containing white and black tickets in infinite number, so that the roortion of white and black tickets is unknown. The extracted tickets are white q black, the robability of drawing out m white and n black tickets in the future is sought. The extracted tickets are suosed cast again into the vessel. Let n be an integer number of tickets which we shall assume infinite afterwards, it is evident the numbers of white and black tickets to be able to be varied, just as the following table reorts: white tickets black tickets white tickets black tickets n n n 2 n n 3 n 4 n 4 n 4 4 Translated by Richard J. Pulskam, Deartment of Mathematics & Comuter Science, Xavier University, Cincinnati, OH. December 20, 2009 See Disquisitio Elementaris circa Calculum Probabilium, Commentationes Societatis Regiae Scientiarum Gottingensis, Vol. XII, 793/4,

2 Under this hyothesis the resective Probabilities of white and q black tickets being drawn roceed, n q n q n 2 2 q n q, 2 n 2 q n q, n q n q,, 3 n 3 q n q,... n 3 3 q n q, and according to the general rule of the calculation of Probabilities, the corresonding term itself divided by the sum of the terms will be the Probability of that given hyothesis. This osed, the Probability of extracting m white tickets, and n black will be according to the individual hyotheses resectively, m n qn n qmn n 3 m 3 qn n qmn, 2m n 2 qn n qmn, 3m n 3 qn n qmn,..., n 2 m 2 qn n qmn, n m qn n qmn. Each and every term must be divided by the common divisor, of course the sum mentioned above. I have omitted moreover here, for the sake of brevity the coefficients of the terms, I shall restore these successively. Therefore the sought Probability will be m n qn 2 m n 2 qn 3 m n 3 qn n 3 m 3 qn n 2 m 2 qn n m n qmn n q 2 n 2 q 3 n 3 q n 3 3 q n 2 2 q n n q However let there be for the sake of brevity, 2 3 n 2 n n q 2 n 2 q 3 n 3 q n 3 3 q n 2 2 q n q n q q n q qq n q 2.2 q... q 2 ± n 3 q 3 qq n In the same manner it will roduce, there will be 2 q... q 2 n q q 2 ± q n q 3 q. m n qn 2 m n 2 qn 3 m n 3 qn n 3 m 3 qn n 2 m 2 qn n m qn n qn m q n n qn m q nq n n qn 2.2 q nq n ± n 2 qmn 2 q n n qmn ±.2 2 m2 qmn

3 But I ut n to infinity, they roduce by known formulas see Jac. Bernoulli ars conjectandi the following equations, m n m m, m2 n m2 m 2 etc. Therefore the first series being divided by n q roduces, n q qq qq q 2 qq ±.2.3 q q ± q q q qq q Next the other series being divided by n qmn roduces, n q n q nq n m m 2.2 m 3 q nq n q n ±.2 q m n q m n ± q m n Moreover let S be the sum of the first series, I suose successively q, 2, 3 etc. it will roduce S n 2,.2.n 3,.2.3.n q.n whence by analogy it roduces in general S q. But the sum of the other series, roduces substituting m in lace of, q m in lace of q.2... q nn m q m n. Now by dividing the other sum by the rior, the sought Probability will be q n 2... q β q m... q m n q q 2... q n 2... q β., m m 2... q m n β being the coefficient which now we will determine. From the theory of combinations it follows the numerator to be multilied by the number of combinations q of the quantities taken according to the exonent q, or by q q,.2... q and by the number of combinations m n of the quantities taken according to the exonent n, or by m nm n m n 3

4 However the denominator must be multilied in such manner by therefore there will be β q q,.2... q m nm n m n Which agrees with the statement of the Celebrated Lalace Mem. des Sçavans Étranges T The series is able to be summed in another way q 2 There exists If therefore the series qq.2 q 3 qq.2 q q q etc. q 0. qq.2 4 ± q. etc. 0 is subtracted from the roosed series, the value of it will not be changed, and it will take this form, q 2 q q q ± q subtracting now from the new series 2 q 2 our series will take this form, qq 2 3 q 2 4 subtracting again from the new series 3 q 2 3 q q 2.2 q 2q 3.2 our series will take this form, qq q q 3 5 q 2q See Mémoire sur la Probabilité des causes ar les évènements. etc. 0, 2 etc. 0, 3 q 3q etc. 6 etc. 4

5 In general if from the series v the series q v v v q v v q v q v.2 q v q v.2 is subtracted the first series will take this form, q v v v q v q vq v v 2.2 Therefore our series will take finally this form as above. qq 2. 2 q v 2 ± q v etc. 0 v 3 v 3. If in the vessel of the white tickets were suosed as many as α in total, the following Table will reresent the various numbers of white and black tickets: white tickets black tickets white tickets black tickets n α n α 2 n 2 3 n 3 4 n 4 Under this hyothesis the resective Probabilities of extracting white and q black tickets arise, n q n q, 2 n 2 q n q, 3 n 3 q n q α n α q n q. Therefore the sum of these terms will exist, using the same method of. n q 2 α qn q 2 α qq n q α 2.2 qq ± n 2 q 2 2 q 2 α q 2.2 q n q 2 q α q ± q 2 q α q Now I ut α to infinity, it will roduce n q α ± qq.2 α2 qn q 2 n 2 αq q q qq q 2 α3 n n αq q ± αq q..

