1 3 4 4 + 1 2 3 $3 to as 2 4 2 X 3 9 n+1 2 3524 8624 2-1+1. On the other hand, the probability of A has of winning once in two games will be 1 the event's not happening so often asl times, mak ; consequently the odds of A's winning ing n–1+1=p, will be expressed by the series bp are three to one. sp 9. If A and B play together, and A wants one game, pa pip, and B wants two of being up; but the chances whereby B 1. 2. SS may win a game are double to ihose of A ; to find the respective probabilities of winning. In this case, the • P+1.8+2.4+3.24 &c. ) continued probability which B has of winning a game is 1.2.3.4.st 3 and the probability of his winning twice tomany terms, exclusive of the common multiplier, 2 as are denoted by the number l. The sum of gether is ; and, therefore, the prothese two probabilities is always equal to unity; 3 9 and, therefore, the first series inay be used when 4 5 ; and 9-1+1 is less than l, and the second when dis bability of A's winning the set is 1 less than n-ait1: in other words, the first or see the odds of A's winning one before B wins cond may be used as I is less or greater than twice, which the question requires, are as five to four. E. gr. let a be=1, b=35, n=24, and 1=1: in this In general, whatever be the number of games case twenty-four terms of the first series would which A and B respectively want of being up, the be necessary to answer the question, but one set will be concluded at the most in so many term of the second will be sufficient; and the pro- games, wanting one, as is the şuin of the games bability of an event's not happening once in wanted between them. For if A wants three twenty-four trials, which has one chance to hapo games, and B wants five of being up, the greatest pen, and thirty-five to fail, will be expressed by number of games which A can win of B, before the determination of the play, will be two, and * 1: this quantity, by the help of logarithms, the greatest number which B can win of a will will be found nearly equal to the decimal 0.50871, be four, and the whole number six : but, supwhich, subtracted from unity, leaves a remain posing they have played six games, the next game der 0.49129, expressing the probability required; will terminate the play; and, therefore, the utand, therefore, the odds against the happening of most number of games that can be played between them will be =5+3-1. the event will be fifty to forty-nine nearly. Again, if 1=2, »=60, and "-/+ 1. = 59, in this being up, and their respective chances of winning a game 10. If A wants three games, and B wants seven of case fifty-nine terms of the first series will be ne. cessary, and only two of the second; and, there are as three to fove; to find the respective probabilities of fore, hy the second series, the probability of an winning the set. The sum of the games is ten, and event's not happening twice in sixty trials will be therefore, A undertakes to win three out of nine tlie set will be concluded in nine games, and, 3559 1 + 3650 36 series of the preceding theorem, n being=9, l=3, ed from 1, leaves the remainder 0.4993, expressing a=3, b=5, and n-1+1, or p=7, the probability the probability required; consequently, the odds which B has of winning the set will be expressed against the event's happening twice in sixty times 21 will be little more than 500 to 499. by 5*(1+ x 484 = 0.28172 7. To find the probability of throwing one ace, and no nearly, which, subtracted from unity, leaves a remore, in four throws. From the unlimited probabi- mainder = 0.71828, expressing the probability ļity of the ace's being thrown in four throws, sub- which A has of winning ; consequently, the odds tract the probability of its being thrown twice in of A's winning the set will be 71828 to 28172, or that number of throws: i. e. from by case 3. nearly as 23 to 9. Mr. de Moivre has also illustrated the prin171 500 ciples of gaining, applied in the preceding cases, subtract, 1296 by case 6, and the remainder in a different arid more general way, by suppos. 