This is a step-by-step reverse engineering of the performance of the trick Richard has put on youtube here. Watch that, and applaud him for its cunning, first. In simple terms, unlike the Whispering Jokers trick, the good professor didn't fool me. However, that doesn't make this a bad trick, just one that is aimed at those that don't have as much of a mathematical pattern-matching mind as I do.
This explanation is written more as a discovery - non-obvious insights are only explained in the order that they're revealed, which means that they may be incomplete at first, and then revisited later when more information is available. However, it is assumed that you've seen the trick from beginning to end so that it's clear what I'm referring to.
The deck clearly has three parts:
Immediately obvious to a mathematician from the video was that when the targets were shown to the mark, all of the numbers were even, and none were divisible by four (a mathematician might say "numbers of the form '4n+2'"). Note that at 3m28s into the video, Richard explains that "they're all different", which does not correspond to the reality we see - there are clearly two 22's and two 26's. Absent from the "4n+2" pattern are 38 and 46; at this stage in the trick it's not known if that is significant, later we'll see it's not.
These are, in order,
8, 7, 9, 10, 11, 1, 6, 3, 2, 12, 4, 5. When dealt
into 4 piles, the 4 piles will be
[8, 11, 2],
[7, 1, 12],
[9, 6, 4],
[10, 3, 5]. When finally dealt out, and spread
apart, they will have the following arrangement:
8 11 _ 1 _ 2 7 12 3 _ 9 6 10 5 4 _As you can see, each row, column, and diagonal has exactly one gap in it, and their sums are as follows:
/ 18+D 8 11 B 1 = 20+B A 2 7 12 = 21+A 3 D 9 6 = 18+D 10 5 4 C = 19+C ------------------- \ 21+A 18+D 20+B 19+C 19+CThe piles of 4, and the para-diagonals, also have the same pattern to their sums.
Those familiar with perfect squares may recognise the following as being a pandigital one:
1 12 6 15 14 7 9 4 11 2 16 5 8 13 3 10and indeed the above is just a rotated version of the template above, with
A, B, C, and Dreplacing
13, 14, 15, and 16.
It is clear therefore what the final section of the deck must begin with in order for the trick to work, namely a quad of numbers in increasing sequence, and in fact with predetermined values such that the sums match the target. As each additional card only affects one pile, one row, one column, and one diagonal, incrementing the starting value of the quad by one will increment all of those sums by one also, therefore any starting value for a quad works.
What may not have been obvious is how this part of the deck is manipulated. Once the first 8 target cards are given to the mark, the target chosen, and the 12 top remaining cards ("X") placed to one side, a seemingly arbitrary number of cards ("Y") are also placed to one side (complete with a clever little feint to continue, making it feel like that stopping point was arbitrary), leaving some cards ("Z") in the magician's hand. However, on collecting those 3 piles, they are placed in the order "XZY". I.e. the seemingly arbitrary number of cards ("Y") are removed from relevance, and the 4 top cards of "Z" are made the ones that will be dealt out into The Grid.
The clear implication is that the targets are marked cards, so that they can be read, and the appropriate size of "Y" can be discarded.
As we saw, the targets were
22, 26, 30, 34, (not 38), 42, (not 46), and 50.
In order to sum to those, the quad raised to the top of "Z" and its top card
in bold must be as follows:
|22||1, 2, 3, 4|
|26||5, 6, 7, 8|
|30||9, 10, 11, 12|
|34||13, 14, 15, 16|
|38||17, 18, 19, 20|
|42||21, 22, 23, 24|
|46||25, 26, 27, 28|
|50||29, 30, 31, 32|
There seems to be no intrinsic reason why
38 and 46 are not
in the set of targets. I conclude that the "all different" claim could be
made true. The false claim made jars more with this revealed (unless I'm
overlooking something). I don't mind magicians lying to me - everything
(important to the trick) that they do is lying, it's that this easily
falsifiable lie was unnecessary (perhaps it was true in the past, and has
just become part of the patter?).
Note that it is completely irrelevant what order these quads appear in in the third section of the deck, all that matters is that you know how many to deal out, or what top card you stop on, in order to get the appropriate quad for the selected target.
The skill involved in this part is that of arranging the 4 cards in a pile without making it too obvious that you're trying to build a magic square. Fortunately, the direction each card needs to be shoved in is identical every time, so it can be done without thinking.
This is a relatively easy flourish, requiring nothing more than the
12 fixed cards having the appropriate words or blanks on the back, and
C card have the appropriate target number.
Even that's not necessary, as the earlier-revealed
A, B, and
D cards could contain earlier words in the phrase.
The "4n+2" pattern was immediately obvious to me, and a give-away
that there was some trivial mathematical pattern involved. This pattern
is not necessary. Any target can be achieved. Were
A, B, C, D
2, 3, 4, 5, then the target 23 would be obtained, for example.
As none of the quads in the "rest of the deck" overlap each other,
they can be independently selected. Having the targets as prime numbers,
for example, would have been a much better decoy - the smart ones who
pattern match such things would have been immediately sent off on a wild
If one is willing to forego the message part of the trick, or at least the word-at-a-time nature of it, then the quads in the "rest of the pack" can in fact overlap each other, as long as the increasing order of any sequence of 4 that you stop on is preserved. This permits more targets to be attained. The "message" can be put back in by, for example, writing the whole of it on the back of one of the 4 (or any fixed position) in the quad, or perhaps writing the words of the message on one of the 12 fixed cards, complete with an arrow pointing to the appropriate "Z" card. In the latter case, I'm pretty sure that with the right selection of targets, one could have the 1-12 card that is turned over selected by the target, rather than being arbitrary (e.g. sum the digits). (More investigation needed...)
Another hastily constructed page by Phil Carmody
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