may undertake to throw three aces twice. The num ber of all the chances on three dice being 216, of which there are 215 against throwing three aces; multiply, therefore, 215 by 1.678, and the product 360.8 will give the number of required throws.

may undertake to throw fifteen points twice. The num Ex. 2. To find in how many throws of six dice one ber of chances for throwing fifteen points is 1666, and the number of chances for missing 44990, see Prob. III. Divide 44990 by 1666, and the quotient will be 27 nearly; therefore, the chances for throwing and missing fifteen points are as 1 to 27 respectively: multiply 27 by 1.678, and the product 45.3 will be the number of required throws.

Prob. 5. To find how many trials are necessary to make it equally probable that an event will happen 3, 4, 5, &c. times. Let a, b, and x, be as before; and a:b:: 1:9; then ( 1 +

And 2002-336=1666 the number required. In
like manner it may be found that the number of
chances for throwing twenty-seven points with
six dice is also 1666. We may observe in general,
that all the points equally distant from the ex-
tremes, i. e. from the least and greatest points
that are upon the dice, have the same number of
chances by which they may be produced: thus,
in finding the number of chances for throwing
twenty-seven points with six dice, let 27 be sub-
tracted from 42, the sum of the extremes 6 and
56, and the remainder being 15, we may infer and
that the number of chances for throwing twenty-
seven points is the same as for throwing fifteen
points.

Ex. 3. To find in how many throws of six dice one may undertake to throw fifteen points precisely. The number of chances for throwing fifteen points being 1666, and the whole number of chances upon six dice being 46656, it follows that the nun ber of chances for failing is 44990; therefore, dividing 44990 by 1666, and the quotient is 27 nearly; multiply 27 by 0.7, and the product 18.9 will shew that the number of throws required will be very near 19.

Prob. IV. To find how many trials are necessary to make it equally probable that an event will happen twice. Let a, b, and x, be as in Prob. II. Then by Case 12, b + x abr is the number of chances whereby the event may fail, and (a+b)* the whole number of chances whereby it may happen or fail, and, therefore, the probability of its b x + xa bx — 1 failing is -; but as the probabilities

(a+b)*

of happening and failing are equal, we shall
bx + x ab*_ 1
have the equation

= 1⁄2, or (a+b)*
(a + b) s
; or making
a: b: 1:9,
Let = 1, and will be

9

3. But if q be infinite, and

=

and z ༡ log. 2+ log. (1 + z), and the value of z will be found 1.678 nearly; therefore the value of x will always be between the limits 3 q and 1.678 q, but will soon converge to the last of these limits; and if q be not very small, x may in all cases be supposed 1.678q: but if x be suspected to be too little, substitute this value in the original

equation (1 + - )2=

9

2x

= 2+ and note the error, if it be worth regarding, increase a little the value of x, and substitute this new value in the aforesaid equation, and noting the new error, the value of x may be sufficiently corrected by the rule of double false position.

Ex. 1. To find in how many throws of three dice one

x

x-1

1+ + x

x) in case of the quadruple

2 393

-X event, &c. Then if 9 in the first equation be supposed 1, x will be 5; if q be infinite, or very large in respect to unity, the aforesaid equation, making z, will become z = log. 2+ log. (1 + 2 + 1 x 2); and z will be found nearly = 2.675, and x will always be between 5 g and 2.675 q. La but if q be infinite, or very large in respect to the second equation, if 9 be= 1, x will be unity, will be

Prob. VI. Two gamesters A and B, each having twelve counters, play with three dice,on condition that if eleven points come up. B shall give one counter to A; if fourteen points come up, A shall give one counter to B;

and that he shall be the winner who shall soonest get all the counters of his adversary; what is the probabi lity that each of them has of winning? Let p be the numbers of counters of each, and a and b the num ber of chances they have respectively for getting lities of winning are respectively as af to b: and a counter at each cast of the dice; their probabip being 12, and a = 27, and b = 15, by Prob. III. the probabilities of winning are respectively

as 9714 to 1512, or as 912, to 512, or as 282129536481 to 244140625. See the demonstration of this rule in Doctrine of Chances, p. 52.

Prob. VII. Two gamesters A and B lay by twentyfour counters, and play with three dice on this condition, that if eleven points come up, A shall take one counter out of the heap; if fourteen, В shall take one; and he that be reputed the winner who shall soonest get twelve

counters.

This problem differs from the preceding, in that the game must necessarily end in twenty-three throws, whereas in the former, the play may be unlimited, on account of the reciprocations of loss and gain, which destroy one another. Let a and represent the proportion of the chance for throwing 11 and 14; and raise a + b to the twentythird power, or to a power whose index is the sumber of all the counters wanting one, and the 19 first terms of that power will be to the 12 last in the same proportion as the probabilities of Winning.

