The Final 3- Amazing Card Trick

[easyrider]

Ok , I really haven't figured this out yet. It does work with a regular deck. Whats the dealio here ?

http://www.flixxy.com/math-magic-card-trick.htm

 

[Karen G]

Wow! That really was amazing. I'm easily baffled and never can figure out magic tricks. I hope someone can explain it.

 

[Elan]

Actually, not that hard to figure out if you write down what the final stack looks like (after removing the 4 cards):

From the top to the bottom:

5 cards
card A
15 cards
card B
15 cards
card C
10 cards

So cards A,B and C reside in positions 6, 22 and 38 (from the top) all even numbers, so they won't be seen flipping the odds face up. After one pass thru, cards A, B and C now reside in spots 6, 14, and 22, remembering that he's reversed (top to bottom) by going thru the cards once. Again, all evens, so they won't be seen. After the 2nd pass thru, cards A, B and C reside in positions 2,6 and 10. Same story. After the 3rd pass they reside in positions 2,4 and 6, etc.

 

[easyrider]

That must be it. I thought it was strange enough to pratice with my own deck and kind of thought it was a numbers deal and not the devil.

 

[Karen G]

Actually, not that hard to figure out if you write down what the final stack looks like (after removing the 4 cards):

From the top to the bottom:

5 cards
card A
15 cards
card B
15 cards
card C
10 cards

But, he says to have the person you're doing the trick for to randomly cut the cards in the piles and then he puts the A, B, & C cards in the deck. It seems like cutting the cards randomly would mess up the number of cards between the A, B, & C cards.

 

[vckempson]

But, he says to have the person you're doing the trick for to randomly cut the cards in the piles and then he puts the A, B, & C cards in the deck. It seems like cutting the cards randomly would mess up the number of cards between the A, B, & C cards.

The cut is random but ends up the same no matter how you cut them. Look at what's happening and you'll see why.

1. you cut pile B and place 5 cards out 15 on to pile C, (leaving 10 cards in pile B)
2 you place your card on top of pile B
3. you place all of pile b onto pile C

When you put pile B onto pile C, you've got the 10 cards under the one you picked plus the 5 you cut, or a total of 15. No matter how many you cut, once you move pile b over to pile C, you've got all 15 cards of pile B under your choicen card which is now all on top of pile C. Does that make sense? While the cut is random, the placement of the chosen card will always be the same in the final pile.