- Snapshots of modern mathematics
- Diderot Mathematical Forum 2013: “Mathematics of Planet Earth”
- Pierre de Fermat and Andrew Wiles in Czech Republic stamps
- Stefan Banach (March 30, 1892 – August 8, 1945)
- Diderot Mathematical Forums
- Guessing the Numbers
- What is mathematics for Ehrhard Behrends
- What is mathematics for Krzysztof Ciesielski
- The Three Ducks Trick
- What is mathematics for Franka Brueckler

# Cheney’s trick

The trick we shall describe here is also known as the best card trick there is. It is more elaborate than most standard mathemagical tricks, but the effort is worthwhile, because the effect is stunning for most audiences. In contrast to other mathematically based tricks, this one needs two performers. Also, a standard deck of 52 cards is needed.

*Franka Miriam Brueckler*

The effect of the trick is that one performer, who communicates with the spectator, leaves 4 cards on the table, while the spectator holds a card in his hand, and when the first performer leaves the room, the second enters and can determine the card the spectator holds just by looking on the 4 cards that were left on the table! There is no communication between the two performers, except the 4 cards. How can 4 cards signalize which of 52 cards the spectator holds? It is simply (mathe)magical!

In the above description we have not said that that the spectator has only a limited free choice. In the beginning, he can choose any 5 cards from the deck – here he really has freedom of choice. But then, the first performer decides which of the 5 cards the spectator is to keep in his hand, and uses the remaining 4 cards to leave the message for the second performer. Still, this limitation of spectator’s freedom cannot explain how the 4 arbitrary cards can give full information on the 5th of the arbitrarily chosen cards (check M. Kleber’s article for a thorough discussion on the numbers of possible choices and informations that can be “sent” by the cards on table).

Obviously, the way the four cards lie on the table is important. There are several methods to achieve the goal, and here we describe one we have discovered to work very well in practice, even in crowded and loud situations.

First, notice that of the 5 cards the spectator chooses, due to the pigeonhole principle at least two will have the same colour (diamonds, hearts, spades or clubs). Secondly, the first and second performer agree beforehand which position of the four cards is to be considered as the leftmost. The four cards will be arranged left to right, according to this agreement.

Now, the card the spectator is to hold will have one of the values 1 = A, 2, 3, …, 10, 11 = B, 12 = Q, 13 = K. Imagine an analogue clock with 13 instead of 12 hours:

Consider the two same-colour cards of the five, as mentioned above. Note that on this “clock” in one direction – clockwise or anticlockwise – they will be less than 8 “hours” apart. The spectator is to hold the one card with value less than 8 “hours” *after* the other, counted in the clockwise direction. E.g., if there is a 3 ♥ and a Q ♥ to be chosen from, the spectator keeps the 3 ♥ as 3 is 4 “hours” after 12 = Q, while the 12 = Q would be 8 “hours” after 3.

We see now that there are two things to be communicated about the card that has been left to the spectator using the remaining four cards: the colour and the difference of values. There are 4 colours and, according to the above agreement, the possible differences are 1 to 7. In the method of signalising these, we use the binary number system: each number 1 to 7 can uniquely be represented as a sum of 1, 2 and 4, each taken 0 or 1 times. The 0 is represented by keeping a card face down, and a 1 by a face up card.

Imagine the two performers agreed on the following meaning of the four cards: leftmost is 0 or 1, depending on 1 appearing in the binary representation of the difference of “hours”; the second is like that, but indicates the 2 in the binary representation, and the third represents the 4. The rightmost card is to represent the colour an will always be open: here the first performer places the second of the two same-colour cards (the one he did not return to the spectator), in our example the Q ♥. As the difference to be indicated is 4, which is 0 times 1 plus 0 time 2 plus 1 times 4, the cards will be placed (left to right): face down (0), face down (0), face up (1), Q ♥. When the second performer enters, he immediately sees that the spectator holds a ♥ card and adds 4 “hours” (binary 001) to Q = 12 to “guess” the spectator holds a 3 ♥!

Here is another example: To communicate a K ♣, the four cards left for the second performer could be like in the bottom row on the picture below (up card, down card, up card, 8 ♣) because from 8 ♣ to K ♣ is 5 and 5 = 1 + 0 + 4.

And that’s all there is to it – (not so) simple, isn’t it?!

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