1296 500 ing two dice, having the same or a different numgives the probability required, and 1 ber of equal faces : it is obvious, that the number 1296 of all the variations which the t'vo dice can undergo 796 the probability of the contrary; and the of the one by the number of faces of the other. will be obtained by multiplying the number of faces 1296 odds against throwing one ace, and no more, in four From hence it follows, that if the faces of each throws, are 796 to 500, or 8 to 5 nearly. die are distinguished into white and black, and the number of white faces on the first is A, 8. If A and B play together, and A wants one game of and that of the black faceş B, and also the number, being up, and B wants trvo, what are their respective of white and black faces 'on the second a and b probabilities of winning the set? It is plain that a respectively, the product of A + B multiplied by wants only to win once in two games, and B a+b, i.e. Á e+ A 6+Ba+B 6 will exhibit all the wants to win twice together : and if both have variations of the two dice: A a will represent the an equal chance, the probability which B has of number of all the chances whereby the two dice 1 winning the first game will be and the may exhibịt two white faces; A b'the number of 2) pro all the chances whereby a white face of the first 1 bability of his winning twice together will be may be joined with a black face of the second; 2 a B the number of all the chances whereby a white face of the second may be joined with a -; and, therefore, the probability which black face of the first; ang B b the bumber of *(1 3%) = 0,5007, which subtract- games, and B to win seven ; then, by the second 1 all the chances whereby a black face of the first respective numbers of white and black faces are may be joined with a black face of the second. represented by the indices of a and b in any given Moreover, as these several quantities may be con- term. nected together several ways, the sum of two or To find the probability of throwing an ace in four more of any of them will answer some question throws with a common die' of six facs. Since the of chance : e. gr. the number of chances for throwing of one die four times successively is the throwing, a white face with the two dice above same thing as throwing fuur dice at once, the mentioned, will be represented by the three parts chance of ihrowing an ace is the same in one case A 2+4 6+ B a, because every one of these com- as in the other. Let the ace, then, in each of prehends a case wherein a white face is concern- the four dice, answer to one white face, and the ed: hut if the question were restricted by a pre- rest of the points to tive black faces, or a be 1. vious stipulation, that, if a person threw iwo and b = 5; raise a + b to its fourth power, and white faces together, the wager would be lost; every one of the terms in a4 + 4 23 6 + 6 2262 +4 a bez the two terms 16+ B a would represent all +64, that has an a, will be a part of the number his chances. Farther, if a third die be added, of chances whereby an ace may be thrown : conwhose white faces are e, and black faces ß, the sequently, a++ 4 a3b + 6 ao b2 + 4 a 63, i. e, 1+20) + preceding product, viz.. A + Ab+Ba+ Bb, mul1511 + 500 = 671, will express the number of tiplied by the whole number of faces of the third chances whereby an ace may be thrown with four die, viz'a + sor daat Aba + Baat Bbc dice, or in four successive throws of a single die; + A a B + Abs+ Baß + B 6%, will exhibit the but the number of all the chances is the fourth number of all ihe variations which the three dice power of a+b or 64=1296; and, therefore, the can undergo : in this case A aa represents the probability required is measured by the fraction number of chances for throwing three white 671 faces, Abe the number of chances whereby both the first and third die may exhibits while 1296 See Case 3. It is evident, that Lt, or the last face, and the second a black one, &c. and by chances wherebytheace may failinevery one of the term of the above power, expresses the number of joining two or more of these terms, some ques ** tion of chance will be answered ; e. gr. the num. dice; and the probability of that failing is ber of chances for throwing two white faces and (+6,4 a black one will be represented by A ba + Baa+ 625 AB, and the odds in favour of him who under- 1296; and, therefore, the probability of not takes to throw them will be Abæ+ Bar+ Acß failing, or of throwing an ace in four throws, is 20 A 2a + Bbre + A 6$ + Ba B+B bß; and the 695 671 probability of the event will be expressed by a as before. When the number fraction, whose numerator is the number of 1296 1296' chances required, and the denominator the whole of dice is n, b", the last term of the power (a+b), number of variations which all the dice can under. will always represent the number of chances go. E. gr. the probability of throwing three white whereby the ace may fail in n times; and, therefaces with the three dice above-mentioned will be fore, is the probability of its failing; and expressed by (a + b)" Aan the probability of throwing an ace, in a number 69 Aaa+ Abu + Baa * daß + Bba + Abß + Baß + Bbb, of throws expressed by n, will be 1 Dr, separating the factors in the denoninator, (a+b) Аа се (a+b)? - 77 (A+B)*(a+b)x(+3) A+B^atb (a + b) a so that the probability of the happening of seve- To find the probability of throwing with one single ral events independently, is the product of all the die two aces in four throws, or of throwing at once tivo particular probabilities whereby each particular aces with four dice. The three terms, u*+ 423 to event may be produced. 