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 au ace four times together with a die, this I consider as the failing of four different events. Now the proba bility of missing the first is the se cond is also 6, the third 1, and the fourth Z.; therefore the probability of missing it four times together is X 1, X f X 2.= ; which being subtract ed from 1, there will remain for the probability of throwing it once or of tener 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 X 5 X a- 5 = which being subtract 216' ed from 1, there will remain for the probability of throwing it once or oftener in three times; therefore 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 1, and of missing it the other three times is iXix it follows that the probability of throwing it the first time, and missing it the other three suc cessive times, is iXiX4Xi 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 three, is 4 x 125 which being sub.
1296 = 1296' being tracted from 1, there will remain 32st. 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 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 proba bility that the event, which has the great er number of chances to come up, does not happen twice before the other hap pens 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 happening fore the second is 4, but the probability of its happening twice before it, is but V w 2 X 2 or 4 ' • therefore it is 5 to 4 se ven 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 T7, and the probability of throwing it twice X 4-?, or In; therefore the probability of not doing it is 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 • therefore the probability of taking out both aces is . X . which being subtracted from 1, there will re main 41.4: therefore the odds against me are 168 to 1.
In cases where the events depend on one another, the manner of arguing is somewhat 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 yet the ace being supposed as taken out already, there will remain only twelve cards in the heap, which will make the probabili ty of taking out the two to be there fore 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 be ing supposed as ll'aving happened, the probability of the other's happening is thereby altered. But the case is not so in the two heaps a cards.
If the events in question be n in num ber, and be such as have the same num ber a of chances by which they may hap pen, and likewise the same number b of chances by which they may fail, raise a -I- 6 to the power n. And if A and B play together, on condition that if either one or more of the events in question hap pen, A shall win, and B lose, the probabili ty of A's winning will be ; and that of B's winning will be _ bn -; for when a b is actually raised to the power 7i, the only term in which a does not occur is the last 6 n ; therefore all the terms but the last are favourable to A.