g a w

.540

G A M

four of being up, ami B 6; but B is so much the belter gamester,that his chance against A upon a single throw would be as three totwo; What is the ratio of their chances r Since A wants four, andB six, the game will he ended at nine throws ; therefore, raisea-\-b to the ninth power, and it will be a J -}-f) <fib -j- 36 <Ob- —84u"Z) :l -j- 120 w>b >-f- 126a'i'-f- 84 aW-\- 36-aW + <jab* + IP: calla 3, and b 2, and you will have the ratio of chances in numbers,viz. 1759077 to 194048. A and B play at single quoits, and A isthe host gamester, so that he can give Btwo in three; What is theratio of their chances at a single throw ? Suppose the chances ass to 1, and raise a -j- 1 to its cube, which will he a 3 -)- 3 i 2 -|~3 z + 1.Now since A could give B two out of three, A might undertaketo win three throws running; and consequently the chances inthis case will be as ; 3 to 3: 2 -f-3 z-j-1. Hence z 3 = 3 -f-3s+1 ;

or 2 a 3 = a 3 -f- 3 a 3 -}- 3 a -j- 1. And therefore z y' 2 = a + 1 !and consequently, a aa: ;i— l • The chances, therefore, ate

l

\/2 — 1, and J, respectively. Again, suppose [ have two wa-gers depending, in the first of which L have three to two the bestof the lay, and in the second, seven to four; What is the proba-bility 1 win both wagers ? f. The probability of winning the firstis-i, that is the number of chances 1 have to win, divided by thenumber of all the chances: the probability of winning the secondis t 7 t : therefore, multiplying these two fractions together, the pro-duct will be 44, which is the probability of winning both wagers.Now, this fraction being subtracted from I, the remainder is 4*,which is the probabilit y 1 do not win both wagers: therefore theodds against me are 34 to 21. II. If I would know what the pro-bability is of winning the first, and losing the second, I argue thus;the probability of warning the first is 4, the probability of losingthe second is-i 4 r : therefore multiplying 4 by T 4 r , Die product 44w ill be the probability of my winning the first, and losing thesecond; which being subtracted from 1, there will remain 4i,which is the probability I do not win the first, and at the sametime lose the second. 111. If l would know what tire probabilityis of winning the second, and at the same time losing the first, I saythus: The probability of winning the second is^Zj.; the probabilityof losing the first is therefore, multiplying these two tractionstogether, the product 4* is the probability 1 win the second, andalsojose the first. IV. if I woidd know what the probability is oflosing both wagers, I say, the probability of losing the first is 4,and tiie probability of losing the second y r : therefore the proba-bility of losing them both is-^: which, being subtracted from 1,there remains 44 : therefore, the odds of losing both wagers is 47 tos. This reasoning is applicable to the happening or fading of anyevents that may fall under consideration. Thus if I would knowwhat the probability is of missing an ace four times together withst die, this 1 consider as the failing of four different events. Nowthe probability of missing the first is 4. the second is also 4, thethird 4. and the fourth 4 1 therefore the probability of missing itfour times together is 4 X 4x 4 X 4 = tzVk; which being sub-tracted from 1, there will remain -/iVs *°r the probability ofthrowing it once or ofteiier in four times: therefore the odds ofthrowing an ace in four times, is 671 to 025. But if the flinging ofan uee was undertaken in three times, the probability of missing itthree times would he^x^X 4 = which being subtractedfrom 1, there will remain^ or theprohability of throwing itonce or ofteiier in three times : therefore the odds against throw-ing it in three times are 123 to 91. Again, suppose we wouldknow the probability of throwing an ace once in four times, and nomore: since the probability of throwing it the first time is 4, andof missing it the other three times, is 4 x 4 X 4 * it follows, thatthe probability of throwing it the first time, and missing it theother three successive times, is 4 x 4 X i X 4 — ! because it

is possible to hit every throw as well as the first, it follows, that theprobability of throwing it once in four throws, and missing it theotherthree, is ! which being subtracted from I, there

will remain for the probability of throwing it ouce, and no"more, in four times. Therefore, if one undertake to throw an aceonce, and 110 more, in four times, he has 300 to 796 the word ofthe lav, or five to eight very near. Suppose two events are such,rliat one of them has twice as many chances to come up as the.other; what is the probability that the event, which has the greaternumber of chances to come up, does not happen twice before the