Prob. VIII. Three gamesters A, B, and C, out of a heap of twelve counters, of which four are while and eight black, draw blindfold one counter at a time, in this Banner: A begins to draw, B follows A, C follows B; then A begins again: and they continue to draw in the same order, till one of them, who is to be reputed the ner, draws the first white: what are their respective probabilities of winning? Let n be the number of counters, a the number of white, and b the number of black, and 1 the stake or sum played for. 1. A has a chances for a white counter, and b chances for a black one; and, therefore, the probability of his winning is ; and his exa+b n pectation on the stake 1, when he begins to draw,

a

=

a

-: now— being taken out of the

n

n

Ex (b-1) ** (n−1)

5. C has a chances for a white counter, and the number of remaining counters is n-2; therefore his probability of winning will be ; and his expectation in the remaining stake will be bx (b-1)xa

*x(n−1) × (n−2)`

a
n-2

4. A may have out of the remainder the sum bx(b-1)x(b−2)× á *X(n−1)× (n−2) × (n−3)' stake be exhausted.

&c. till the whole

Q, R, S, &c. denote the preceding terms, and take as many terms of this series as there are units in b+1, (for b representing the number of black b+1), then the sums of the first, fourth, seventh, counters, the number of drawings cannot exceed &c. terms, of the second, fifth, eighth, &c. terms, and of the third, sixth, &c. terms will be the respective expectations of A, B, C; or as the stake is fixed, these sums will be proportional to their respective probabilities of winning. Let n, then, in the case of this problem, 12, a 4, and b=8; and the general series will become

7

10

-

5

-Q+ -R+ S+

9

Y; or, multiplying the whole by 495 in order to take away the fractions, the series will be 165+ 120 +84 +56 + 35 +20 + 10+4+1; and A will have 165 +56 +10=231, B, 120+35+4=159, and C, 84 +20+1=105; therefore their respective probabilities of winning will be proportional to the numbers 231, 159, and 105, or 77, 53, and 35.

Prob. IX. A and B having twelve counters, four of them white, and eight black; A wagers with B, that taking out seven counters, blindfold, three of them shall be white: what is the ratio of their expectancies? 1. Seek how many cases there are for seven counters to be taken out of twelve; they will be found from the doctrine of combinations to be 792.

2. Set aside three white ones, and find all the cases wherein four of the eight black ones may be combined therewith; they will be found to be 70. 8 7 6 5 4

=70.

And since there are four cases, in which three white may be taken out of four; multiply 70 by 4: thus the cases, wherein three whites may come out with four blacks, are found to be 280.

3. By the common rules of gaming, he is reputed conqueror who produces an effect oftener than he undertook to do, unless the contrary be expressly agreed on; and therefore, if A take out four whites with three blacks, he wins. Set aside four whites, and then find all the cases wherein three of the eight blacks may be combined with four whites: these cases will appear to be 56.

4. A therefore has 280+56=336 chances, whereby be may win; which, subtracted from the whole number of chances 792, leaves 456, the number of chances wherein he may lose. Divide 336 14 336 by 792, and the quotient will ex 792 33 press the probability of A's winning; and 114 19 the probability of his losing; and there33 33 fore the odds against taking three white counters are 19 to 14.

From the solution of this problem it appears, that, if a B E = the number of white counters, b the number of black, n the whole number=a+b, c the number of counters to be taken out of Pn, and the number of white counters to be found precisely in c, then the number of chances R+ S, &c. in which P, for taking none of the white, or one single white, or two white and no more, or three white and no

ratio of their expectancies? This problem M. Ber
noulli soives analytically. Here, calling the
number of gamesters n+1, he finds that the pro-
babilities of any two immediately following
each other in the course of playing, are in the
ratio i+2" to 2"; and therefore the expectancies
of the several gamesters, A, B, C, D, F, &c. are
in a geometrical progression 1+2": 2# : a: c
::cd::d: e, &c. Hence it is easy to deter-
mine the state of the probabilities of any two
gamesters, either before the game, or in the course
thereof. If, e. gr. there be three gamesters, A, B,
C, then n=2 and 1 + 2" : 2* : : 5 : 4 :: a: c; that
is, their several probabilities of winning, before
A have overcome В, or B, C, are as the num-
bers 5, 5, 4; and therefore their exp ctancies are
5 5 4
: for all of them taken together must
14' 14' 14
make 1, or absolute certainty. After A has over-
come B, the probabilities for A, B, and C, will be
2 4

