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

Gaming has at all times been looked upon as a thing of pernicious consequence to the commonwealth; and is therefore severely prohibited by law. It is considered as a practice generally intended to supply, or retrieve, the expenses occasioned by LUXURY: it being a kind of tacit confession, that the company engaged therein do, in general, exceed the bounds of their respective fortunes; and therefore they cast lots to determine upon whom the ruin shall at present fall, that the rest may be saved a little longer. But taken in any light, it is an offence of the most alarming nature, tending by necessary consequence, to promote public idleness, theft, and debauchery, among those of a lower class; and, among persons of a superior rank, it hath frequently been attended with the sudden ruin and desolation of ancient and opulent families, and abandoned prostitution of every principle of honour and virtue, and too often hath ended in self-murder. To restrain this pernicious vice among the inferior sort of people, the statute 33 Hen. VIII. c. 9. was made; which prohibits to all but gentlemen, the games of tennis, tables, cards, dice, bowls, and other unlawful diversions there specified, unless in the time of Christmas, under pecuniary pains and imprisonment. And the same law, and also the statute 23 Geo. II. c. 24. inflict pecuniary penalties, as well upon the matter of any public house wherein servants are permitted to game, as upon the servants themselves, who are found to be gaming there. But this is not the principal ground of modern complaint; it is the gaming in high life that demands the attention of the magistracy; a passion to which every valuable consideration is made a sacrifice, and which we seem to have inherited from our ancestors, the ancient Germans; whom Tacitus describes to have been bewitched with the spirit of play to a most exorbitant abitant degree. "They addict themselves (says he) to dice (which is wonderful) when sober, and as a serious employment, with such a mad desire of winning or loosing, that, when strip'd of every thing else, they will stake at last their liberty, and their very selves. The loser goes into a voluntary slavery; and, though younger and stronger than his antagonist, suffers himself to be bound and sold. And this perseverance in so bad a cause they call the point of honour: ea est in re prava peruvicacia, isth fidelis vacant." One would almost be tempted to think Tacitus was describing a modern Englishman. When men are thus intoxicated with so frantic a spirit, laws will be of little avail: because the same false sense of honour that prompts a man to sacrifice himself, will deter him from appealing to the magistrate. Yet it is proper that laws should be, and be known publicly, that gentlemen may consider what penalties they wilfully incur, and what a confidence they repose in sharpers; who, if successful in play, are certain to be paid with honour, or, if unsuccessful, have it in their power to be still greater gainers by informing. For, by flat. 16 Car. II. c. 7. if any person by playing or betting shall lose more than 10l. at one time, he shall not be compelled to pay the same; and the winner shall forfeit treble the value, one moiety to the king, the other to the informer. The statute 9 Ann. c. 14. enacts, that all bonds and other securities, given for money won at play, or money lent at the time to play withal, shall be utterly void: that all mortgages and encumbrances of lands, made upon the same consideration, shall be and endure to the heir of the mortgager: that, if any person at one time loses 10l. at play, he may sue the winner, and recover it back by action of debt at law; and, in case the loser does not, any other person may sue the winner for treble the sum to lost; and the plaintiff in either case may examine the defendant himself upon oath: and that in any of these suits no privilege of parliament shall be allowed. The statute farther enacts, that if any person cheats at play, and at one time wins more than 10l. or any valuable thing, he may be indicted thereupon, and shall forfeit five times the value, shall be deemed infamous, and suffer such corporal punishment as in case of wilful perjury. By several statutes of the reign of King George II. all private lotteries by tickets, cards, or dice, (and particularly the games of faro, basset, ace of hearts, hazard, passage, rolly poly, and all other games with dice, except backgammon), are prohibited under a penalty of 200l. for him that shall erect such lotteries, and 50l. a-time for the players. Public lotteries, unless by authority of parliament, and all manner of ingenious devices, under the denomination of sales or otherwise, which in the end are equivalent to lotteries, were before prohibited by a great variety of statutes under heavy pecuniary penalties. But particular descriptions will ever be lame and deficient, unless all games of mere chance are at once prohibited; the invention of sharpers being swifter than the punishment of the law, which only hunts them from one device to another. The flat. 13 Geo. II. c. 19. to prevent the multiplicity of horse races, another fund of gaming, directs, that no plates or matches under 50l. value shall be run, upon penalty of 200l. to be paid by the owner of each horse running, and 100l. by such as advertise the plate. By statute 18 Geo. I. c. 34. the statute 9 Ann. Gaming. is farther enforced, and some deficiencies supplied: the forfeitures of that act may now be recovered in a court of equity; and, moreover, if any man be convicted, upon information or indictment, of winning or loosing at any fitting rel. or 20l. within 24 hours, he shall forfeit five times the sum. Thus careful has the legislature been to prevent this destructive vice: which may shew that our laws against gaming are not so deficient as ourselves and our magistrates in putting those laws in execution.

