Magic Trick With 27 Cards

[Jarrell Campbell]

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Hello friends! Every teacher should have one or 2 really good jokes, 1 really good card trick (mine is pretty amazing) and 1 or 2 math magic tricks for their students. This trick is one of my favorite and I thought I would share it with you.

I am Sara Van Der Werf, a 24-year mathematics teacher in Minneapolis Public Schools. I have taught math in grades 7-12 as well as spent several years leading mathematics at the district office. I currently teach Advanced Algebra at South High School and I'm also the current President of the Minnesota Council of Teachers of Mathematics (MCTM). I am passionate about encouraging and connecting with mathematics teachers.I'd love to connect via twitter. Join the community. Tweet me @saravdwerf.

The first thing to note is that the authors are both respected mathematicians, so it is perhaps not surprising to learn that the mathematics involved is actually non-trivial. In my undergraduate course on Knots and Surfaces I do a few knot and rope tricks to enliven the lectures and to demonstrate some of the ideas in the course, but these are generally sleight-of-hand tricks unlike the tricks in this book which all have some interesting mathematics underlying them.

The next sneaky part is that these 32 cards were arranged in an order such that the permutation of red and black cards amongst five consecutive cards uniquely identifies the five cards. The repeated cutting of the deck does not alter this cyclic order.

The final sneaky part is that the mathematician, being a mathematician, rather than memorising the order of the cards, has a cunning algorithm for turning the five bits of information into the suits and values of the five cards.

All you need to do now to perform the card trick is take a pack of cards, throw out all of the nines, the tens and the royals, then order the remaining the cards as above. Next you need to memorize the order of these 32 cards(!!). Then you can pass round the cards allowing the audience to cut the deck, thus not altering the cyclic order. When the sequence of red and black cards is revealed to you, you can decode it to learn the first card, and your knowledge of the order allows you to say what the other four cards are as well.

Now we can incorporate the linear feedback shift register into the trick. When the order of red and black cards are revealed to you, for instance as BRBBR, you first translate to binary, 0100101001, and then to a card value, ace of spades, as before. The difference is that now you can calculate the next card rather than having to remember it.

You add the first and third bits together modulo 22, append that to the end of the string and forget the first bit. This gives you 1001010010 which is the two of diamonds, and you can check that this is indeed the next in the card sequence listed above! You repeat this to obtain the other three cards that are being held by members of the audience.

The astute amongst you will note that our linear feedback shift register algorithm only gives a punctured de Bruijn sequence, so would not include the substring 0000000000, which corresponds to the eight of clubs. There are two ways around this.

Alice shuffled a pack of cards, and asked me to take five. I looked at them. She put the rest of the pack down on the table. Alice asked for my cards. She gave four of them to Bob (he was across the table), and the fifth back to me. Bob looked at the four cards for a while. Then Bob looked at me, and named the card I was holding. He was right. I'm quite sure he couldn't have seen it (we weren't sitting by a mirror).

They did the trick again later to someone else. I watched for funny business. Alice didn't say anything to Bob, so I don't think they have a code. Also Alice is famously clumsy, so I doubt it was sleight of hand.

Edit to answer a question: I learnt the trick from a maths magazine several years ago. I don't know who invented it. "Michael Kleber. The best card trick. Mathematical Intelligencer 24 #1 (Winter 2002)"

I'm going to assume Alice looked at the cards and chose which one to give back to you. The key to the puzzle is then to encode a single card's suit and value in 4 cards without the luxury of choosing those 4 cards arbitrarily from the whole deck.

The suit is easy. In 5 cards there must be a double of at least one suit. So the first (or last, but I'll choose arbitrarily) card in the bunch she passes is the same suit as yours. Now there are three cards left to encode a number from 1 to 13. However Alice chose which card of your suit to pass to Bob and which to return to you. She can choose according to a rule that gets the number of possible cards down significantly.

The three passed cards can be designated small medium and large according to their number, and then breaking ties by suit order (clubs smallest, diamonds, hearts, spades as in bridge.) This gives six possible numbers to be represented by the 3 passed cards based on their order: SML, SLM, MLS, MSL, LSM, LMS.

