RECHERCHES sur ue questio relative au calcul des probabilités M. JEAN TREMBLEY Mémoires de l Académie royale des scieces et belles-lettres, Berli 1794/5 ages 69 108. Oe fids at the ed of the secod volume of the aalytic Works of Mr. Euler, prited i etersburg i 1795, a Memoir etitled: Solutio quarudam quaestioum difficiorum i calculo probabilitatum. 1 These questios revolve o the lottery of Geoa, which has bee sice imitated at Maheim, at aris, at Berli & i other cities of Europe, i which out of 90 tickets marked with umbers 1,, 3,... 90, oe draws 5 of them at some fied periods. Mr. Euler supposig ay umber tickets, of which oe draws at each time a umber p, & which oe puts back agai ito the ur which cotaied them, seeks the probability that i a give umber of coups oe will have brought forth all umbers, the probability that oe will have brought forth 1 umbers at least, the probability that oe will have brought forth umbers at least &c. Has igitur quaestioes, says Mr. Euler, utpote difficillimas hic e pricipiis calculi probabilitatum iam pridem usus receptis resolvere costitui. Qui i huiusmodi ivestigatioibus elaborarut, facile perspiciet resolutioem harum quaestioum calculos maime itricatos postulare, quos autem mihi beeficio certorum caracterum superare licuit. 3 The aalysis of which Mr. Euler makes use i this Memoir is very igeious & worthy of this great geometer, but as it is a little idirect & as it would ot be easy to apply it to the geeral problem of which this oe is oly a particular case, I have udertake to treat the thig directly accordig to the theory of combiatios, & to give to the questio all the etet of which it is susceptible. 1. Let a regular prism of which the umber of faces are, amog which there Traslated by Richard J. ulskamp, Departmet of Mathematics & Computer Sciece, Xavier Uiversity, Ciciati, OH. December 1, 009 Read to the Academy 18 Jue 1795. 1 Opuscula Aalytica Vol. II, 1785, p. 331-346. Traslator s ote: The word used is uméro. This sigifies a umber used i the sese of a label. 3 Traslator s ote: We have here i the first setece a etract from the first paragraph of Euler s paper E600. Therefore here I have decided to resolve these questios, iasmuch as they are the most difficult oes, from the priciples of the calculus of probability, which have log sice bee accepted by practice. It is followed by most, but ot all of the first setece of the secod paragraph. Everyoe who has carefully labored i ivestigatios of this sort, easily will observe the solutio of these questios to require the most itricate calculatios, but which by beefit of certai characters, it is permitted by me to have the upper had. 1
are a marked 1, a marked, a marked 3,... a marked, oe demads the probability that at the ed of a umber coups oe will have brought forth all the kids of faces. The probability that 1 will ot come at all is =, therefore the probability that 1 will come with or without is 1. The probability that, will ot come at all is = the probability that either 1 or will come is =. The first of these probabilities gives the umber of combiatios by a where all the umbers are foud ecept, the secod gives the umber of combiatios where either 1 or is foud. If oe subtracts this umber of combiatios from the precedig, oe will have the umber of combiatios where 1 is foud without ; thus a epresses the probability that 1 will come without. If oe subtracts this probability from that which 1 will come with or without, there will remai for the probability that oe will brig forth 1 &, a a a a 1. This formula epresses thus the probability that 1 & will come with or without 3, ow epresses the probability that 3 will ot come at all, that is to say it gives the combiatios where all the umbers will come ecept 3; a epresses the probability that either 1 or 3 will come, that is to say it gives the combiatios where all the umbers will come ecept 1 & 3; subtractig this umber from the precedig oe has the combiatios where 1 is foud & ot 3, therefore a epresses the probability that 1 will come without 3, et a epresses the probability that either or 3 will come, that is to say it gives the combiatios where either or 3 is foud; a a epresses the probability that either 1, or, or 3 will come, that is to say it gives the combiatios where either 1 or or 3 is foud. Subtractig this umber from the precedig oe has the combiatios where 1 is foud without or 3, therefore a epresses the probability that 1 will come without or 3; ow a a subtractig from the umber of combiatios where 1 is foud & ot 3, the oe where 1 is foud without or 3, there will remai the umber of combiatios where 1 & is foud without 3, therefore a a a a a a a a epresses the probability that 1 & will come without 3; if oe subtracts this probability from the oe that 1 & will come with or without 3, there will remai the