6 Let α n q θ, it roduces α n q θ, omitting therefore the coefficient n q which will vanish in the final fraction by aid of the similar coefficient in the denominator, as we have seen. the series will roduce, q θ ± qq.2 q q θ 2 2 q q θ q q θ ± qq.2 q 2 q θ q qq.2 q q q θ q θ 3 3 q θ q ± q q 2 q q θ qq q θ q q q q θ q θ q ± q. Let there be made, for the sake of brevity, q θ v, while however let there be v q qv qq.2 v 2 etc. let there be made S from that series let be subtracted it will roduce by suosing S q v 2 3 from the same series let 2 q v 2 qv qq qv qq 2 3 S q q 2.2 be subtracted, it will roduce by suosing S 3.2 v 2.2 v 2 qq q v 3 etc. 4 qq q v 3 vq etc. vq qv S q q 2.2 S q 2 v 2 2 vq 2 v 4 v 2 4 q... q q... q q v 2 S, q 2q 3.2 v 3 2 v 2 5 etc. v 3 5 etc. etc. vq 2

7 from this series let 3 q 2 be subtracted, it will roduce by suosing S 4 Therefore it will roduce S vq qv S vq vq v 3 S q 3 q 2q 3.2 vq 2 3 v 5 v 2 3 q 2v 3 S q 3q 4.2 qv vq qq v2 2 2 S qv vq 2 qq v2 v q 2 3 etc. vq 2 3 v 2 6 etc. Therefore it will be in general as it is known from analogous series, S vq qv vq 2 qq v2 v q qq... v q 2... q [ v q q 2 3 q... q 2 v... 4 v v v q q q... q 2v S. qq... 3 v v v ] q v by suosing and q enormous numbers [ v q q 2 ] 3 v 2 q2 v v 3 q3 v v 4 etc. v v q qv v q v v q v q v q q v q. However there is v q θ q θ, q v q qθ, q 7 2

8 therefore there exists S The sought sum will be therefore vq qθ q q θ q qθ q θ q θ q qθ namely the first term of the series which the Celebrated Lalace reorts Mem. sur les Prob. Mem. de Paris If it is necessary to aroach nearer to the true sum, the following method shall be alied. Let in general the series be qα 2 qq α 2 q... q 2α qq... α q... the sum of which is sought through aroximation, by suosing and q enormous numbers. The following formulas will aear by division 2 3 ; ; ; ; ; ; etc.; and hence our series will take this form, 2 qα qq qα q 3 3q 2 2q q 4 6q 3 q α 4 7 q 5 0q 4 35q α 5 8 q 6 5q 5 85q α 6 etc. 9 qα. 7 3 qq qα q 3 3q 2 2qα q 4 6q 3 q 2 α q 5 0q 4 35q 3 α q 6 5q 5 85q 4 α etc. However there is 9 7αq 3 25α2 q α3 q α4 q α5 q α6 q 6 8 9,. α 3 6 etc.

9 2αq αq 2αq 5 αq ; 5 25α 2 q 3.65α3 q α4 q α5 q α6 q 5 8 etc. 9 α 2 q αq α2 q α3 q α4 q 4 0 etc. 25α2 2 q20α 3 q 2 αq 7 ; 2.65α 3 q 6.40α4 q α5 q α6 q 4 9 etc. α 3 q αq α2 q α3 q 3 2 etc. 30α 3 q 3 370α 4 q α 5 q 3 αq 9 ; 2α 3 q α4 q α5 q α6 q 4 6 α 3 q 2 5 αq 35α2 q α3 q 3 7 etc. 8 etc. 7 2α3 qα 4 q 2 2.0α 3 q.5α4 q α5 q α6 q 4 7 etc. 8 α 3 q αq 35.2α2 q α3 q 3 9 etc. 0 αq 5 ; 20α 3 q 2 25α 4 q 2 αq. Our series will have therefore the following form, 7 qα2 qα qα 2 2αq 6qα2 2α 3 q α 4 q 2 qα 4 25α2 2 q 20α 3 q 2 20α 3 2 q 25α 4 q 2 qα 6 30α3 3 q 370α 4 2 q 2 30α 5 q 3 αq 8 etc. etc. But in the following manner, I have obtained the sum of these series. In the first series I seek the difference of the coefficients which the following table exhibits: Differences of the fourth order are constants, so that the general term of the series is of known form 6n 2nn.2 0nn n nn n 2n This general term contains five kinds of terms multilied by, n, n 2, n 3, n 4. Therefore each and every term will deend on the five receding terms and by taking A, B, C, D, E 9

10 the coefficients to be determined I establish the following equation 2nn 0nn n 2 3nn n 2n 3 6n An n 2 0An n 2n 3 3An n 2n 3n 4 A 6An Bn 2n 3 0Bn 2n 3n 4 3Bn 2n 3n 4n 5 B 6Bn Cn 3n 4 0Cn 3n 4n 5 3Cn 3n 4n 5n 6 C 6Cn Dn 4n 5 0Dn 4n 5n 6 3Dn 4n 5n 6n 7 D 6Dn En 5n 6 0En 5n 6n 7 3En 5n 6n 7n 8 E 6En This equation remains, as it is well known if n is written in lace of n, in which case the order of each and every term in the series will be greater by unity. If n 2 is written in lace of n, the order will be greater by two units; and hence any one term in the same manner will deend on the revious five terms, and the scale of the relation will be, Aαq Bα2 q 2 2 Cα3 q 3 3 Dα4 q 4 4 Eα5 q 5 5. The same roof succeeds in all cases in whatever kind of series the dislayed differences become constants, so that if the differences of the v th order become constants, each and every term deends on the v receding terms, whence a recurrent series will aear of which the scale of the relation will contain v 2 terms. In the resent case, by aid of the general term I seek the terms 750, 55, 75. Now I set u the following five equations, 70 55A 750B 462C 266D 40E A 462B 266C 40D 65E A 266B 40C 65D 25E A 40B 65C 25D 7E A 65B 25C 7D E By eliminating the quantity E, the following four equations aear, A 35B 0C 24D A 359B 485C 0D A 3763B 359C 35D A 8350B 3038C 74D By eliminating the quantity D, the following three equations aear, A 07B 230C A 297B 678C A 4085B 938C 0

