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goochelen New user Hexagonia 76 Posts |
Hi,
I am looking for a self-working trick I once came across but I can't remember the details nor the name of the author. Perhaps someone can help as I remember it being a great selfworking trick. Steps as far as I recall : 0. setup: the forced card is at a specific location in the deck (position X) 1. magician predicts a that a card Y (selected by the spectator) will be at a specific position (Z) at the end of the trick 2. spectator is asked to remember a number from 1 to 20 and take that many cards from the deck and put them aside (out of sight). 3. magician takes 20 cards and shows them all one-by-one to the spectator who must remember the card that corresponds to his selected number (from step 2). 4. magician puts his 20 cards back together with the cards that remained on the table 5. the spectator's "selected" card is now always at a position Z in the deck regardless of the number he had chosen in step 2. As mentioned I'd appreciate if someone can point me to the origin of this trick. Thanks, |
magicthree Special user 619 Posts |
Paul Gordon, I think. If I get a chance later I'll see if I can find it, can't remember the name off hand.
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goochelen New user Hexagonia 76 Posts |
Hi magicthree,
Thanks in advance : looks promising I am curious if you can find the reference. |
Ferry Gerats Regular user the Netherlands 190 Posts |
Looks like "Where it has to go" from Rick Lax. Method is similar to Paul Gordon's Unfathomable Location. For more variations search for Automatic Placement here at the Café.
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goochelen New user Hexagonia 76 Posts |
Hi Ferry,
Thanks for those pointers and the reference to the underlying principle. Exactly what I was looking for. Excellent. |
stickmondoo Veteran user 306 Posts |
Vietnam Card trick ?
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LoveKey1988 Elite user 443 Posts |
You can find tricks like this in The Last Word on Cards by "Rufus" Steele also..
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The Magic Cafe Forum Index » » All in the cards » » Can't find trick where thought off card is at predetermined position at the end of the trick (0 Likes) |
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