The people who shared this magic trick ask, in the video, for an explanation of how it works. Here it is:

First, it should be fairly clear that the last part of the trick, where the cards are repeatedly divided into two piles, will always yield the same set of cards relative to their position in the initial pack. Suppose we number the cards in the pack 1-52 with 1 at the top. Card 1 becomes the bottom card in the “Up” pile and card 2 becomes the bottom card in the “Down” pile. By the end of the first round, the “Down” pile contains all the even-numbered cards in reverse order – so counting from the top, it has cards 52,50,48,…,2.

The second round has the same effect – we retain every other card, and reverse the order, so after this the down pile contains cards 2,6,10,14,18,22,26,30,34,38,42,46 and 50. The next round gives us 46,38,30,22,14 and 6 and the final round leaves us with 6, 22 and 38.

So, the question is, how come the chosen cards always end up in these positions in the pack? It looks like they should be at random positions. This is where the “illusion” part of the trick comes in; it relies on making us think we’ve cut the cards at random when in fact we’ve just moved a counted pile of cards from one place to another.

In the trick, there are two occasions when we move a random number of cards. Let’s call the number of cards we move x and y, and call the three cards we’ve chosen at the start of the trick cards A, B and C. To start with, we’ve got the following arrangement of cards:

After we position card A, we have

So, once we’ve positioned the other two cards, and put the nine card pile on top of the final card to give us three piles, we have:

Because x and y could be any number of cards, it looks like we can’t tell the position of A, B and C. But look what happens when we combine the three piles into a single pack:

Where we have 15-y cards on top of y cards, clearly we’ve just got 15 cards, and the same goes for x; so the final arrangement is just

That is, the cards A, B and C are in completely predictable positions. The last step is to move four cards from the top of the pack to the bottom, giving us

So now, C is in position 6 (ie is the 6th card from the top of the pack), B is in position 22 and A is in position 38 – the three positions that will be retained in the “up-down” rounds which follow.

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