An Incomprehensible Divination
UNDER this title I have pleasure in describing what is, in my opinion, a very clever non-sleight-of-hand trick. It has never yet been published, and is, in fact, absolutely
unknown. It is so simple that a child can learn it in five minutes, yet it can be exhibited to experts for hours without the slightest fear of detection. It is one of the very few
tricks that can be repeated ad lib.
In effect it is as follows:
Eleven cards are placed face downward on the table in a heap. The performer is now securely blindfolded, and, if desired, can be led into another room. In his absence a spectator
cuts the cards, and after counting the lower portion replaces them on what was originally the upper portion.
The performer now enters the room, still blind-folded, waves his hands over the cards, and immediately picks one of them out of the packet, which upon examination proves to have
the number of pips or spots that corresponds with the number of cards counted.
For instance, suppose four cards were removed, the performer would unhesitatingly turn up, say, the four of clubs. Supposing that no cards were re-moved, and the packet was left
in its original state, the knave would be turned up by the artiste. Not only can the performer be blindfolded, but a thick cloth or handkerchief can be covered over the packet, and
yet the card with the proper number of pips is produced.
This trick can be repeated as often as desired without any rearrangement of the pack, and it is this part of the experiment that mystifies conjurers unacquainted with the modus
operandi.
For the performance of this excellent drawing-room trick, the performer must previously arrange eleven cards as follows:
Place a knave face downward, and on the top an ace, then a deuce, then a tray, and so on to the ten, and the cards must be kept in this order throughout the trick. They can,
however, be cut as often as desired before you start the experiment; but as the packet is placed on the table the performer must manage to catch a glimpse of the bottom card, as this
forms the key to the whole mystery.
Supposing the bottom card is the four spot, the cards will be arranged one on the top of the other, as in first table:
Now, no matter how many cards the spectator moves, all that the performer has to do is to show the fourth card down from the top, and this will have the number of spots that
corresponds with the number of cards removed.
We will suppose a spectator moved seven cards, which would leave the cards as in second table:
The performer now shows the fourth card from the top, which is a seven-spot. Now, without rearranging the cards, or even looking at the bottom one, the trick can be repeated.
The artiste remembers that the card just shown -a seven-spot - was fourth from the top, and he, therefore, knows that the third card from the top must be the eight-spot, the
second the nine-spot, and the top card the ten-spot, leaving the jack at the bottom.
A certain number of cards are again moved, and, as in all cases when the knave is at the bottom of the packet, all the performer has to do is to turn the cards over, and on the
bottom card will be found the correct number of spots.
Should the ace be at the bottom of the pack, then the top card will always denote the answer.
If the two-spot be at the bottom, the second card from the top will denote the answer, and so on all through.
If no cards are moved, say, when the cards are in the position shown in the first table, it is apparent that the performer would turn up the knave, this being fourth from the top,
so that if no cards are shifted the knave always turns up.
After five minutes' study, this trick will be found quite easy of accomplishment.
|
|
 |
|
|