1

as there are units in e, for a denominator. E. gr. Resume the supposition of the problem, only that of the seven counters drawn, there shall not be one white; and let p=0, and c-p=7b1; then taking 1 of the first series and 7 terms of the second, the number of chances will be 1 x 8; the ratio of which to all the 7's that can be taken out 8 1 of 12 is :therefore the odds that there 792 99 shall be one or more white counters among the 7 which are drawn are 98 to 1. The probability of drawing one white counter and no more will be 112 14 , or the odds 85 to 14: the probability 792 99 of drawing all the 4 white among the 7 will be Prob. XII. Three gamesters, A, B, and C, whose ; or the odds 92 to 7. If n and c dexterities are equal, deposit each one piece, and engage were large, the foregoing method would be im- upon these terms; that two of them shall begin to play, and that the vanquished party shall give place to the practicable; and, therefore, the following theorem third, who is to take up the conqueror; and the same may be used. Let n, c, and p be as before; and condition to go round: each person, when vanquished, formake n-c=d. The probability of taking pre-feating a certain sum to the main stake; which shall be

=

56 7 792

99

cisely the number p of white counters will be cxc-1xc-2, &c. xd x d−1 x d -2, &c. x a-l a-2

777, as in the answer above. If there be four gamesters, A, B, C, D, their probabilities from the beginning will be as 81, 81, 72, 64. After A has beat B, the several probabilities of B, D, ter A has beat B and C, the probabilities of C, B, C, A, will be as 25, 32, 56, 56, respectively. AfD, A, will be as 16, 18, 28, 87.

all swept by the person who first beats the other two successively. How much, now, is the chance of A, and B, better or worse than that of C? 1. If the forfeiture be to the sum each person first deposited, as 7 to X-Xn-Xn-3×n-4Xn-5Xn-6Xn 6, the gamestors are upon an equal footing. 2. If

X

3

-7x-8, &c.

&c.

Note, The first series of the numerator contains as many terms as there are units in p; the second as many as there are units in a-p; and the third as many as there are units in p; and the denominator as many as there are units in a.

To avoid tiresome prolixity in this article, we must desist from farther investigations; which, in the following problems, grow very long, and more perplexed. In the rest, therefore, we shall content ourselves to give the answer, or result, without the process of arriving at it; which may be of use, as it furnishes so many data, from whence, as standards, we may be enabled occasionally to judge of the probability of events of the like kinds; though without letting the mind into the precise manner and reason thereof,

Prob. X. A and B play with two dice on this condition, that A shull win, if he throws six; and B, if he throws seven: A to have the first throw, in lieu of which B to have two throws; and both to continue with two throws each turn, till one of them wins: What is the ratio of the chance of A to that of B? Ans. As 10355 to 12276.

Prob. XI. If any number of gamesters, A, B, C, D, E, &c. equal in point of dexterity, deposit each one piece of money; and engage,on these conditions, that two of them, A and B, beginning the game, whichever of them shall be overcome, shall give place to the third C, who is to play with the c conqueror ; and the conqueror here to be taken up by the fourth man D; and thus øn, till some one, having Congeered them all round, draws the stake: what is the

be

1

the forfeiture be in a less ratio to the deposit, A
and B are on a better footing than C; if in a
greater ratio, the advantage is on,the side of C.
3. If there were no fines, the probabilities of win-
ning would be proportional to the expectations;
and the expectations, after the first game, would
12 6 2
the first belonging to B, the se-
7' 7' 7
cond to C, and the third to A; and, therefore,
dividing the sum of the probabilities belonging to
A and B into two equal parts, the probabilities of
winning would be proportional to the numbers 5,
4, 5; and it is 5 to 2 before the play begins that
either A or B win the set, or 5 to 4 that one of
thein who shall be fixed upon win it. 4. If three
gamesters, A, B, and C are engaged in a poule,
and have not time to play it out, but agree to di
vide (s) the sum of the stake and fines, in propor
tion to their respective chances;

share of B, who has got one game;

who should next come in, and

A, who was last beat.

4 -s will be the

7
2

- that of C,

7

M. Bernoulli gives an analytical solution of the same problem, ouly made more general; as not being confined to three gamesters, but extending to any number at pleasure.

