The Slit Envelope

Various forms of slit envelopes have been used by mind
readers as a means of quick access to the writing within. As the
title indicates, a secret slit, or opening is made in the envelope
with a knife. The slit may be at the bottom edge, the side edge,
or at some point on the face or back of the envelope, depending
upon the routine to be followed.
In working the one ahead billet switch wherein the thumb tip is
employed (Thumb Tip Steal, One Ahead) as previously
described, the idea of using a slit envelope instead of a thumb
tip was suggested. Some experimenting revealed a simple and
very practical method which is here described.
Cards and drug envelopes are used, the latter having the usual
end flaps with gum for sealing. One envelope is prepared by
slitting open the bottom crease along its entire length. The
others are unprepared. The prepared envelope is at the bottom of
a stack of say a dozen envelopes, all held in one hand along
with the cards. The cards are passed out first, and the spectators
are instructed to write their names, questions, drawings, etc.
The envelopes are then distributed until you get down to the last
(prepared) envelope which you hold by the slit end and while
directing one of the spectators to turn his card over, you reach
for it and insert it in the fake envelope. You are talking to the
whole audience as you do this--in other words, all of them are
cautioned to let no one see what they have written--they are to
turn the written side downwards, insert card in envelope and
seal it. You seal the faked one by way of illustration and
proceed to collect the others, putting them underneath that one.
Holding the pile in one hand as you return to the stage, it is an
easy matter to secretly extract the card from the slit envelope
and place it on top of the envelope, using the fingers of the other
hand for this move. The written side of the card is exposed to
you, and faces you as you hold the stack in left hand.
The bottom (unprepared) envelope now faces the audience, and
this one is removed by the right hand and raised to the level of
your eyes. You pretend to read the contents, but, of course, read
the card facing you on the pile of envelopes.
After completing your answer, the raised envelope is placed
between the first and second fingers of the left hand, being thus
in full view of the audience. The pile of envelopes is still held in
left hand, being gripped between thumb and first finger. The
right hand picks up a pair of scissors and snips off the end of the
separated envelope. The scissors are put down, and the right
thumb and index finger extract the card which you hold with
writing towards you as you read it "to confirm" your
impressions, doing the regular one ahead stunt. Now a simple
switch of cards must be made, and the move is screened by the
envelopes held in left hand.
The right hand, holding the card between the thumb and first
finger, approaches the left hand for the Ostensible purpose of
getting the separated envelope, and that is all that seems to take
place. However, the card in right hand is placed on the pile, and
the other card removed as the separated envelope is gripped
between the first and second fingers of right hand, and both the
card and envelope are drawn away in the right hand at the same
time and may be handed to the writer.
A little practice on this exchange move will show how easy it is,
and familiarity will perfect its quick and smooth execution. The
same routine is followed with the remaining envelopes which
you take from the bottom until but one remains--the slit
envelope--with the last card on it. This is treated just as though
the card were inside, you raising it to your forehead, giving the
answer, and then cutting off the end (slit end). Insert the first
and second fingers as though to get card and with thumb on
back, the card is "extracted" in a move that is perfectly natural.

Stealing Folded Billet from Tray

With a full understanding of the foregoing, it will be obvious
that no confederate is needed if the performer can secretly gain
possession of any one question and ascertain its contents before
he starts to give his answers.
One of the ways to do this is as follows:-- The performer carries
a small opaque tray and some blank slips of paper about 1-1/2"
x 3-1/2" down into the audience. He passes out the slips on
which the spectators are requested to write their questions. After
this is done, he explains the necessity for secrecy and requests
each spectator to fold his slip in half with the writing inside,
then fold again in half the other way.
This done, the performer collects the questions on the tray. All
the time the performer has held a folded blank slip on the under
side of the tray, concealed by the fingers of the left hand. As he
returns to the stage, the performer changes the tray from the left
hand to the right, at the same time performing two simple secret
moves. As the right hand takes hold of the tray, the right thumb
slides off one of the question slips lying near the edge of tray.
This question is slid over the edge of tray and falls into the right
hand where it lies concealed between the right hand and the
bottom of the tray.
The left hand at the same moment has done the reverse with the
fake billet. As the tray is withdrawn from the left hand, the
blank folded billet remains in the left hand until the edge of the
tray passes over it when a slight motion of the left fingers will
flip the billet in with the others on the tray. This fake billet has a
bent corner or other secret mark by which it can be easily
distinguished, otherwise the performer will have to follow it
with his eyes so as to avoid picking it up until the end.
The tray is emptied on the table, the stolen question being
secretly held in the palm of the right hand at the roots of the
fingers. On the table is a crystal gazing ball resting on a
cushion. The performer sits down back of the table, picks up a
question slip from the pile, holds it to his forehead and gazes
into the crystal. In the meantime, the right hand is back of the
cushion quickly opening the stolen billet so it can be read, and
announced as the one held at the forehead. The one ahead
system is followed throughout, the stolen billet being switched
for the blank billet after the performer has pretended to read it at
the finish.

