G A M 
G A M 
SI 1 
win ; what is the ratio of their chances ? Since 
there is but one case wherein an ace may turn 
up, and five wherein it may not, let a — l, and 
f— 5. And again, since t&ere are eight throws 
ttf the die, let n = 8 ; and you will have a -j~ bi 
_ b' 1 — nab’ 1 — ] , to b* -j- nab’' — 1 : that is, 
the chance of A will be to that of 13, as 663991 
to 10156525, or nearly as 2 to 3. 
A and B are engaged at single quoits, and 
, after playing some time, A wants 4 of being up, 
and B G ; but B is so much the better gamester, 
that his chance against A upon a single throw 
would be as 3 to 2 ; what is the ratio of their 
chances? Since A wants 4, and B 6, the game 
will be ended at nine throws ; therefore, raise 
| a _1_ b to the ninth power, and it will be a' -j- 
9 a*b 4- 36V ’bb 4- 84TA ! -f- 126 VA 3 -f 126TA’, to 
* g4,p//_j_ ?,C,aab' -\- Gab 1 -f- b ‘ ; call a 3, and b 2, 
I and you will have the ratio of chances in num- 
bers, viz. 1759077 to 194048. 
A and B play at single quoits, and. A is the 
ji best gamester, so that he can give B 2 in 3, what 
i is the ratio of their chances at a single throw ? 
I Suppose the chances as z to 1, and raise z -f- 1 
i to its cube, which will be -s’ -j- 3z" -{- 3s 1. 
I Now, since A could give B 2 out of 3, A might 
undertake to win three throws running ; and, 
consequently, the chances in this case wsil be as 
s ! to 3z 2 -J- 3z -j- 1 . Hence s J = 3s 2 -f- 3s -f 1 ; 
or, 2s 3 = z 5 -f- 3 z 2 3z 4- 1. And there- 
fore, z^/2 = z -j- 1 *, and, consequently, z = 
1 1 
. The chances, therefore, are - 
:; % /2-l V--- 1 
j and 1, respectively. 
Again, suppose I have two Wagers depend- 
1 ing/io the first of which I have 3 to 2 the best 
I of the lay, and in, the second 7 to 4, what is the 
j probability I win both wagers ? 
1. The probability of winning the first is 4, 
that is, the number of chances l have to win, 
divided by the number of all the chances : the 
probability of winning the second is _ 7 _ : there- 
| fore, multiplying these two fractions together, 
! the product will be 2A, which is the probability 
of winning both wagers. Now, this fraction 
j being subtracted from 1, the remainder is f-f> 
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 
j of winning the first, and losing the second, I 
argue thus : the probability of winning the first 
is A, the probability of losing the second is AL. ; 
j therefore multiplying A by T 4 T , the product 14 
will be the probability of my winning the first, 
I and losing the second ; which being subtracted 
! from 1, there will remain 4 A, which is the pro- 
bability I do not win the first, and at the same 
j time lose the second. 
3. If I would know what the probability is 
of winning the second, and at the same time 
I losing the first, I say thus: the probability of 
; winning the second is _ 7 _. ; the probability of 
1 l-osing the first is |. : therefore, mutiplying these 
i two fractions together, the product ii is the 
1 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 probability of 
Rising the first is A, and the probability of losing 
the second _A_; therefore, the probability of 
losing them both is JL ; which being subtracted 
from 1, there remains AA : 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 
G A M 
times together with a die, this I consider as the 
failing of four different events. Now the pro- 
bability of missing the first is A, the second is 
also 4, the third A, and the fourth A ; therefore 
6 . . ^ , . .6 
the probability of missing it four time3 together 
is A X A X A X A — _AlA_ : which being siib- 
6 6 ^ 6^6 1296 . b 
tracted from 1, there will remain J>_ 7 A. for the 
129s 
probability of throwing it once or oitener 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 undertaken 
in three times, the probability of missing it three 
times would be A X A X' A — A 2 A ; which be- 
0 6 6 216 
mg subtracted from 1, there will remain JLA. 
for the probability of throwing it once or oftener 
in three times; therefore the oddsagi’nst throw- 
ing it in three times arc 125 to 91. 
