Share page |

226 BACKGAMMON.
adversary must take two throws before he enters his man in vour table, and two throws more be-fore he puts that man into his own table, and three throws more to put into his own table the men which he has abroad—in all, seven throws: and as you have twelve men to bear, these probably will take seven throws in bearing, because you may twice be obliged to make an ace, or a deuce, before you can bear all.
N.B. No mention is made of doublets on either side, that event being equal to each party.
The foregoing case shows it is in your power to calculate very nearly the odds of saving or winning a gammon upon most occasions.
2. Suppose you have three men upon your adversary's ace-point, and five points in your table, and that the adversary has all his men in his table, three upon each of his five highest points ; what is the probability for a gammon ?
Answer. Points.For his bearing three men from his 6 point, is 18
Ditto......................from his 5 point, is 15
Ditto......................from his 4 point, is 12
Ditto......................from his 3 point, is 9
Ditto......................from his 2 point, is 6
Total.... 60
To bring your three men from the adversary's ace-point to the six-point in your table, being for each 18 points, makes in all......54
The remainder is.... 6
And as, besides the six points in your favour, there is a further consideration, that your adversary may make one or two bolts in bearing, you have greatly the probability of savingyour gammon. |
||