Three of a Kind

This trick seems to have been invented independently by Martin Gardner and Karl Fulves. A blindfolded magician asks a spectator to lay three pennies on a table, in any arrangement of heads and tails. The magician’s goal is to put all three coins into the same state, all heads or all tails.

If the three coins already match, then the trick is done. If not, then the magician gives three instructions: Flip the left coin, flip the middle coin, flip the left coin. After each step he asks whether the three coins now match. By the third flip, they will.

“It’s no surprise that the magician can eventually equalize all the coins,” writes MIT computer scientist Erik Demaine, “but it’s impressive that it always takes at most three moves.” The technique exploits a principle used in Gray codes, which are used to reduce errors when using analog signals to represent digital data. Demaine relates a similar trick involving four coins in the November-December 2010 issue of American Scientist.

See Lincoln Seeks Equality.

All Roads

A self-working card trick by New York magician Henry Christ:

Shuffle a deck thoroughly and deal out nine cards in a row, face down. Choose a card, look at it, and assemble the nine cards into a stack face down, with the chosen card at the top. Add this stack to the bottom of the deck.

Now deal cards one at a time from the top of the deck into a pile, face up, counting backward from 10 as you do so. If at some point the card’s rank matches the number said, then begin dealing into a new pile at that point, counting again backward from 10. If you reach 1 without a match occurring, then “close” that pile by dealing a face-down card onto it, and start a new pile.

Keep this up until you’ve created four piles. Now add the values of any face-up cards on top of the piles, count down through the remaining cards until you’ve reached this position, and you’ll find your chosen card.

This works because it always leads to the 44th card in the deck, but it takes some thinking to see this. You can put a sealed deck into a stranger’s hands and direct him to perform the trick himself, with mystifying results.

Two in One

Look at this image closely and you’ll see the features of Albert Einstein.

But look at it from across a room and you’ll see Marilyn Monroe.

It’s a “hybrid image,” created using a technique developed by Aude Oliva of MIT and Philippe Schyns of the University of Glasgow. The image combines the low spatial frequencies of one picture with the high spatial frequencies of another, so that it’s processed differently at different viewing distances.

See their paper for the details, and this gallery for more examples.

Star Power

A puzzle by A. Korshkov, from the Russian science magazine Kvant:

It’s easy to show that the five acute angles in the points of a regular star, like the one at left, total 180°.

Can you show that the sum of these angles in an irregular star, like the one at right, is also 180°?

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Image: Wikimedia Commons
  • The clock face on the Marienkirche in Bergen auf Rügen, Germany, has 61 minutes. Does this mean time moves more slowly there — or more quickly?
  • To ensure quiet, poet Amy Lowell hired five rooms at every hotel — her own and those on either side, above, and below.
  • A perplexing sentence from a letter by Dorothy Osborne, describing shepherdesses in Bedfordshire, May 1653: “They want nothing to make them the happiest people in the world but the knowledge that they are so.”
  • OVEREFFUSIVE is a palindrome in Scrabble — its letter values are 141114411141. (Discovered by Susan Thorpe.)
  • The sum of the digits of every multiple of 2739726 up to the 72nd is 36. (E.M. Langley, Mathematical Gazette, 1896)
  • I’ll bet I have more money in my pocket than you do. (Of course I do — you have no money in my pocket!)
  • In 1996 a model airplane enthusiast was operating a remote-controlled plane in Phoenix Park in Dublin when the receiver died and the plane flew off on its own. It flew five miles to the northeast, ran out of fuel, and glided to a landing … on the taxi-way to Runway 28 at Dublin Airport.

(Thanks, Brian and Breffni.)

Say Red

Cornell mathematician Robert Connelly devised this intuition-defying card game. I shuffle a standard deck of 52 cards and deal them out in a row before you, one at a time. At some point before the last card is dealt, you must say the word “red.” If the next card I deal is red, you win $1; if it’s black you lose $1. If you play blind, your chance of winning is 1/2. Can you improve on this by devising a strategy that considers the dealt cards?

Surprisingly, the answer is no. Imagine a deck with two red cards and two black. Now there are six equally likely deals:


By counting, we can see that the chance of success remains 1/2 regardless of whether you call red before the first, second, third, or fourth card.

Trying to outsmart the cards doesn’t help. You might resolve to wait and see the first card: If it’s black you’ll call red immediately, and if it’s red you’ll wait until the fourth card. It’s true that this strategy gives you a 2/3 chance of winning if the first card is black — but if it’s red then it has a 2/3 chance of losing.

