Mathemagic with a Deck of Cards - Department of …colm/cambshort.pdfCentre for Mathematical Sciences, University of Cambridge Mathemagic with a Deck of Cards “Card Colm” Mulcahy

Centre for Mathematical Sciences, University of Cambridge

Mathemagic with a Deck of Cards

“Card Colm” Mulcahy@CardColm

Spelman College, Atlanta, Georgia

18th July 2013

Inspiration

The scene: an old people’s home in Norman, Oklahoma.

The height exceeds the circumference, right? No!

Demonstrating a classic illusion with a scrap of paper and a tallskinny glass: our failure to appreciate the value of π.

Martin Gardner (1914-2010)Prince of Recreational Mathematics

–standing by every word he ever wrote

Follow @WWMGT on Twitter.

Three Scoop Miracle

Consider this demonstration of mathemagic:

A quarter of a deck of cards is handed to a spectator, who isinvited to shuffle freely. She is asked to call out her favouriteice-cream flavour; let’s suppose she says, “Chocolate.”

Take the cards back, and mix them thoroughly, before dealing cardsinto a pile, one card for each letter of “chocolate,” before scoopingthem up and dropping the remainder on top (as a topping!).

This spelling/scooping/topping routine is repeated twice more (forthree times total).

Three Scoop Miracle

Emphasize how random the dealing was, since the cards wereshuffled and you had no control over the named ice-cream flavour.

Have the spectator press down hard on the card that ends up ontop, requesting that she magically turn it into a specfic card, saythe four of diamonds.

When that card is turned over it will indeed be found to be thedesired card.

Published as ”Low Down Triple Dealing” and dedicated to MartinGardner, on the occasion of his 90th birthday, 21 October 2004.

Low Down Triple Dealing

The underlying mechanism behind this is a reversed transfer ofsome fixed number of cards in a packet—at least half—from top tobottom, done three times in total.

The dealing out (and hence reversing) of k cards from a packetthat runs {1, 2, . . . , k − 1, k, k + 1, k + 2, . . . , n − 1, n} from thetop down, and then dropping the rest on top as a unit, yields therearranged packet {k + 1, k + 2, . . . , n − 1, n, k, k − 1, . . . , 2, 1}.

When k ≥ n

2, doing this three times brings the bottom card(s) to

the top. Why?

Does this result generalize?

Is there a corresponding move which brings the top card(s) to thebottom if done three times?


Given an ice cream flavour of length k and a number n ≤ 2k . . .

Given a packet of n cards and k ≥ n

2, the packet naturally breaks

symmetrically into three pieces T ,M,B (top, middle and bottom),of sizes n − k, 2k − n, n − k, respectively, such that the basiccount-out-and-transfer operation (of k cards each time) is

T ,M,B → B ,M,T ,

where the bar indicates a complete subpacket reversal.

Using this approach, the Bottom to Top (with three moves)property can be proved. Actually . . .


Low Down Triple Dealing Generalized?

The Bottom to Top property is only 75% of the story.

Here’s the real scoop:

The Period 4 PrincipleIf four reversed transfers of k cards are done to a packetof size n, where k ≥ n

2, then every card in the packet is

returned to its original position.

What about mint chocolate chip?

Additional Certainties

Shuffle a deck of cards a few times, and have two or threeselections made by different spectators.

They share the results with each other and announce the sum ofthe chosen card values.

You promptly announce what each individual card is!

Secret Number 1:

When several numbers are added up, each may bedetermined with certainty from the sum.

Secret Number 2:

Totally free choices are indeed offered, but from acarefully controlled small subset of the deck.

The possibilities may be narrowed down by having half a dozen keycards at the top of the deck at the start, in any order, and keepingthem there throughout some fair-looking shuffles.

Secret Number 3:

You have memorized the suits of the half dozen key cardswhich start at the top.

The Value of a Good Education

Secret Number 1 Revisited:

Every natural number has a unique representation as asum of (distinct) non-consecutive Fibonacci numbers(1, 2, 3, 5, 8, 13, 21, 34, . . . ).

This Zeckendorf representation was apparently not first noticeduntil 1939, by Edouard Zeckendorf from Belgium, an amateurmathematician—but a real (army) doctor, and later dentist. Hedidn’t publish it until 1972, after he’d retired.

For instance, 6 = 5 + 1 (we don’t allow 3 + 2 + 1 or 3 + 3),and 20 = 13 + 5 + 2.

Zeckendorf–Fibonacci

Given any natural number, it’s easy to find the appropriatedecomposition, by first peeling off the largest possible Fibonaccinumber, and repeating for the difference, until we are done.

