These things will be easily apprehended if it be considered that the word probability includes a double idea; first, of the number of chances whereby an event may happen; secondly, of the number of chances whereby it may either happen or fail. If I say that I have three chances to win any sum of money, it is impossible from the bare assertion to judge whether I am likely to obtain it; but if I add that the number of chances either to obtain it or miss it, is five in all, from this will ensue a comparison between the chances that are for and against me, whereby a true judgment will be formed of my probability of success; whence it necessarily follows that it is the comparative magnitude of the number of chances to happen, in respect of the whole number of chances either to happen or to fail, which is the true measure of probability. To find the probability of throwing an ace in two throws with a single die. The probability of throwing an ace the first time is 1/6; whereof 1/ is the first part of the probability required. If the ace be missed the first time, still it may be thrown on the second; but the probability of missing it the first time is 5/6, and the probability of throwing it the second time is 1/6; therefore the probability of missing it the first time and throwing it the second, is 5/6 X 1/6 = 5/36 and this is the second part of the probability required, and therefore the probability required is in all 1/6 + 5/36 = 11/36. To this case is analogous a question commonly proposed about throwing with two dice either six or seven in two throws, which will be easily solved, provided it be known that seven has 6 chances to come up, and six 5 chances, and that the whole number of chances in two dice is 36; for the number of chances for throwing six or seven 11, it follows that the probability of throwing either chance the first time is 11/36, but if both are missed the first time, still either may be thrown the second time; but the probability of missing both the first time is 25/36, and the probability of throwing either of them on the second is 11/36; therefore the probability of missing both of them the first time, and throwing either of them the second time, is 25/36 X 11/36 = 275/1296, and therefore the probability required is 11/36 + 275/1296 = 671/1296, and the probability of the contrary is 625/1296.