11 By eliminating the quantity C the following two equations aear A 3098B A 798B By eliminating the quantity B the following equation aears A, thence results A 5, B 0, C 0, D 5, E. The scale of the relation will be therefore 5αq 0α2 q 2 2 0α3 q 3 3 5α4 q 4 4 α5 q 5 5 αq 5. But the roosed series is able to take this form, However, there exists 5 7αq 6 25α 2 q α 3 q α 4 q α 5 q α 6 q 6 etc. αq 5 5 5αq 6 5α 2 q α 3 q α 4 q 4 9 etc. Let this series be multilied through by a 6αq. and it will roduce α 5 5aαq 6 5aα 2 q aα 3 q aα 4 q 4 9 b 5b 5b 35b etc. Comaring terms, there is a, 5a b 7, or b 2, and thus it haens sufficient to the other conditions. Therefore the sum will be 2αq αq as above. 5 Other series are able to be summed in the same manner. But it should be observed how we have managed more easily to have been able to sum some series, by the figurate number taken of order five, and multilied by the coefficient to be determined a, with the same numbers multilied again by the other coefficient to be determined b, and by the same with the revious added, but advanced by one order thus as the first with the second, the second with the third, and thus added in order. While each and every term of the second series is multilied by the ower of the quantity, by two units higher than in the revious case, I add the figurate number of order seven, and while the second series is able to take this form, α 2 q αq α 2 q α 3 q α 4 q 4 etc. I shall form a new series, namely α 7 7aαq 8 28aα 2 q aα 3 q aα 4 q 4 b 7b 28b 84b etc.

12 Comaring terms, there exists a 25, 7a b 95, or b 20, and thus it haens sufficient to the other conditions. Therefore the sum will be as above. The third series is able to take this form 25α 2 2 q 20α 3 q 2 αq 7 α 3 q αq α 2 q α 3 q 3 2 etc. But from the figurate numbers taken of order nine, I shall form a new series, α 9 9aαq 0 45aα 2 q 2 65aα 3 q aα 4 q 4 3 b 9b 45b 65b c 9c 45c etc. Here I have added the third coefficient c, because the first two were not sufficient. Now comaring terms, there exists a 30, 9ab 540 or b 370, 45a9bc 930 or c 30, and thus it haens sufficient to the other conditions. The sum is therefore as above. The fourth series is able to take this form, 30α 3 q 3 370α 4 q α 5 q 3 αq 9 α 3 q2 5 αq 6 35α 2 q α 3 q 3 8 etc. From the figurate numbers taken of order five I shall form a new series α 5 5aαq 6 5aα 2 q aα 3 q aα 4 q 4 0 b 5b 5b 35b etc. By comaring terms, there exists a 2, 5a b, or b, and thus it haens sufficient to the other conditions. Therefore the sum will be as above. The fifth series is able to take this form, 2α 3 q 4α 4 q 2 αq 5 α 3 q αq α 2 q α 3 q 3 0 etc. From the figurate numbers taken of the seventh order, I shall form a new series, α 7 7aαq 8 28aα 2 q aα 3 q aα 4 q 4 b 7b 28b 84b etc. 2

13 By comaring terms, there exists a 20, 7ab 65, or b 25, and thus it haens sufficient to the other conditions. Therefore the sum will be 20α 3 q 2 25α 4 q 2 αq 7 as above. However the order of the figurate numbers which must be used, will deend on the order of differences which become constants, the first order surasses the second by unity. Substituting now in lace of α its value αq qθ q θ ; v v qα 2 q q θ2 q θ2 q 2 θ 2 q θ q θ, it will roduce 2 2 q θ q q θ q 2 θ 2 q qθ2 q q 2 θ 2 q q2 θ 2 q 3 θ 2. The formula will become q θ q θ q qθ [q q2 θ 2 ] q 3 θ 2 etc. This formula is what the famous Lalace reorts 8. of the cited memoir, and 8. Memoir year The series of the receding q 2 v v 2 qq v etc v converges if v < v, that is, if v < 2. Therefore this series serves as long as the selected Probability is demanded from 0 to a fraction inferior to the fraction 2. If the selected Probability is demanded from 0 to a fraction greater to 2 the following method should be used. The Probability is taken from to this fraction greater than 2. To discover this Probability it will be sufficient in the formula of the receding to write in lace of q, q in lace of, and n α in lace of α, or to write v in lace of v and v in lace of v, whence the series will arise, [ 2 v q v v v etc.] q q 2 v q... q 3 v vq v qv, if and q are enormous numbers q q θ q q θ, qθ 3

