GAMING. 
probability that, taking one card at a ven- 
ture 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 J- ; there- 
fore the probability of taking out both aces 
is x x tj = ts 7 > wfiich being subtracted 
from 1, there will remain iff : therefore the 
odds against me are 168 to 1. 
In cases where the events depend on one 
another, the manner of arguing is some- 
what altered. Thus, suppose that out of 
one single heap of thirteen cards of one 
colour I should undertake to take out first 
the ace ; and, secondly, the two : though the 
probability of taking out the ace be + and 
the probability of taking out the two be 
likewise T ’ 5 ; yet the ace being supposed 
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 
T 'j ; therefore the probability of taking out 
the ace, and then the two, will be x 
In this last question the two events have 
a dependence on each other, which consists 
in this, that one of the events being sup- 
posed as having happened, the probability 
of the other’s happening is thereby 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 u + 5 to the 
power n. 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 
a-\-E\ n —b n 
will be ; and that of B’s win* 
a-f-£| 
b n 
nmgwill be for when a + b is ac- 
tually raised to the power n, the only term 
in which a does not occur is the last bn ; 
therefore all the terms but the last are fa- 
vourable to A. 
Thus, if n = 3, raising a -(- b to the cube 
<r + 3 « 2 6 + 3 « b 1 -|- b 3 , all the terms but 
b 1 will be favourable to A ; and therefore 
the probability of A’s winning will be 
a 3 +- 3a 2 b — |— 3 ah 1 ^ a ■+ 6| 3 — b 3 
« + + ’ a — j— 6l 3 ’ 
the probability of B’s winning will be 
and 
But if A and B play on con- 
dition, 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 
- ; for the only two 
terms in which a a does not occur, are the 
two last, viz. nab ” ~ 1 and b ", See Simp- 
son’s “ Nature and Laws of Chance.” We 
shall now add a table that may be useful 
to persons not skilled in mathematics, and 
which is applicable tp many subjects ; - 
TABLE, 
Shewing the Odds of Winning in any Game, when the Number of Games wanting 1 
does not exceed Six, and the Skill of the Contenders is equal. 
Games 
wanting. 
Odds of 
winning. 
Games 
wanting. 
Odds of 
winning. 
Games 
wanting. 
Odds of 
winning. 
1, 2 ... 
.. 3:1 
2, 3 ... 
.. 11: 5 
3, 5 .... 
.. 99: 29 
1, 3 ... 
.. 7:1 
2, 4 ... 
.. 26: 6 
3, 6 .... 
..219: 37 
1, 4 ... 
.. 15:1 
2, 5 ... 
.. 57: 7 
4, 5 .... 
..163: 93 
1, 5 ... 
2, 6 ... 
4, 6 .... 
..382:130 
1, 6 ... 
.. 63:1 
3, 4 ... 
.. 42 ; 22 
5, 6 .... 
The above proportions are found by the 
binomial theorem in a very easy way. 
Suppose the games wanting 1 and 5 raise 
a -(- 6 to the 5th power being the number 
of games which must determine the bet. 
a = b in this case, as the skill is equal ; 
a 5 -)- 5, a 4 6 -)- 10, a 3 b 1 + 10, a 2 5+ 5, a i> 4 + 
A’,~the first five coefficients are tlie chances 
of him who has l game to get, viz. 1 -J- 
+ 10 + 10 + 5 = 31, and the other, viz. 
% the chance of him who has five to get. 
Suppose the games wanting are 2 and 5, 
hi* 
A 
