GAMING. 
chances, therefore, arc c> j"— , and 1, re- 
spectively. 
Again, suppose I have two wagers de- 
pending, in the first of which I have 3 to 2 
the best of the lay, and in the second, 7 to 
4, what is the probability I win both wa- 
gers ? 
1, The probability of winning the first is 
|, that is the number of chances I have to 
win, divided by the number of all the. 
chances : the probability of winning the se- 
cond is ^ : therefore, multiplying these two 
fractions together, the product will be §f, 
which is the probability of winning both 
wagers. Now, this fraction being subtract- 
ed from 1, the remainder is ||, which is the 
probability I do not win both wagers: 
therefore the odds against me are 34 to 21. 
2. If I would know what the probability 
is of winning the first, and losing the second, 
I argue thus : the probability of winning 
the first is f, the probability of losing the 
second is £ : therefore multiplying f by ft, 
the product if will be the probability of my 
winning the first, and losing the second ; 
which being subtracted from 1, there will 
remain ff, which is the probability I do not 
win the first, and at the same time lose the 
second. 
S. If I would know what the probability 
is of winning the second, and at the same 
time losing the first, I say thus : the proba- 
bility of winning the second is ■£. ; the pro- 
bability of losing the first is f ; therefore 
multiplying these two fractions together, 
the product if is the probability I win the 
second, and also lose the first. 
4. If I would know what the probability 
is of losing both wagers, I say, the probabi- 
lity of losing the first is f , and the probabi- 
lity of losing the second ft. ; therefore the 
probability of losing them both is ; which 
being subtracted from 1, there remains f|; 
therefore, the odds of losing both wagers is 
47 to 8. 
This way of reasoning is applicable to the 
happening or failing of any events that may 
fall under consideration. Thus, if I would 
know what the probability is of missing an 
ace four times together with a die, this I 
consider as the failing of four different 
events. Now the probability of missing 
the first is f, the second is also f, the third 
f, and the fourth f ; therefore the probabi- 
lity of missing it four times together is f X 
4 X f X | = r$rs ; which being subtracted 
from 1, there will remain for the proba- 
bility of throwing it once or oftener in four 
times ; therefore the odds of throwing an 
ace in four times, is 671 to 625. 
But if the flinging of an ace was under- 
taken in three times, the probability of 
missing it three times would be 4 X 4 X 4 — 
iff : which being subtracted from 1, there 
will remain ^ for the probability of throw- 
ing it once or oftener in three times ; there- 
fore the odds against throwing it. in three 
times are 125 to 91. Again, suppose we 
would know the probability of throwing an 
ace once in four times, and no more : since 
the probability of throwing it the first time 
is 4, and of missing it the other three times 
is 4 X f X |) it follows that the probability 
of throwing it the first time, and missing 
it the other three successive times, is i X 4 
X 4 X | = but because it is possible 
to hit it every throw as well as the first, it 
follows, that the probability of throwing it 
once in four throws, and missing the other 
4 V 1 25 500 
three, is ; which being sub- 
tracted from 1, there will remain T ff| for 
the probability of throwing it once, and no 
more, in four times : therefore, if one un- 
dertake to throw an ace once, and no more, 
in four times, he has 500 to 796 the worst 
of the lay, or 5 to 8 very near. 
Suppose two events are such, that one of 
them has twice as many chances to come 
up as the other, what is the probability that 
the event, which has the greater number of 
chances to come up, does not happen twice 
before the other happens once, which is the 
case of flinging 7 with two dice before 4 
once ? Since the number of chances are as 
2 to 1, the probability of the first happen- 
ing before the second is §, but the probabi- 
lity of its happening twice before it, is but 
f X f or | ; therefore it is 5 to 4 seven does 
not come up twice before four once. 
But if it were demanded what must be 
the proportion of the facilities of the com- 
ing up of two events, to make that which 
has the most chances come up twice, be- 
fore the other comes up once ? The an- 
swer is 12 to 5 very nearly ; whence it fol- 
lows, that the probability of throwing the 
first before the second is -if, and the proba- 
bility of throwing it twice is if X ff, or iff ; 
therefore the probability of not doing it is 
4H : therefore the odds against it are, as 145 
to 144, which comes very near an equality. 
Suppose there is a heap of thirteen cards 
of one colour, and another heap of thir- 
teen cards of another colour, what is the 