Chance, or Hazard, in GAMING. Hazard, or chance, is a matter of mathematical consideration, because it admits of more and less. Gamesters either set out upon an equality of chance, or are supposed to do so. This equality may be altered in the course of the game, by the greater good fortune or address of one of the gamesters, whereby he comes to have a better chance, so that his share in the stakes is proportionably better than at first. This more and less runs through all the ratios between equality and infinite difference, or from an infinitely little difference till it come to an infinitely great one, whereby the game is determined. The whole game, therefore, with regard to the issue of it, is a chance of the proportion the two shares bear to each other.

The probability of an event is greater or less, according to the number of chances by which it may happen, compared with the number of all the chances by which it may either happen or fail.

M. 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 gamesters, 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 dice, 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 \( a^b + b^n - n a b^{n-1} \), to \( b^n + n a b^{n-1} \); that is, the chance of A will be to that of B as 663991 to 156725, 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 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 \( a+b \) to the ninth power, and it will be \( a^9 + 9 a^8 b + 36 a^7 b^2 + 84 a^6 b^3 + 126 a^5 b^4 + 126 a^4 b^5 \) to \( 84 a^3 b^6 + 36 a a^2 b^7 + 6 a b^8 + b^9 \): call a 3, and b 2, and you will have the ratio of chances in numbers, viz. 1759077 to 194948. G A M A and B play at single quoits, and A is the best gamester, so that he can give B 2 in 3: What is the ratio of their chances at a single throw? Suppose the chances as \( x \) to 1, and raise \( x+1 \) to its cube, which will be \( x^3 + 3x^2 + 3x + 1 \). 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 \( x^3 \) to \( 3x^2 + 3x + 1 \). Hence \( x^3 = 3x^2 + 3x + 1 \); or \( 2x^3 - x^2 + 3x^2 - 3x + 1 \). And therefore \( x \sqrt{2} = x + 1 \); and, consequently, \( x = \frac{1}{\sqrt{2} - 1} \). 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 \( \frac{4}{7} \): therefore multiplying \( \frac{3}{5} \) by \( \frac{4}{7} \), the product \( \frac{12}{35} \) will be the probability of my winning the first and losing the second; which being subtracted from 1, there will remain \( \frac{23}{35} \), 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 \( \frac{2}{5} \): therefore, multiplying these two fractions together, the product \( \frac{14}{55} \) 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 \( \frac{4}{7} \): therefore the probability of losing them both is \( \frac{8}{35} \): which, being subtracted from 1, there remains \( \frac{27}{35} \): 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 \( \frac{5}{6} \), 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 \( \frac{91}{216} \) 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 \( \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \), it follows, that the probability of throwing it the first time, and missing it the other three successive times, is \( \frac{1}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{125}{1296} \); but because it is possible to hit every throw as well as the first, it follows, that the probability of throwing it once in four throws, and missing it 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 is 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 \( \frac{1}{3} \times \frac{1}{3} \) or \( \frac{1}{9} \); therefore it is 5 to 4, even does not come up twice before 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 \( \frac{12}{17} \times \frac{12}{17} \), or \( \frac{144}{289} \); 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 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 \( \frac{1}{13} \), 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 \( \frac{a+b^n}{a+b} \); 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 \( \frac{a^3+3a^2b+3ab^2}{a+b} \), or \( \frac{a+b^3}{a+b} \); 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; the probability of A's winning will be \( \frac{a+b^n-nab^{n-1}}{a+b^n} \); for the only two terms in which \( a \) does not occur are the two last, viz. \( nab^{n-1} \) and \( b^n \).