The Math Behind Magic

THE MATH BEHIND MAGIC

By Kristen Berish

PERCI DIACONIS • Professor of Mathematics and Statistics at Stanford University • Main focus was on magic through the use of cards • Most famously know for his finding in 1992, along with Dave Bayer, that it takes about seven ordinary riffle shuffles to randomize a deck. • Although the proof of this took years to discover and is of long length, 7 shuffles completely randomizes the deck and therefore every arrangement of the 52 cards is equally likely or that any card is as likely to be in one place as in another.

♦ THE ART OF SHUFFLING CARDS ♦ • The Perfect Shuffle of a deck of 2n cards • Cutting the deck exactly in half and then riffling them so that they alternate perfectly. • This will not completely randomize the deck! • Let the deck be labeled (1,2,…,n-1,n,….2n-1, 2n)

• The in-shuffle; • Leaves the original card second from the top • After an in shuffle, the order is (n+1,1,…,2n,n) • In-shuffling an even number n of cards n times results in the original card order. •

Example: Deck of 52 cards • Takes 52 shuffles to return to the original

♠ THE ART OF SHUFFLING CARDS ♠ •   • The out-shuffle: • Leaving the original card on top • After an out shuffle, the order is (1, n+1, 2,…n, 2n) • Out-shuffling an even number n cards n-2 times when n-1 is prime results in the original order • For any n-1 that is not prime, amount of shuffles, r, can be written as • Example: Deck of 52 cards •  8 perfect shuffles

WHAT THE PERFECT SHUFFLES ARE USED FOR: • If a magician wants to bring the top card to a given position in the deck • Suppose he wants to move the top card to the 13th position Step 1: Subtract 1 from the position wanted --- 13–1= 12 Step 2: represent that number in binary 12 = 1100 Step 3: let 0 = out-shuffle and 1= in-shuffle 1100 = in, in, out, out

COLM MULCAHY • Professor of Mathematics at Spelman College, in Atlanta, Georgia, where he's been teaching since 1988. • He’s been publishing original mathematical card trick principles bi-monthly at MAA.org since 2004. • Other puzzles of his have appeared in the New York Times.

♣ THE THREE SCOOP MIRACLE ♣ • The magician hands about a quarter of a deck of cards to a spectator, and ask her to shuffle freely. Taking those cards back, and mixing them further the magician asks the spectator what her favorite ice cream flavor is. Let’s suppose she says, “Chocolate.” Dealing from the packet to the table, the magician puts down one card for each letter of “chocolate,” then scoop those up with one hand and drops the remainder on top. This process is repeated twice more, for a total of three times. The magician then now ask her to press down hard on the top card of the packet on the table, requesting that it be miraculously turned a specific card, which in this case we’ll say the Four of Diamonds. When the card is turned over, it is seen by all to be the desired card. Congratulate the spectator on a job well done.

♣ THE THREE SCOOP MIRACLE ♣ • There are two secrets here: • (1) A key relationship between the number of letters in the word being spelled out and the size of the “quarter” deck being used • The size of the deck must be at least as big as the kind of ice cream the person chooses but no larger then twice that number

• (2) The you must know the identity of the bottom card at the start of the spelling and dropping

• The question now is how can the magician get that bottom card to the top of the deck??

Bottom to Top Principle The original bottom card of the packet ends up on top after three such reverse transferrings of k cards from n, provided that k ≥ n/2 • This is considered a reversed transfer of some fixed number of cards in a packet from top to bottom, done three times in total. • The dealing out of k cards from a packet that runs {1, 2,...,k − 1, k, k + 1, k + 2,...,n − 1, n} from the top down, and then dropping the rest on top as a unit, yields the rearranged packet {k + 1, k + 2,...,n − 1, n, k, k − 1,..., 2, 1}. Picture shows the order when k = 9

Top - - - - - - - - - - - - - - - - - - > Bottom

Low-Down Deal Packet Separation Principle If k cards are reverse transferred from the top to the bottom of a packet of size n, and k ≥ n/2 , then the top and bottom halves switch places, subject to some internal reordering. Picture shows the order when k = 7 Patterns to be aware of: If the deck has an odd number of cards, then the middle card will always be fixed. If the deck has an even number, the middle two cards will switch places.

SYMMETRY OF A DECK OF CARDS • When reverse transferring the same number of cards over and over, there are three sections of the deck to keep an eye on. • First, note that since k ≥ n /2 , we have 2k − n = k − (n − k) ≥ 0 • Let n = k + (n − k) = [(n − k)+(k − (n − k))] + (n − k) = (n − k) + (2k − n)+ (n − k) • Packet of n cards naturally breaks symmetrically into three pieces T,M,B, (top, middle, and bottom) of sizes n−k, 2k −n, n−k.

T  n−k M  2k−n B  n−k

T  n−k M  2k−n B  n−k

• Example: n = 13, and k = 8, and therefore the deck is ordered: {1, 2,..., 12, 13} T = {1, 2, 3, 4, 5} M = {6, 7, 8} B = {9, 10, 11, 12, 13} Counting out eight cards results in: {9, 10, 11, 12, 13, 8, 7, 6, 5, 4, 3, 2, 1}, that is B followed by M reversed followed by T reversed. The middle card, in position 7 in this case, remains fixed throughout.

•  Reminder: the goal is to get the last card to be on the top of the deck • In general, counting out the first k cards (always at least T and M together) and dropping the rest on top leads to: T,M,B → B,, • The second round yields: B,,→ , M, • The third round yields: , M, → , T • Hence, the original bottom card will now be on top.

‘THINGS ARE MEANT TO BE’ TRICK • You claim that some things are simply meant to be, and set about proving it. You need your friend to help you create a ‘random’ target number. But before she even picks the number, you predict that the total is 1089. • To prove, you get your friend to secretly write down any three-digit number. • The digits must all be different and the biggest digit must be at the front. • Then you get them to reverse their selected number and write it underneath the first number. • They should then subtract this lower number from their first number. • Finally, have them write their answer backwards and add this new reversed number to their answer. • After all of this, you now have a random number even your friend couldn’t have predicted in advance. But as predicted the final number is 1089

• Say your friend chose 742: 742 - 247 495 +594 1,089

WHY DOES THIS WORK? • Let ABC represent the chosen number: Subtracting: ABC ― CBA

 Since we know A>C we cant to subtraction without borrowing

Lets try this instead: If you add 100 to ABC and then take away 100, then you’re still left with the same number, ABC. Representing adding 100 as 10+90 and subtraction of 100 as A-1: ABC = (A-1) (B+9) (C+10) We now have:

―

C

(A-1)-C

(A-1) (B+9) (C+10) B (B+9)-B

+ (C+10)-A 9

18

A______ (C+10)-A (B+9)-B 9

(A-1)-C___ 

A=9 , B=18, C=9

 1089

REFERENCES • Magical Mathematics: The mathematical ideas that animate great magic tricks • By Persi Diaconis and Ron Graham

• Mathematical Card Magic Fifty-Two New Effects • By Colm Mulcahy

• http://www.mathematicalmagic.com/docs/mathsmagic_full.pdf