The One-Ahead Principle

This is one of the oldest, and, at the same time, one of the
simplest methods ever devised for secretly reading questions. It
has been used by mind readers, pseudospiritualists and others
with great success for many years before both large and small
audiences. As originally performed, it was not a one-man effect;
hence, we shall first describe it that way and then give the
variations used by the single performer.

Original One-Ahead Method
The spectators are requested to write their questions on slips of
paper and to let no one see what they have written. These slips
are then folded by the writers so the writing is concealed on the
inside. Envelopes may be furnished and the questions sealed
within, if desired. This is a fair sized audience where some write
questions and some do not.
In any event, a total stranger is directed to collect the questions
and dump them on the table on the platform. The performer
picks up one of the sealed envelopes, holds it to his forehead, as
though to "see" with supernatural vision and shortly announces
that he "gets" the initials M. C.--"is M. C. present?" M.C.
acknowledges that she wrote a question, whereupon the
performer proceeds to give a suitable answer to her question
about a trip to Boston, etc. As he finishes the answer, the
performer tears open the envelope and reads aloud the question,
"Will I go to Boston? M. Clark."
That is what he appears to do, but all is not what it seems. M.
Clark wrote a question all right, but M. Clark is a confederate of
the performer's, and wrote that question by agreement, secretly
bending over a corner of her envelope so as to distinguish it
from the others, identifying it as the one to be avoided until the
last. The performer has picked up and opened some other
envelope, first pretending that it belongs to M. C. Then, as he
looks at the open slip, he orally reads "Will I go to Boston? M.
C.", but, in reality, he is at the same time mentally reading and
remembering the question written on the genuine slip before
him--let us assume this slip hears the question, "Will father get
well?--signed John Jones."
The performer has thus apparently confirmed his divination of
the first question, and now knows a genuine question which he
pretends to be the one contained in the next (second) envelope,
which he reads while held at the forehead as before. The same
procedure is followed throughout, each new envelope torn open
furnishing the data for the next reading.
The performer can stop at any time but if he desires to read all
of the questions, he leaves the M. Clark envelope till last, and
after it is torn open to apparently confirm the last test, it is
tossed among the others, and they may all be returned to the
writers.

Time Zones By Edward Marlo

Set-up: As in .Time For Si Stebbins,. the deck is previously set in Si Stebbins order.

Method: Table the deck and request that Spectator A cut the pack several times, giving
it straight cuts. As this is done, turn your back to the audience.
Next ask Spectator A to think of any hour on the clock and to remove an equal number
of cards off the top of the deck and so on, following the procedure already outlined
several times in this manuscript.
When he has finished following your instructions, turn around and pick up the talon. In
squaring the cards or in the process of handing them to Spectator B, glimpse the bottom
card. Remember it for-it keys your first force card--the one Spectator A will eventually
select.
Instruct Spectator B to repeat the same actions of Steps 2 and 3. Once again you turn
your back. Turn around and pick up the remaining cards and place them aside or in your
pocket. Use any ruse for logically handling the talon as you glimpse the bottom card. As
in Step 4, you now know the card Spectator B will eventually select.
Pick up each 12-card packet in front of each spectator (in turn) and deal them into a pair
of clock configurations, placing them side by side. At this point you can go into some
business about .different time zones.. and that Spectator A.s clock dial represents
Eastern Standard Time and Spectator B.s clock dial represents Central Standard Time.
Turn your back and ask each spectator to note the card at his chosen hour. Request that
both spectators assemble all the cards, having one volunteer shuffle them to further mix
them. After some byplay, name both selections. Several patter approaches should
suggest themselves. Also the student can work out several handlings on the glimpses
and when and how they are taken.

Best Of Nine Card

Effect
It’s Open Evening and the hapless parents are being dragged round the Maths Is Fun
Department by their goggle-eyed offspring to play with all the games and puzzles on
display. I pick the parent who most clearly would prefer to be at home right now and
ask him to look at a pack of 9 cards which I have just dealt out from my shuffled
deck. I then ask him to take out his favourite and return the others face down to the
table. By now there is a little crowd forming around my desk, so he knows there is no
backing out now. Finally I ask him to show his card to a few others before placing it
on top of the other face down cards. I give the remainder of the deck a quick shuffle
and complete the pack. I thank the parent for his efforts, and promise that he has done
all the hard work. The rest of the trick will be done by the cards and some
devastatingly devious algebra.
Picking up the pack, I deal out the first card face up, saying “Ten!”. On top of that I
deal the second card “Nine!” and so on down to “One!”. I then place a face down
“lid” on that pile with one other card and repeat the process three more times, making
4 piles altogether.
If a face up card appears with the same number as the one I am saying then I stop and
move on to the next pile, starting again from “Ten!”. “These cards seem to be telling
me something” I mutter mysteriously.
When the last pile is complete, I have some cards in my hand. On the table in front of
me are some face up cards, let’s say they are a 3 and a 5. (“They are a 3 and a 5!”) I
now add these numbers together and count down to the eighth card in my hand. It is,
of course, the parent’s card. Pumping his hand vigorously, I thank him for his time,
and explain that Maths really does have many surprising uses.
Method
Johnny Ball named this trick as his favourite card trick of all. It is completely selfworking,
and the underlying algebra is certainly accessible to school children.
When I place the balance of the deck on top of the spectator’s pile of nine, it makes
his card 44th from the top (with 8 below it). The fancy counting is just doing 4 x 11.
If there are no matches, then the final “lid” is the spectator’s card, but this rarely
happens. If I stop part way through, then the number on display tells me how many
cards are missing from the intended 11.
If there are n cards in the pile then I need (11-n) to complete it.
As I deal the nth card I am saying the number (11-n), and the number (11-n) is on
display if I get a match.
After dealing four piles in this way, the cards needed to complete each pile are still in
my hand. Adding the face up cards is equivalent to placing them back on their piles,
and the final card is therefore the 44th card. I usually ask the spectator to name his
card first. “Seven of hearts” he says. “Not this seven of hearts by any chance?” I ask,
as I turn over the final card.
Most packs of cards in school have a few cards missing. If this is the case, then just
subtract the number of missing cards from the 9 in the introduction.