probability that 1 will come with & 3 = a a a a a 1 a a a a a a a. This formula epresses the probability that 1, & 3 will come with or without 4; ow iv epresses the probability that 4 will ot come at all, that is to say it gives the combiatios where all the umbers ca be, ecept 4; a iv epresses the probability that either 1 or 4 will come, that is to say it gives the combiatios where all the letters ca be, ecept 1 & 4; subtractig this umber from the precedig, oe has the combiatios where 1 will be foud, but ot 4, therefore iv a iv epresses the probability that 1 will come without 4, et a iv epresses the probability that either or 4 will come, that is to say it gives the combiatios where it will be watig & 4; a a iv epresses the probability that either 1 or or 4 will come, that is to say it gives the combiatios where 1, & 4 will be watig. Subtractig this umber from the precedig, oe will have the combiatios where 1 will be foud without or 4, therefore epresses the a iv a a iv probability that 1 will come without or 4; subtractig these combiatios from those where is foud 1 without 4 with or without, oe will have those where 1 is foud with without 4, therefore a iv a a iv a a iv a a a iv epresses the probability that 1 will come with without 4; moreover a iv epresses the probability that either 3 or 4 will come, that is to say it gives the combiatios where 3 & 4 will be watig; a a iv epresses the probability that either 1 or 3 or 4 will come; subtractig this umber from the precedig, oe has the combiatios where 1 will be foud without 3 or 4, therefore a iv a a iv epresses the probability that 1 will come without 3 or 4; fially a a iv epresses the probability that either or 3 or 4 will come, that is to say it gives the combiatios where, 3 & 4 will be watig; a a a iv epresses the probability that either 1 or or 3 or 4 will come. Subtractig this umber from the precedig, oe will have the combiatios where 1 comes without, 3 or 4; therefore epresses the probability a a iv a a a iv that 1 will come without, 3 or 4. Subtractig this probability from that which 1 will come without 3 or 4, oe will have the probability that 1 will come with without 3 or 3
4 = a a iv a a a iv. a a a iv a a a a iv Subtractig this probability from that which 1 will come with without 4, oe will have the probability that 1 will come with & 3 without 4 = a iv a a iv a a iv a a iv a a a iv a a a iv a a a iv a a a a iv Subtractig this probability from that which 1 will come with & 3 with or without 4, oe will have the probability that 1,, 3 & 4 will come together = a a a a a a iv a a a a iv 1 a a a a a a iv a a a iv a a a iv a iv a a a a a iv a a iv a a iv Mr. de Moivre is arrived to the same results i his Doctrie of chaces. The aalogy is ow evidet, & oe sees that if oe calls A the sum of the terms which oe obtais by subtractig from the combiatios of a, a, &c. a oe by oe, A the sum of the terms which oe obtais by subtractig from the sum of the combiatios of the same quatities take two by two, A the sum of the terms which oe obtais by subtractig from the sum of the combiatios of the same quatities take three by three, & i geeral A the sum of the combiatios of the 4
same quatities take by, oe will have for the probability that 1,, 3,... will have eited at the ed of coups 1 A A A A iv &c. ± A.. If oe makes ow a = a = a = a = 1, the precedig formulas will give us the solutio of the problem that Mr. Euler treats i first place & that Mr. de la lace has treated i T. 6. of the Mémoires des Savas étrages preseted to the Academy of Scieces of aris. 4 Oe has i this case here, by virtue of the first priciples of the doctrie of combiatios, the probability that all the umbers of the lottery will have eited at the ed of coups = 1 1 1 3 1 &c.±. 1. 1..3 It is ecessary ow to substitute ito this formula for 1, &c. their values. Now 1 epresses the probability that i coups a umber for eample o will ot eit; as oe draws p tickets at each coup, the umber of all possible cases results from the combiatio of thigs take p by p, of which the result must be raised to. 1 p1 1..3...p This umber is therefore by the doctrie of combiatios = & the umber of cases where 1 is ot foud, results from the combiatio of 1 thigs take p by p. This umber will be by the same rule = 1 p. 1..3...p Therefore by the geeral rule of probabilities, the probability that i coups the umber 1 will ot eit will be = 1 p 1 p1 = p. Likewise epresses the probability that i coups, two umbers, for eample 1 &, will ot eit; ow the umber of possible cases beig always the same, the umber of cases where 1 & will ot be foud, results from the combiatio of thigs take p by p, this umber is therefore = 3 p1, 1..3...p therefore the sought probability will be = 3 p 1 p p 1 =. 1 p 1 1 Net 3 epresses the probability that the three umbers 1,, 3 will ot eit; ow the umber of cases where 1,, & 3 are ot foud, results from the combiatio, of 3 thigs take p by p amely therefore the sought probability will be = 3 4 p = 1 p 1 34 p 1..3...p Oe will fid likewise 4 4 5 p 3 = = 1 p 1 p p 1 p. 1 p p 3. 1 3 4 Mémoire sur les suites récurro-récurretes et sur leurs usages das la théorie des hasards, Mém. Acad. R. Sci. aris Savats étragers 6, 1774, pages 353-371. 5
