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# You must do it three times!

**Take a deck of n cards where n might be arbitrary (usually 10-20)…**

*by Ehrhard Behrends*

*Step 1:* It will be necessary to know the card at the bottom. There are several possibilities to achive this

(for example by a secret glimpse when presenting the deck).

*Step 2:* One of your spectators chooses a number $k$. The only condition: k must be larger than n/2 and not greater than n.

*Step 3:* Then t h r e e t i m e s the following happens: The magician counts k cards one by one from the top of the deck on the table, and then the rest of the deck is put on top of the cards on the table.

After these operations one knows the card on top: it is that card that in the original deck was at the bottom. It is rather surprising that this fact is true regardless which n and k had been chosen.

One can use this information in various ways. For example, after step 3, your spectator can have a look at the top card (and show it to the other spectators) before he or she shuffles it into the deck. The magician puts the cards face-up on the tabel and finds the spectators card.

**The explanation**

Call (from top to bottom) the first n-k cards block A, the next 2k-n cards block B and the remaining n-k cards block C.

Thus the original deck can be abbreviated ABC. The first of the operations in step 2 transforms ABC to CB’A’, where the prime indicates that the order is reversed. After the second step we arrive at A’BC’ and after the third at C’B’A. This explains that the top card is now the bottom card of C, that is, the bottom card of the original deck. (By the way, a further operation would yield ABC, the original order: After f o u r operations the deck is restored!)

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