For instance, 50 = 34 + 16 = 34 + 13 + 3.

Zeckendorf–Fibonacci Card Trick

Now, consider any Ace, 2, 3, 5, 8 and King (value 13), for instance.

A♣, 2♥, 3♠, 5♦, 8♣, K♥ (suits in CHaSeD order).

If two (or three) cards are selected from these, then they can bedetermined from the sum of their values.

In other words, any possible total can only arise in one way.

Even better, it’s easy to decompose a given sum into its pieces.

Generalized Fibonacci and . . . ?

The Lucas sequence 2, 1, 3, 4, 7, 11, 18, ..., which is a kind ofgeneralized Fibonacci sequence, also has the desired property, if weomit the 2 at the start.

When all is said and done, we don’t even need segments ofgeneralized Fibonacci sequences to pull off the above kind of effect.

The set 1, 2, 4, 6, 10 works.

So does 1, 2, 5, 7, 13.

Can you generalize?

The Birthday Card Match Principle

How many cards, picked randomly from a standard deck of 52,are required so that there is a greater than 50% chance ofgetting at least one pair with the same value?

Let’s call this desirable phenomenon a birthday card match—that’s “birthday” card match and not “birthday card” match.

This problem can be tackled using a common approach to theclassic Birthday Problem, which concerns the number of peoplerequired to ensure a greater than 50% chance of having at leastone birthday match.

The surprisingly small answer there is 23 people!

The Birthday Card Match Principle

The key to estimating such probabilities is to turn things around,and focus on the chances of there being no match, noting that

Prob(at least one match) = 1 - Prob(no match).

If k cards are picked at random, then since a deck contains fourcards of each value, it’s clear that for 2 < k < 14 we get:

1 − 5252 ×

4851 ×

4450 × . . . × 52−4k+4

52−k+1 .

It turns out that we need to pick at least 6 cards to be at least50% sure of a birthday card match.

Given 8 or 9 cards, there is a high probability (89% or 95%) of amatch, and with 10 cards, it’s very likely (98%) to occur.

Better Poker Hands with Bill Simon

When you glance at the faces of the cards you are given, assumingthat there is at least one matching pair, the basic idea is to ensurethat by dropping clumps of cards casually, the “winning cards” areamong the bottom four.

If you play your cards right, you can ensure that later on, you getthose winning cards.

Forget about the first two cards for now, and suppose you havejust eight cards (WLOG, 10 = 8 . . . ).

We need a wonderful observation from Bill Simon’s 1964 bookMathematical Magic, which included an effect called “The FourQueens.”

Bill Simon’s Sixty-Four Principle

It is possible to give the illusion of multiple free choicesto a spectator, while controlling how to split up a packetof eight cards into two piles of four.

In fact, you retain control of the division in one key sense: the topfour cards all end up in the first pile.

You can even have the spectator handle the cards throughout,after you appear to have shuffled them.

Hence, if you start with 4 red cards of top of 4 black ones, thepiles maintain that colour separation, with the reds in the first pile.


We now describe how Simon’s separation scheme works in practice:an effective way to follow along is to work with a face-up packet of4 reds followed by 4 blacks.

The spectator is given the choice of putting the top card on thetable to start Pile A, or tucking the top card underneath the rest ofthe packet. The second card then goes wherever the first one didnot (under the packet if the first one went to the table, and viceversa).

Overall, one of the first 2 cards starts Pile A, and the other goes tothe bottom of the packet.

Give the spectator the exact same free choice for the second pairof cards.


Unsuspected by most is the fact that Pile A now contains 2 redcards, and the retained packet consists of 4 blacks followed by 2reds.

Next, the spectator is asked to make similar choices to determine 2cards for Pile B. Of course, the result is that 2 blacks start thatpile and the retained packet consists of 2 reds followed by 2 blacks.

(Note that at this stage we have a scaled down version of theoriginal packet.)

Now, the spectator uses the same procedure to pick just 1 card forPile A, and finally 1 for Pile B, unwittingly maintaining the colourseparation.


Pause to recap what has happened, claiming “Six times, I gave youcompletely independent free choices. That’s two to the power ofsix, or sixty-four, different things that could have happened so far.”

The last 2 cards are a red followed by a black, and you must havethe first added to Pile A, and the second to Pile B.

This can be done either by casually asking the spectator to dealthem that way, or you can come up with some magicians force toachieve the same result (Simon suggested a specific one).

Of course, this can also be applied to packets of size 16 (or 32), ifsuitable modifications are made.

Given Any Five Cards

Aodh gives a deck of cards to a spectator and asks for fiverandomly chosen cards.

