Question

I am interested in magic tricks whose explanation requires deep mathematics. The trick should be one that would actually appeal to a layman. An example is the following: the magician asks Alice to choose two integers between 1 and 50 and add them. Then add the largest two of the three integers at hand. Then add the largest two again. Repeat this around ten times. Alice tells the magician her final number $n$. The magician then tells Alice the next number. This is done by computing $(1.61803398\cdots) n$ and rounding to the nearest integer. The explanation is beyond the comprehension of a random mathematical layman, but for a mathematician it is not very deep. Can anyone do better?

Answer 1

If you are not mathematically inclined, this game can drive you crazy. http://www.transience.com.au/pearl.html

Answer 2

I hope this is contribution is appropriate; I think that a nice puzzle based on Hamming codes discussed a little here: http://ocfnash.wordpress.com/2009/10/31/yet-another-prisoner-puzzle/

is the following:

A room contains a normal 8×8 chess board together with 64 identical coins, each with one “heads” side and one “tails” side. Two prisoners are at the mercy of a typically eccentric jailer who has decided to play a game with them for their freedom. The rules are the game are as follows.

The jailer will take one of the prisoners (let us call him the “first” prisoner) with him into the aforementioned room, leaving the second prisoner outside. Inside the room the jailer will place exactly one coin on each square of the chess board, choosing to show heads or tails as he sees fit (e.g. randomly). Having done this he will then choose one square of the chess board and declare to the first prisoner that this is the “magic” square. The first prisoner must then turn over exactly one of the coins and exit the room. After the first prisoner has left the room, the second prisoner is admitted. The jailer will ask him to identify the magic square. If he is able to do this, both prisoners will be granted their freedom

Answer 3

A variant on Anton Geraschenko answer above- say you are in a fourth grade school that for some reason let these poor kids use calculators. you ask them to pick for themselves a 3 digit number say abc. Tell them to write it twice in their calculator ,i.e., abcabc and then divide by 77. Then by 13. What did you get? do it again with 143 and then by 7? What did you get. again with...

It teaches them about prime decomposition, about the decimal structure, about consecutive division etc.

I learnt it from Avraham Arcavi.

Answer 4

Here is a trick much in the spirit of the original number-adding example; moreover I'm sure Richard will appreciate the type of "deep mathematics" involved.

On a rectangular board of a given size $m\times n$, Alice places (in absence of the magician) the numbers $1$ to $mn$ (written on cards) in such a way that rows and columns are increasing but otherwise at random (in math term she chooses a random rectangular standard Young tableau). She also chooses one of the numbers say $k$ and records its place on the board. Now the she removes the number $1$ at the top left and fills the empty square by a "jeu de taquin" sequence of moves (each time the empty square is filled from the right or from below, choosing the smaller candidate to keep rows and columns increasing, and until no candidates are left). This is repeated for the number $2$ (now at the top left) and so forth until $k-1$ is gone and $k$ is at the top left. Now enters the magician, looks at the board briefly, and then points out the original position of $k$ that Alice had recorded. For maximum surprise $k$ should be chosen away from the extremities of the range, and certainly not $1$ or $mn$ whose original positions are obvious.

All the magician needs to do is mentally determine the path the next slide (removing $k$) would take, and apply a central symmetry with respect to the center of the rectangle to the final square of that path.

In fact, the magician could in principle locate the original squares of all remaining numbers (but probably not mentally), simply by continuing to apply jeu de taquin slides. The fact that the tableau shown to the magician determines the original positions of all remaining numbers can be understood from the relatively well known properties of invertibility and confluence of jeu de taquin: one could slide back all remaining numbers to the bottom right corner, choosing the slides in an arbitrary order. However that would be virtually impossible to do mentally. The fact that the described simple method works is based on the less known fact that the Schútzenberger dual of any rectangular tableau can be obtained by negating the entries and applying central symmetry (see the final page of my contribution to the Foata Festschrift).

Answer 5

This is a trick that I designed years ago and I have used it in many different occasions for amusement only or educational purpose or both. It is indeed the finial difference method to find a polynomial. Ask the person to write down a polynomial without you knowing the polynomial and even the degree of the polynomial. To keep your life easy, it would be better to keep the degree less than or equal 3. (It wouldn't be hard to let a layman know what a polynomial is just by giving two or three examples). Then you ask for some information that is essentially the value of the polynomial for 0, 1, 2, 3. As soon as you take one of the value you should calculate the difference. And in a few seconds after taking the last information, you announce not only the degree of the polynomial but also the exact polynomial.