Prob. XIII. A and B, two gamesters of equal o terity, play with a given number of balls; a er some

one below the proslambanomenos: the latter he called hypo-proslambanomenos, and denotwhich, being at the foot of the scale, imparted ed it by the letter G or the Greek r, gamma, a name to the whole. The gammnt was divided into three coluinns, the first called molle (flat), the second natural, and the third durum (sharp). It consisted of twenty notes, viz. two octaves and a major sixth. The first ocG, A, B, &c. The second by small letters, ave was distinguished by capital letters, as

Prob. XIV. Two gamesters, A and B, of equal dexterity, are engaged in play, on this condition, that as often as A exceeds B he shall give him one piece of money; and that B shall do the like, as often as A exceeds him; and that they shall not leave off till one has won all the other's money: and now having four pieces, tuo by-standers, R and S, lay a wager on the number of turns in which the game shall be finish, a, b, &c. and the supernumerary sixth by ed; viz. R that it shall be over in ten turns; what double letters, as gg, aa, bb, &c. By the 560 35 word gammut we now generally understand is the value of the expectancy of S? or 1024' the whole present existing scale; and to learn 64 the names and situations of its different notes is of the wager; or it is to that of R as 560 to 464: to learn the gammut.

35

29 64

and subtracting from 1, the remainder will

64 express the probability of the play ending in ten ganes; and it is 35 to 29, if two equal gamesters play together, that there will not be four stakes lost on either side in ten games.

If each player has five pieces, and the wager were, that the game shall end in ten turns, and the dexterity of A were double that of B, the expectof S would be

ancy

3800

6561*

If each gamester have four pieces, and the ratio of the dexterities be required to make it an even wager, that the game shall end in four turns, it will be found that the one must be to the other as

5.974 to 1. And if the skill of either be to that of the other as 13.407 to 1, it is a wager of 3 to 1, that the play will be ended in four games.

If each gamester have four pieces, and the ratio of their dexterities be required to make it an even lay, that the game shall be ended in six turns, the

answer will be found to be as 2.576 to 1.

Prob. XV. Two gamesters, A and B, of equal dexterity, having agreed not to leave off playing till ten games are over: a spectator, R, lays a wager with another, S, that by that time, or before, A shall hate beat B by three games: what is the value of the 350 11 expectancy of R? of the same wager; 1024' 32 or it is to that of S, as 352 to 672. See BasSET, HAZARD, LOTTERY, PIQUET, QUADRILLE, RAFFLING, WHIST, &c.

or

GAMMARUS, in the Fabrician system of entomology, a tribe of the cancer genus. See CANCER.

GAMMER. s. The compellation of a woman corresponding to gaffer.

GA'MMON. s. (gumbone, Italian.) 1. The buttock of a hog salted and dried; the lower end of the flitch (Dryden). 2. A kind of play with dice (Thomson).

GAMMUT, GAMUT, or GAM-UT, in music, the name given to the table or scale laid down by Guido Aretinus, to the notes of which he applied the syllable, ut, re, mi, fa, sol, la. See Guido.

The gammut is also called the harmonical hand, because Guido first arranged his notes upon the figure of a hand.

This scale is an improvement upon the dia gram of the ancients, which was indeed conined within too narrow limits. Guido added four notes above the note hyperboleon, and

'GAN, for began, from 'gin, for begin (Spenser).

GANACHES, in the manage, the two bones on each side of the hinder part of a horse's head, opposite its onset, or neck, which form the lower jaw, and give it motion. Here it is that those glands are placed which are chiefly affected in the strangles or glanders.

To GANCH. v. a. (ganciare, Italian.) To drop from a high place upon hooks, by way of punishment; a practice in Turkey.

GANDER. S. (gaudra, Saxon.) The male of the goose. See ANAS.

GANDERSHEIM, a town of Lower Saxony, in Germany. Lat. 51. 54 N. Lon.

18. 20 E.

To GANG. v. a. (ganghen, Dutch.) To go; to walk an old word not now used, except ludicrously (Spenser. Arbuthnot).

GANG. s. (from the verb.) A pumber herding together; a troop; a company; a tribe (Prior).

GANGES, a river of Asia, which rises by two branches from the mountains of Kentaisse, in the country of Thibet; these two branches take a westerly direction, inclining to the north, for the course of about 300 miles in direct distance, when meeting the great chain or ridge of mount Himmaleh, which extends from Cabul along the north of Hindustan, and through Thibet, the rivers are compelled to turn to the south, in which course they unite their waters, and form what is properly termed the river Ganges. This body of water now forces a passage through the ridge of mount Himmaleh, at the distance, possibly, of 100 miles below the place of its first approach to it, and sapping its foundation, rushes through a cavern, and precipitates itself into a vast bason, which it has worn in the rock, at the hither foot of the mountains. From this second source (as it may be termed) of the Ganges, its course becomes more eastwardly than before, through the rugged country of Sirinagur, until, at Hurdwar, it finally escapes from the mountainous tract, in which it has wandered for about 800 British miles. At Hurdwar it opens itself a passage through mount Sewallick which is the chain of mountains that borders on the level country, on the north of the province of Delhi. After entering Hindustan, it passes