You take out a deck of cards and shuffle it, before dealing twelve cards into a packet on the table. Pick it up, and ask somebody to shout out any one of "two," "three" or "four." Deal the packet into the appropriate number of piles, left to right, in the usual way. For instance, if "three" is called for, you will end up with three piles of four cards each. Now reassemble the packet by collecting the piles in order, and repeat, first requesting that a new number be shouted out to determine how many piles to deal into this time. Reassemble the packet as before.
Emphasize how mixed up "these randomly selected cards" must now be. "Just to make sure," you add, "Let's do this." Split the packet in two and riffle shuffle the halves together. "That should do it," you say triumphantly.
Split the packet in two again and invite a spectator to take either half. "Please look at the faces of the cards you have in your hand, and arrange them in numerical order. An Ace counts as one, a Jack as eleven, and so on. The lowest value card should be at the top of the face-down packet, the highest one at the bottom."
Once that has been accomplished, announce, "I'm now going to deal my cards into a magical circle, and I'd like you do exactly as I do. Please stand beside me with your packet of cards ready, like mine, face-down." Slowly deal your cards into a face-down clockwise circle, starting at the 12 o'clock position, as indicated below, allowing the spectator time to duplicate your actions with a second such circle beside yours.
Pause, and comment, "It would be wonderful if we could generate some synergy between our respective circles of cards, despite the completely random methods used to determine them." Continue, "Something's not quite right. I think we need to combine our efforts. Let's try again. Once again, please copy my actions exactly."
Pick up the six cards in your circle, making four "NE to SW" diagonal sweeps, starting at the bottom right corner and ending at the top left corner, as indicated in the following picture. Place the first card you pick up on top of the second, and those two on top of the third, etc.
When the spectator has done the same, place those six cards on top of yours. Say, "I believe one larger circle would be better, I was confusing circumference and area before. Also, I should deal starting from the 3 o'clock position, and in the other direction, going through the quadrants in the usual order." Deal out this twelve-card packet into a single face-down larger circle, starting at the 3 o'clock position this time, but going anti-clockwise.
"Maybe we've got a perfect clock here," you say hopefully. Turn over the card in the 3 o'clock position, which is indeed a 3. Express satisfaction at this. Next, turn over the card in the 2 o'clock position. Alas, it's not a 2. Look crestfallen, and say, "I guess that was too much to hope for. Maybe it's because of my anti-clockwise approach." Turn over the spectator's remaining four cards, still going anti-clockwise. None of them match the corresponding hours on a clock.
"But I have not given up hope. Remember you copied my every move a little while back, with the random cards we both obtained after all of the earlier mixing and shuffling," you remind the spectator. "Perhaps, by doing so, you turned your cards into matches of mine. Then mine, of course, will match yours! The cards I just exposed were yours, let's take a look at mine now." Turn over the next card, in the 9 o'clock position. It it a 3, just like the spectator's first card. Turn over the following one, in the 8 o'clock position. It matches the corresponding spectator's card (in the 2 o'clock position). Continuing, it will be found that all six of your cards match the spectator's. Moreover, when the spectators card is red, in most cases yours is black, and vice versa. The situation will resemble the following image.
The trick appears to be over. An extraordinary six-fold match has occurred with seemingly random cards which were certainly mixed and shuffled in many ways earlier. Now comes the kicker: draw attention to the first six exposed values. "Look: three---in the correct position---then one, four, one, five, nine...does that ring a bell?"
Assuming your audience is well educated and alert, somebody should perk up and exclaim, "It's the start of the decimal expansion of π." "Yes, indeed," you reply radiantly. "Note, we have two copies here, your upper semi-circle and my lower semi-circle. It's inevitable if you think about it: after all, a full circle corresponds to 2π!"
