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# A trick for three

While most mathemagical tricks involve only the performer and one spectator, here is an example of a trick that involves three spectators. To perform it, you need a standard deck of 52 cards, which are given to some spectator for shuffling.

#### Franka Miriam Brueckler

While most mathemagical tricks involve only the performer and one spectator, here is an example of a trick that involves three spectators. To perform it, you need a standard deck of 52 cards, which are given to some spectator for shuffling.

Now, the three spectators – Anne, Bob and Chris – are asked to take one card each, remember it and hold it in their hands so that the performer cannot see the three cards chosen. The performer holds the remaining cards as a pack, face down, and deals them into three pack – one with 10 cards in it, and two with 15 cards in each of them. The remaining cards the performer keeps in his hand.

Ann is asked to place her card face down onto the first pack (the one with 10 cards). Then she is asked to take some cards from the second column and put them on top of her card “to hide it”. Now, Bob is to place his card onto the remainder of the second column and hide it with some cards from the third one. Finally, Chris is to put his card on the remainder of the third pack and his card is “hidden” by the cards that remained in the performer’s hands.

The performer now puts the second column onto the first one and the third one on top of these joined columns to have just one pack. He may comment on how the three cards are well hidden, and chatter about how the spectators had free choice in their actions, etc., etc.

Now, the performer announces: “First, let us remove the top 4 cards,” (he deals them face up), “I guess you all noticed that I held more than 4 cards in my hand, so yours cannot be among them. Now, I will deal the cards from the pack, alternating one face up, one face down. You are all to keep track of the face up cards. If any of you sees his or her card dealt face up, please intervene!”

So, the performer deals: one up, one down, one up, one down, etc., until all the cards have been dealt. The chosen cards will not appear face up, so they must be in the other half of the deck. He repeats the dealing with this half-deck, and again the cards will not appear. He repeats this until there are only three face down cards – these will be the chosen cards, Ann’s will be the bottom one, Bob’s the middle one and Chris’s the last one!

Why does this work? Simply because it was not true that the cards are well hidden! Let us denote by a1, a2, …, a52 the cards of the pack, with a1 the bottom card. Then, cards a1 to a10 were the ones from the first column, onto which Ann’s card was placed as a11. Then she placed some cards from the second colum (of 15 cards) on top of it and Bob placed his card onto the remaining ones. Note that when collecting the columns, this second column was placed onto the first one, so independently on Ann’s choice there are 15 cards between her and Bob’s. Thus, Bob’s card is at position a27. Analogously, Chris’s card will land on position a43. Consequently, when the performer removes the top 4 cards, he is left with a1 to a48 and Ann’s, Bob’s and Chris’s card are a11, a27 and a43, respectively. Note that 11, 27 and 43 have one thing in common: when divided by 16, they leave a remainder of 11, and these are the only three such numbers between 1 and 48. The repeated dealing of face up and face down cards will essentially classify the card according to the remainders their indices (original positions) have when divided by 16.

In the first dealing, a48 is dealt face op, a47 face down, etc.: all even cards are dealt up, i.e. after the first step we eliminated all even remainders of division by 16 and are left with 24 cards in the order (bottom to top) a47, a45, .., a3, a1.

The second dealing deals a1 face up, a3 face down, a5 face up, etc.: The cards with indices having remainders 1, 5, 9 or 13 now appear face up, and we are left with 12 cards in order (bottom to top) a3, a7, a11, a15, a19, … , a43, a47.

The third dealing is “a47 face up, a43 face down, a39 face up, a35 face down, …, a7 face up, a3 face down” – the cards with remainders 7 and 15 have now appeared face up, and only those 6 with remainders 3 and 11 are now face down: a43, a35, a27, a19, a11, a3. Obviously, the last deal eliminates the remainder 3, and the three cards of our spectar end up in order, bottom to top, a11, a27, a43, i.e. in the order: Ann, Bob and Chris!

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