TO GUESS CARDS SELECTED BY SEVERAL PERSONS.
Show as many cards to each person as there are persons to choose; that is to say, 3 to each, if there are three persons. When the first has thought of one, lay aside the three cards in which he has made his choice. Present the same number to the second person, to think of one, and lay aside the three cards in the like manner. Having done the same in regard to the third person, arrange all these cards in three rows, with their faces turned downwards, and then put them together in order. If you take the 3 first, and present them successively to the different persons, and do the same thing with the others, you may easily guess the cards, by observing, that the card thought of by each person will have the same place among the cards as the person has in regard to the other two; that is to say, the card thought of by the first person will be first of that packet in which he discovered it; that thought of by the second will be the second in the packet where he recognized it; and that of the third will be the last and in the last packet.
The operation is exactly the same when the number of persons is greater. If, instead of 3, there are 4 or 5 persons, four or five cards must be presented to each.
A NEW THREE-CARD TRICK.
As it is necessary that the cards presented should be distinguished, we shall call the first A, the second B, and the third C. Let the persons, whom we shall distinguish by first, second, and third, choose privately whichever of the cards they think proper, and when they have made their choice, which is susceptible of six variations, give the first person 12 counters, the second 24, and the third 36: then desire the first person to add together the half of the counters of the person who has chosen the card A; the third of those of the person who has chosen B; and the fourth part of those of the person who has chosen C; and ask the sum, which must be either 23 or 24, 25 or 27, 28 or 29, as in the following table:
| First. | Second. | Third. | Sums. |
|---|---|---|---|
| 12 | 24 | 36 | |
| A | B | C | 23 |
| A | C | B | 24 |
| B | A | C | 25 |
| C | A | B | 27 |
| B | C | A | 28 |
| C | B | A | 29 |
This table shows, that if the sum is 25, for example, the first person must have chosen the card B, the second the card A, and the third the card C; and that, if it be 28, the first person must have chosen the card B, the second the card C, and the third the card A; and so of the rest.
TO TELL THE SPOTS ON ALL THE BOTTOM CARDS OF SEVERAL HEAPS.
Arrange each heap of cards in such a manner that the spots on the bottom one, added to the cards above it, may always amount to twelve; continue to make as many heaps as possible, in the manner above prescribed, and place the remaining cards on one side. Then separate in your mind four heaps, and multiply the heaps which remain, after these are deducted, by 13; this product, added to the number of cards, will be that of the spots required. We give the solution of this problem by an analysis in “To Guess the Spots on a Card,” p. [253].