GAMING. 
239 
if A throw two or more aces at eight throws, he fliall 
win ; otherwife B fhall win ; What is the ratio of their 
chances ? Since there is but one cafe wherein an ace 
may turn up, and five wherein it may not, let i , and 
^—5. And again, fince there are eight throws of the 
die, let w=8 ; and you will have — nab'- —i, 
to b"-\-7iab '‘—I : that is, the chance of A will be to that 
of B as 663991 to 10156523, or nearly 2 to 3. 
A and B are engaged at fingle quoits ; and, after 
playing feme time, A wants 4 of being up, and B 6 ; 
but B IS fo much the better gamefter, that his chance 
againft A upon a fingle throw would be as 3 to 2 ; What 
i °the ratio of their cltances ? Since A wants 4, and B 6, 
the game will be ended at nine throws ; tliereiore, raife 
fl-p-Ao the ifinth pov.’er, and it will be 
o7W-l-84iz5i3xm6 a^b'^ a‘*b^ , to + z(> aab'' f 
f,ab^-\-b'^ ; call a 3, and 2, and you will have the ratio 
of chances innumbers, viz. 1759077 to 194048. 
A and B play at fingle quoits, and A is the befi: game- 
■fter, fo that he can give B 2 in 3 : Vv'hat is the ratio of 
their chances at a fingle throw ? Suppofe the chances 
as z'o I, and raife z+i to its cube, which will be 
_ 23 ^^z^+ 3 z-i-i. Now fince A could give B 2 out of 
3, A might undertake to win three throws running ; and 
confequently the chances in this cafe will be as z^ to 3z3 
-^3z+i. Hence 23—2^2; or 2Z^z=z^-]-zz -—32 
-bi ■ And therefore 2 ^\/2=z+1 ; and, conlequently, 2= 
3|/2 —I. The chances, therefore, are ^3/^—^ttd i 
refpeiffively. 
Again, fuppofe I have two wagers depending, in the 
firft of which 1 have 3 to 2 the befi; of the lay, and in the 
fecond 7 to 4 ; What is the probability I win both wagers? 
I. The probability of winning the firfi is|., that is the 
number of chances I have^o win, divided Ijy the num¬ 
ber of all the chances: the probability of winning the 
fecond is .^5-: therefore, multiplyingthefe two fradtions 
together, the product will be -fi, which is the probabi¬ 
lity of winning both wagers. Now, this fradfion being 
fubtradled from i, the remainder is-|A, which is the pro¬ 
bability I do not v/in both wagers : tiierefore the odds 
againfi me are 34 to 21. 2. If I would know what the 
probability is of winning the firfi, and lofing the fecond, 
I argue thus : the probability of winning the firfi is 
the probability of lofing the fecond is -"L: therefore mul¬ 
tiplyingby^, the product w'ill be the probability 
of my winning the firfi, and lofing the fecond ; which 
being fubtradted from i, there will remain -H, which is 
the probability I do not win the firfi, and at the fame 
time lofe the fecond. 3. If 1 would know what thepro- 
bability is of winning the fecond, and at the fame time 
lofing the firfi, I fay thus : The probability of winning 
the fecond is ; the probability of lofing the firfi is |: 
therefore, nuiltiplying thefe two fradtions together, the 
produdi is the probability I win the fecond, and alfo 
lofe the firft. 4. If I would know what the probability 
is of lofing both wagers, I fay, the probability of lofing 
the firfi is f, and the probability of lofing the fecond^^ : 
therefore the probability of lohng them both is ; 
w hich, being fubtradled from i, there remains ; there¬ 
fore, the odds of lofing both wagers is 47 to 8. 
1 his way of reafoning is applicable to the happening 
or failing of any events that may fall under ebnfideration. 