Subtle Glimpse of Folded Billet

This method of ascertaining the contents of a genuine billet is
subtle because of its very boldness. No one would suspect that
the performer would do this bare faced deed, yet it is
accomplished with the greatest ease. It can be done before small
groups, but is best suited to audiences of thirty, forty, or more.
We have witnessed a performance of
this method, the billets, or slips of
paper measuring about 2-3/4" x 2-
3/4", or a trifle less. This particular
performer used paper of a Golden
Rod color but white would do as
well. Prior to the show, the
performer had folded each of these
slips three times as in Fig. 1 (Note
A, B, C, and D).

After which the billets were just oneeighth
of the original size, and they
were numbered consecutively on the
outside from 1 to 40, or 150 or more,
according to the size of the audience.
In his vest pocket the performer has a number of soft lead
pencils, and with the folded billets in his left hand, on top of a
few opened billets (but with creases), he is ready to proceed. He
makes the usual opening talk about mind reading, etc., and
explains that he has the spectators write their questions the
better to concentrate on them. He steps down in the audience,
and as he passes out the billets and pencils, he calls attention to
the fact that the slips are numbered--each spectator is to
remember his number, then open his billet and write his name at
the top, the question underneath, and finally fold the slip in the
original creases. No one is to show or tell what he has written
but must concentrate his thoughts on the question.
The performer now and then opens a billet before handing it to a
spectator. He moves about from one side of the aisle to the
other, and sometimes steps back and forth, always keeping on
the move, as he talks and passes out the slips. He finally has
only two folded billets remaining in his left hand, on top of the
open billets. One of these folded billets, say, No. 5, was
originally given a bent corner when folded by the performer so
he can distinguish it from the others. He has carefully withheld
this No. 5 billet and now hands it to a spectator who is requested
to hurry up "as the others are all ready."
Performer steps across the aisle, says "anybody else?" hands out
remaining folded billet to a spectator, and at the same time
watches No. 5 to see when he has finished writing, then speaks
out "All ready, we will now collect the questions, who will
volunteer? Anybody?" (man offers his services). "All right,
thank you, sir, go down to the front, and start there--take your
hat for the purpose." The performer turns and looks toward the
rear, and acknowledges an imaginary request for a slip, by
saying, "Yes sir, just a moment, I'll be right there." Performer
quickly turns to No. 5, and extends his hand, quietly saying,
"Are you ready?" and takes the No. 5 question, turns away
quickly and walks toward the rear.
Now comes the bold move. Performer's hands are brought
together and he quickly and secretly opens the No. 5 question as
he walks rapidly to the rear, inquiring as he goes, "Who was it
that wanted a slip ?--Where is that party?" Somebody will
respond. "Oh, there you are, here's a pencil and paper--and
hurry please." Performer takes the bottom, opened slip from left
hand and gives it to the spectator, in the meantime getting a
good chance to read the No. 5 question which is now lying open
on top of the other open slips. This only takes a second, you
need not get all the details if hard to read-merely the name, or
initials and an idea of the subject matter, such as, "J. C. Wilson,
sick father," or anything you can glimpse.
The performer keeps moving about and talking, he quickly
refolds No. 5, and turns around, walking toward stage, and as he
passes the volunteer collector, remarks, "That's right, get them
all," and quietly places his right hand over the hat for a second
as sort of a gesture to pull the hat down to see how many
collected--but the right hand has the No. 5 billet gripped at the
roots of the fingers, and the billet is allowed to drop in the hat
with the others. This casual move will not be noticed, and even
if it were, you would appear to be dropping in some stray billet
that had been handed to you. Nothing is said about it, 'you just
do it. If desired, the performer can stop just before he passes the
volunteer assistant, and, as though to speed up things, turn to
one row and say, "Pass your questions over, please," he taking
them and tossing them along with No. 5 into the hat.
In any event, the performer returns to the stage, requests the
collector to kindly come upon the stage and dump the folded
slips out on the table, the collector returning to his seat. The
performer has spotted the No. 5 question in the pile, either by
seeing the number visible, or by the bent corner--he gives the
pile a stir with finger if necessary to bring the No. 5 question
into view. With right hand, the performer picks up a billet, or
what to the audience appears to be one billet, but he really picks
up two, the No. 5 along with some other billet, for example No.
9. The actual picking up is done with the right hand and the left
hand immediately comes to the right hand, so the two billets are
momentarily held as one, between the tips of the right and left
fingers. The right fingers instantly let go and the right hand is
withdrawn, leaving the billet (really two of them) in full view at
tips of left fingers which then raise the billet to your forehead,
and you pretend to divine the question--the one by "J. C.
Wilson, who has a sick father, etc." This is done in the usual
manner, announcing first the name, then the nature of the
question, and finally giving the answer.
The left hand
has been
lowered, the
right fingers
approach the
left, and with
the aid of the
thumbs,
quickly and
secretly slide
the No. 5
billet into the
right hand
where it is
held or palmed against the roots of the right fingers. See Fig.
1A.