6 a 12 of the fourth power of the binomial a +b, In the preceding case of two dice, if A be=a, in which the indices of a equal or exceed the and B=b, A 2+ 8b+ a B+B b will become = aa number of times that the ace is to be thrown, + 2ab+bb; and 2 a b=Ab+a B will express the will denote the number of chances whereby two number of variations, whereby with two dice of aces may be thrown; so that a being=1, and be the same respective number of white and black =5, 4*+ * 23 6+ 6 aa , will be = 1 + 20+ 150) = faces, a white face and a black one may be 171, and the whole number of chances, viz. thrown. And in the case of three diçe, A ba + Back (a + b)+= 1296 ; consequently the probability re+A . ß will become =S a ab;or, one term of the 171 See Case 6. binomial a +b, raised to its cube, will express the quired will be expressed by 1296 number of variations, whereby three dice of the 1125 same kind would exhibit two white faces and a The probability of the contrary will be 1296 black one. 11. From the above reasoning the following 13 + 14, a is found only once in the first, and not in the two terms of the fourth power of a + 1, 4 general rule may be deduced : viz. in a number of dicę = n, each of which is distinguished into the number of chances against throwing two at all in the second ; therefore, these will express white and black faces, represented respectively by term of that power will express the number of bability of succceding will be 1 e and b, if a+b be raised to the power n, the first aces, which is = 500 + 625 = 1125. And the pro 1125 171 chances wherebyn white faces may be thrown; 1296 1296 the second term will express the number of as before. And universally, the last term of any chances whereby n-1 white face and 1 black power (a + ?)' being b", and the last but one na in face may be thrown; the third term will express , in neither of which a: enters, it follows that the number of chances whereby *-2 white faces the two iast terms of that power express the Lad 2 black faces may be thrown, &c. and the number of chances that are contrary to the throw А CC x ate ; for . 9 Games Odds of Gamies Odds of Games Odds of eng of two aces, in any number of throws deno 5 minated by n ; and that the probability of throw. 3, b=5, therefore q=; and the quantity nab"-1 + bn 19+9a 18 +35 ao ? ing two aces will be l 1 + 9 + 369 becomes (a+b)" (a + b) (1+9) (a + b)* — nabi -1-" And likewise, in thethree q?x (32 + 99 +36) - but the factor qe +99 + 36 is = (a+b)? (1+4) last terms of the power (a+ li)", every one of the 25 45 484 indices of a will be less than three, and conse 9 + 36 and its logarithm is tlie 3 quently those last terms will shew the number of chances that are contrary to the throwing of an logarithm of 484 log. 9. = 1.7506029 ace three times in any number of trials n; and the To which add the lcg. of q', 1.5529409 same rule will hold perpetually. If the chances or 7 x log. 5 log. 3. for happening and failing in any particular trial And from the sum 3.2535438 be respectively represented by a and b, the pre- Subtract the log. of (1+y)" ceding rules may be applied to the happening or or 9 x (log. 8.-log. 3.) 3.8337183 failing of any other kind of event in any number and the remainder 1.4498255 of times. will be the logarithm of B's chance, viz. 0.281725, 12. If A lays a wager that a certain event and the complement of this to unity, or 0.718275, will happen i times in n trials, and B lays to will be the chance of a. See another method of the contrary, and the number of chances of reducing ratios in larger numbers to their least happening and failing in one trial be respectively exact terms, under Ratio. a and l, the number of chances whereby B may win his wager will be determined by as many of The following Table shews the odds of winning, the last terms of the power (a + b)" expanded, as when the number of games wanting dces are represented by And the number of chances not exceed six, and the skill of the contenders whereby A may be a winner, will be expressed is equal. by as many of the first terms of the