April 2009
In Sheer Luck, the December 2004 Card Colm, we highlighted a wellknown card control based on an elimination deal credited to Stewart Judah in the 1930s. It hinges on the fact that the 22nd card from the top is the last one dealt when the following procedure is applied to a full 52card deck, alternately dealing cards to two people (e.g., yourself and a spectator) in several rounds:
The Yummie Deal: Deal a packet of cards to two piles, first to a spectator (pile A),
and then to yourself (pile B), saying "You, me," silently to yourself over and over. Then,
pick up pile B and deal again, first to the spectator, thereby adding to the existing pile A,
and then to yourself, forming a new pile B. Repeat, picking up the diminished pile B,
and dealing "You, me" as before. Eventually, just one card remains in pile B.
Note that "Yummie" sounds a bit like "you, me" said fast. Also, "Yummie ABs" may help in remembering that you deal to A (spectator) before own B(eing).
If the last card is placed on top of pile A, we get a permutation (i.e., a rearrangement) of the whole packet; this is implicit in the mathematical analysis that follows. Also, after the first choice, of whichequally sized, all things being evenpile to pick up for the second deal, it's always the smaller pile that is picked up for additional dealing.
Given a packet of k cards, and a particular welldefined deal such as the Yummie Deal, there exists a unique position i such that the card starting in that position ends up as the last card dealt. We can easily determine i as follows. Order the cards from top to bottom in some specific order, such as: A♣, ..., K♣, A♥, ..., K♥, A♠, ..., K♠, A♦, ..., K♦so that, e.g., the 15th card is 2♥, omitting the cards past the kth one if fewer than 52 cards are used. Next, perform the deal and observe what card is dealt last.
For instance, with a full deck of 52 cards, ordered as above, the last card is 9♥, which was originally the 22nd card. So, k = 52 does indeed imply that i = 22, as explored in Sheer Luck.
Here's a twist on the elimination idea, which uses some old ideas of Charles Jordan and others. It's a 2001 cocreation with Gordon Bean.
The Either/Or Trick
"Here's a trick based on pure logic, which I call The Either/Or Trick." A spectator is handed the deck, and invited to cut it repeatedly. Take the cards back, and deal them into two piles, inviting the spectator to decide which cards to continue with by pointing at each pile in turn and saying, "Either we discard this pile, or we discard this pile. It's your choice." The twentysix discarded cards are set aside, and the rest are dealt out into two new piles, and the same choice offered. The thirteen discarded cards are placed on top of the previously rejected ones, and the dealing and chosing continues as before, until only one card remains, along with a pile of fiftyone discarded cards.
Pick up the solo remaining card, saying, "This is your chosen card, determined with no input from me." Have it noted and then plunged into the middle of remainder of the deck. The cards are riffleshuffled and then cut. Take the deck back, and remark that you had no control over what card was chosen or what subsequently happened to it. Glance through the cards quickly, saying, "Recall that at each deal you either discarded one pile or the other. Actually, the real reason this is called the Either/Or Trick is because either I find your card, or you won't be at all impressed." Pause while that sinks in, and then produce the chosen card triumphantly.
The deck is setup in advance using the CHaSeD order: the suits cycle ♣, ♥, ♠, ♦. When you get the deck back, start as in the Yummie Deal. The first deal separates the deck into black and red packets of twentysix cards each, the second deal separates one of those into two suits. As the spectator makes choices, place each rejected pile on top of the previous ones. The third deal leads to six or seven cards of the chosen suit being added to the growing pile of rejected cards, and so on. The final cardthe chosen oneneed only be returned to the deck at least a quarter of the way down to be very easy to find, despite the last riffle shuffle and cut: simply look for threads for each suit, and you should be able to locate the one card which is out of place relative to the rest.
(It's a good idea to shuffle the deck before anybody else inspects it! A simple ruse is to locate the chosen card, cut it to the top of the deck, and then hide all of the evidence by doing several riffle shuffles while maintaining the top card. This also makes the revelation more interesting; you can add a little showmanship to conclude the trick.)
Good Deals on a Smaller Scale
The Yummie Deal applied to a whole deck numbered 152 leads to the cards being in order 22, 38, 6, 14, 30, 46, 50, 42, 34, 26, 18, 10, 2, 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 51, 49, 47, ..., 5, 3, 1. As a permutation, this is the product of a 49cycle, a 2cycle and a 1cycle; specifically, using standard cycle notation, it's: (1 52 26 10 12 16 17 44 24 19 43 31 37 34 9 48 25 40 23 41 32 21 42 8 15 45 30 5 50 7 49 28 20 18 11 47 29 38 2 13 46 6 3 51 27 39 33 36 22) (4 14) (35). It follows that the original 35th card is fixed, and a second Yummie Deal brings to the top the card that started in position 36.
Any permutation, if repeated sufficiently often, restores a packet to its original order, but there is little hope of locating spectators who will let you do the Yummie Deal on a whole deck to the point where it is back as it started, even assuming that this feat would impress them. The permutation in question has period 49 x 2 = 98, and while that number of Yummie Deals will indeed restore order to the deck, your audience may prove less compliant. It's time to consider working with much smaller packets.
For Yummie Deals applied to packets of size k (up to 28, a perfect number to stop at), we now give the cycle decompositions of the resulting permutations, as well as the periods of the dealsnamely, the least common multiple of the lengths of the constituent (disjoint) cyclesand a list of the fixed points, i.e., the positions of the cards which effectively do not change. Transpositions, that is to say pairs of positions which are merely interchanged, are highlighted in bold font for ease of reference.