14 if its value q θ is written for v. This formula will reresent the Probability taken from to q θ. Therefore the sum of the Probabilities taken from 0 to q θ and from θ to is q q θ q θ q q θ q θ q qθ Therefore let K be the Probability taken from 0 to, the selected Probability will roduce from q θ to q θ [ q ] q q θ q q θ q θ q q θ K qθ as the famous Lalace reorts in the Memoir year If q, and then α, by formula 4. it will haen, etc. By suosing, 2, 3... etc. the following Table of will arise, q Now it is well known by analogy for to be the sum By multilying this formula by 2 because in this case is v 2 2 it will be q But the entire robability is. Therefore the robability, the number of white tickets not exceeding the number of black tickets is 2 as it is clear a riori. 7. The sum of the Probability taken from 0 to the given quantity β and of the Probability taken from β to must be equal to the Probability taken from 0 to. Therefore by the calculation I begin near the receding and by reserving the same denominations, the following equation will aear v v q qα 2 qq α qq... α 9... q... v v q q α q q 2 α 2 q... q 3 α q... q.2... q... q 4

15 and hence α qα 2 qq α qq... α q... q q α q q 2 α 2... q... q 3 α q... q.2... q... q v v q.2... q α q... q α.2... α q q... q α. If α, the equation will be.2... q... q 2q q 2 qq... 3 q q q q q 2 q... q 3... q... q or because of.2... q... q.2... q... q.2... q... q 2 Therefore it is.2... q... q q q 2q [ q qq.2 q q q qq.2 q...q2.2...q q qq.2 q q q qq.2 q q q q q q2 ] q...q [ q qq.2 q q q qq2 ] q...q3...[ q 2] q 2 qq... 3 qq q.2... q... q 2q 2 q.... q 2 q... q q 5 q q 2 q... q 3 q...q q.

16 By suosing α, let v 2, v 2, now our formula will be multilied by v v q.2... or by 2. Dividing next by q q...q it will show the Probability the number of white tickets will not be exceeding the number of black tickets, and it will be 2 q... q 2 q... q q q 2... [ q 2] q... q 3... q q. 8. This last formula is able to serve in the cases where q is a small number, in which case the first formula frequently diverges. Of this kind the Celebrated Lalace handles an examle in the Memoir year There exists in Burgundy a little city named Viteaux, in which are born during one year 45 infants, 203 of male 22 of female kind. Let 22, q 203 be made, they roduce q 45, q 46. The series q 2... [ q 2] q 2q 3 q... q q Now there will be from the theorem of Stirling q... q.2... q q 3 2 2π 2 q q 2 e e is the number of which the hyerbolic logarithm is unity. Thus the calculation roceeds, 46 2 log log log log l l 2π l e Therefore the Probability of the number of girls not being in excess of the number of boys will be , and the Probability of the number of girls being in excess of the number of boys will be 0.674, just as the Celebrated Lalace reorts. 6

17 9. By the quantity α remaining indeterminate, the general equation is able to take the following form.2... q α q... q α.2... q α q q α q... q q q q α q q α q 2.2 α q q q α.2... q q... q q α q.2... q 2 q q α α q α q The method of summation of these various series by aroximation requires contemlations, which here it was set forth at greater length, and which deserve to be set forth aart. 0. Let there be some vessel containing an infinite number of white and black tickets, there are extracted from that vessel white q black tickets, the Probability is sought m white and 2a m black tickets will be exiting in a following oeration. From. the Probability of m white, n black tickets exiting from the vessel whence already exited there are white tickets, q black, is m m 2... m n q q 2... q n 2... q.2... n m m 2... q m n by substituting the quantity 2a m in lace of n, this formula is m m a q q 2... q 2a m 2... q a m m m 2... q 2a a.2... q 2a m.2... q.2... m.2... m a m.2... q q 2a.2... a.2... q 2a.2... q q.2... q 2a m.2... m a m.2... m Now in order to obtain the Probability of the number of white tickets to not be in excess of the number of black tickets, the sum of all these values should be taken from m to m a. However constant is the coefficient a q.2... q 2a q, therefore it suffices to consider the formula.2... q 2a m.2... m a m.2... m 2a m 2a m 2... q 2a mm m 2... m 7 m. 7

18 Let the following term be 7 m there will exist the equation 7 m 2a mm 2a m m 7 m. However there is 7 m 7 m 7 m, is the sign of a finite difference. Hence it roduces, m q 2a m m q 2a m 7 m 2a m m 2a m m 7 m, or 7 m In a similar way the equations will exist m q 2a m 2a q m q 7 m. mq 2a m 7 m 2a m q 7 m, m q 2a 2 m 7 m 2 2a m 2 q 7 m 2, etc. If it haens 7 m α 7 m, 7 m 2 α 7 m 2 etc. by α roving to be constant it will roduce or 7 m 7 m 2 7 m 3 etc. α 7 m 7 m 2 7 m 3 etc. α7 m Σ7 m mq 2a m 2a m q 7 m Σ is the sign of the summation of finite differences but by suosing m a, it makes Σ7 a aq a a q 7a. But it makes, if the formula of the aroximation of Stirling is alied, and it is suosed m n a, it makes I say, γq a qa 2 a a 2 q q q q q 2 q 2a q2a 3 2 γ aq a q 3 2 q a qa a a q q 3 q q 2a 2 q 2a q2a q q qa γ aq a q 3 2 q2a qa q2a a q q2a 3 q q 2a 2 q 2a q q γ a q a q 3 qa a 2 aq a aq a q a q 2a q q 2a 3 2 q q 2a q 2a γ a q a q 3 qa 2 aq a q 2a q q 2a 3 2 a a q q q 2a a q q 2a a 8