In the meantime, the No. 9 billet is being visibly opened, the
fingers concealing the No. 9 on the slip. The same pretense is
made, confirming the answer just given, and mentally
ascertaining a new question, as previously described. The No. 9
billet is refolded, and in the operation, the positions of No. 5
and No. 9 are reversed--No. 9 being now palmed in the right
hand, and No. 5 in full view at left finger tips. No. 5 may now
be returned to "J. C. Wilson," or may be tossed out into the
audience, or simply thrown upon the floor. It would be well to
use all three of these methods of disposal during the
performance, as was the habit of John Slater in his
demonstrations.
Of course, the readings are continued in the same manner, and
all of the billets can be returned to the writers, if desired, but it
becomes a bit tiresome to the balance of the audience, and it is
much more dramatic to vary the final disposition, as suggested.
The performer we witnessed, repeatedly thrust his right hand
into his trouser's pocket, leaving the billet there so he could
show his right hand empty but he had to dive into the pocket
again to get the billet so he could make the final switch, and this
pocket procedure did not look so good.

Magic Squares, Any Total

Effect
The 30-second “Countdown” theme is played (or similar) while the pupils quickly
pass a Teddy round the class. When the music stops, Sameer is left holding Teddy. I
ask Sameer to tell me the number of his house. He tells me that it is 46. Immediately I
draw on the board the following square:

Quickly we add up each row: 46! And each column! The two diagonals as well!
The magic total 46 is obtained in every direction!
But then clever old Suraj and Emily have been adding in other ways. They point out
that the corners add up to 46 too, and the 4 middle numbers! Before long the class has
found that each corner 2x2 square totals 46, as well as the top/bottom half middle
four. Later it is noticed that 1 + 12 + 27 + 6 = 11 + 5 + 2 + 28 = 46, and then Lateral
Lisa pipes up with 1 + 11 + 28 + 6 = 12 + 2 + 5 + 27 = 46.
Method
In the square above, the four numbers in the twenties are the only numbers which are
altered to make the trick work. I only need to learn this square:

When I first performed this trick in a classroom, it was
back in the days of blackboards and chalk. I had used (cunningly, I thought) a pencil
outline on the board which I was then planning to write over with the chalk during
performance. Unfortunately for me, the graphite in the pencil was reflective enough to
catch the sunlight and be perfectly visible to my audience, thus explaining my surprise
as they called out the numbers before I had even chalked them in. Andrew Jeffrey has
subsequently given me the far more professional and useful tip that this square could
be stuck on the barrel of the whiteboard pen. It could even be memorised!
I don’t go on to reveal this trick to my students for two reasons. Firstly it is in the
working repertoire of several professional magicians (I first saw it done by Paul
Daniels), but secondly and more importantly, the impact of the apparently endless
totals is immediately lost. I prefer to leave them with that sense of wonder.