YUMMIE DEALS 

k 
period 
cycle decomposition 
fixed points 
2 
2 
(1 2) 
 
3 
3 
(1 2 3) 
 
4 
3 
(1 2 4) 
3 
5 
5 
(1 2 4 3 5) 
 
6 
5 
(1 4 5 3 6) 
2 
7 
6 
(1 4 7) (3 6) 
2, 5 
8 
7 
(1 6 5 7 3 4 8) 
2 
9 
15 
(1 6 7 5 9) (3 4 8) 
2 
10 
20 
(1 8 5 10) (2 4 6 9 3) 
7 
11 
11 
(1 8 7 9 5 10 3 2 4 6 11) 
 
12 
24 
(1 6 12) (2 10 5 8 9 7 11 3) 
4 
13 
24 
(1 6 12 3 2 10 7 13) (5 8 11) 
4, 9 
14 
14 
(1 8 13 3 4 2 12 5 6 10 9 11 7 14) 
 
15 
6 
(1 8 15) (2 12 7 14 3 4) (5 6 10 11 9 13) 
 
16 
28 
(1 6 8 16) (2 14 5 4) (3 10 13 7 12 9 15) 
11 
17 
17 
(1 6 8 16 3 10 15 5 4 2 14 7 12 11 13 9 17) 
 
18 
120 
(1 8 14 9 18) (2 16 5) (3 12 13 11 15 7 10 17) 
4, 6 
19 
55 
(1 8 14 11 17 5 2 16 7 10 19) (3 12 15 9 18) 
4, 6, 13 
20 
180 
(1 6 4 10 20) (2 14 13 15 11 19 3 18 5) (7 8 12 17) ( 9 16) 
 
21 
21 
(1 6 4 10 20 3 18 7 8 12 19 5 2 14 15 13 17 9 16 11 21) 
 
22 
18 
(1 8 10 18 9 14 17 11 22) (2 16 13 19 7 6) (3 20 5 4 12 21) 
15 
23 
60 
(1 8 10 18 11 22 3 20 7 6 2 16 15 17 13 21 5 4 12 23) (9 14 19) 
 