19 as the Celebrated Lalace refers in the Memoir year Here for the sake of brevity I have suosed γ.2...2a a. 2 Let the equation m q 2a m 7 m 2a q m q 7 m, be considered again and by successively attributing to the quantity m the values 0,, 2... a, the following equations exist: q 2a 7 0 2a q 7 0, q 2a 2a 2q 7, 7 2 4q 2a 3 2a 3 4q 7 aq a a a aq 7 a. Certainly q 2a a q 7 0 2q 2a a 2q 7 3q 2a a 2 3q 7 2 or, q 2a 7 0 2a q 7 0, a But there is 2a 2 3 2a 2 3q 7 2, 7 4 a a a aq 7 a a 2a q 7 0, 3q 2a 2 2a 2 3q 7 2, 2a 2 2a 2q 7 ; 2a 2 3 2a 2 3q 7 2 2a 2a q 7 2a 2 0, 7 2 2a 2q 7, 2a 3 4 2a 3 4q a 2 2a 2q 7 2a 2a q 7 0. Therefore Likewise Therefore 2a 2a q 7 22a q 0 2a 2q a 2q 2a.22a q2a q a2a 2.22a q2a q

20 In addition Therefore and hence a 2 3 2a 2 3q 7 2a2a a q2a q a2a 2.22a q2a q 7 32a q 2 0 2a 2 3q 7 2, 7 2 2a 2 3q 2aa a q2a q 2a q 2 7 0, 7 3 2a2a 2a a q2a q 2a q 2 7 0, 7 0 q 2a 2a q 7 2a 0, 7 2a q 7 2a2a 2 0, a q2a q 7 0, 2a2a 2a a q2a q 2a q a2a 2a 22a a q2a q 2a q 22a q a Therefore 2a2a... a a a a2a q2a q 2a q 2... a q a q 2a 2a q a 2a2a 2 2a q a q 2a2a 2a a q 2a q 2 2a2a 2a 22a a q 2a q 22a q 3 2a2a... a a a a2a q 2a q 2... a q This inverted series will take the following form, aa q aa a q a q 2 7 a aa a a a a 2 aa a 2a q a q 2a q 3 a a a 2a a 2a 3 etc. 20

21 if and q are suosed enormous numbers aa q 7 a aa a2 a q 2 a 2 a 2 a3 a q 3 a 3 a 3 etc. 7 a aa 7 aaq a aa aa q 7 aa a a q. aa. If I wish to rogress further, I would observe the series anew, and there will roceed aa q aa a q a q 2 aa a a a a 2 etc. a a q 2 a a 2 aa q aa aaq aa q2 aa 2 aa 2aa q a a 2 a 2 aa q aa 2aq aaq aa a 2 a 2 q 2 aa by neglecting quantities in which the sum of the set forth quantities a,, q is smaller than unity. In the same manner a 2a q 3 aa q 2aa 2q aaq a 2a 3 aa a 2 a 2 ; a 3a q 4 aa q 3aa 2q aaq a 3a 4 aa a 3 a 3, and so forth. Let, for the sake of brevity, A aa q aa 2aq aaq, B aa a 2 a 2, and by neglecting squares of the quantity B, the sought series will be A AA B AA BA 2B AA BA 2BA 3B etc. A AA A 3 A 4 A 5 etc. AB 3A 6A 2 0A 2 etc. A AB A 3 AB A A 2 etc. a a a2 a q a 2q aq a aq a aq a 2 a 2 etc.. The Celebrated Lalace reorts a similar series 20, and the higher terms of this series can be discovered, the laborious calculation will become only rolix. 2

22 2. The Celebrated Lalace in the Mémoires Etrangers T. 6 attacks the following Problem. Two gamesters A and B of whom skill is unknown, between themselves on the same condition that he who first will have won v games, obtains the sum. However the gamesters are forced to cease the game, while gamester A lacks f games, gamester B h games. With this ut it is sought in what way the sum is divided between the gamesters. If the skill of the gamester A was, of the gamester B, and the quantity had been known, the Problem was able to be reorted to the second Problem among the most general Problems which the Celebrated Lagrange has handled, Mem. Acad. Berlin 775 and which I have demonstrated by an elementary method in the memoir transmitted in the Royal Society. The formula reorted at the end of this Problem is this a a aa aa... b... a b a b a b. Here there must be a f h, b h. But I rove in, that if there are white tickets q black drawn from a vessel containing an infinite number of white and black tickets, the Probability of extracting in the future m white tickets, n black, will be q q 2... q n 2... q. m m 2... q m n Here there must be v f, q v h, thus the formula becomes v h v h 2... v h nv f v f v f h. v f m v f m v f h m n Now into the receding series, in lace of the first term a this formula is substituted by suosing m 0, n f h,whence the term will roduce v h v h 2... v f v f v f v f h v f v f v v h v h 2... v f 2v f f 22v f h v. In the lace of the second term a this same formula is substituted by suosing m, n f h 2, whence the term will roduce v h v h 2... v f 2v f v f v f h v f 2v f v which is the term of the first case multilied by v f vf. In the lace of the third term a 2 2 this same formula is substituted by suosing m 2, n f h 3, whence the term will roduce v h v h 2... v f 3v f v f v f h v f 3v f v 22