Aodh glances at the cards, and hands one back to the spectator,who hides it.

Aodh places the remaining four cards in a face-up row on thetable.

Bea, who has witnessed nothing prior to this display, enters theroom, glances at the cards on the table, and after a suitable pause,promptly reveals the identity of the hidden card.

This is a two person card trick, although the second person—Aodh’s mathematical accomplice Bea—only participates in thedramatic finale.

How can this be done, without any verbal or physical cues?

Fitch Cheney’s Five-Card Trick

This superb effect is usually credited to mathematician WilliamFitch Cheney Jnr (1904-1974), who in 1927 received the first PhDin Mathematics awarded by MIT.

A keen magician all his life, Cheney was also Editor of the PuzzleSection of the American Mathematical Monthly from 1930 to 1940.

Note that Aodh gets to choose which card to hand back, and then,in what order to place the remaining four cards.

(The first condition can be worked around actually . . . ask howlater if this interests you!)


Three main ideas make this magic possible:

1. The pigeonhole principle guarantees that (at least) two ofthe five cards are of the same suit.

WLOG Aodh has two Clubs.

One Club is handed back, and by placing the remaining four cardsin some particular order, Aodh effectively tells Bea the identity ofthe Club handed back.

2. Aodh can use one designated position (e.g., the first) ofthe four available for the retained Club—which determinesthe suit of the hidden card

Aodh used the other three positions for the placement of theremaining cards, which can be arranged in 3! = 6 ways.


If Aodh and Bea agree in advance on a one-to-one correspondencebetween the six possible permutations and 1, 2, . . . , 6, then Aodhcan communicate one of six things.

What can one say about these other three cards? Not much—forinstance, some or all of them could be Clubs too, or there could beother suit matches!

However, one thing is certain: they are all distinct, so with respectto some total ordering of the entire deck, one of them is LOW, oneis MEDIUM, and one is HIGH.

Asssume suits are in CHaSeD order.

This permits for an unambiguous and easily remembered way tocommunicate a number between 1 and 6.


But surely 6 isn’t enough?

The hidden card could in general be any one of 12 Clubs!

This brings us to the third main idea:

3. Aodh must be careful as to which card he hands back.

Considering the 13 card values, 1 (Ace), 2, 3, . . . , 10, J, Q, K, asbeing arranged clockwise on a circle, we see that the two suitmatch cards are at most 6 values apart, i.e., counting clockwise,one of them lies at most 6 vertices past the other.

Aodh gives this “higher” valued Club back to the spectator tohide. Aodh will then use the “lower” Club and the other threecards to communicate the identity of the hidden card to Bea.


For example, if Aodh has the 2♣ and 8♣, then he hands back the8♣.

However, if he has the 2♣ and J♣, he hands back the 2♣.

In general, he saves one card of a particular suit and needs tocommunicate another of the same suit, whose numerical valueis k higher than the one he makes available, for some integer kbetween 1 and 6 inclusive.


Put this total CHaSeD linear ordering on the whole deck:

A♣, 2♣, . . . , K♣,A♥, 2♥, . . . , K♥,A♠, 2♠, . . . , K♠,A♦, 2♦, . . . , K♦.

Mentally, he labels the three cards L (low), M (medium), and H(high) w.r.t. this ordering.

The 6 permutations of L,M,H are always ordered by rank, i.e., 1 =LMH, 2 = LHM, 3 = MLH, 4 = MHL, 5 = HLM and 6 = HML.


Finally, he orders the three cards in the pile from left to rightaccording to this scheme to communicate the desired integer.

For example, if he is playing the J♣ and trying to communicatethe 2♣ to Bee, then k = 4, and he plays the other three cards inthe order MHL.

Bee knows that the hidden card is a Club, decodes the MHL as 4,and mentally counts 4 past the visible J♣ (mod 13) to get the 2♣.


A weakness in the method as described above is the invariant useof the first position in the pile as the “suit giver,” this is soonspotted by alert audiences if the trick is repeated.

Here is a better idea:

Since everybody gets to see the display four cards, Aodh first sumstheir values and reduces mod 4 (using 4 if 0 is obtained). He nowuses that number for the position in the display of the suitdetermining card. Bea, upon seeing these cards, first sums mod 4also to figure out which card is special and which three tell herhow far to count up from the value of that suit giver card.

E.g., a Jack, 8, 2 and 7 gives 11 + 8 + 2 + 7 = 0 (mod 4), so thefourth slot would be used for the suit determining card.


Suppose the cards Aodh is handed are 2♣, 2♥ 8♠, 7♦, and J♣.