The law is ow rather clear. Substitutig therefore these values, our formula will become 1 p 1 3 p 1 1 1 p 1 1 p 1 1 3 4 p &c..3 1 p 1 This is the formula that Mr. de la lace fids. Accordig to that which we just saw, oe ca set it uder the followig form which is much simpler. p 1 p p 1 1 1 1 p p 1 p 1..3 1 1 3 p p 1 p p 3 &c. 1..3.4 1 3 For that which regards the arithmetic calculatio of these formulas, oe will cosider that A p 1 =, A 1 p 1 A =, 1 A p A =, 3 A iv 3 p 3 = A 4 3 & i geeral Therefore A λ λ = A λ1 λ 1 p λ. λ l A = l l p, l A = l A l 1 l p1 1, l A = l A l 3 l p &c. l A λ = l A λ1 l l λ λ1 pλ λ Let as i the lottery of Berli = 90, p = 5, & make = 100, we will have A = 0.9640, A = 0.04066, A = 0.00344, A iv = 0.0000, therefore 1 A A A A iv &c. = 0.7410. Make = 00, we will have A = 0.000976, 1 A = 0.99904. There are therefore early odds of three agaist oe that all the umbers will have eited at the ed of 100. 6
drawigs, & early odds of oe thousad agaist oe that all the umbers will have eited at the ed of 00 drawigs, as Mr. Euler fids it. If oe would wish to kow what is the umber of coups where oe ca wager eve that all the umbers will have eited, oe would fid this umber betwee 85 ad 86, so that there is advatage to wager that all the umbers will have eited at the ed of 86 drawigs, & disadvatage to wager that all the umbers will have eited at the ed of 85 drawigs. The eact umber is earer 85 tha 86. 3. Supposig the same thigs as i 1. oe demads the probability that at the ed of coups oe will have brought forth at least 1 kids of faces. Reasoig always i the same maer, I say: a is the probability that i drawigs either 1 or will come, therefore 1 a epresses the probability that of the two umbers there will eit at least oe, a a a epresses the probability that will come without 1 or 3, a a a epresses the probability that 1 will come without or 3. Subtractig these two probabilities from the first, oe will see that 1 a a a a a a a a a epresses the probability that of the three umbers there will eit at least two of them, sice 1 & must ecessarily eit together or separately, & sice 1 must come with & 3 together or separately, & with 1 & 3 together or separately. Oe fids likewise that a iv a a iv epresses the probability that 3 will come without or 4; that a iv a a iv epresses the probability that will come without 1 or 4; that a iv a a iv epresses the probability that 1 will come without 3 or 4. Oe fids agai that a a iv a a a iv epresses the probability that 3 will come without 1 or or 4; that a a iv a a a iv epresses the probability that will come without 1, 3 or 4; that a a iv a a a iv epresses the probability that 1 will come without, 3 or 4. Therefore by subtractig these probabilities two by two, oe will have the followig epressios: a a iv a a a iv a a a iv a a a a iv = 7
the probability that will come with 3 without 1 or 4; a a iv a a a iv a a a iv a a a a iv = the probability that 3 will come with 1 without or 4; a a iv a a a iv a a a iv a a a a iv = the probability that 1 will come with without 3 or 4. Ideed, if the probability that will come without 1 or 4 with or without 3, oe takes off the probability that 3 will come without 1, or 4, there remais the probability that will come with 3 without 1 or 4, & it is the first of our three ew formulas. The two others are obtaied precisely i the same maer. If oe subtracts ow these three formulas from the first which epresses the probability that at least two of the three umbers 1,, 3 will eit, it is easy to see that which remais. These three umbers furish three combiatios two by two 1, ; 1, 3;, 3. Now 1, ca ot come without 3 or 4, by virtue of subtractig it from the third formula; 1, 3 ca ot come without or 4, by subtractig it from the secod;, 3 ca ot come without 1 or 4, by subtractig it from the first. Therefore the formula a a a a a a a a a iv 1 3 a a a a a iv a a iv a a a iv a a a a a iv a a iv a a iv epresses the probability, that of the umbers 1,, 3, 4 there will eit at least three of them. These four umbers take three by three form the followig four combiatios, 1,, 3; 1,, 4;, 3, 4. It is ecessary to seek the probability that each of these combiatios will come without the two other umbers, for eample that 1,, 3 will come without 4 & 5, & subtractig these four probabilities from the probability that we just 8
