art of playing or practising any game, particularly those of hazard, as cards, dice, tables, &c.

Mr de Moivre, in a treatise de Mensura Sortis, has computed the variety of chances in several cases that occur in gaming, the laws of which may be understood by what follows.

Suppose \( p \) the number of cases in which an event may happen, and \( q \) the number of cases wherein it may not happen, both sides have the degree of probability, which is to each other as \( p \) to \( q \).

If two gamblers, A and B, engage on this footing, that, if the cases \( p \) happen, A shall win; but if \( q \) happen, B shall win, and the stake be \( a \); the chance of A will be \( \frac{p}{p+q} \), and that of B \( \frac{q}{p+q} \); consequently, if they sell the expectancies, they should have that for them respectively.

If A and B play with a single die, on this condition, that, if A throw two or more aces at eight throws, he shall win; otherwise B shall 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 \( a = 1 \), and \( b = 5 \). And, again, since there are eight throws of the die, let \( n = 8 \); and you will have \( \frac{a^n + b^n - nab^n}{n} = \frac{1}{8} \), to \( b^n + nab^n - 1 \); that is, the chance of A will be to that of B, 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 6; but B is so much the better gambler, 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 \( a + b \) to the ninth power, and it will be \( a^9 + 9a^8b + 36a^7b^2 + 84a^6b^3 + 126a^5b^4 + 126a^4b^5 + 84a^3b^6 + 36a^2b^7 + 6ab^8 + b^9 \); call \( a_3 \), and \( b_2 \), and you will have the ratio of chances in numbers, viz., 1759077 to 194048.

A and B play at single quoits, and A is the best gambler, so that he can give B 2 in 3, what is the ratio of their chances at a single throw? Suppose the chances as \( z \) to 1, and raise \( z + 1 \) to its cube, which will be \( z^3 + 3z^2 + 3z + 1 \). 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 will be as \( z^3 \) to \( 3z^2 + 3z + 1 \). Hence \( z^3 = 3z^2 + 3z + 1 \); or, \( 2z^3 = z^3 + 3z^2 - 3z + 1 \). And, therefore, \( z^3 = \frac{1}{\sqrt{2}} \); and, consequently, \( z = \frac{1}{\sqrt{2} - 1} \). The chances, therefore, are \( \frac{1}{\sqrt{2} - 1} \), and 1, respectively.

Again, suppose I have two wagers depending, 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 wagers?

1. The probability of winning the first is \( \frac{3}{5} \), that is the number of chances I have to win, divided by the number of all the chances; the probability of winning the second is \( \frac{7}{11} \); therefore, multiplying these two fractions together, the product will be \( \frac{21}{55} \); which is the probability of winning both wagers. Now, this fraction being subtracted from 1, the remainder is \( \frac{34}{55} \), 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 \( \frac{3}{5} \), the probability of losing the second is \( \frac{4}{11} \); therefore multiplying \( \frac{3}{5} \) by \( \frac{4}{11} \), the product \( \frac{12}{55} \) will be the probability of my winning the first, and losing the second; which being subtracted from 1, there will remain \( \frac{43}{55} \), which is the probability I do not win the first, and at the same time lose the second.

3. If I would know what the probability is of winning the second, and at the same time losing the first, I say thus: the probability of winning the second is \( \frac{7}{11} \); the probability of losing the first is \( \frac{3}{5} \); therefore, multiplying these two fractions together, the product \( \frac{21}{55} \) 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 probability of losing the first is \( \frac{2}{5} \), and the probability of losing the second \( \frac{4}{11} \); therefore, the probability of losing them both is \( \frac{8}{55} \); which being subtracted from 1, there remains \( \frac{47}{55} \); 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. deration. 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 $\frac{5}{6}$, the second is also $\frac{5}{6}$, the third $\frac{5}{6}$, and the fourth $\frac{5}{6}$; therefore the probability of missing it four times together is $\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{625}{1296}$; which being subtracted from 1, there will remain $\frac{671}{1296}$ for the probability 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 undertaken in three times, the probability of missing it three times would be $\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{125}{216}$; which being subtracted from 1, there will remain $\frac{93}{216}$ 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 $\frac{1}{6}$, and of missing it the other three times is $\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{125}{216}$; 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 $\frac{4 \times 125}{1296} = \frac{500}{1296}$; which being subtracted from 1, there will remain $\frac{796}{1296}$ 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 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 happening before the second is $\frac{2}{3}$, but the probability of its happening twice before it is $\frac{2}{3} \times \frac{2}{3} = \frac{4}{9}$; 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 throwing the first before the second is $\frac{12}{17}$, and the probability of throwing it twice is $\frac{12}{17} \times \frac{12}{17} = \frac{144}{289}$; therefore, the probability of not doing it is $\frac{145}{289}$; 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 $\frac{1}{13}$, the probability of taking the ace out of the second heap is $\frac{1}{13}$; therefore the probability of taking out both aces is $\frac{1}{13} \times \frac{1}{13} = \frac{1}{169}$, which being subtracted from 1, there will remain $\frac{168}{169}$: therefore the odds against me are 163 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 $\frac{1}{13}$, and the probability of taking out the two be likewise $\frac{1}{13}$; 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 $\frac{1}{12}$; therefore the probability of taking out the ace, and then the two, will be $\frac{1}{13} \times \frac{1}{12}$.

In this last question the two events have a dependence on each other, which consists in this, that one of the events being supposed 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 $a+b$ 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 will be $\frac{(a+b)^n - b^n}{a+b}$; and that of B's winning will be $\frac{b^n}{a+b}$; for when $a+b$ is actually raised to the power $n$, the only term in which $a$ does not occur is the last $b^n$: therefore all the terms but the last are favourable to A.

Thus if $n=3$, raising $a+b$ to the cube $a^3 + 3a^2b + 3ab^2 + b^3$, all the terms but $b^3$ will be favourable to A; and therefore the probability of A's winning will be $\frac{a^3 + 3a^2b + 3ab^2}{a+b} - b^3$, or $\frac{a^3 + 3a^2b + 3ab^2}{a+b} - b^3$; and the probability of B's winning will be $\frac{b^3}{a+b}$. But if A and B play 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 $\frac{(a+b)^n - na^nb^{n-1}}{n(a+b)}$; for the only two terms in which $aa$ does not occur, are the two last, viz. $na^nb^{n-1}$ and $b^n$.