Thus if I would know what the probability isof mifiino- 
an ace tour times together with a die, tliis I confider a*! 
the tailing of four ditferent events. Now the orobabili- 
ty of milling the firfi is f, the fecond is alfo S-,‘the third 
and the fourth A; therefore the probability of milfino- 
It tour times together ii -Ay ?-X ?-X A —. which hein° 
’ r ° I 6 ^ 6 ^ 66-1 296 > " lllCII OeUlg 
Uihtradtea from i, there .viu remain for tlie proba¬ 
bility of tl*rowing it once or ofteuer in lour times : iliere- 
lore the odds of throwing an ace in four times, is 671 
£0 625. ’ ' 
But if thii^ flinging of an ace were undertaken in three 
times, the probability of mifiing it three times would 
which being fubtradled from I, there 
will remain for the probability of throwing it once 
or oftener in three times : therefore the odds againft 
throwing it in three times are 125 to 91. Again, fuppofe 
we would know the probability of throwing an ace once 
in four times, and no more : fince the probability of 
throvdng it the firfi time is i, and of miffing it the other 
three times, is -bXfX l, it follows, that the pr®bability 
of throwing it the hrfi time, and miffing it the other tliree 
liiccelfive times, is-g-X ® Xf X ®=2 ^a? 6 i but becaufe it is 
poffible to hit every throw as well as the firft, it follows, 
that the probability of throwing it once in four throws, 
and milfing it the other three, is ^ which 
1296 1296 
being fubtradled from I, there will remain for the 
probability of throwing it once, and no more, in four 
times. Therefore, if one undertake to throw an ace 
once, and no more, in four times, he has 500 to 796 the 
worft of the lay, or 5 to 8 very nearly. 
Suppofe two events are Aich, that one of them has 
twice as many chances to come up as the other ; what is 
the probability, that the event, which lifts the greater 
number of chances to come up, does not happen twice 
before "the other happens once, which is the cafe of 
flinging7 with,two dice before 4once ? Since the num¬ 
ber of chances is as 2 to I, the probability of the firft 
happening before tlie fecond is I-, but the probabilitv or 
its happening twice before it is but -Ix-I or : therefore 
it is 5 to 4 feven does not come up tv/ice before four 
once. But, if it were demanded, what muftbe the pro¬ 
portion of the facilities of the coming up of two events, 
to make that which has the mofichances come up twice, 
before the other comes up once? Theanfweris, i2t0 5 
very nearly : wlience it follows, that the probability of 
throwing the firfi before the fecond is t^t^d the proba¬ 
bility of throwing it twice is ■j-yX'f|'> o'" ’ therefore 
the probability of not doing it is ; therefore the odds 
againft it are as 145 to 144, which comes very near an 
equality. 
Suppofe there is a heap of thirteen cards of one co¬ 
lour, and another heap of thirteen cards of another co¬ 
lour; What is the probability, tliat, taking one card at 
a venture out of eacli heap, I fhall take out the two aces ? 
The probability of taking rhe ace out of the firft heap is 
■jfj, the probability of taking the ace out of the fecond 
heap is therefore the probability of taking out both 
aces is i’3XT3=T5-qj whicli being fubtradled from i, 
there will remain : therefore the odds againft me are 
168 to I. 
In cafes where the events depend on one another, the 
manner of arguing is fomewhat varied. Tlius, fuppofe 
tiiatout of one fingle heap of thirteen cards of one cOr 
lour, I Ihould engage to take out firfi the ace ; and, fe- 
condly, the two : though the probability of taking out 
the two he likewife-Jj ; yet, the ace being fuppofed as 
taken out already, there will remain only twelve cards 
in the heap, which will make the probability of taking 
Out the two to be-jJg-; therefore the probability of taking 
out the ace, and then the two, will be-J^X-^- Ht this 
laft queftion the two events have a dependence on each 
other; which confilts in this, that one of tlie events be¬ 
ing fuppofed as having happened, the probability of the 
other’s happening is thereby altered. But the cafe is not 
fo in the two heaps of cards. 
If the events in queftion be n in number, and be fuch as 
have the fame number a of chances by which they may 
happen, and likewife the fame number b of chances by 
which they may fail, raife a-^^b to the power n. And if 
A and B play together, on condition that if either one Oj- 
more of the events in queftion happen, A lhaii win, an^j 
B lofe, the probability of A’s winning will be”^^^'*—; 
c-PZ]‘ 
and that of B’s winning will be —-—; for when a-^b is 
I adtually 