So how does she choose which of the suit cards to pass and which to return? Bob will add the encoded number to the passed card (going around K-A-2 if need be) to get the returned card. Alice passes whichever card is within an add of 6.

Say you have the 2 and 4 of spades in your 5. She can't pass the 4 because no number between 1 and 6, added to 4, will wrap around to the 2. So she passes the 2 and encodes 2 in the other three cards by ordering them SLM. If you have the Q and K she passes the Q and encodes 1. If you have the 7 and K she passes the 7 and encodes 6. But 6 and K, she passes the K and encodes 6.

I've tried to find a set of cards you could choose, knowing this algorithm, that would make it impossible to perform the trick, and I can't. Not from a proper deck that doesn't have any duplicate cards.

This is a great card trick known as Fitch Cheney's Five Card trick. Alice selects the fifth card to be identified and arranges the remaining four cards in a specific sequence. Then Bob interprets the four card sequence and identifies the fifth card.

Wanted to point out that the classic solution is just the most "human" way of doing this and while we humans tend to appreciate things that are easy for us, the classic technique is wasteful and there is another generic solution that came to me while observing the problem.

Of course this is harder to do cause it is not human friendly but it is much more efficient and generic so it should work with any set of integers and any number of integers where such a bijection exists.

You shuffle the cards any way you want. Then you take the first card from the top of the deck and put it down on the table face up. Now, if it's a king you put it back in the deck and try again. If it's anything else you read the value on the card. Lets say you picked a 3. Now you have to put down cards from the deck in your hand on top of the 3 until you have counted 13. So 13 - 3 = 10. So you place 10 cards from the deck face up on the 3 on the table. It does not matter what the cards you place on top have in terms of values.

Now you move on to make a new set of cards on the table after the same principle. Again you cannot start on a king. If you reach a point where you can no longer put down 13 cards you keep the cards in your hand.

Now you take three of the decks, turn them around so they lie face down and then you pick up the rest of the cards to your hand. Now you choose 2 decks of cards and make the first card turn around so that they are face up. I have a 3 and a 10 and I have 24 cards on my hand. With 3 and 10 facing up you put that together which is 13. You have to add an extra 10 to whatever number you get so it will become 23. Now I put down 23 cards from my hand which leaves me with one card. This means that the last card currently facing down of three sets of cards on the table will be an ace.

They should give back all the cards with their number on it. The magician immediately knows which number the volunteer chose and announces it. The volunteer is baffled: the number is correct! The audience cheers, the magician takes a bow, and the volunteer? Still amazed.

How did the magician know? Did they memorize all the combinations? No, they added up the numbers in the top left corner of the cards they got back from the volunteer. That's it. That's the entire trick! But why does it work?

Let's have a look at the distribution of the numbers on the cards. I'll number the cards 1 to 6 (sorted by the number in the top left corner) and will analyse which numbers occur on which cards. For this, I'll use a table for the first few numbers:

So by handing back the cards, the volunteer describes a binary string, representing the number they chose. All the magician needs to do is converting it back to decimal by adding up the bits. Neat!

That's where the name of the trick comes from: A computer stores numbers as binary strings. But those six cards are by no means Turing complete... In my opinion, this trick should have been called the Magic Number Base Converter - that would be a lot more accurate.

Of course, you can also use these in a comedy routine.

Do a few "amazing flourishes" with the deck, then "lose control" and let the cards fall- they are obviously attached to each other!

Hilarious!

Please note: Because of the way this deck is made, you cannot do regular card tricks with this deck.

This gimmicked deck is made up of 26 cards.

The attaching string takes up room, so using a full deck of cards would make this way too thick.

It is NOT a full deck- and it is not supposed to be.

It is a VISUAL effect.

We also carry an economy priced Electric Deck here.

What's the difference?

The Electric Deck sold on this page made of heavier-weight playing cards that have a geometic back design.

The Economy Electric Deck is less expensive because it is made with lighter weight playing cards.

Also, the back design on those cards are images, like flowers.

BOTH DECKS WORK EXACTLY THE SAME WAY.

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