same power as are equal to pip being assumed equal to nu-l +1; because B, by laying against A's winning 1 wanting, winning. wanting. winning wantinz. winning times,does in effect lay that he will not win above 1,2 3,1 2,3 11,5 3,5 99.29 1-1 times; but the whole number of winnings and 1,3 7,1 2,4 26,6 3,6 219,97 losings being n, (n-1)-1, or n-1+1, will be the 15,1 12,5 57,7 4,5 163,93 number of times which В himself undertakes to 1,5 31,1 2,6 120,8 4,6 382,130 win. 1,6 63,1 3,4 42,225,6 638,380) 13. If A and B being at play, respectively want l and p games of being up, and their respective 14. To assign the probability of throwing one ace, and chances for winning any one game be as a to b; no more, with four dice at one throw. From the numraise the binomiala + b to a power whose index is lt ber of chances whereby one ace or more may be p-1, and the number of chances,whereby they may thrown, subtract the number of chances whererespectively, win the set, will be in the same pro- by two aces or more may be thrown, and the reportion as the sum of so many of the first terms as mainder will be the number of chances for throware expressed by p, to the sum of so many of the ing one ace, and no more. Having raised the bilast terms as are expressed by l; for when A and nomial a + to its fourth power, which is at #4a3 B respectively undertook to win ? games and 6 + 64?l2 + 4ab3 + 14, it is evident, that the four p games, and n represented the whole number of first terms express the number of chances for games, in the preceding case, p was = 1-1+1, throwing one ace or more; and the three first and, therefore, 1+p=n+ 1, and n=l+ p - 1, terms express the number of chances for throwing which will, consequently, represent the power to two aces or more ; and, therefore, the single which a+b must be raised. E. gr. Let l=3,p= term 4a 13 expresses the pumber of chances for 7, a=3, and b=5; then raise a + b to the power throwing one ace and no more; and the probarepresented by 1+P-1,i. e, to the ninth power, 500 125 and the sum of the first seven terms will be to the bility required will be See (a + b)* 1296 sum of the three last, in the proportion of the re 2.24 Case 7. spective chances whereby A and B may win the set. The chances of B, or the three last terms, When it is required to assign the chances for are 19 + 9 a 18+ 36 a al 37812500, which, dia throwing any number of aces, and no more than that vided by (a+b)' = 134217728, gives the fraction number, one single term of the power(+1)* will 37812500 9453125 always answer the question; and this term may be expressing the probabi- expeditiously found, supposing n to be the num134217728 33554432' lity of B's winning; and this fraction, subtracted ber of dice, andl the precise number of aces to be thrown, when l is less than n, by writing as 9453215 24101307 from unity, or 1 will n m - 3, 33554432 33554432 miany terms of the series express the probability of A’s winning; and the T' S odds in favour of A'are in the proportion of n-3, &c, as there are units in l; and when I is 24101307 to 9453125, or nearly as 23 to 9. See 4 Case 10. Examples often occur, as in this case, greater than 1.n, by writing as many of them as in which it is required to sum up several terms of there are units inn-1; then multiplying all those & high power of the binomial a tu, and to divide terms together, and multiplying the product by their sum by that power : the most convenient al 6* - ; and this last product will exhibit the method of doing this is to write 1 and q for a and term expressing the number of chances required. b, having taken q:1:: 6:2, and to use a table of E. gr. Let l be =3, and n=10; since l is less than logarithms. Thus, in the preceding example, a= n, take three terms of the series, which will be 4 a 13 n 3 9 9 (1+1) 1 1 + + &c. = 2 + 1 Z х 10 9 8 differs from unity, substitute in the room of Ta' z; their product will be 120:a! 14 ), a being = 1, 1 = 9, &c. will be = 6042969; and log. its value expressed in a series, 6042069x120=725156250, which expresses the viz. number of chances required. Divide this by 1 1 1 (a + b) = 10000000000, and the quotient &c. and we 0.0725156280, will express the probability of 2q2 393 49+ 5qs 690? throwing precisely three aces with ten dice: subtract this quotiene from unity, and the remainder shall have the equation log. 20 0.9274843720 will express the probability of the 9 contrary; and the odds against throwing pre- the first term of the series will be suficient; and If be infinite, or large in respect to unity: cisely three aces with ten dice, are 9274843720 