foud, oe will have the probability that of the umbers 1,, 3, 4, 5 there will eit at least four of them, sice there must eit at least three of the first four, & sice ay of the combiatios three by three of these first four ca ot come without brigig forth at least oe of the two other umbers. Now, i order to fid the probability that 1,, 3 will come without 4 & 5, it is ecessary to seek the probability that 1, will come without 4 or 5 with or without 3, & by subtractig the probability that 1, will come without 3, 4 or 5. Now the probability that 1, will come without 4 & 5 is by that which precedes, a iv a v a a iv a v a a iv a v a a a iv a v. I order to have the probability that 1, will come without 3, 4 or 5, it is ecessary to seek the probability that 1 will come without 3, 4 or 5 with or without, & by subtractig the probability that 1 will come without, 3, 4 or 5. Now the probability that 1 will come without 3, 4 or 5 is by that which precedes a iv a v, & the probability that 1 will come without, 3, 4 or 5 is 5 is a a iv a v a a a iv a v. Therefore the probability that 1, will come without 3, 4 or a a iv a v a a a iv a v a a a iv a v a a a a iv a v & the probability that 1,, 3 will come without 4 or 5, is = a a iv a v a iv a v a a iv a v a a a iv a v a a a a iv a v a a iv a v a a a iv a v a a iv a v a a a iv a v Oe will have likewise the probability that 1,, 4 will come without 3 or 5 = a a v a a a v a a a a v a a a a iv a v a a a v a a a iv a v a a iv a v a a a iv a v 9
Oe will have likewise the probability that 1, 3, 4 will come without or 5 = a a v a a a v a a a a v a a a a iv a v a a a v a a a iv a v a a iv a v a a a iv a v Oe will have likewise the probability that, 3, 4 will come without 1 or 5 = a a v a a a v a a a a v a a a a iv a v a a a v a a a iv a v a a iv a v a a a iv a v The sum of these four probabilities will be therefore a a v a a a 4 a a a a v a a a a iv a v 3 4 a a v a a a v a a a iv a v 3 a a v a a iv a v a a a iv a v 3 3 a iv a v a a iv a v a a a iv a v 3 a a iv a v 10
& by subtractig it from the precedig formula, oe will have the formula, 1 a a a 3 a a a iv 4 a a a iv a v a a a iv 3 a a a v a iv a a v 3 a a iv a v a v a a iv 3 a a iv a v a a a v a iv a iv a v a v a a iv a iv a a v a v a iv a v iv a v a iv a v which epresses the probability that of the umbers 1,, 3, 4, 5 there will eit at least four of them. The aalogy is ow evidet, & oe sees that by coservig the deomiatios of 1 oe will have for the probability that i coups oe will have brought forth at least 1 kids of faces = 1 A A 3A iv 4A v 5A vi 6A vii &c. 4. If a = b = c = d =&c.= 1 oe will have by the theory of combiatios, 5 the probability that out of umbers oe will have brought forth at least 1 of them i coups= 1 1 3 1 1. p 1..3 3 1 3 4 4 1 4 5 &c. 1..3.4 1... 5, & by settig for 3, 4 p &c. the values foud. oe will have this probability = 1 p p 1 1 1. 1 1 p p 1 p 1..3 1 3 1 3 1..3.4 p p 3 1 3 &c. 5 Traslator s ote: a = a, b = a, c = a, d = a iv,... Trembley repeatedly uses this alterate otatio. 11
This is that which Mr. Euler fids. 5. Supposig always the same thigs as i 1. oe demads the probability that at the ed of coups oe has brought forth at least faces. I reaso always i the same maer, & I say: a a is the probability that there will come either 1 or or 3 i drawigs, therefore 1 a a epresses the probability that of the umbers 1,, 3 there will eit at least oe of them. I order to have the probability that of the umbers 1,, 3, 4 there will eit at least two of them, it is ecessary to subtract from the precedig probability the probability that oe of the umbers 1,, 3 comes without oe of the three others. Now the probability that 1 will come without, 3 or 4 is by that which precedes a a iv 1, 3 or 4 is without 1,, or 4 is a a iv a a a v a a iv ; the probability that will come without a a a iv ; the probability that 3 will come a a a iv. Therefore a a a a a a a iv 1 3 a a a iv a a a iv a a a iv epresses the probability that of the umbers 1,, 3, 4 there will eit at least of them. I order to have the probability that of the umbers 1,, 3, 4, 5 there will eit at least three of them, it is ecessary to subtract from the precedig probability the probability that each of the si combiatios two by two of the umbers 1,, 3, 4 will come without the three other umbers. Now by that which precedes, the probability that 1, will come without 3, 4 or 5 is a a iv a v a a a iv a v a a a iv a v a a a a iv a v. The probability that 1, 3 will come without, 4 or 5 will be a a iv a v a a a iv a v a a a iv a v a a a a iv a v. 1
The probability that 1, 4 will come without, 3 or 5 will be a a a v a a a a v a a a iv a v a a a a iv a v. The probability that, 3 will come without 1, 4 or 5 will be a a iv a v a a a iv a v a a a iv a v a a a a iv a v. The probability that, 4 will come without 1, 3 or 5 will be a a a v a a a a iv a a a iv a v a a a a iv a v. The probability that 3, 4 will come without 1, or 5 will be a a a v a a a a v a a a iv a v a a a a iv a v. Oe will have therefore by subtractig these si probabilities from the precedig, 6 1 a a 3 a a a iv a a a iv a v 3 a a iv a a v a a iv a a v a iv a v a a iv 3 3 a a v a iv a v a iv a v a a a v a a iv a v a a iv a v 13