23 which is the term of the first case multilied by v fv f2 vf v f 2. In the lace of the fourth term a 3 3 this same formula is substituted by suosing m 3, n f h 4, whence the term will roduce v h v h 2... v f 4v f v f v f h v f 4v f v which is the term of the first case multilied by v fv f2v f3 vf v f 2vf 3. In general in lace of the term a b, which is b b a b the same formula is substituted by suosing m a b, n f h a b, whence the term will roduce v h v h 2... v f a b v f v f v f h v f a b v f a b v which is the term of the first case multilied by v f v f 2... v f a b v f v f 2... v f a b. But it is a f h, b h, and also this multilier will be v f v f 2... v v f v f 2... v by substituting values mentioned above the sought Probability will be v h v h 2... v f 2v f h 22v f h v f h v f v f f h f h 2 v f v f 2.2 v f v f 2 f h... h n f n f 2... n f f n f n 2... n as the Celebrated Lalace reorts. In the same manner all Problems are able to be solved which the Celebrated Lagrange has treated and which are contained in the receding dissertation, and in general in the same manner, all Problems of this kind are able to be solved, by reserving the coefficients and whatever terms, and in the lace of the terms themselves by substituting the terms which result from the formula demonstrated in. 3. The method which we are using is able to be reduced to the method of the Celebrated Lalace the other in an easy manner, it is sufficient to aly the integral calculus in the summation of series. Thus the series q 2 qq.2 3 qq q ± q 23

24 is able to take this form x x 2 q 2 qq.2 x 3 3 qq q x 4 4 etc. S here is S 0 where x 0, and after integrating the series there must become S. Therefore there is [ ] ds dx qq x qx x 2 qq q 2 x 3 etc. x x q, ds x x q dx, S x x q dx, by being integrated from x 0 to x, as the Celebrated Lalace reorts. In the same manner the series n q α is able to take this form Let it will roduce ds dx x n q α x [ qα n x qα n α2 qnq 2 qα n 2 x2 x 2 2 qq α 2 x.2 n 2 x2 qq q 2 α3 n.2 3 etc. qq.2 qq.2 α 2 x 3 α 2 x etc..2 3 etc. S, q... q 2 α 3 ].2.3 n 3 x3 etc. x α q n x, therefore ds x α n x q dx, S x α n x q dx. Now let there be it will roduce S α n q θ, q x [ x] q θ dx, by being integrated from x 0 to x, and dividing the result by n q, the series becomes q q θ x [ x] q θ dx. But it is, by being integrated from x 0 to x x vx q dx x x q dx v 24

25 by being integrated from x to x v, because in both cases it roduces the series qv 2 qq v 2 etc..2 Therefore the series will be q θ x x q dx, by being integrated from x 0 to x q θ. In like manner the calculation will be in other cases. 4. According to the aforesaid it follows the Probability originating from the causes of effects, whatever method to require corresonds to two arts. In the first art the formulas are being assigned which reresent this Probability, in the other art they indicate aroximations which deliver ossible uses of these formulas where enormous numbers are resent. We see in the revious integral formulas used by the men the most celebrated Lalace, Condorcet etc. to be nothing other than abbreviated formulas which coincide with the common method of combinations. In truth x x q dx by being integrated from x 0 to x is nothing other than the sum of the terms which are roduced from the unfolding of the formula x x q, by substituting successively for x of the values n, 2 n, 3 n,... n n and suosing the quantity n infinite. This conclusion is deduced from the very beginnings themselves of the integrals of the calculus. Demonstrated once by this analogy, the integral formulas are able to be used for the sake of brevity, and chiefly these abridgments are useful while the quantities arrive more comosite, for the sake of an examle, x x q x x q, and the quantities n, 2 n, 3 n,... x are substituted for x and the quantities n, 2 n, 3 n,... n n for x. In this case it will be sufficient to integrate the formula x x q x x q dx from x 0 to x x, and from x 0 to x. I shall add only one examle lest this inquiry become longer. The Celebrated Lalace suoses to have been born in Paris between certain times boys q girls, but in London within a different interval of time boys q girls, and he seeks the Probability, the cause which roduces boys to be more efficient than in London. From the aforesaid it follows this Probability to be reresented by the formula x x q x x q dxdx x x q x x q dxdx with the numerator being integrated from x 0 to x x, and from x 0 to x, but the denominator from x 0 to x, and from x 0 to x. In this first art no difficulty is involved, and by itself its general rule of the calculation of robabilities is being alied for its own sake. However the other art which is concerned with the aroximation manner, is subject to enormous difficulty, as it surassed all things, the Celebrated Lalace has discovered the more sublime Analysis. Now we see in the receding by what manner the common calculus of series were able to lead to analogous 25

26 conclusions. In this case the integral of the denominator ermits no difficulty. And indeed x x q x q x q q x 2 q... q 2 x 3 x q x x q dx x q x 2 2 q q x ± x q q a by being integrated from x 0 to x q 2 q q.2 3 q... q ± q. Therefore x x q x x q dxdx q 2 q q.2 3 q... q q qq q... q q... q... q... q. 4 ± q 4 q However the integral of the higher numerator is exlored as seen from the following. I return to formula a which is made, suosing x x, x x q dx x q x 2 2 q q.2 Therefore x x q x x q dxdx x q q x 2 2 q q.2 qq x x 2.2 x 3 x 3 3 ± x q q. 3 ± x q q q... q 2 x 3 ± x q