to 725156280, or nearly 64 to 5. we shall have the equation Having thus given as compendious a view as =log. 2, or 1=9* 9 possible of the principles and rules of gaming, log. 2. Then taking the hyperbolic logarithm of from Ms. De Moivre's Introduction, we shall se- 2, which is 0.693, &c. or 0.7 nearly, * will be lect a few problems, for the farther illustration equal to 0.7q nearly. Thus the limits of the ratio and exercise of these principles and rules. of x to q are assigned; for it begins with unity, and terminates at last in the ratio of 7 to 10 very Prob. I. If A and B play with single bowls, and A nearly. This value of x may be assumed in all can give 5 two games out of three; what is the propor- cases, whatever be the value of q. tion of their skill, or what are the odds, that A may get any one game assigned? Let the proportion of odds take with an equality of chance to throw two aces with Ex. 1. To find in how many throws one may underbe as z to l; now since A can give B two games out of three, A may, upon an equality of play being 36, of which there is but one chance for two dice. The number of chances upon two dice undertake to win three games together ; but the two aces, the number of chances against it is 35; probability of his winning the first time is. multiply, therefore, 35 by 0.7, and the product 24.5 will shew that the number of required throws and the probability of his winning three times to- will be between 24 and 25. gether is 1 But, Ex. 2. To find in how many throws of three dice one 2+1 *+1 (2+1) may undertake to throw three aces. The number as A and B are supposed to play on equal terms, of all the chances on three dice being 216, of the probability which A has of winning three which there is but one chance for three aces, games together ought to be expressed by and 215 against it, multiply 215 by 0.7, and the ; 2 product 150.5, will shew that the number of re quired throws will be 150 nearly. and, therefore, 금, -, or 2:3 = (2+ 1)3; (5+1)3 Prob. III. To find how many chances there are upon and, extracting the cube root on both sides, 23/2 any number of dice, each of them having the same number of faces, to throw any given number of points. Let p+ os+1; and :} 2–5 = 1; therefore : = 34-1 i i be the given number of points, n the number of and the odds in favour of A's getting any game dice, and f the number of faces in each dice: let p-f=9,9-f=r,r-f=s, s s, s-f=t, &c. T'he numassigned, are as V-1 as 50 to 13 very nearly. PyP-1 |- &c. Prob. II. To find in how many trials any event will probably bappen, so that A and B may lay a wager on een terms. Let the number of chances for its happening in any one trial be a, and the number of chances for its failing be b; and the number of &c. XX trials x. By Case 12. bv will represent the number of chances of the event's failing 1 times successively, and (a + b be the whole number of ? &c. chances for happening or failing; and, therefore, ba represents the probability of the event's (a + b)* 3 failing : times together, which is equal to the And these serieses ought to be continued till probability of its happening once at least in that some of the factors in each product become either number of trials; and either of these may be ex- =0, or negative: and so many factors are to be bx pressed by the fraction 4 ; consequently P it taken in each of the products P(a + b)* 1 ,&c. = h, or (a+b)x = 2 b*; therefore x, x logarithm (a+b) * * logar. 9 + logar. 2. and I = &c. as there are units in n-1. See the log. 2. log.(a+b) – log. 6. Again, suppose a : 6::1:9, demonstration of this rule in De Moivre's Doc trine of Chances, p. 41. then the equation (a+b)* =2 bt, will be changed Ex. 1. To find how many chances there are for throwa in ing sixteen points with four dice. Here p+ i=16 and p=15; and the foregoing series will be, 42. Therefore, * * log. 2. If, 15 14 13 9 + 455 in this equation, q = 1, * will also be=1 ; but if 1 212=to 1, or as 1 to 1/2–1, i.e. ber of chances required wil be, + X х 1 X Х х 3 2 3 +--XXXX I 2 3 1 2 14 10. X 4 Х 5 1 that are upon the dice, have the same number of 4:61:1:9; then ( 17) in + + 1-2 х Х 3 * x may undertake to throw three aces fovice. The num. -336 ber of all the chances on three dice being 216, of which there are 215 against throwing three 1 4 +6. And 455 aces ; multiply, therifore, 215 by 1.670, and the product 360.8 will give the number of required 336 + 6 125 is the number of chances re, throws. quired. Ex. 2. To find in kow many throws of six dice are Ex. 2. To find the number of chances for throwing may undertake to throw fifteen points twice. The sunAfteen points with six dice. The series will be, Let of chances for throwing tifteen points is 13 11 1666, and the number of chances for missing = + 2002 44990, see Prob, 111. Divide 44990 by 1666, and 2 3 5 the quotient will be 27 nearly; therefore, the 8 ។ 6 5 6 Х chances for throwing and missing fificen points 336. 