this is the probability that out of the umbers 1,, 3, 4, 5 there will eit at least three of them. I order to have the probability that of the umbers 1,, 3, 4, 5, 6 there will eit at least four of them, it is ecessary to subtract from the precedig probability the probability that each of the te combiatios three-by-three of the umbers of the first five umbers will come without the three other umbers. Now these probabilities result easily from the precedig calculatios; here is a eample of it for the probability that 1,, 3 will come without 4, 5, 6. The probability that 1, will come without 4, 5, 6 is foud by that which precedes = a iv a v a vi a a iv a v a vi a a iv a v a vi a a a iv a v a vi. The probability that 1, will come without 3, 4, 5, 6 is = a a iv a v a vi a a a iv a v a vi a a a iv a v a vi a a a a iv a v a vi. Therefore subtractig the secod formula from the first, the probability that 1,, 3 will come without 4, 5, 6 is = a iv a v a vi a a iv a v a vi I. a a a iv a v a vi a a a a iv a v a vi a a iv a v a vi a a a iv a v a vi a a iv a v a vi a a a iv a v a vi Oe will have likewise the probability that 1,, 4 will come without 3, 5 & 6 by echagig i the precedig c ito d & d ito c, a a v a vi a a a v a vi II. a a a a v a vi a a a a iv a v a vi a a a v a vi a a a iv a v a vi a a iv a v a vi a a a iv a v a vi 14
Oe will have the probability that 1,, 5 will come without 3, 4, & 6 by echagig i the precedig d ito e & e ito d, III. a a iv a vi a a a iv a vi a a a a iv a vi a a a a iv a v a vi a a a iv a vi a a a iv a v a vi a a iv a v a vi a a a iv a v a vi Oe will have the probability that 1, 3, 4 will come without, 5, 6 by echagig i the secod b ito c & c ito b, IV. a a v a vi a a a v a vi a a a a v a vi a a a a iv a v a vi a a a v a vi a a a iv a v a vi a a iv a v a vi a a a iv a v a vi Oe will have the probability that 1, 3, 5 will come without, 4, 6 by echagig i the precedig d ito e & e ito d, V. a a iv a vi a a a iv a vi a a a a iv a vi a a a a iv a v a vi a a a iv a vi a a a iv a v a vi a a iv a v a vi a a a iv a v a vi Oe will have the probability that 1, 4, 5 will come without, 3, 6 by echagig 15
i the precedig c ito d & d ito c, VI. a a a vi a a a a vi a a a a iv a vi a a a a iv a v a vi a a a iv a vi a a a a v a vi a a a v a vi a a a iv a v a vi Oe will have the probability that, 3, 4 will come without 1, 5, 6 by echagig i formula II a ito c & c ito a, VII. a a v a vi a a a v a vi a a a a v a vi a a a a iv a v a vi a a a v a vi a a a iv a v a vi a a iv a v a vi a a a iv a v a vi Oe will have the probability that, 3, 5 will come without 1, 4, 6 by echagig i the precedig formula d ito e & e ito d, VIII. a a iv a vi a a a iv a vi a a a a iv a vi a a a a iv a v a vi a a a iv a vi a a a iv a v a vi a a iv a v a vi a a a iv a v a vi Oe will have the probability that, 4, 5 will come without 1, 3, 6 by echagig 16
i the precedig formula c ito d & d ito c, IX. a a a vi a a a a vi a a a a iv a vi a a a a iv a v a vi a a a iv a vi a a a a v a vi a a a v a vi a a a iv a v a vi Oe will have the probability that 3, 4, 5 will come without 1,, 6 by echagig i the precedig formula b ito c & c ito b, X. a a a vi a a a a vi a a a a iv a vi a a a a iv a v a vi a a a iv a vi a a a a v a vi a a a v a vi a a a iv a v a vi Subtractig these te formulas from the precedig, oe will have the probability that out of the umbers 1,, 3, 4, 5, 6 there will eit at least four of them, = 17
1 3 a a a a a iv a a iv a v a vi 6 a a a iv a v 10 a a iv 3 a a a v 6 a a a iv a vi a a v 3 a a a vi 6 a a a v a vi a a vi 3 a a iv a v 6 a a iv a v a vi a a iv 3 a a iv a vi 6 a a iv a v a vi a a v 3 a a v a vi 6 a a iv a v a vi a a vi 3 a a iv a v a iv a v 3 a a iv a vi a iv a vi 3 a a v a vi a v a vi 3 a iv a v a vi a a iv 3 a a iv a v a a v 3 a a iv a vi a a vi 3 a a v a vi a a vi 3 a a v a vi a iv a v 3 a iv a v a vi a iv a vi 3 a iv a v a vi a v a vi a iv a v a iv a vi a v a vi iv a v a vi. The aalogy is ow evidet, & oe sees that by coservig the same deomiatios as above, oe will have for the probability that out of faces oe will brig forth at least 3 of them, 1 A 3A iv 6A v 10 vi &c. 6. If a = a = a = a iv &c. = a = 1, oe will have by the theory of combiatios, 1 3 3 3 4 1 1..3 1... 4 6 4 5 10 5 6 &c. 1... 5 1... 6 18
for the probability sought, & by puttig for 3 foud above, oe will have this probability, = 1 p p 1 1..3 6 4 p p 4 &c. 1... 5 4 This is the result that Mr. Euler fids., 4, 5 &c. the values 3 3 1... 4 p p 3 3 7. Supposig always the same thigs, we demad the probability that at the ed of coups oe will have brought forth at least 3 faces. I cotiue to reaso likewise, & I say: a a a iv is the probability that there will come either 1 or or 3 or 4 i coups, therefore 1 a a a iv epresses the probability that of the umbers, 1,, 3, 4 there will eit at least oe of them. I order to have the probability that of the umbers 1,, 3, 4, 5 there will eit at least two of them, it is ecessary to subtract from the precedig probability the probability that oe of the umbers 1,, 3, 4 will come without the four others. Now a a a iv a v a a a a iv a v = the probability that 1 will come without, 3, 4, 5. a a a iv a v a a a a iv a v = the probability that will come without 1, 3, 4, 5. a a a iv a v a a a a iv a v = the probability that 3 will come without 1,, 4, 5. a a a a v a a a a iv a v = the probability that 4 will come without 1,, 3, 5. Therefore 1 a a a iv 4 a a a v a a iv a v a a iv a v a a iv a v = a a a iv a v 19