27 by integrating from x 0 to x 2 q qq ± q 2 q 3 3 q qq ± q 3 q q q qq ± q 4 q...q q qq ± q 5 ± q q 2 q qq q 3.2 q 4 ± q q 2 by summing this series by the method of q 2... q 2 q 2 q q q q q q q q 2... q q q 2... q 2 q 2 2 q q 3... q 3...q q q 5... q 5 q q.2 2 q 3 3 q 3 q q 3... q q ± q q 3... q q 2 The series 2 q 2 2 q 3 etc. rovides, by substituting successively for q the values 0,, 2, 3 etc. q 0 q 2 q.2 q 2 2 q q 2 3 q3 qq2 3 q3 q q q3 3qq2 3 4 q3 q4 qq2q3 4 q3 q4 q

28 Now it is clear by analogy, and hence there exists for q q.2... q... q q q 2... q q q q q q q 3 q q q... q q q 3... q 5 q 2q q q... q q 3 q 2... q q 3... q q q q q... q q 2 q... q q 3... q q q q... q q q... q q 3... q q q... q q q q 3... q q 2 q... q q q 3... q q 2 q q q q q q 2 q q q q q q 2 q q q q q q q q q q q q 2 q q 2... q q q q q q q q 2 q q... q 2 q q... q 3 q q 2... q q q q q q 3 q... q 3 q... q q 2... q 3 q q q q... by emloying the first for aroximation q... q q q 3... q q 2 q q q q q q q q q 2 q q 2 q q 2 q q 2 etc. q... q q q 3... q q q q aq q... q q q q q 3... q q 2 q q. 28 qq

29 By multilying by the comlete integral will aear,.2... q 2... q 2... q q q q 2... q q 2 q q 2π 3 2 q q qq 3 2. q q q q qq 2 If it be necessary to rogress further, let for the sake of brevity, α q, β q, γ q q, δ q q 2, whence it will roduce the series α β δ ββ δδ γ αα γγ αα α 2 ββ β 2 δδ δ 2 γγ γ 2 αα α 2α 3 ββ β 2β 3δδ δ 2δ 3 γγ γ 2γ 3αα α 2α 3α 4 etc. α β δ γ α α α 2 α 3 β γ β δ α δ γ γ 2 γ 3 α 2 3 α 3 7 α 4 β γ ββ 3β 2 δ α δδ 3δ 2 α 3 6 α 4 25 β γ β3 6β 2 β α 5 γ 3 6 γ 4 25 γ 5 δ 3 6δ 2 δ γ 2 3 γ 3 7 γ 4 α 4 0 α 5 65 α 6 etc. 29

30 α βδ α 2 γ β2 δ 2 α 3 γ 2 β3 δ 3 α 4 γ 3 β4 δ 4 α 5 γ 4 etc. β δ γ βδ3β 3δ α 3γ βδ α 3 γ 2 α 4 γ 3 β2 δ 2 6β 6δ 3α 6γ α 5 γ 4 β3 δ 3 0β 0δ 6α 0γ α 6 γ 5 etc. βδαγ γ2 βδβ δα 3δ α 4 γ 3 α 4 γ 3 βδ2β2 9βδ 2δ 2 α 4 γ 3 β2 δ 2 α 2 3αγ 7γ 2 α 5 γ 4 3β2 δ 2 β δ3α 6γ α 5 γ 4 β2 δ 2 β 2 36βδ δ 2 α 5 γ 4 β3 δ 3 7α 2 8αγ 25γ 2 α 6 γ 5 6β3 δ 3 β δ6α 0γ α 6 γ 5 β3 δ 3 35β 2 00βδ 35δ 2 α 6 γ 5 β4 δ 4 25α 2 60αγ 65γ 2 α 7 γ 6 0β4 δ 4 β δ0α 5γ α 7 γ 6 β4 δ 4 85β 2 225βδ 85δ 2 α 7 γ 6 The sum of the first series is α βδ γ etc. γ αγ βδ q q q q q q. The other series has taken this form α 3 3βδ α 4 γ 6β2 δ 2 α 5 γ 2 0β3 δ 3 β δ α 6 γ 3 etc. βδ γ 2 γ βδ γ 3 βδ γ 3 [β δγ βδ γ2 ] α βδ γ q q q 2 2q q q q q 3. βδβ δγ βδ γ2 3 αγ βδ 3 30