Thumb Tip Billet Steal

All magicians are familiar with the fake known as the thumb
tip, which is a hollow shell shaped and painted flesh color to
resemble the first joint of the thumb over which it fits. As used
in this effect, the thumb tip should be long enough to cover the
full first joint of the thumb, and fit easily so there is room to
hold a folded billet which lies under the ball of the thumb, and
concealed within the tip.
The slips of paper used should measure about 1" x 2-1/2". When
folded twice, the billet may easily be concealed within the tip,
as described. The thumb tip is used to get secret possession of
one of the questions.
One method is to have the blank slips in the left vest pocket
along with the thumb tip, the latter being nearest the body with
the open end up. An envelope, size 6-3/4, is in your left side
coat pocket. After passing out a few slips, allowing time for
these spectators to finish writing their questions, the performer
withdraws a slip from his pocket and at the same time brings out
the thumb tip in position on the right thumb. He uses this slip to
show how the spectators should fold them. With this sample
folded billet in his left fingers he thrusts his left hand in his coat
pocket, leaving the sample billet there and getting the envelope.
He spots a spectator whose question is written and slip folded.
This spectator and one or two more are allowed to drop their
billets in the envelope.
The performer then quickly inserts his right thumb in the
envelope as though to open it, withdraws his thumb, leaving the
thumb tip inside the envelope through which it is lightly grasped
and held upright by the fingers of the left hand. While doing
this, the performer requests a spectator to "just put your slip in
envelope," however performer does not let this spectator drop it
in--performer simply reaches for the spectator's billet, taking it
in his right hand between first finger and thumb which go into
envelope. Actually, the billet is put into the thumb tip, the right
thumb going in with it and the hand quickly withdrawn, thus
secretly bringing out the billet in thumb tip. In getting more
slips from pocket, thumb tip containing stolen billet is left in
pocket.
The envelope is then handed to another spectator to drop in his
billet, and he passes envelope along to others, meanwhile
performer has been handing out slips to others, and in talking
and moving about, has plenty of opportunity to get thumb tip
with stolen billet out on his thumb and quickly extract stolen
billet and open it. He must not look at it, just get slip opened out
and placed underneath slips in left hand. The next spectator is
handed a slip and instructed to "write briefly and plainly and be
sure to sign your name--then fold writing inside like this" this
giving performer a chance to take stolen slip and secretly read it
as he folds it.
Some spectator looks after the collection envelope and brings it
to the stage, and is directed to dump the billets out on the table.
In the meantime, performer has gotten thumb tip with stolen
question in it on his right thumb, and he has also secretly gotten
from his left coat pocket, the blank sample billet that he first
folded down in the audience, and this blank billet is secretly put
on the edge of the pile and used as in previous methods. The
one ahead principle is employed, but the use of the thumb tip
provides an easy and most natural switch whereby the question
just answered may be returned at that moment to the writer.
The right hand, wearing thumb tip containing stolen question,
picks up a slip from the pile, and after giving the answer,
switches the one ahead billet for the stolen one in this manner:
If you take hold of thumb tip with left fingers and thumb, you
should be able to withdraw right thumb and billet both at once
from thumb tip. With palms towards you, try it before a mirror.
The tip remains concealed behind the left fingers, while the
billet appears to be taken by the right fingers from the left hand.

Now, with the loaded tip
on right thumb, and
billet No. 2 held openly
between tips of right
fingers and thumb (with
tip on) you have just
completed answering the
stolen billet, and you
open billet No. 2 to
confirm (really to read
the one ahead). You
refold No. 2 with both
hands and finish with it
in left hand. To
exchange the No. 2
billet for the stolen one
in tip, you merely bring
the hands together,
palms toward you, and
put right thumb (with tip
on) on No. 2 billet
behind the left fingers,
grip thumb tip with left
thumb and fingers, slide
out stolen billet as right
thumb is withdrawn
from thumb tip, and it
will appear to be the No.
2 billet just seen in left
hand. See Fig. 2.
Try this before a mirror,
and the deceptiveness of
the move will be
apparent. The stolen billet is now returned to its writer by an
usher, or voluntary assistant.
Your left hand holds concealed, the No. 2 billet against the
fingers, the thumb tip against the billet, and the left thumb
against the thumb tip. The fingers are curled inwards in a
natural position and no one suspects anything in the hand.
The next billet (we will call it No. 3) is now picked up by the
right hand which is raised to the forehead and an answer given
(to No. 2). The hands are brought together to open No. 3 to
verify, and is refolded and finally held in right hand which
pushes it into the thumb tip along with the right thumb, and the
No. 2 billet is brought into view at the same instant, being
grasped between fingers and thumbs of both hands for a second,
and may then be returned to its writer. You are again prepared
with one ahead for the next reading.
It will be noted that the second move, or switch, is the reverse of
the first, and both should be practiced before a mirror until you
can make the moves with rapidity and certainty, without looking
at your hands. It should be done while you are addressing the
audience with some remark, such as, "Where is Miss White, I'll
return your question, etc."
Also note that both hands are seen to be "empty" as you answer
the first question, and likewise on every alternate billet. You
make no comment about it, but the "emptiness" of the hands
permits you to make open handed gestures so frequently that no
one will suspect that anything is, or could be, concealed in the
hands at any time.

The no card, card trick

Presentation and Effect: The performer starts his patter
"You have seen or heard of many ways in which a performer can ascertain what cards have been
selected by his audience. I have developed a science which gives me 100% accuracy without seeing a
card, touching a card, in fact, without using any cards at all.
I would like you to participate in this demonstration, and I hope this will entertain and give you something
to think about!
Now Simply follow my instructions.
Ready?
I want you to think of any card in the deck, got one?
Good now double its value
If it's an ace count it as one, two as two, etc., up to ten. If you think of a jack the value is eleven,
Queen is twelve, and King is thirteen.
Now, after doubling its value, add one and multiply this total by five
You done that?
Good!
Right now do this
If the card you are thinking of is
a spade, add nine to the total;
if it is a club, add six;
if a heart, add eight;
if a the diamond add seven.
As soon as you have your answer tell me the number you have in your mind but of course not the card
you are thinking of, Ill do that part hopefully."
The Spectator says “74” You say, "Your total is 74; then the card you thought of was the 6 of spades.
That is correct, you say? They answer “YES”
Another spectator has 131, and he thought of the queen of clubs.
Another called out 82 was thinking of a 7 of diamonds.
You are correct everytime without failure, in fact if you get two or more thinking of the same card, really
milk it for all it’s worth!!! It does happen, a lot more than you would think it does!