other happens once, which is the case of flinging seven with twodice before four once? Since the number of chances is as two toone, the probability of the first happening before the second is 4,hut the probability of its happening twice before it, is hut 4 X 3or $: therefore it is five to four, seven does not come up twice be-fore four once. But, if it were demanded, what must be the pro-portion of the facilities of the coining np of two events, to makethat which has the most chances come up twice, before the othercomes up once. 5 The answer is 12 to 5 very nearly: whence itfollows, that (he probability of throwing the first before the secondis 44, and the probability of throwing it twice is 44 X )y, or 4f4 5therefore the probability of not doing it is 4J& : therefore the oddsagainst it areas 143 to 144, which comes very near to an equality.Suppose there is a iieap of 13 red cards, and another heap of 13black cards, What is the probability, that, taking one card at aventure out of each heap, I shall take out the two aces ? Theprobability of taking the av.e out of the first heap is-/?, the proba-

A ’ J V' .

bilily of taking the ace out of the second heap is 4,; therefore the

probability of taking out both aces is t 7 x t‘t = 7 tpt> which beingsubtracted from 1, there will remain 4?K! : therefore the oddsagainst me are 168 to 1. In cases w here the events depend on oneanother, the maimer of arguing is somewhat altered. Thus, sup-pose that out of one single neap of !3 cards of one colour 1 shouldundertake to.take out first the ace; and, secondly, the two: thoughthe probability of taking out the ace he-^, and the probability oftaking out the two be likewise ^ : yet, the ace being supposed astaken out already, there will remain only 12 cards in the heap,which will make the probability of taking out the two to be -fj:therefore the probability of taking out the ace, and then the two,will he X T ' T . In this last question the two events have a de-pendence on each other; which consists in this, that one of theevents being supposed as having happened, the probability of theother’s happening is thereby altered, But the case is not so in thetwo heaps of cards. If the events in question ben in number, andbe such as have the same number « of chances by which they mayhappen, and likewise the same number b of chances by which theymay fail, raise a -j- b to the power », And if A and B play toge-ther, on condition that if either one or mere Of the events in ques-tion happen, Ashall win, and B lose, the probability of A’swin-<t -f b\»—b»

ning will be ~^~^Tb \“ ; and that of B’s winning will be 44.71]'!

for wheno + i is actually raised to the power 11; the only term hiwhich « does not occur is the la-t b»; therefore all the terms butthe last are favourable to A. Thus if h = 3, raising a -f- b to thecube « 3 -|- 3a*Ji-4 3rfo 2 + A 1 , all the terms hut A 3 will be favourableto A ; and therefore tlu» p robability of A’s winning w ill he

a 3 + 3 (I'b -f 3 ah- a + Aj 3 —■A 3

“ ——— . an d the probability of B’sw in-

a + b\

, or

A 3

a+b | 3 ’

ning will b But if A a,ul B ^ 6n condl "™> thateither two or more of the events in question happen, Ashall win*but in case one only happen, or none, b shall win ; the piobabm ,

of A’s winning will he a -4- A|“— nab a ^~ , —A“

0 - —prp r—■-; for the only two terms

n + b\

in which aa does not occur are the two last, viz nab a— 1 ami A n -See Chancks. Tire reader may also consult “Clarkes laws 0Chance,” and “ Simpson's Nature and Laws ol Chance,”

Gaming, Laws against. By slat. !G. Car. II. c. 7, if anyperson by playing or belling shall lose more than look at ontime, lie shall not be compellable to pay the same; and the winn®shall forfeit treble the value, one moiety to the king, tlmother to the inforntfer. The statute 0 Ami. e. 14, enacts, that abonds and other securities, given for money won at play, 0money lent at the time to play withal, shall be utterly void: that amortgages and incumbrances of lauds, made upon the same ecu'"sideration, shall be and enure to the heir of the mortgager: thanifanv person at one time loses 101. at play, he may sue the winnt'6and recover it hack by action of debt at law; and, in case m®loser does not, any other person may sue the winner tor treble I' 11stun so bwt; and the pktintilf in either ease may examine the de-fondant himself upon oath and that In any of these suits no ] ,rl "

vii'-'g®