31 The third series has taken this form βδ γ 2 γ 3 α 4 βδα2 3αγ 7γ 2 α 5 γ β3 δ 3 25α 2 60αγ 65γ 2 α 7 γ 3 βδβ δ γ 3 α 3γ 3βδ3α 6γ α 4 α 5 γ 0β3 δ 3 0α 5γ α 7 γ 3 etc. βδ γ 3 β2 δ 2 7α 2 8αγ 25γ 2 α 6 γ 2 etc. 6β2 δ 2 6α 0γ α 6 γ 2 2β 2 9βδ 2δ 2 α 4 βδβ2 36βδ δ 2 α 5 γ β2 δ 2 35β 2 00βδ 35δ 2 α 6 γ 2 β3 δ 3 85β 2 225βδ 85δ 2 α 7 γ 2 etc. This series is able to be resolved in the following manner, β 2 δ 2 α 2 γ 4 α 5 7βδ α 6 γ 25β2 δ 2 α 7 γ 2 65β3 δ 3 α 8 γ 3 40β4 δ 4 α 9 γ 4 266β5 δ 5 α 0 γ 5 etc. However of the numbers the fourth differences are constants, therefore by emloying the method exosed in the above, the denominator of the generating fraction of the series will be δ α βδ γ α 5βδ 5 α 6 γ 5β2 δ 2 α 7 γ 35β3 δ 3 2 α 8 γ 3 Bβδ αγ α 5 5βδ α 6 γ 5β2 δ 2 α 7 γ 35β3 δ 3 2 α 8 γ 3 B 5B 5B I have multilied this series by Bβδ αγ, hence B being constant term and by comaring terms it roduces B 5 7, therefore B 2, and thus it satisfies the other conditions. Therefore there exists a art of the series α 2 β 2 δ 2 α 2βδ αγ γ 4 5 αβ2 δ 2 αγ 2βδ α βδ αγ βδ 5. γ The sequence of members is 3αβ 2 δ 2 γ 3 α 5 6βδ α 6 γ 20β2 δ 2 α 7 γ 2 50β3 δ 3 α 8 γ 3 05β4 δ 4 α 9 γ 4 96β5 δ 5 α 0 γ 5 etc. 3

32 However of the numbers the fourth differences are constants, likewise therefore I enjoy by the method α 5 5βδ α 6 γ 5β2 δ 2 α 7 γ 35β3 δ 3 2 α 8 γ etc. 3 Bβδ αγ α 5 5βδ α 6 γ 5β2 δ 2 α 7 γ 35β3 δ 3 2 α 8 γ 3 B 5B 5B etc. it roduces B, therefore this member is 3αβ 2 δ 2 α βδ αγ γ 3 5 3β2 δ 2 γαγ βδ α βδ αγ βδ 5. γ The sequence of members is αβδ γ α 5 7βδ α 6 γ 25β2 δ 2 α 7 γ 2 65β3 δ 3 α 8 γ 3 etc. The sequence of members is α2 βδβ δ γ 3 However of the numbers βδγ3 αγ 2βδ αγ βδ 5. α 5 9βδ α 6 γ 36β2 δ 2 α 7 γ 2 00β3 δ 3 α 8 γ 3 225β4 δ 4 α 9 γ 4 etc the fourth differences are constants, likewise I enjoy by the method α 5 5βδ α 6 γ 5β2 δ 2 α 7 γ 35β3 δ 3 2 α 8 γ etc. 3 Bβδ α 5 αγ Cβ2 δ 2 α 2 γ 2 5βδ α 6 γ 5β2 δ 2 α 7 γ 35β3 δ 3 2 α 8 γ 3 B 5B 5B C 5C it roduces B 4, C, therefore this member becomes α 2 βδβ δ 4βδ αγ β2 δ 2 α 2 γ 2 γ 3 5 βδβ δα2 γ 2 4αβγδ β 2 δ 2 α βδ αγ βδ 5 γ etc. 32

33 The sequence of members is 3αβδβ δ γ 2 α 5 6βδ α 6 γ 20β2 δ 2 α 7 γ 2 50β3 δ 3 α 8 γ 3 3αβδβ δ α βδ αγ γ 2 α βδ γ etc. 5 3βγ2 δβ δαγ βδ αγ βδ 5 The sequence of members is 2αβ 3 δ γ α 5 2 βδ 35 α 6 γ 2 β2 δ 2 α 7 γ β3 δ 3 α 8 γ 3 etc. However of the numbers 2,, 35, 85 etc. the fourth differences are constants, likewise I enjoy by the method, α 5 5βδ α 6 γ Bβδ αγ α 5 5βδ α 6 γ 5β2 δ 2 α 7 γ 2 35β3 δ 3 α 8 γ 3 etc. 5β2 δ 2 α 7 γ 2 35β3 δ 3 α 8 γ 3 etc. B 5B 5B it roduces B 2, therefore it haens this member 2αβ 3 δ γ 3 α βδ 2αγ α βδ γ 5 β3 γδ2αγ βδ αγ βδ 5. The sequence of members is α 2 βδ γ 2 α 5 9βδ α 6 γ 36β2 δ 2 α 7 γ 2 00β3 δ 3 α 8 γ 3 etc. α 2 βδ γ 2 α 4βδ αγ β2 δ 2 α 2 γ 2 α βδ γ 5 βγδα2 γ 2 4αβγδ β 2 δ 2 αγ βδ 5 The last member is 2αβδ 3 γ 3 α 5 2 βδ 35 α 6 γ 2 β2 δ 2 α 7 γ 2 etc. α 2 βδ 3 α βδ αγ γ 2 5 βγδ3 2αγ βδ α βδ αγ βδ 5 γ 33

34 The sum of these values, [by suosing for the sake of brevity is T αβδ γ 3 αγ 2βδ 3βγδ βγ 2 γ 2 δαγ βδ β δ γα 2 γ 2 4αβγδ β 2 δ 2 β 2 γ γδ 2 αγ βδ] βδt αγ βδ. The integral of the numerator will be therefore 5 γ βδβγ γδ βδ γ2 βδt αγ βδ γαγ βδ 2 γαγ βδ 4 etc. Likewise it will be ermitted to roceed by the method of another. This series, which is similar to the series which the Celebrated Lalace reorts, requires many considerations, which here, lest this dissertation be longer, I am forced to omit; it is sufficient here to indicate the method. Still other worthy examles to note may be given which, lease God, I will develo in a following dissertation. 34