2 3 4 are as I to 27 respectively: multiply 27 by 1.678, And 2002–336=1666 the number required. In and the produci 45.3 will be the number of relike manner it may. be found that the number of quired throws. chances for throwing twenty-seven points with Prob. 5. To find hotu many trials are necessary to six dice is also 1666. We may observe in general, make it equally probable that an event will happen S, 4, that all the points equally distant from the ex- 5, &c. times. Let a, b, and x, be as before; and tremes, i. e. from the least and greatest points 1 + %3D1x 9 chances by which they may be produced : thus, -1 in finding the number of chances for throwing in the case of the triple event ; 1 twenty-seven points with six dice, let 27 be cube 242 tracted from 42, the sum of the extremes 6 and and 1 + 36, and the remainder being 15, we may infer 1 222 that the number of chances for throwing twenty- x-1 seven points is the same as for throwing fifteen ī in case of the quadruple 39 points. event, &c. Then if q in the first equation be Ex. 3. To find in how many throws of sir dice one supposed=1, will be 5; if q be infinite, or way undertake to throw fifteen points precisely. The very large in respect to unity, the aforesaid equanumber of chances for throwing fifteen points being 1666, and the whole number of chances tion, making -=s, will become ::=log. 2 + log. upon six dice being 46656, it follows that the 9 number of chances for failing is 44990; therefore, (1++1=2); and will be found nearly = 2.675, dividing 44990 by 1666, and the quotient is 27 and x will always be between 59 and 2.675 ቀ g. In nearly ; multiply 27 by 0.7, and the product 18.9 the second equation, ifq be = 1, will be will shew that the number of throws required but if q be infinite, or very large in respect to will be very near 19. unity, s will be log. 2 + log: +:+ izat Prob. IV. To find how many trials are necessary to make it equally probable that an event abill happen Case 12. Br + xa bx is the number of chances s will be between 7 q and 3.6719 g. whereby the event may fail, and (a+b)x the whole number of chances whereby it may hap A Table of the Limits. pen or fail, and, therefore, the probability of its The value of * will always be bir + xabx-. I For a single event, between -; but as the probabilities for a double event, be:ween and 0.693 9. failing is (a + b)* 39 1.6789 of happening and failing are equal, we shall For a triple event, between 54 bx + xab-1 For a quadruple event, between 79 have the equation =j, or (a + b)* For a quintuple event, between 99 (a+b)* For a sextuple erent, between 119 *26* +2.ra br -?; or making a : b;:1:9, &c. 2 * 1 + = 2 + Let q=1, and a will be And if the number of events contended for, as 9 well as the number q, be pretty large in respect to 3, But if o be infinite, and = s; and . = unity; the number of trials required for those 2n-1 log. 2+ log. (1+s), and the value of : will be events to happen n times will be -q.or bare 2 found = 1.678 nearly; therefore the value of x lyn. will always be between the limits 3 g and 1.678 q, but will soon converge to the last of these limits; Prob. VI. Two gamesters A and B, each having and if q be not very small, x may in all cases bé twelve counters, play with three dice, en condition that if supposed = 1,678 q: but if x be suspected to be cleven points come up, B shall give one counter to A; if too little, substitute this value in the original fourteen points come up, A shall give one counter to Bi 23 and that he shall be the winner who shall soonest get and note the er- all the counters of his adversary; what is the probabirop; if it be worth regarding, increase a little the lity that each of them has of winning? Let o be the value of a, and substitute this new value in the number of counters of each, and a and 6 the numaforesaid equation, and noting the new error, ber of chances they have respectively for getting the value of may be sufficiently corrected by the lities of winning are respectively as af to b: and a counter at each cast of the dice; their probabiTule of double false position. p being = 12, and a = 27, and b = 15, by Prob. Ex. I. To find in how many throws of three dice ene 1. the probabilities of winning are respectively |