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A Mathematical Magic Trick – The Chatty Jokers

by Dr. Tao Chen


The mechanism behind the magic trick "The Chatty Jokers"

video credts: youtube channel STEMMathsMagic

is very straightforward and can be explained with basic algebra. As its title indicates, it is the two jokers that help the magician to locate two cards, which two participants select. Let us try to show how it works.

First the magician prepares a full deck of cards including jokers, a total of 54 cards. The two jokers are placed upside down so the magician could find them easily during the performance. Suppose that the red joker is Xth card from the top and the black joker is the Yth. For the performance, X is less than 18, and Y is between 19 and 36. That is, the red joker is in the first one third and black joker is in the second third. The magician needs to remember the numbers X and Y. Regardless of what the participants do, the position of two cards they found will only rely on X and Y.

The magician asks the first volunteer and the second volunteer to grab about a third of the deck each, and to show their bottom cards to the audience. Suppose the first volunteer grabbed M cards and the second volunteer grabbed N cards. Note that the red joker is in the first pack and the black joker is in the second pack. It can be shown as the following figure.


Then the magician asks the volunteers to return their thirds to the deck. This is first done by the first volunteer and followed by the second.The black Joker becomes the (Y-M)th card and the red Joker becomes the (N+X)th card. The card that the first volunteer selected is the (M+N)th card, and the card the second audience selected is the Nth card. This is illustrated in the following figure.


Note that there are Y-M-1 cards on top of the black Joker, N-(Y-M) cards between the black Joker and the card that the second volunteer selected, X-1 cards between the card the second volunteer selected and the red Joker, M-X cards between the red Joker and the card the first volunteer selected. Also there are 54-(M+N) cards below the card that the first volunteer selected, as indicated by the blue section.

The magician finds the two jokers and asks for their help. This is the key part of the performance. As he finds the two jokers, he pulls out the cards between them and puts them to the bottom of the deck. This can be done inconspicuously if desired by setting aside each pack of cards leading up to each joker and then simply returning the first to the top of the remaining deck and the second to the bottom. The result is illustrated in the following figure.


Therefore the card that the first volunteer selected is positioned at (Y-M-1)+(M-X)=(Y-X-1), and the card the second volunteer selected is positioned at (Y-X-1)+(54-M-N)+(N+M-Y)=(53-X). For example, if X=10 and Y=30, then the first card will be the 19th card, and the second card will be the 43th card.

Because the magician knows X and Y, no matter what the two volunteers do, he will immediately know where to find these two cards.