24 
42 
(1 14 21 7 4 18 13 23 3 22 5 10 16 17 15 19 11 20 9 12 24) (2 6) 
8 
25 
90 
(1 14 23 5 10 16 19 13 25) (2 6) (3 22 7 4 18 15 21 9 12 24) (11 20) 
8, 17 
26 
153 
(1 16 21 11 18 17 19 15 23 7 2 8 6 4 20 13 26) (3 24 5 12 22 9 10 14 25) 
 
27 
140 
(1 16 23 9 10 14 27) (2 8 6 4 20 15 25 5 12 22 11 18 19 17 21 13 26 3 24 7) 
 
28 
429 
(1 14 28) (2 6 10 12 20 17 23 11 16 25 7) (3 22 13 24 9 8 4 26 5 18 21 15 27) 
19 

This table reveals many interesting facts which can be adapted to performance. For instance,

For packets of size eight to twentythree, either the original 6th or 8th card ends up as the last remaining card dealt. Moreover, there is a simple way to predict which. This observation was applied in Temple Patton's book Card Tricks Anyone Can Do (Castle Books, 1968).

For a packet of size nineteen, three cards are effectively unmoved by the whole experience.

For a packet of size twentyfive, six cards will end up in their original positions after two applications of the Yummie Deal.
The Cat's Deal
A small modification to the Yummie Deal leads to:
The Cat's Deal: Deal a packet of cards to two piles, first to yourself (pile B),
and then to the spectator (pile A), saying "Me, you," silently to yourself over and over.
Pick up pile B and deal again, first to yourself, forming a new pile B, and then to the
spectator, thereby adding to the existing pile A. Repeat, picking up the diminished pile B,
and dealing "Me, you" as before. Eventually, just one card remains in pile B.
Note that "me, you" said fast sounds a little like "meow"hence the cat connotation. Also, while cats and sheep don't have that much in common, we suggest thinking "BA" to aid in remembering the fact that this time you deal to your own B(eing) before dealing to A (spectator). Again we assume, for mathematical reasons, that the last card is placed on top of pile A, so that we get a permutation of the whole packet, and once more, after the first choice of which pile to pick up for the second deal, it's always the smaller pile that is picked up for additional dealing.
If we apply the Cat's Deal to a full deck of 52 cards, the last card is the one that started in position 35 in the deck. Curiously, this is the very card which remained fixed under the Yummie Deal; invent your own trick here to exploit this happy coincidence! When the Cat's Deal is applied to a deck numbered 152, assuming that the last card is placed on top of the others at the end, the cards end up in order 35, 3, 51, 19, 11, 27, 43, 47, 39, 31, 23, 15, 7, 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, 45, 49, 52, 50, 48, 46, ..., 6, 4, 2. This permutation is the product of an 8cycle, a 27cycle, a 15cycle and a 2cycle. Specifically, it's (1 14 46 30 38 34 36 35) (2 52 27 6 50 28 39 9 16 45 25 20 43 7 13 17 18 44 31 10 48 29 21 19 4 51 3) (5 15 12 47 8 49 26 40 33 22 42 32 37 23 11) (24 41). Hence, the Cat's Deal requires even more tedious repetition to get the whole deck restored to its initial state: the permutation in question has as period 1080, the least common multiple of 8, 27, 15 and 2. Don't try this at home (or in public)! Here is the table for Cat's Deals applied to packets of size k (up to 28):


CAT'S DEALS 

k 
period 
cycle decomposition 
fixed points 
2 
1 
( ) 
1, 2 
3 
3 
(1 3 2) 
 
4 
3 
(1 3 4 2) 
 
5 
2 
(2 5) 
1, 3, 4 
6 
4 
(2 5 4 6) 
1, 3 
7 
6 
(1 3) (2 7) (4 5 6) 
 
8 
10 
(1 3) (2 7 4 5 8) 
6 
9 
6 
(1 9 2) (3 5 7 6 8 4) 
 
