men abroad, although he should lose one point or two in his putting his men into his tables, it is his interest to leave a man upon his adversary’s ace point, because it will prevent his adversary from bearing his men to the best advantage, and at the same time the player will have a chance of the adversary’s making a blot which he may chance to hit. However, if a player finds, upon a throw, that he has a probability of saving his gammon, he should never wait for a blot, as the odds are greatly against his hitting it, but should embrace that opportunity. The following are directions for calculating the odds of saving or winning the gammon:—Suppose the adversary has so many men abroad as require three throws to bring them into his tables, and at the same time that the players’ tables are made up, and that he has taken up one of the adversary’s men, in this case it is about an equal wager that the adversary is gammoned. For, in all probability, the player has borne two men before he opens his tables, and when he bears the third man, he will be obliged to open his size or cinque point. It is then probable, that the adversary is obliged to throw twice before he enters his men in the player’s tables, twice before he puts that man into his own tables, and three throws more to put the men which are abroad into his own tables, in all seven throws. Now the player having twelve men to bear, he may be forced to make an ace or a deuce twice before he can bear all his men, and consequently will require seven throws in bearing them; so that, upon the whole, it is about equal whether the adversary is gammoned or not. Suppose a player has three men upon his adversary’s ace point, and five points on his own tables, and that the adversary has all his men in his tables, three upon each of his five highest points, has the player a probability of gammoning his adversary or not? POINTS. For bearing three men from his sixth point is 18 From his fifth point 15 From his fourth point 12 From his third point 9 From his second point 6 —— In all 60 Bringing his three men from the adversary’s ace point, to his size point in his own tables, being eighteen points each, and making together 54 —— There must remain 6