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The Magic Cafe Forum Index » » Magical equations » » A Touch of Tarot (0 Likes) Printer Friendly Version

Payner44
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I'm not sure that this is the correct forum, but math is used to accomplish the trick. Problem is, following the instructions to the letter does not provide the expected outcome.

Basically, you have five envelops with a Tarot card placed in each by the spectator. The magi then fans the envelopes in front of the spectator and instructs him/her to select an envelope by it's location counting from top to bottom or as they see it from right to left, and remember that number. The magi then mixes the envelopes in such a way as to not disturb their order. He then gives an example of how the spectator is to try and find their envelope by using #2 as the possible selection and placing the top envelope to the bottom and then the #2 envelope to the bottom. The spectator now does the same thing using his/hers remembered number. Now, here's the issue I need help with. All of this moving of the envelopes supposedly places the spectator's selected envelope in the middle position of the 5. You then use theatrics to eliminate 4 envelopes while getting to the target envelope.

Any help would be much appreciated!!
"I'm not arguing with you, I'm just explaining why you're wrong!"
Russo
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One I have a lot of fun with - is - " pick a number between 2 - 10 // multiply buy 9 // add the 2 numbers // take away 5 // think of the corresponding alphabet (1=a--2=b -- 3=c etc.) //think of a Country that starts with that letter // think of a large animal that starts with the 2nd letter of that Country // think of its color //"there's gray elephants in Denmark" ???????????
leonard
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Payner,
If I understand you post, I think it works. I will use the notation (1,2,3,4,5) to mean 1 on top and 5 on the bottom.
Start with (1,2,3,4,5).
After your demonstration (which must use two moves) you end up with (3,4,5,1,2).
There are now five cases, depending on the spectators choice. I consider all five.

Case #1 = move one card, yields (4,5,1,2,3).
Case #2 = move two cards, yields (5,1,2,3,4).
Case #3 = move three cards, yields (1,2,3,4,5).
Case #4 = move four cards, yields (2,3,4,5,1).
Case #5 = move five cards, yields (3,4,5,1,2). Note that this is the same as moving zero cards.

If you look down the third (or middle) column of the results, you will see the spectators choice. Therefore, their card is in the center of the spread, as desired.
Payner44
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Quote:
On Apr 5, 2020, leonard wrote:
Payner,
If I understand you post, I think it works. I will use the notation (1,2,3,4,5) to mean 1 on top and 5 on the bottom.
Start with (1,2,3,4,5).
After your demonstration (which must use two moves) you end up with (3,4,5,1,2).
There are now five cases, depending on the spectators choice. I consider all five.

Case #1 = move one card, yields (4,5,1,2,3).
Case #2 = move two cards, yields (5,1,2,3,4).
Case #3 = move three cards, yields (1,2,3,4,5).
Case #4 = move four cards, yields (2,3,4,5,1).
Case #5 = move five cards, yields (3,4,5,1,2). Note that this is the same as moving zero cards.

If you look down the third (or middle) column of the results, you will see the spectators choice. Therefore, their card is in the center of the spread, as desired.


Thanks so much for your example. It does work just fine. This made me realize the mistake I was making in the handling!
"I'm not arguing with you, I'm just explaining why you're wrong!"
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