box together with two unfaked pieces of white and green chalk.

Before presentation you will require looking slate similar in appearance to the prepared one and three pieces of adhesive tape two inches in width.

Place a rubber band round the Jumbo pack and then seal it inside an envelope. This packet will lie on the table together with the two slates.

The faked slate comes in for a little attention. On the lower part of the slate proper, i.e., the part with the appendage ;write in yellow chalk " C REVERSED 6 P.M." Adjust the slide over this so that the slate appears blank except for the white chalk lines. Incidentally the slate should lie faked side down so that the lines are not seen.

Easily at hand are the pieces of tape the normal sized pack of cards and the box containing the chalks.

The Presentation

The faked slate is taken, the " lined " side being kept away from the audience. Holding it in the left hand, the right hand takes a piece of white chalk from the box and appears to write lines across the slate. The audience have seeming proof of this when you turn it round showing the lines that actually were there before you started. The white chalk is placed down and the stick of green chalk is taken. Saying that you intend writing a prediction for spectator ' A,' casually toss the piece of green chalk into the air and catch it. There is a reason for this. The spectators see a piece of green chalk and nothing else. When they see you handling another piece of coloured chalk later on they will take it that this too is just as unprepared. With the green chalk in view of the audience, write ' A,' and then turn the slate so that the surface is hidden from the audience. Write another word which can be any word. The chalk is dropped into the box and one of the lengths of tape is picked up and placed across the frame and slate, covering the message. It is essential, should you use tape, that the adhesive surface should not be allowed to touch the slate surface. It must be stretched across the frame and leave the surface of the slate free.

Spectator 'A' is now asked to name any colour, and when he does so you take the other slate and with a piece of white chalk record it as previously mentioned.

With the remark that you will now write a prediction for another spectator ' B,' but this time with red chalk, you reach into the chalk box and remove the piece of red-green chalk, masking the green part with the fingers and allowing the red part to be seen. Taking the 'taped' slate and keeping the ruled side towards you, reverse the chalk you are holding end for end, and in the second space write in green "A . . . Yellow" (the colour named by ' A'). The chalk is returned to the box, care being taken that no green chalk can be seen by the wary eyed spectator, and the second length of adhesive tape picked up and placed over the second (?) prediction.

' B' now names a multiple number and this is recorded on the unfaked slate.

The last prediction. Taking the piece of red-yellow chalk and showing it as yellow, you write in the third space in red " B . . . 4567 " (the number given by B). After replacing the chalk into the

box the same hand stretches forward to pick up the third length of tape, whilst the thumb of the hand holding the faked slate, slides the slide part upwards, so that all the predictions are in the correct position. The tape is then placed over the third prediction and the slate is placed in a standing position against a chair back or some similar rest.

The pack of cards is taken, fanned and handed to ' C' with a request that he shuffles the cards and asks his neighbour to take any one. When this is done, the card is named and it is recorded on the unfaked slate as " C . . . ace of diamonds " (the chosen card).

The unfaked slate should now be placed so that it too rests against some object and the writing can be seen by the audience.

The faked slate is picked up, the top tape is removed, and the first prediction is found to be correct.

The middle tape is also removed. The second prediction is correct! Finally the third tape is taken off to show the message, " C . . . Reversed 6 p.m."

Ripping open the envelope the pack of Jumbo cards is removed, the rubber band is taken off and fanning the cards faces towards the audience you leaf through the pack until you find the group of three the middle one of which is the reversed duplicate of the chosen card. Split the roughened assembly allowing the back of the duplicate to be seen. Then slowly turn round the pack and the face of the duplicate card is apparent.



Continued from June issue


PIS prime. In addition to the case of a pack of 2x cards, there is one other case when a an immediate solution of the equation is possible. This is when P is a prime number, then 2P-1—1 is divisible by P. This is Fermat's theorem, a theorem in the theory of numbers. Applied to the weave shuffle equation it means that when P is prime, P cards will return to their original order in P-l shuffles. Incidentally, this is not necessarily the smallest number of shuffles necessary for the cards to return to their original order.

For example, fifty-two cards in-shuffled are equivalent to an odd pack of fifty-three. Fifty-three is a prime, and therefore fifty-three cards return to the same order after fifty-two shuffles. Hence, fifty-two cards return to the same order after fifty-two in-shuffles.


General method. This is the quickest all-round method. It uses the binary system, which would take too long to explain here, so I shall assume it is known. Express P (which of course is odd) in the binary system, and write it down. Write it down a second time underneath, so that the final * 1' of the lower expression of P comes under the farthest right ' O' in the upper expression. Add the two together. If you get nothing but ' IV stop right there. Otherwise, write P down again under the sum with the final 1 in P under the farthest right '0' in the sum. Again add. Carry on until you get a sum containing nothing but' IV The number of ' l's' in the final sum is the necessary number of shuffles to bring P cards back to the original order.