the probability that of the umbers 1,, 3, 4, 5 there will eit at least two of them. Oe will fid by cotiuig the operatio, & coservig always the same deomiatios, that 1 A iv 4A v 10A vi 0A vii 35A viii &c. epresses the probability that out of umbers there will eit at least 3 of them. 8. The aalogy is ow evidet, the umeric coefficiets beig the figurate umbers, & oe will have the followig formula, which is geeral: 1A λ λ 1A λ1 λ 1λ A λ 1. λ 1λ λ 3 A λ3 λ 1 λ 4 A λ4 1..3 1... 4 λ 1 λ 5 A λ5 &c. = 1.... 5 the probability that of umbers there will eit at least λ of them. 9. If ow we make a = a = a = a iv &c. = 1, we will have A = 1, 1 A 1 =, 1. A 1 3 =, 1..3 A iv 3 4 = 1... 4 A λ λ 1 λ = 1.... λ Oe will have therefore by substitutig these values, λ 1 λ λ λ 1 1 λ 1 1.... λ 1.... λ 1 λ 1λ λ 1 λ 1. 1... λ λ 1 λ 3 λ λ 3 1..3 1... λ 3 λ 1 λ 4 λ 3 λ 4 &c. = 1... 4 1... λ 4 0
λ 1 1 1.... λ λ 1 λ 1 1... λ 1 λ 1λ 1. λ 1λ λ 3 1..3 λ 1 λ 4 1..3.4 p p λ 1... λ 1 p p λ... λ λ 1 p p λ 1 1... λ... λ 1 λ p p λ 1... λ 3... λ λ 3 p p λ 3 &c. 1... λ 4... λ 3 Mr. Euler fids a aalogous formula. Oe sees that the direct use of the doctrie of combiatios has led us quite simply to the solutio of the geeral problem of which the oe of Mr. Euler is oly a particular case, sice a, a, a... a which i the problem of Mr. Euler are = 1 ca be aythig i ours. The geeral problem would have place i a lottery where there would be a tickets marked 1, b marked, c marked 3 &c. & where oe would demad the probability that there would have eited a certai umber of kids of them. Oe will fid always the value of the epressios, a &c. by way of the followig formula which epresses the probability that by takig t thigs out of, of which u are of a certai kid, there will be foud m of this kid of them. This formula is, as oe kows, t t 1 t u m 1tt 1 t m 1 uu 1 u m 1 1 u 1 1..3... m 10. It is evidet by the same ature of thigs, that i problem 1, if the umber of drawigs is smaller tha p, it will be impossible that all the umbers have eited at the ed of a umber of drawigs; thus i all the cases where < p, the foud probability must be = 0, & the miimum of where the probability is ot ull, has place whe = p or = p. Mr. Euler remarks that i this case the sum of the series which gives the probability, ca be epressed by some products. This ca be demostrated by the itegral calculus, by the followig method which is quite simple. 11. The series that the questio is to sum is i this case here p p pp 1 p pp p 1 1p p 1. pp 1 pp 1p p p p p 1..3 p... p p p 3 p p p p 3 &c. 1... 4 p... p 3 I see first that if = 1, all the series = 1, because all the terms vaish with the eceptio of the first. 1
If =, oe has the series p pp 1 pp 1 1p p 1. pp 1 pp 1p pp 1p 1..3 pp 1p pp 1p p 3 1..3.4 = 1 p p 1 1. pp 1p p 3 pp 1p p 3 p p 1 pp 1 1 p p 1 p 1..3 pp 1p 1 p p 1 p p 3 1..3.4 pp 1p p 3 &c. &c. I order to fid the sum of this series, I form the summatory sequece of that there, & I have by makig successively p = 1,, 3, 4 &c. the sequece 1 3 4 1 1. 3.4 1..3 4.5.6 1..3.4 5.6.7.8 &c. The geeral term of this sequece, that is to say the sum of the sought series whe 1..3...p =, is therefore p1p p. If = 3, oe has the series, 1 p3 3p 1 1. p 3 p 1 3 3p 3p 1 1 p 3 p 1 3 p 3 1..3 3p 3p 1 3p 1 p 3 p 1 3 p 3 p 3 3 1..3.4 3p 3p 1 3p 3p 3 &c. Make successively p = 1,, 3, 4 & we will have the summatory series, 1 3 4 of which the geeral term is evidetly If = 4, oe has the series, 1 3p4 4p 3 1 1. 1. 1..3.4 1..3.4.5.6 1..3.4.5.6.7.8 3 5.6 7.8.9 9.10.11.1 &c. 1..3...p p1p 3p. 3p 4 3p 1 4 4p 3 4p 1 3 1 3p 4 3p 1 4 3p 4 1..3 4p 3 4p 1 3 4p 3 1 3p 4 3p 1 4 3p 4 3p 3 4 1..3.4 4p 3 4p 1 3 4p 3 4p 3 3 &c. Make successively p = 1,, 3, 4 & we will have the summatory sequece, 1 3 4 1..3 4 3 1..3.4.5.6 7 3.8 3 1..3.4.5.6.7.8.9 10 3.11 3.1 3 1..3...1 13 3.14 3.15 3.16 3 &c.
1..3...3p of which the geeral term is evidetly 3p13p 4p. The aalogy is ow evidet, & oe sees that the proposed series will be 3 = by makig p = 1..3... 1p 1p 1 1p p 1 = 1..3... 1p 1 p 1 1 = 1..3... p 1 p 1 = 1..3... 1 p 1. 1. Oe ca demostrate also that the sum of our series is = 0 i the case where > p. Let = p 1, the series will become p p 1 1p 1 pp 1 1p 1 p 1 1. 1p 1p 1 1p 1p 1..3 If = 1, this series becomes 1 p pp 1 1. If =, this series becomes p p 1p 1p pp 1p 1..3 &c. &c. = 1 1 p = 0. 1 p 3p 1 p p 1 1 p p 1 p 1. 3p3p 1 1..3 3p3p 13p 1 p p 1 p p 3 &c. 1..3.4 3p3p 13p 3p 3 = 1 p p pp 1 p p 1 1 3p 1. 3p 3p 1 pp 1p 1..3 p p 1 3p 3p 1 p 3p &c. If oe forms the summatory sequece of this series by makig p = 1,, 3, 4, &c. oe will have each term = 0, the geeral term of the series is therefore = 0, & the sum sought = 0. If = 3, this series becomes 1 3p 1 3p 3p3p 1 4p 1. 3p3p 13p 1..3 3p 4p 3p 3p 1 4p 4p 1 3 3p 1 4p 1 3p 4p &c..