The Secret:
The final total announced by the spectator will give the performer the clues to both suit and value
of the selected cards.
If the total ends in 1, remember that it will be a club;
if it ends in 2, it will be a diamond;
if it ends in 3 will be hearts,
and if it ends in 4 will be spades.

The first number, or the first 2 numbers minus 1, determines the value of the card.
For example, the number that the first spectator called was 74, and this proves to be the 6 of Spades
(doubling its value made it 12, adding 1 made it 13, and multiplying by 5 gave him 65. Adding 9 for
spades gave a final total of 74.)
The last number being a 4 indicates that the selected card is a spade.
A further example: the value of the second card called was 131.
The final 1 indicates a club;
The first 2 numbers are 13(minus 1 makes 12, and 12 means the Queen);
therefore, the card is the queen of clubs.
To be certain, when you start your demonstration, impress upon your audience with the idea of
concentrating on the selected card and also to listen to your instructions.
It is also important to make it plain that the ace counts as 1, 2 as 2, etc., up to 10, after which the jack
counts as 11, queen as 12, and the king as 13.
You can vary the presentation in a mixed group by asking the ladies, for example, to think of hearts and
diamonds, and the men, spades and clubs.
Using this system--if a lady calls out a number, you will know that the selected card will be either hearts or
diamonds. This effect can be done with as many people as you like.
You can perform it with many people at the same time getting them to call out their numbers and as they
do you just tell them their thought of card. It is the speed when doing it for a group of people that is
impressive, it is also useful for leading into a mind reading demonstration, you could indeed even state
after performing this trick then that maybe if could be solved with maths, then go into another mentalist
routine that would make this seem even more impossible , thus shelving the idea that it could have been
maths…..
The history of this trick is really unknown but a mention of Blackstone would be disrespectful if we did not
mention his name as a credit.

I hope you like this and it gives you much fun performing it whether it be in person via
the phone or even via a text message or an email
If the instructions seem a little complex, simply write or print out the formula and the
code for the suits and glue it to the side of a card box.
In fact why not glue it if you can’t remember it to the side of a card box that contains
an Invisible deck and you can then say you knew that they would have picked the
card they did as there is one card face down in the pack…then take out the invisible
deck and BANG! You nail them on 2 levels…..Believe me this kills.
Think outside the box.

Mentalists Translucent Clipboard

The trend of having a see-through clipboard has introduced all kinds of clever products
of late, many of which use old principles that range from waxed paper and graphite
powder to a wide range of other interesting and exciting ideas.
So with this in mind I thought about it and came up with this idea for creating this
simple, effective, practical and inexpensive clipboard.
Requirements
1) A clear plastic clipboard
The clipboard must have absolutely no coloring and can be purchased from your local
office supply store, such as Staples/ Office World
2) A box of white chalk
3) Some white paper
Method
Lie the chalk on its side and thoroughly saturate at lest two-thirds of one surface of the
white paper and ensure that you work the chalk in deep into the paper. You'll end up
with a lot of chalk powder so I recommended that you get a container and collect it as
you can use it later whilst treating other sheets.
The next step is to carefully dust off the back of the paper (this will take some
experimenting) so that when you rub your hand across the chalked surface, you aren't
getting it on your hands. This way, the spectator can fold the slip and place it into an
envelope without noticing its condition.
With the chalk side of the paper place against the clipboard's surface, anything written
on the front side of the sheet will be transferred to the plastic surface but will essentially
go unseen until you lay the board on top of or move it to some kind of dark/black
background or a manila folder etc.
Handling Advice
In my handling of it I leave a manila folder under the folder which is where I remove the
paper from and as I replace the clipboard onto the folder after the paper has been
removed from the clipboard I glimpse the info, and just casually set both the clipboard
and folder aside.
Notes
Clearing the board is super easy, just hold it against your leg or chest and give it a little
rub and it's automatically ready for the next sheet of paper. Also don’t wear black or
real dark colors if you use this cleaning method as it will show!
Believe me this is just one of the best kept secrets there is!!! Use the knowledge well.

