Page:Magic oracle, or, Conjuror's guide.pdf/19

 by ten, and to that product add the number of the third dice. From the total let there be subtracted 250, and the figures of the number that remain will answer to the points of the three dice, as they stand on the table

Suppose the points of the three dice thrown on the table to be Then the double of the first dice will be To which add

That sum multiplied by five will be To which add the number of the middle dice

And multiply the sum by

To that product add the number of the third dice

Subtract

The number of the dice and the order in which they stand.

4, 6, 2. 8 5 — 13 5 — 65 6 — 71 10 — 710 2 — 712 250 — 462

In the first place we shall suppose that the dice are of the ordinary kind, namely, having six faces marked with the numbers 1, 2, 3, 4, 5, 6; and we shall analyze some of the first cases of the problem, that we may proceed regularly to those that are more intricate.

I. It is proposed to throw a determinate point, 6 for instance with one die.

Here it is evident, that as the die has six faces, one of which is only marked six, and as any one of them may as readily come up as another, there are five chances against the person who wishes to throw a six at one throw, and only one in his favour.