which oe will fid i the same maer = 0. If = 4, this series becomes 1 4p 3 4p 4p4p 1 1 5p 1. 4p4p 14p 1..3 4p 5p 4p 4p 1 5p 5p 1 3 4p 1 5p 1 4p 5p which oe will fid likewise = 0. Therefore the series itself which oe ca put uder this form 1 p 1 p pp 1 p 1 1p 1. 1p pp 1p p p 1 1..3 1p 1p 1 3 &c.. p 1 1p 1 p 1p is = 0. Let ow = p, & oe will prove likewise that the series p 1 1p p p p 1 1. = 1 1p 1p p 1 1p 1p 1 1. p 1 p 1p p 1p 1p 1 p 1..3 p 1 1 &c. p p 1 1 p 1 1 &c. p p 1 1p 1p 1 p p 1 1 &c.. 1p &c.. p = 0. Therefore fially if = p m, m beig ay whole umber, oe will fid i the same maer that the series m 1p 1 1 m 1p mp 1 m 1p m 1p 1 1. m 1p mp m 1p m 1p 1 m 1p 1..3 m 1p m 1p mp m 1p 1 mp 1 mp 1 m 1p 1 &c. mp 1 &c. = 0. 13. If & are very great umbers, & p a small umber, the calculatio would become impractical by its legth, but the aalysis furishes i this case the way to 4
abbreviatio. Oe will have uder this assumptio, p = 1 p p = 1 p p &c. = e & likewise p 1 = e p 1 1 p λ λ = e p λ. We will have therefore for the probability that all the umbers have eited, 1 p 1 1 e e p 1 p 1. 1 e p p 1 p 1..3 3 e p 3 p p 1 p 1... 4 3 = 1e p = e e p p e = e q.3 e &c., 3p 4 4p 1.... 4 e &c. by makig for brevity q = e p. If oe wishes to push the approimatio further, oe will have 1 p p = 1 p 3.3 p3 3 4.3.4 p4 4 &c. p p 1 p 3.3 p3 3 &c. = e p Oe will have likewise 1 p 1 1 p 1 p λ = e p 1 1 = e p 1 1 = e p λ 1 p 1 p 1 p λ 1 p,,.. 5
We will have therefore for the probability that all the umbers have eited, 1e p 1 p 1 1 p 1. p 1 1 e p p 1 e 1 1..3 1 p e p e p 1 e p 1 1 3 1..3.4 1 p 1 = 1e p = 1e p p 1 1 e p e p 1 e p e p 3 p 1 1 p 1 e 3p 1..3 1 1 3 1..3.4 1 p e 3 3p 1..3 e p e =by makig e p = q 1 p 1 p p e 1 p 1 1. 1 p e 4p p 3 4.3.4 e 1 e p e q 1 p q 1 p e q e p q q 1 p e q e p q q 1 p p p 1 e p 3 1 p 1 p 4p 1 p 4 &c. 1 p 1 p 3 p.3 e 1 p p 3 &c. 4 &c. 3 &c. e q 1 p = e q e p q = e q e p q. 6
Now e p q = 1 p q &c. Therefore the sought probability will be e q 1 p q q e q = e q 1 p q q. 1 14. If oe cosiders this process with attetio, oe will see that oe ca put 1 1 i place of 1, &c. without committig a error greater tha that of the terms where q 3 eters. It is ecessary to ecept the terms e p 1, e p &c. Oe has 1 1 = 1 1 1, = 1 1,... λ = 1 λ. We will have therefore e p 1 = e p e p, e p = e p e p, e p 3 = e p e 3p,... e p λ = e p e λp. Therefore takig the precedig formulas, & applyig these correctios to the first part of the formula aloe, it will become Now 1e p 3.3 e 4.3.4 e 1 p 3p 3p 1 p e 3 4p 6p 1 p e p e 3p e 6p =1 p, =1 3p, p p 1 p 4 &c. =1 6p &c. whece it follows that it is ecessary to add to the foud formula, p p e p p e p q e q. 3 3p 3p.3 e 4 1 e p e 4p.3.4 e 6p &c. = p 3 3p.3 e &c. = Thus the probability sought will be e q 1 p q p q. 7
If oe makes p = 1, oe will have e q 1 q q. It is that which Mr. de la lace fids i the Mémoires de l Académie des Scieces de aris for 1783 6 by a very profoud aalysis, draw from the itegral calculus i fiite differeces. It is remarkable that oe ca arrive to the same coclusios without itegral calculus & by some quite simple cosideratios. 15. Let for eample, as Mr. de la lace supposes it = 10000, p = 1, & let oe demad the umber of coups at the ed of which oe ca wager eve that all the umbers will have eited. We will have i this case to resolve the equatio e q 1 q q = 1. Now 7 q = e, therefore = l q, = l q. For first approimatio I make e q = 1 l, therefore q = l, q = = 0.693147 10000 = 0.0000693147, l q = 4.1591745, l q = 4.1591745. Multiplyig this quatity by.30585, oe will have l q = 9.57679. Therefore = l q = 95767.9 first value. I substitute ow these values of & q ito the formula 1 q q & I have l q =5.840855 l q =1.6816510 l =4.981173 l =4.73313 l q =0.8048 l q =6.404973 l =4.3010300 q =0.0005408 l q =6.51018 q =0.00033190 q =0.0005408 q q =0.0000778 Therefore 1 q q = 1.0000778. Therefore e q 1 = 1.0000778, q = l.00015564 = 0.0000693548, l q = 4.159154. Multiplyig this quatity by.305851, oe will have l q = 9.576740. Therefore = l q = 95767.401. The sought umber is therefore a little earer to 95767 tha to 95768. Mr. de la lace arrives to the same coclusio at the ed of his Memoir. Oe sees et that i this case 6 Suite du mémoire sur les approimatios des Formules qui sot foctios de très-grads ombres, Mém. Acad. R. Sci. aris 1783 1786, p. 43-467. 7 Traslator s ote: Where Trembley writes log. hyp., I have writte l. 8