Again, suppose we 'would know the probabi- 
lity of throwing an ace once in four times, and 
no more : since the probability of throwing it 
the first time is A and of missing it the other 
three times is A X — X — , it follows that the 
6 6 . 6 . 
probability of throwing it the first time, and 
missing it" the other three successive times, is 
AX 5 y 5 X s — _AAA_; but because it is 
possible to hit it every throw as well as tne first, 
it follows, that the probability of throwing it 
once in four throws, and missing the other three, 
. 4 X 125 .500 , . . . 
is — : which being subtracted 
1296 1296 
from 1, there will remain „ LAA. for the proba- 
bility 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 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 fling- 
ing 7 with two dice before 4 once ? Since the 
number of chances are as 2 to 1, the probability 
of the first happening before the second is 
but the probability of its happening twice be- 
fore it, is but x j, or A ; 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 coming up of 
two events, to make that which has the most 
chances come up twice, before the other comes 
up once, the answer is 12 to 5 very nearly : 
whence it follows, that the probability of throw- 
ing the first before the second is A 2 ., and the 
probability of throwing it twice is AA x AA> or 
A. 44 . ; therefore, the probability of not doing it 
is A 4 A : 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 thirteen cards 
of another colour, what is the probability that, 
taking one card at a venture out of each heap, 
I shall take out the two aces ? 
The probability of taking the ace out of the 
first heap is the probability of taking the 
ace out of the second heap is —A.; therefore the 
probability of taking out both aces is x -A 
— _A_ , which being subtracted from 1, there 
16 9 ° 
will remain AAA : therefore the odds against me 
16 9 
are 168 to 1. 
In cases where the events depend on one an- 
other, the manner of arguing is somewhat al- 
tered. Thus, suppose that out of one single 
heap of thirteen cards of one colour, 1 should 
undertake to take out first the ace; and, second- 
ly, the two : though the probability of taking 
out the ace be _A_, and the probability of taking 
1 J 5 K t 
out the two be likewise —A. ; yet, the ace being 
supposed a3 taken out already, there will remain 
only twelve cards in the heap, which will make 
the probability of taking out the two to be • 
therefore the. probability of taking out the ace, 
and then the two, will be _X_ X 
In this last question the two events have a de- 
pendance on each other, which consists in this, 
that one of the events being supposed as having 
happened, the probability of the other’s happen-* 
ing is tl'.ereby altered. But the case is not so in 
the two heaps of cards. 
If the events in question be n in number, and 
be such as have the same number a of chances 
by which they may happen, and likewise the 
same number b of chances by which they may 
fail, raise a -)- b to the power «, And if A and 
B play together, on condition that if either one 
or more of the events in question happen, A 
shall win, and B lose, the probability of A’s 
winning will be 
winning will be 
and that of B’s 
b K 
— — ; for when a -J- b is ac- 
tually raised to the power the only term in 
which a does not occur is the last b ■■ therefore, 
all the terms but the last are favourable to A. 
Thus,’ if n — 3, raising «-f 4 to the cube 
a' _j_ Sa 2 b -(- 3:1 b 1 4- b\ all the terms but A 3 will 
be favourable to A ; and therefore the probabi- 
a* 4- 3 a 2 b -f- 3ab 2 
lity of A’s winning will be — 
v+b) 3 
a b> 1 
a -j- b I 
and the probability of B’s 
winning will be — — . But if A and B play 
a -j- A] 3 
on condition, that if either two or more of 
the events in question happen, A shall win ; 
but in case one only happen, or none, B shall 
win ; the probability of A’s winning will be 
Z+$ n -nab a ~ 1 -A" , . 
— ; for the only ,tw® 
r »« 
n -j- b ) 
terms in which aa does not occur, are the tw<s 
last, viz. nab” and A . See Chance. 
GAMMONING, among seamen, denotes 
several turns of rope taken round the bow- 
sprit, and reeved through holes in knees oi 
the head, for the greater security of the bow- 
sprit. 
GAMMUT, inmusic, the name given to the 
table or scale laid down by Guido, and to the 
notes of which he applied the monosyllables 
at, re, mi, fa, sol, la. Having added a note 
below the proslambanomenos, or lowest tone 
of the antients, he adopted for its sign the 
gamma, or third letter of the Greek alphabet ; 
and hence his scale was afterwards called 
gammut. This gammut consisted of twenty 
notes, viz. two octaves and a major-sixth. 
The lirst octave was distinguished by capital 
letters, as G, A, B, &c. the second by small 
letters, g, a, b, &c. and the supernumerary 
sixth by double letters, as gg, aa, bb, &c. 
By the word gammut, we now generally un- 
derstand the whole present existing scale; 
and to learn the names and situations of its 
different notes is to learn the gammut. It, 
however, sometimes simply signifies the low- 
est note of the Guidonian or common com- 
pass. 
/ 