10 
10 
(1 9 4 3 5 7 8 6 10 2) 
 
11 
14 
(1 11 2 3 7 10 4) (6 9) 
5, 8 
12 
12 
(1 11 4) (2 3 7 12) (6 9 8 10) 
5 
13 
30 
(1 9 10 8 12 4 5 3 13 2) (6 7 11) 
 
14 
36 
(1 9 12 6 7 11 8 14 2) (3 13 4 5) 
10 
15 
24 
(1 11 10 12 8 13 6 5) (2 3 15) (4 7 9 14) 
 
16 
14 
(1 11 12 10 14 6 5) (2 3 15 4 7 9 16) (8 13) 
 
17 
12 
(2 17) (3 9 15 6) (4 13 10 16) (8 11 14) 
1, 5, 7, 12 
18 
56 
(2 17 4 13 12 14 10 18) (3 9 15 8 11 16 6) 
1, 5, 7, 11 
19 
55 
(1 3 11 18 4 15 10 17 6) ( 2 19) ( 5 7) (8 9 13 14 12 16) 
 
20 
66 
(1 3 11 20 2 19 4 15 12 18 6) ( 5 7) (8 9 13 16 10 17) 
14 
21 
10 
(2 17 10 15 14 16 12 20 4 21) (3 9 11 19 6 5 13 18 8 7) 
1 
22 
60 
(2 17 12 22) (3 9 11 19 8 7) ( 4 21) (5 13 20 6) (10 15 16 14 18) 
1 
23 
14 
(1 3 11 17 14 20 8 5 15 18 12 21 6 7) (2 19 10 13 22 4 23) 
9, 16 
24 
110 
(1 3 11 17 16 18 14 22 6 7) (2 19 12 21 8 5 15 20 10 13 24) (4 23) 
9 
25 
198 
(1 9 7 5 21 10 11 15 22 8 3) ( 2 25) (4 17 18 16 20 12 19 14 24) (6 13 23) 
 
26 
126 
(1 9 7 5 21 12 19 16 22 10 11 15 24 6 13 23 8 3) (2 25 4 17 20 14 26) 
18 
27 
140 
(1 11 13 21 14 25 6 15 26 4 19 18 20 16 24 8) (2 27) (5 23 10 9) (12 17 22) 
 
28 
133 
(1 11 13 21 16 26 6 15 28 2 27 4 19 20 18 22 14 25 8) (5 23 12 17 24 10 9) 
3, 7 

This table also reveals several interesting facts, e.g., for packets of size seventeen or eighteen, four cards are fixed.
Meaty Music
Combining Yummie and Cat's Deals leads to all sorts of interesting possibilities. For instance, if you are confident in your abilities to split a full deck exactly in halfit's a nontrivial skill but one worth acquiringhave two matching cards, say the two red Kings, in positions nine from the top, and eleven from the bottom. Split the deck roughly in half and riffle shuffle, being careful to drop a lot from the bottom part at first and a lot from the top last, so as not to disturb the planted cards. Then split the deck exactly in half, and give the top half to Spectator C and the bottom half to Spectator D. Guide C through the Cat's Deal, until one card is left. Have it set aside face down. Then, guide D through the Yummie Deal, until one face down card is left. Remark that it would be really amazing if their cards, like their methods of selection, somehow complemented each other. Have both cards turned over at the same time; they should be the two red Kings.
Alternatively, if the precise deck splitting requirement seems too risky, one can arrange the two desired matching cards so that when twentysix, or some other specific number, are casually (and silently) counted into a pile, thereby reversing their order, the resulting two packets can have elimination deals (perhaps of the same type) done on them to tease out the matching cards. As long as you do the counting out, the rest will follow in the spectator's hands.
One can also insert Jokers into the packet at strategic locations, with the goal of those being the two last cards; this is one reason we list all relevant information for packets up to size 28. Have fun experimenting.