the first operatio sufficed, so that the correctio was superfluous, this which cofirms the goodess of the method. 16. Oe ca fid likewise a approimatio for the secod of the formulas foud above, whe & are very great umbers. The probability that at the ed of drawigs there will have eited at least 1 umbers is p p 1 1 1 1. 1 1..3 3 3 1... 4 1 p p p 1 1 p 1 1 p p p 3 3 by beig cotet with a first approimatio & makig q = e p 1 q 3 1..3 q3 34.3.4 q4 &c. by makig q = e p. Let q = oe will have the sought probability Therefore S = 1 3.3 34.3.4 &c. &c. = ds d = 1 1 3 4.3 &c. = 1 1 3 3 &c. = e. Therefore ds = d e, S =e e C. Whe = 0, oe has S = 1, therefore C = 0. Therefore S = e 1 = e q 1 q. Thus istead as i the precedig case it was ecessary for first approimatio to resolve the equatio e q = 1, it will be ecessary to resolve the equatio eq 1 q = 1. I order to come to ed, I suppose first e q = 1, whece I draw as above q =.0000693147. Therefore l q =5.840855 l =4.0000000 l q =9.840855 q =0.693147 1 q =1.693147 9
Oe has ow e q = 1 1.693147, q = l 3.38694 = 0.59747.305851 10000 l 0.000059747 =5.740503 l.305851 =0.36156 l q =6.086659 l =4.0000000 = 0.000059747.305851 l q =0.086659 q = 0.00011974, q = 1.1974, 1 q = 7.1974, e q = 1.1974, q = l4.43948 l 0.0000647331 =5.811171 l.305851 =0.36156 l q =6.173347 = 0.0000647331.305851 q =0.000149054. Therefore q = 1.49054, 1 q =.49054, e q = 1.49054, q = l4.98108 l 0.0000697336 =5.8434344 l.305851 =0.36156 l q =6.056500 = 0.0000697336.305851. q =0.000160565. Therefore q = 1.60565, 1 q =.60565, e q = 1.60565, q = l5.11130 l 0.00007093659 =5.8508703 l.305851 =0.36156 l q =6.130859 = 0.00007093659.305851. q =0.0001633375. Therefore q = 1.633375, 1 q =.633375, e q = 1.63338, q = l5.6675 l 0.000071547 =5.85860 l.305851 =0.36156 l q =6.04776 q =0.0001661415. = 0.000071547.305851. 30
Therefore q = 1.661415, 1 q =.661415, e q = 1.661415, q = l5.3830 l 0.000076146 =5.861019 l.305851 =0.36156 l q =6.3375 = 0.000076146.305851. q =0.000167. Therefore q = 1.67, 1 q =.67, e q = 1.67, q = l5.344 l 0.0000778664 =5.860518 l.305851 =0.36156 l q =6.4674 = 0.0000778664.305851. q =0.0001676. Therefore q = 1.676, 1 q =.676, e q = 1.676, q = l5.35 l 0.0000785161 =5.864391 l.305851 =0.36156 l q =6.46547 = 0.0000785161.305851. q =0.000167747. Therefore q = 1.67747, 1 q =.67747, e q = 1.67747, q = l5.35494 l 0.0000787546 =5.865813 l.305851 =0.36156 l q =6.47969 = 0.0000787546.305851. q =0.0001678, l q = 3.775031 l 3.775031 =0.5769404 l.305851 =0.36156 0.9391560 l q =8.6973 = l q = 8697.3. 31
Oe would have bee able to fid the limit much more rapidly by supposig q a little too great, but I have preferred to give the direct approimatio without makig ay error. 17. Oe will fid likewise for the probability that at least umbers will have eited, 1 3 1..3 q3 34 1..3.4 q4 65 1..3.4.5 q5 &c. = by makig q = Therefore 1 3 1..3 34 1..3.4 65 1..3.4.5 106 &c. = S. 1..3.4.5.6 ds = d 1. 33 1..3 64 1..3.4 105 1..3.4.5 &c. = 1 = e ds = d e, S = e d e = 1 e 1 1. 3 1..3 4 1..3.4 &c. =e q 1 q 1 q = 1. Fially to shorte the gropigs, as q must be greater tha i the precedig case, I make q = 0.001, & I have q = 10, q = 50, 1 q q = 61, e q = 1 1, q = l 1 = 0.00008636.305851, l 0.00008636 =6.3193893 1 6 l.305851 =0.36156 l q =6.6816049 q =0.00048040 I make q = 0.0004 & I have q = 4, q = 8, 1 q q = 13, e q = l 6, q = = 0.00014149733.30851. l 0.00014149733 =6.1507453 l.305851 =0.36156 l q =6.519609 q =0.0003581 3
I make q = 0.000 & I have q =, q =, 1 q q = 5, e q = 1 l 10 10, q = = 0.000305851. I make q = 0.0006 & I have q =.6, q = 3.38, 1q q = 6.98, e q = 1 l 13.96 13.96, q = = 0.00011448854.305851. l 0.0001144885 =6.0587619 l.305851 =0.36156 l q =6.409775 q =0.000636 18. Oe approaches thus quite ear to the value of q, but the approimatio is less eact, because it is already raised to above the value of 1. 19. Oe would fid likewise for the probability that there will have eited at least 3 umbers 3 S =.3 1 e by makig = q, & i geeral for the probability that there will have eited at least λ umbers λ S = 1... λ λ1 1... λ 1 1 e but ca ot be eact as log as q is of order 1. There would be may cosideratios to make o the later approimatios, but I myself could ot deliver the detail which they require without legtheig this memoir too much. 33