Similarly, it would seem that if the first two cards are black then you have a sure thing — the next card must be red. This is true, but it will happen only once in six deals; on the other five deals, calling red at the third card wins only 2/5 of the time — so this strategy has an overall success rate of (1/6 × 1) + (5/6 × 2/5) = 1/2, just like the others. The cards conspire to erase every seeming advantage.

The same principle holds for a 52-card deck, or indeed for any deck. In general, if a deck has r red cards and b black ones, then your chance of winning, by any strategy whatsoever, is r/(b + r). Seeing the cards that have already been dealt, surprisingly, is no advantage.

(Robert Connelly, “Say Red,” Pallbearers Review 9 [1974], 702.)

Six by Six

The sestina is an unusual form of poetry: Each of its six stanzas uses the same six line-ending words, rotated according to a set pattern:

This intriguingly insistent form has appealed to verse writers since the 12th century. “In a good sestina the poet has six words, six images, six ideas so urgently in his mind that he cannot get away from them,” wrote John Frederick Nims. “He wants to test them in all possible combinations and come to a conclusion about their relationship.”

But the pattern of permutation also intrigues mathematicians. “It is a mathematical property of any permutation of 1, 2, 3, 4, 5, 6 that when it is repeatedly combined with itself, all of the numbers will return to their original positions after six or fewer iterations,” writes Robert Tubbs in Mathematics in Twentieth-Century Literature and Art. “The question is, are there other permutations of 1, 2, 3, 4, 5, 6 that have the property that after six iterations, and not before, all of the numbers will be back in their original positions? The answer is that there are many — there are 120 such permutations. We will probably never know the aesthetic reason poets settled on the above permutation to structure the classical sestina.”

In 1986 the members of the French experimental writers’ workshop Oulipo began to apply group theory to plumb the possibilities of the form, and in 2007 Pacific University mathematician Caleb Emmons offered the ultimate hat trick: A mathematical proof about sestinas written as a sestina:

emmons sestina

Bonus: When not doing math and poetry, Emmons runs the Journal of Universal Rejection, which promises to reject every paper it receives: “Reprobatio certa, hora incerta.”

(Caleb Emmons, “S|{e,s,t,i,n,a}|“, The Mathematical Intelligencer, December 2007.) (Thanks, Robert and Kat.)

Image: Wikimedia Commons

A puzzle by Princeton mathematician John Horton Conway:

Last night I sat behind two wizards on a bus, and overheard the following:

A: I have a positive integral number of children, whose ages are positive integers, the sum of which is the number of this bus, while the product is my own age.

B: How interesting! Perhaps if you told me your age and the number of your children, I could work out their individual ages?

A: No.

B: Aha! AT LAST I know how old you are!

“This is an incredible puzzle,” writes MIT research affiliate Tanya Khovanova. “This is also an underappreciated puzzle. It is more interesting than it might seem. When someone announces the answer, it is not clear whether they have solved it completely.”

We can start by auditioning various bus numbers. For example, the number of the bus cannot have been 5, because in each possible case the wizard’s age and the number of his children would then uniquely determine their ages — if the wizard is 3 years old and has 3 children, then their ages must be 1, 1, and 3 and he cannot have said “No.” So the bus number cannot be 5.

As we work our way into higher bus numbers this uniqueness disappears, but it’s replaced by another problem — the second wizard must be able to deduce the first wizard’s age despite the ambiguity. For example, if the bus number is 21 and the first wizard tells us that he’s 96 years old and has three children, then it’s true that we can’t work out the children’s ages: They might be 1, 8, and 12 or 2, 3, and 16. But when the wizard informs us of this, we can’t declare triumphantly that at last we know how old he is, because we don’t — he might be 96, but he might also be 240, with children aged 4, 5, and 12 or 3, 8, and 10. So the dialogue above cannot have taken place.

But notice that if we increase the bus number by 1, to 22, then all the math above will still work if we give the wizard an extra 1-year-old child: He might now be 96 years old with four children ages 1, 1, 8, and 12 or 1, 2, 3, and 16; or he might be 240 with four children ages 1, 4, 5, and 12 or 1, 3, 8, and 10. The number of children increases by 1, the sum of their ages increases by 1, and the product remains the same. So if bus number b produces two possible ages for Wizard A, then so will bus number b + 1 — which means that we don’t have to check any bus numbers larger than 21.

This limits the problem to a manageable size, and it turns out that the bus number is 12 and Wizard A is 48 — that’s the only age for which the bus number and the number of children do not uniquely determine the children’s ages (they might be 2, 2, 2, and 6 or 1, 3, 4, and 4).

(Tanya Khovanova, “Conway’s Wizards,” The Mathematical Intelligencer, December 2013.)

Two similar puzzles: A Curious Conversation and A Curious Exchange.