He plays the J♣, in the 4th slot, and communicates k = 4 (hencethe necessity of counting around to 2♣) using the other threecards as follows:

In standard LMH order they are 7♦, 2♥, 8♠, so in MLH order theyare 2♥, 7♦, 8♠,

So Aodh lays out the cards in this order: 2♥, 7♦, 8♠, J♣.


Martin Gardner mentions this trick briefly in his 1956 bookMathematics Magic and Mystery (Dover), and also in theScientific American “Mathematical Games” column collection TheUnexpected Hanging and Other Mathematical Diversions, citingW. Wallace Lee’s book Math Miracles (1950).

There, it is featured as “Telephone Stud” and attributed toWilliam Fitch Cheney, Jnr, Chairman, Department of Mathematics,University of Hartford, Hartford, CT.

Thanks to Art Benjamin for providing this source, and Paul Zornfor alerting us to the existence of the trick in the first place.

Fitch Four Glory

A volunteer from the crowd chooses any four cards at random andhands them to Aodh. He glances at them briefly, and hands oneback, which the volunteer then places face down to one side.

Aodh quickly place the remaining three cards in a row on the table,some face up, some face down, from left to right.

His confederate Bee, who has not been privy to any of theproceedings so far, arrives on the scene, looks at the cards ondisplay, and promptly names the hidden fourth card—even in thecase where all three cards are face down!

Fitch Four Glory

How can the pigeonhole principle help this time?

Redefine the pigeonholes first!

Partition the stardard deck into three new suits of 17 cards each,leaving one special card aside.

Each of these new suits consists of the one of the standard suits ♣,♦, ♥ supplemented with four ♠’s.

Specifically,

Suit Alpha is A♣, 2♣, . . . , K♣, 2♠, 3♠, 4♠, 5♠

Suit Beta is A♦, 2♦, . . . , K♦, 6♠, 7♠, 8♠, 9♠

Suit Gamma is A♥, 2♥, . . . , K♥, 10♠, J♠, Q♠, K♠

Fitch Four Glory

Note that A♠ has been left out in the cold: if this special card isamong those handed to Aodh, he simply hands it back—after asuitable pause—and plays the other three all face down!

Otherwise, the pigeonhole principle guarantees that (at least) twoof the four cards are from one of the three redefined suits, withoutloss of generality Suit Alpha.

He hands one back, and by placing the remaining three cards onthe table in some particular fashion, reveals to Bee confederate theidentity of the hidden card.

As before, the basic strategy is to save the “lower” card from SuitAlpha, and communicate the “higher” one, whose numerical valueis k past the one he holds on to, where this time k is an integerbetween 1 and 8 inclusive.

Fitch Four Glory

In the convention we explain below, at least one card will be faceup, so once more he can use a face up card (the first such if thereare two) to communicate the suit.

The placements UDD, DUD, DDU (one U in 1st, 2nd or 3rdposition) and DUU, UDU, UUD (one D in 1st, 2nd or 3rdposition), respectively, can be used to tell Bee that k is 1, 2, 3, 4, 5or 6.

This time he also needs a way to communicate 7 or 8 . . .

Note: he also has the UUU option at his disposal!

Fitch Four Glory

If we agree to use one particular U (say, the middle one) to givethe suit, there are two ways to play the other two: Low-High (toconvey k = 7) or High-Low (for k = 8) w.r.t. some total orderingof the deck, such as lining up Suits Alpha, Beta, Gamma in thatorder.

Fitch Four Glory

Mike Trick at Carnegie Mellon kindly put together a websiteillustrating this, er, trick in action, it’s athttp://mat.gsia.cmu.edu/CARD/.

http://mat.gsia.cmu.edu/CARD/ Mike Trick page

Follow @WWMGT on Twitter.

ReferencesMathematical Card Tricks, Feature Column, AMS Online(October 2000)

Card Colm, MAA Online (bimonthly since October 2004)

Chapters in MAA books:

Expeditions in Mathematics (The Second Book ofBAMA Talks) (2011)The Edge of the Universe—Celebrating Ten Yearsof Math Horizons (2006)

Chapters in A.K. Peters books:

Mathematical Wizardry for a Gardner (2009)Homage to a Pied Puzzler (2009)A Lifetime of Puzzles (2008)Puzzlers’ Tribute: A Feast for the Mind (2002).

References

Mathematical Card Magic: Fifty-Two New Effects

Hardback, full colour, 380 pages

AK Peters/CRC Press (August 2013)

Documents

Mathemagic with a Deck of Cards - Department of …colm/cambshort.pdfCentre for Mathematical Sciences, University of Cambridge Mathemagic with a Deck of Cards “Card Colm” Mulcahy