You`re putting one card on top of a stack of 10 cards (pile1).

When you cut pile 2 and put it on top of the first card, then place the second car on top of pile 2, then placing that stack on top of the other, you are simply putting all of pile 2 on top of your first card and then placing the second card on top.

You are cutting the piles and then putting them back together. You are just doing it out of sequence.

Works like this: cutting the pile is useless. if you pay attention, you will place the cards like this: 10 first card 15 second card 15 third card 9 cards Then you take 4 cards from the top and put them at the bottom. The amount of cards in between of the "special" cards it always one below 2^x when you show the cards, which garantuess, that you will never show the "special" cards. Finished...

The cutting is a misdirection... if you watch carefully you`ll notice he doesn`t put the second card on pile #1 after pile #2 has been cut, he puts it on TOP of pile 2. So the full pile 2 will still be on top of card #1. Ditto for pile 3. Pretty clever though!

Move four cards to the bottom? I`m not sure why folks are having trouble with this trick, but I can tell you that it`s much easier to make the first pile fourteen instead of ten and skip the "four cards to the bottom" part.

Imagine the top card of a deck as card 1 and the bottom as card 52. For example the 3rd card from the top is 3. This is very important.

In this case the cards are placed with 15 cards in between them,as long as protocol is followed, no matter what. In the position of the deck they are always cards 6,22,and 38 check me if I`m wrong.

Then using alternation we eliminate the odd numbered cards and reverse the order of remaining even cards so now starting with the top, the cards are 52,50,48...2.

Repeating the alternation method removes cards 52,48,44...4. not 38,22,or6 because of sequence. now the order is reversed again. the cards remaining in order are now 2,6,10...50.

Repeat alternation to remove cards 2,10,18...50. reversing the order the cards from top to bottom remaining are 46,38,30,22,14, and 6.

what is left is a simple alternation that removes 46,30,and 14 leaving the last three cards 38,22,and 6 yet to be unturned.

- If you`re a mathematician! Can the brainiacs of I-A-B explain this one?
When you cut pile 2 and put it on top of the first card, then place the second car on top of pile 2, then placing that stack on top of the other, you are simply putting all of pile 2 on top of your first card and then placing the second card on top.

You are cutting the piles and then putting them back together. You are just doing it out of sequence.

cutting the pile is useless. if you pay attention, you will place the cards like this:

10

first card

15

second card

15

third card

9 cards

Then you take 4 cards from the top and put them at the bottom. The amount of cards in between of the "special" cards it always one below 2^x when you show the cards, which garantuess, that you will never show the "special" cards.

Finished...

odds and even

In this case the cards are placed with 15 cards in between them,as long as protocol is followed, no matter what. In the position of the deck they are always cards 6,22,and 38 check me if I`m wrong.

Then using alternation we eliminate the odd numbered cards and reverse the order of remaining even cards so now starting with the top, the cards are 52,50,48...2.

Repeating the alternation method removes cards 52,48,44...4. not 38,22,or6 because of sequence. now the order is reversed again. the cards remaining in order are now 2,6,10...50.

Repeat alternation to remove cards 2,10,18...50.

reversing the order the cards from top to bottom remaining are 46,38,30,22,14, and 6.

what is left is a simple alternation that removes 46,30,and 14 leaving the last three cards 38,22,and 6 yet to be unturned.