For example, eleven in the binary system is 1011.

Going through the procedure described, we have

10 11 10 11

110 111 10 11

10 0 111111 10 11


There are ten ' l's ' in the final sum, so eleven cards return to the same order after ten shuffles. Note that we get the same result from the last section, since eleven is a prime.

To find out how many shuffles are necessary for the cards to reverse their order, we write P down each time with its final 4 1' under the 41,' excluding the final 4 1,' furthest to the right in the sum, or expression of P, above, until we get a sum of the form 1000 --- 001. The necessary number of shuffles is one more than the number of ' 0's.' It is advisable to solve for S before trying for R, for although there is always a solution for S, there is not always a solution for R. When there is, of course, S=2R.

For example, with eleven cards as before— 10 11 10 11

There are four ' 0's,' so eleven cards reverse their order in five shuffles.


Suppose, in an odd pack of P cards, we have a set of cards stacked " a " apart, counting downwards through the pack. As we shuffle, the distance apart of the cards in the stack becomes successively 2a, 4a, 8a and so on. Suppose after x shuffles, the distance apart of cards in the stack (which is 2xa) is less than P but greater than half P. If we now count in the opposite direction, 1. 2. upwards through the pack, the distance apart of cards in the stack is P-2xa. This may give an interesting transformation of one stack to another. At each transformation the order of the cards in the stack reverses.

With an even pack, the calculations are made as for the equivalent odd pack.

For example, fifty-two cards out-shuffled are equivalent to an odd pack of fifty-one. Therefore, a stack of cards seven apart in a full pack will move in successive out-shuffles through the following distances apart 7, 14, 28 or 51—28=23, 46 or 51-46 = 5.

Thus a stack seven apart becomes a stack five apart after three out-shuffles. There were two transformations, so the stack is in its original order. If after shuffling, the distance between two cards of the stack has to be counted round the ends of the pack, there will be an error due to the fact that we are using an even pack.

Other transformations with a pack of fifty-two out-shuffled are: —

10 to 11 in two shuffles, reversing their order.

11 to 7 in two shuffles, reversing their order.

Any multiple of three to itself in four shuffles, reversing their order.

With fifty-two cards in-shuffled we get: —

10 to 13 in two shuffles, reversing their order.

11 to 9 in two shuffles, reversing their order.

7 to 3 in three shuffles, retaining their order.

3 to 5 in four shuffles, reversing their order.

The 10-13 transformation means that cards ten apart can be brought together by four in-shuffles.

Here, incidentally, is a poker deal which uses the 7-3 transformation.


Take out the four royal flushes from the pack, and arrange them in the pack as follows: — The cards of one flush at positions 2- 4- 6- 8-10 The cards of one flush at positions 7-14-21-28-35 The cards of one flush at positions 9-18-27-36-45 The cards of one flush at positions 5-17-29-41-46

You can now have any number of hands suggested by a spectator, shuffle and deal that number of hands, and get for yourself a royal flush. When the number of hands is named, you make the number of in-shuffles shown below, and then deal normally.

For two hands, shuffle not at all.

For three hands, shuffle three times.

For four hands, shuffle once.

For five hands, shuffle twice.

For six hands, shuffle four times.

For seven hands, shuffle not at all.

For eight hands, shuffle twice.

For nine hands, shuffle not at all.

For ten hands, shuffle three times.

The same flush that is set to come out on seven hands turns up after three shuffles set for three hands.


Suppose, in an odd pack, P=ab. We can stack a set of ' b' cards 4 a' apart through the pack, so that by dealing out4 a ' hands of4 b' cards each, all the stacked cards will fall in the same hand.

Taking any card of the stack as a reference card, the position from it of any other card in the stack will be a multiple of 4 a,' i.e., No=xa.

After one shuffle, Nl=2xa or Nl=2xa—P =2xa—ab = a (2x—b)

In either case, N, is also a multiple of 4 a,' i.e., the 4 b' cards are still stacked 4 a' apart through the pack after one shuffle, and hence after any number of shuffles. The cards will not, however, be in the same order in the stack.

For example, suppose we discard all the diamonds from a full pack, leaving 39 = 3 x 13 cards. We set these up in the order clubs, hearts, spades, clubs, hearts, spades and so on throughout. We now have three separate stacks of thirteen cards, three apart. If after any number of shuffles (with cuts in between, if you like) we deal the cards into three hands, each hand will consist of a full suit.

To be concluded


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