The first secret underlying the 3.14159 effect is that the original twelve cards dealt off the "shuffled" deck are far from random. They are in fact A♠, A♣, 3♠, 4♠, 5♣, 9♣, 9♥, 5♦, 4♦, 3♥, A♦, A♥, and care must be taken to maintain this palindromic stack at the top of the deck, while genuinely shuffling the rest. You can even riffle shuffle, to give the illusion of fair mixing, provided that the hand holding the top part of the deck drops the desired stack last.
The second secret is that the repeated dealing into piles, of various sizes, and the final riffle shuffle and even split into two piles of six cards, turns out to ensure that you and the spectator each end up with two Aces, a 3, 4, 5 and 9. (Just why this is the case, we will see in due course.)
When the spectator arranges her cards in ascending numerical order, you of course do the same with yours, but without drawing attention to it. At this point, nobody suspects that your cards already match. (The colours will compliment each other too, but it's not easy to arrange is the Aces come out in the appropriate order later on, unless the spectator asks, "What do I do if I have two cards of the same value?" and you answer accordingly.) When you deal into a small circle, "clockwise from the top," and do the subsequent "anti-diagonal pickups," both actions being copied by the spectator, it's merely a ruse to go from ascending numerical order to digits of π order; in other words, it's the necessary set-up for the final large circle deal.
To understand the effect of dealing a palindromic stack into piles, we first step back and consider dealing general packets into piles.
We restrict ourselves to the case of dealing packets into piles of equal size, which is equivalent to dealing into a rectangular array, like a matrix. For such an array with r rows and c columns--obtained by dealing out a packet of numbered 1 to rc from left to right, building up each column one card at a time--we can track what is going on by means of the table below, which indicates the positions of the cards in the original packet.
The ith row and jth column are shown, and the sole card in both is numbered (i - 1)c + j. For instance, if r = 5 and c = 9, then the card in the 4th row and 3rd column is numbered (4 - 1)9 + 3 = 30. Note that this card is 3 rows from the top of the array, and 2 columns from the left. Now consider its "twin", namely, the card which is 3 rows from the bottom of the array, and 2 columns from the right. It's in row 5 - 3 = 2 and column 9 - 2 = 7, and is numbered (2 - 1)9 + 7 = 16. Observe how the sum of the numbers for the card and its twin is 30 + 16 = 46 = 4x9 + 1, which is also the sum of the numbers for the very first card and its twin, the last card. This is no accident; we claim:
As already remarked, the card in the ith row and jth column is numbered (i - 1)c + j = ic - c + j. Its twin is in the r - (i - 1)th row and the c - (j - 1)th column and thus is numberws ([r-(i-1)]-1)c+[c-(j-1)]=rc-ic+c-j+1. Summing those yields ic-c+j+rc-ic+c-j+1=rc+1, as desired.
It's clear that the above also gives us a way to identify a card's twin: check if the two numbers sum to rc + 1. For instance, if r = 5 and c = 9 as before, then the cards with numbers 11 and 35 must be twins.
Suppose that the packet to be dealt out is a palindromic stack containing an even number of cards 2n, namely a packet for which the last n cards are in some sense "the same" as the first n cards, only in reverse order. We can think of the second half as matching the first half, card by card, but in the opposite order. For instance, A♣, 2♣, 3♣, 4♣, 5♣, 6♣, 6♥, 5♥, 4♥, 3♥, 2♥, A♥. One could be sloppier with the suits, in order not to make a later outcome look so planned, such as in the cards suggested for the digits of π effect (A♠, A♣, 3♠, 4♠, 5♣, 9♣, 9♥, 5♦, 4♦, 3♥, A♦, A♥).
There is a wonderful property of palindromic stacks that we need: dealing such a twelve-card packet into 2, 3, 4 or even 6 piles, and reassembling the piles in order, results in a packet which is still palindromic. This is the so-called Stay Stack principle, which has been known and studied, usually for just 2 pile deals (c = 2 below), since at least the late 1950s. Its 1957 incarnation, as published by Rusduck (J. Russell Duck), was in the context of perfect/Faro shuffles. In general we assert:
It's a fun exercise to verify that this holds, using the row and column table displayed earlier and the fact that any card's number added to that of its palindromic twin always yields rc + 1.