Note 1: Finding the degree is a very important part of this trick since it convinces more knowlegable persons that you are not just solving a simultaneous equation quickly.

Note 2: I used this trick in my Calculus classes to give this seemingly paradoxical idea that "if you don't know what the function is, try to figure out how it changes."

Note 3: Of course, one can use it in many different classes for different purposes.

Note 4: I've just search the internet to see if Martin Gardner ever introduced this trick. Damn it! The answer was yes, here: "The calculus of finite differences". However, I still love to keep the credit of telling the degree for my self :)

Answer 6

How about the "Flash Mind Reader"

Answer 7

Lay out 21 cards face up in three vertical lines. Have a friend pick out any card without telling you which card he/she has chosen. Have your friend tell you which line of cards the selected card is in, and make three stacks of cards, each stack being made from each line of cards. stack the three stacks on top of each other, placing the stack with the selected card between the other two stacks (IMPORTANT!). lay out the cards again in the exact same set up (3 lines of 7 all face up) but here is the trick: when laying out the cards, flip them face up in a line every time. In other words, don't make one line at a time, but put a card in every line one at a time. Have your friend again tell you which line has the selected card. Stack the cards again, the exact same way you did the first time. One more time, lay out the cards the exact same way as the last time, one card per line, and again have your friend tell you which line has the selected card. Stack all the cards again one last time, again placing the line with the selected card between the other stacked cards. now lay out all the cards face down, one at a time. while you're doing this, remember to count, because the 11th card you place down is the selected card. from this point you can do whatever you can think of to make the trick "magical" and shock your friend by suddenly coming up with his/her card.

Answer 8

I think I learned this from a Martin Gardner article:

You are going to take a deck of cards and place them one at a time, face up, on a table (at a rate of about one per second). The person you are performing the trick for is to choose (secretly) one of the first 5 or 6 cards to start with. Whatever the rank of that card, they count that far to choose a new card, and repeats until the deck is exhasuted. Thus they have arrived at a (emphapsize this) "random" card near the bottom of the deck. You then tell them what that card is.

The trick is that it doesn't matter where you start counting, there is a pretty good probability that any two sequences of chosen cards will eventually coincide. The chances are better for larger decks (also if you say that every face card counts as 10).

Answer 9

Start with a deck of 32 cards. Then the player should take a card and tell a number $n$ between 1 and 32 then you divide the stack in 2 smaller stacks and the player has to tell which of the stacks contains his chosen card. according to a rule dependend on that number you put that stack above or below the other stack. After repeating this 5 times the chosen card should be exactly at position $n$. The rule has to depend on the way you want to deal cards (whether you turn around the deck and start dealing from the bottom, or you deal from the top and turn each single card around or you deal at first and then turn bost stacks around). In one of the cases the rule was take $N-11$, find the representation in the system with base $-2$ and revert that presentation. ($0$ tells you to put the stack containing the chosen card on top, etc.). I dont remeber this trick properly, it should not be too difficult to express the final position depending on the choices in some formula; but it is the only situation I know, in which the $-2$-system is useful.

Answer 10

The "casting out nines" sanity check of calculations is dead simple to use (a small child can do it), but the proof requires a deeper knowledge of mathematics (more precisely of arithmetic ; my own students don't have access to it even though they know what series are and can diagonalize matrices!).

Answer 11

Destination Unknown is a magic trick that makes use of Combinatorics. It really fools people.

See http://themagicwarehouse.com/cgi-bin/findit.pl?x_item=SP2453&keyword=DESTINATION

Answer 12

A deck of cards is shuffled, the audience member selects a number $n_0 < 10$ but keeps it private. The cards are then dealt out, face up, one at a time. When the $n_0$th card is turned up (but the magician does not which it is), the value of this card becomes $n_1$. Again, when the $n_1$th subsequent card appears, the value of that card becomes $n_2$, and so on.

Without knowing $n_0$, the magician can still predict the final "secret value". The way to do so is that the magician choose any value of $n_0'$ at random, and performs the corresponding process. If at any time, the magician's and audience member's values agree, then they will continue to agree for the rest of the deal. This in fact occurs with high probability, and so with high probability the magician the magician correctly determines the secret final $n$.

This works because the deal of the card deck behaves as a random function, and random functions under repeated iterations tend to coalesce. The mathematics of the random function mapping under iteration is quite deep. It is interesting that although it would require a quite difficult analysis to determine, for example, the probability that the magician's prediction is correct, to the audience none of this math is needed to be impressed by the trick!