Flipping More Coins At Once

Returning to the magic trick of flipping n coins to become all the same, another generalization is to allow
the magician the additional flexibility of flipping more than one coin at once. The number of coins flipped
per move might be a constant value (as considered in the next section), or might change from move to move
. In either case, we let k denote the number of coins flipped in a move.
In this section, we consider what happens when the spectator gets to choose how many coins the magician
must flip in each move. Obviously, if n is even, then the spectator must choose odd values for k, or
else the magician could never get out of the odd parity class. But even then the magician is in trouble. We
provide a complete answer to when the magician can still succeed:
Lemma 6 If n _ 5, the magician is doomed.
Proof: The spectator uses the following strategy: if the distance between the current configuration and the
all-heads or all-tails configuration is 1, then the spectator tells the magician to flip three coins. Otherwise,
the spectator tells the magician to flip one coin. Because n _ 5, being at distance 1 from one target
configuration means being at distance at least 4 from the other target configuration, and 4 > 3, so the
magician can never hit either target configuration. The spectator always says odd numbers, so this strategy
satisfies the constraint when n is even. 2
Lemma 7 If n = 3 or n = 4, the magician can succeed.
Proof: As mentioned above, flipping k or n − k coins are dual to each other. For n = 3 or n = 4, the
spectator can only ask to flip 1 or n − 1 coins. Thus the magician effectively has the same control as when
flipping one coin at a time. More precisely, if the spectator says to flip 1 coin, the magician flips the next
coin in the k = 1 strategy. If the spectator says to flip n−1 coins, the magician flips all coins except the next
coin in the k = 1 strategy. This transformation has effectively the same behavior because the two targets are
bitwise negations of each other. 2
Despite this relatively negative news, it would be interesting to characterize the sequences of k values for
which the magician can win. Such a characterization would provide the magician with additional flexibility
and variability for the equalizing trick. In the next section, we make partial progress toward this goal by
showing that the magician can succeed for most fixed values of k.