Another wonderful property of palindromic stacks is that when they are split in the middle, and the two halves are riffle shuffled, then some "order" remains, in the sense that one last split into two halves preserves representation of the values involved. For instance, if we start with the stack A♣, 2♣, 3♣, 4♣, 5♣, 6♣, 6♥, 5♥, 4♥, 3♥, 2♥, A♥, then after any amount of deals (into possibly different numbers of piles), and a single riffle shuffle, the top half (and likewise the bottom half) will contain an Ace, 2, 3, 4, 5 and 6, in some order. (In the case of the recommended stack for the 2π effect, each half ends up containing two Aces, a 3, 4, 5 and 9, with appropriate colour switching. Recall, the rest of the trick is contrived to ensure that each of the two piles ends actually up in the order of the start of the decimal expansion of π.) The explanation for this "set preservation under a riffle shuffle" property is the second Gilbreath principle explored here in August 2006. There, one usually thinks of cards dealt from a packet consisting of repeated stacks being riffle shuffled into the remainder of the packet. Of course, an even-sized palindromic stack split in the middle, in readiness for riffle shuffling, is the same as a duplicated stack dealt off the top (reversing the card order) first.
The inspiration for considering the not-so-randomizing deals above at this time derives from British magician Stewart Murray of KR Professional Magic, author of the upcoming book Palindromic Magic Revealed - Theory Techniques Tricks. Special thanks to him a lively correspondence in recent weeks. Palindromic principles have been around for a very long time, needless to say, and are perhaps worthy of additional exploration in future Card Colms. They are the focus of an entire book by Phil Goldstein, naturally enough entitled RedivideR (2003).
When dealing with π, circles are hard to avoid, as we discovered above. We now offer additional circular π trick possibilities, based on a 7- or 8-digit approximation to that important irrational ratio, i.e., 3.1415926(3). As before, cards with those values--eight of them this time--can be stacked at the top of a deck and maintained there through some shuffles. Deal these into a face-down circle, anti-clockwise starting at the 3 o'clock position. Unknown to any but you, the cards--note the CHaSeD suit order--are as shown in the image on the right below.
There are several properties of this display which can be exploited for other tricks. (1) The "compass points" are 2, 3, 4 and 5 (and black). (2) The sums of pairs of opposite numbers are 6, 7, 8 and 10. (3) The sums of three adjacent numbers (around the circle) are 8, 6, 10, 15, 16, 17, 11 and 10. (4) The products of three adjacent numbers (around the circle) are 12, 4, 20, 15, 90, 108, 36 and 18, which are distinct. Hence, this "pi bracelet" is almost sum-rich, in the sense explored in the June 2006 Card Colm, and is most definitely product-rich. The duplicate 10 for triple sums can be avoided by dealing the eight cards into a row instead of a circle, thus removing wrap-around sums from consideration. Knowing the sum or product basically allows one to deduce the identity of the three cards used to compute it.
Riffle shuffling two piles of six cards, as required earlier, is not so easy, as there is "so little to grasp" with each hand. An alternative procedure, which has the same effect, is the rosette shuffle often employed by Swedish magic maestro Lennart Green. It works with two piles of any size.
Start with the piles side by side, as shown in the first image, then use the fingers to "twirl" the left pile into a so-called rosette; the second image shows the result. Repeat for the right pile, and push the rosettes together. Square up the packet: it has effectively been riffle shuffled.
"(A) Pi Evolved Set" is an anagram of "Steve Loved P(a)i" and "Harmonic Split Drill" is an anagram of "Palindromic Thrills." "Need A Pi List Lingo" and "One Aligned Pi List" are anagrams of "Dealing Into Piles." Finally, "Shiny Portabilities" is an anagram for "Another πssibility."