Flipping Exactly k Coins at Once

In this section, we characterize when the magician can equalize n coins by flipping exactly k coins in each
move. Naturally, we must have 0 < k < n, because both 0-flip and n-flip moves cannot equalize a notalready- equal configuration. Also, as observed in the previous section, we cannot have both n and k even, because then we could never change an odd-parity configuration into the needed even parity of an all-equal configuration. We show that these basic conditions suffice for the magician: Theorem 8 The magic trick with k-flip moves can be performed if and only if 0 < k < n and either n or k is odd. The optimal solution sequence uses exactly 2n−1 − 1 moves in the worst case. A lower bound of 2n−1 − 1 follows in the same way as Section 2. Again we can view the trick on the n-dimensional hypercube, where 0 represents a bit unchanged from its initial configuration and 1 represents a changed bit. The difference is that now moves (edges) connect two configurations that differ in exactly k bits. The lower bound of 2n−1 − 1 follows because we need to visit every bit string or its complement among 2n possibilities. Our construction of a (2n−1 −1)-move solution is by induction on n. If k is even, we can consider only odd values of n. The base cases are thus when n = k + 1 for both even and odd k. The n = k + 1 case has k = n − 1, so it is effectively equivalent to k = 1 from Section 2. We will, however, need to prove some additional properties about this solution. It seems difficult to work with general solutions for smaller values of n, so we strengthen our induction hypothesis. Given a solution to a trick, we call a configuration destined for heads if the solution transforms that configuration into the all-heads configuration (and never all-tails), and destined for tails if it transforms into all-tails (and never all-heads). (Because our solutions are always optimal length, they only ever reach one all-heads configuration or one all-tails configuration, never both, even if run in entirety.) We call a transformation destiny-preserving if every configuration on n coins has the same destiny before and after applying the transformation. A transformation is destiny-inverting if every configuration on n coins has the opposite destiny before and after applying the transformation. Now the stronger inductive statement is the following: 1. for nk even, flipping the first j coins for even j < k preserves destiny, while flipping the first j coins for odd j < k inverts destiny; and 2. for nk odd, flipping the first j coins for even j < k inverts destiny, while flipping the first j coins for odd j < k preserves destiny, and flipping coins 2, 3, . . . , k preserves destiny. To get this stronger induction hypothesis started, we begin with the base case: Lemma 9 For any k > 0, the k-flip trick with n = k + 1 coins has a solution sequence of length 2k − 1
such that flipping the first j coins for even j < k preserves destiny, while flipping the first j coins for odd j < k inverts destiny. Proof: The construction follows the Gray code of Section 2. That flip sequence, ignoring the first coin, can be described recursively by Gk = Gk−1, “flip the (n − k + 1)st coin”, Gk−1. To flip n − 1 coins in each move, we invert this sequence into ¯G k = ¯Gk−1, “flip all but the (n − k + 1)st coin”, ¯Gk−1. In the base case, G0 = ¯G0 = ;. Validity of the solution follows as in Section 2; indeed, for any starting configuration, the number of moves performed before the configuration becomes all-heads or all-tails is the same in sequences Gk and ¯Gk. Every move flips the first coin, so the destiny of a configuration is determined by its parity and the parity of n: if n and the number of coins equal to the first coin (say heads) have the same parity, then the configuration is destined is that value (heads); and if n and the (heads) count have opposite parity, then the destiny is the opposite value (tails). To see why this is true, consider the hypercube viewpoint where 0s represent coins matching the initial configuration and 1s represent flipped coins in an execution of Gk (not ¯Gk). Then, at all times, the number of 1 bits in the configuration has the same parity as the number of steps made so far. At the same time, every move in ¯Gk flips the first coin, so the first coin in the current configuration matches its original value precisely when there have been an even number of steps so far. Thus, when we reach a target configuration of all-heads or all-tails, it will match the original first coin precisely if there have been an even number of steps so far, which is equivalent to there being an even number of 1 bits in the Gk view, which means that the initial and target configurations differ in an even number of bits. In this case, the initial and target configurations have the same parity of coins equal to their respective first coins; but, in the target configuration, all coins match the first coin, so in particular n has the same parity as the number of coins equal to the first coin. We have thus shown this property to be equivalent to the target configuration matching the initial first coin. It remains to verify the flipping claims. Flipping the first j coins for even j < k preserves the parity of the number of heads as well as the number of tails, but inverts the first coin, so inverts the destiny. Flipping the first j coins for odd j < k changes the parity of the number of heads as well as the number of tails, and inverts the first coin, which together preserve the destiny. 2 With this base case in hand, we complete the induction to conclude Theorem 8. In the nonbase case, n > k + 2. There are three cases to consider:
Case 1: Both n and k are odd. By induction, we obtain a solution sequence _0 of length 2n−2 − 1 for
n0 = n − 1 satisfying the destiny claims. We view _0 as acting on only the last n − 1 of our n coins. Then
we construct a solution _ for n as follows:
_ = _0, “flip the first k coins”, _0.
This solution has length |_| = 2 |_0| + 1 = 2n−1 − 1.
Next we prove that sequence _ solves the trick. Consider any configuration on n coins, and assume by
symmetry that the last n − 1 of its coins are destined for heads in _0. If the first coin is also heads, then the
magician arrives at the all-heads configuration within the first _0 prefix of _. If the first coin is tails, then
the _0 prefix will not complete the trick, at which point the magician flips the first k coins. This move has
the effect of flipping the first coin to heads as well as flipping the first k − 1 of the _0 subproblem, which is
destiny-preserving because k−1 is even and (n−1)k is even. Therefore, during the second _0, the magician
will arrive at the all-heads configuration.
Now we verify the destiny claims. Note that the destiny of a configuration in _ equals the destiny of
the last n − 1 coins in _0, so we can apply induction almost directly. Flipping the first j coins for even
j < k flips the first j − 1 of the last n − 1 coins, which inverts destiny by induction because j − 1 is even
and (n − 1)k is even. Similarly, flipping the first j coins for odd j < k preserves destiny by induction.
Finally, flipping coins 2, 3, . . . , k flips the first k − 1 coins of the last n − 1 coins, which preserves destiny
by induction because k − 1 is even.
Case 2: For n even and k odd, by induction we again obtain a solution sequence _0 of length 2n−2 − 1
for n0 = n − 1, viewed as acting on only the last n − 1 coins. We construct a solution _ for n as follows:
_ = _0, “flip coins 1, 3, 4, . . . , k + 1”, _0.
Again _ has length 2n−1−1. Flipping coins 1, 3, 4, . . . , k+1 has the effect of flipping the first 2, 3, . . . , k of
the last n − 1 coins, which by induction is destiny-preserving because (n − 1)k is odd. Thus, if the destiny
of _0 does not match the first coin, then it will match the newly flipped first coin during the second _0. As
before, destiny in _ matches destiny in _0 on the last n − 1 coins. Flipping the first j coins for even j < k
flips the first j − 1 of the last n − 1 coins, which preserves destiny by induction because j − 1 is odd and
(n − 1)k is odd. Similarly, flipping the first j coins for odd j < k inverts destiny by induction.
Case 3: For n odd and k even, by induction we obtain a solution sequence _0 of length 2n−3 − 1 for
n0 = n − 2, which we view as acting on only the last n − 2 coins. Then we construct a solution _ for n as
follows:
_ = _0, “flip the first k coins”, _0, “flip coins 1, 3, 4, . . . , k + 1”, _0, “flip the first k coins”, _0.
This solution has length |_| = 4 |_0|+3 = 2n−1 −1. Restricting attention to the first two coins, _ first flips
both coins, then flips the first coin only, then flips both coins again. Together these enumerate all possibilities
for the first two coins. Restricting to the last n − 2 coins, these moves correspond to flipping the first k − 1
coins, coins 2, 3, . . . , k, and again the first k − 1 coins. By induction, all three of these operations preserve
destiny because k − 1 and (n − 2)k are odd. Therefore all three executions of _0 produce the same target
configuration (all-heads or all-tails) which will eventually match one of the combinations of the first two
coins. Flipping the first j coins for even j < k flips the first j − 2 of the last n − 2 coins, which preserves
destiny by induction because j − 2 is even and (n − 1)k is odd. Similarly, flipping the first j coins for odd
j < k inverts destiny by induction.
This concludes the inductive proof of Theorem 8 .