Estimation Bonus

You have purchased the book "Estimation," and we thank you! Below are additional ideas related to this principle.

New Idea:

NUMEROLOGY

After talking about numerology, the mentalist asks a spectator to write the numbers from 1 to 9 on a piece of paper, but in descending order: 9 8 7 6 5 4 3 2 1

Then, he explains that she must eliminate four of the nine digits (by crossing them out), so that only five remain.

Example: 86531

The mentalist asks her to write the same five digits below, but in reverse order (13568)

and subtract them from the first number.

Example: 86531-13568 = 72963

The mentalist states that he does not know any of the eliminated digits, let alone the remaining digits, and therefore the result of the subtraction is completely unknown to him.

That's somewhat true... but not entirely.

Indeed, the middle digit will always be a 9, the sum of the two outer digits will always be 10, and the sum of the second and fourth digits will always be 8.

From there, the mentalist will make it seem like he knows the five digits that are the result of the operation.

To do this, he will deliver a completely esoteric speech, but one that will hold up because everything will turn out to be accurate.

He asks the spectator to hold his hands, then says:

"Focus on the second and fourth digits... Yes, that's right... Can you mentally add those two digits? Don't tell me anything. I perceive the number 8. 8 symbolizes the balance between the material and spiritual worlds.

Do you know that if these two digits surround the number 9, we have a benevolent conjecture... Is that right, is your middle digit indeed a 9?"

The spectator nods. The mentalist continues:

"Now focus on the two outer digits: the first and the last. Again, mentally add them up... Good... I perceive a two-digit number, 10.

10, the symbol of benevolence and open-mindedness that seems to characterize you.

Do you have 10 in mind? Yes? Thank you."

You only knew the middle digit (9), and yet everyone will believe that you have the gift of divination.

Idea 1

Ideas often come unexpectedly, and sometimes even when one thinks they have exhausted an effect. After finishing this work, the following idea came to me, which I believe greatly enhances the impact of the trick. Here it is.

After estimating the number of cards lifted, the magician offers to do something even more difficult. To do this, he takes the deck in his hands and arranges to place a two and a seven at the bottom of the deck, then places the deck on the table.

The magician asks the spectator to cut the deck into three piles of roughly the same thickness. Under the spectator's rightmost pile will be the two cards (2 and 7).

The spectator will follow the previously described steps to have 9 cards in the first pile, 18 in the second, and 25 in the last. The magician asks the spectator to put the first two piles into the card case (the one containing 9 cards with the one containing 18 cards). So, there will be 27 cards in the case (a multiple of 9).

On the table, the last pile of 25 cards will remain. The magician takes this pile and, by double-cut, moves the bottom card to the top. As he throws this last pile from his right hand to his left hand, the magician will hold the top and bottom cards in his right hand, while the pile will end up in his left hand.

The magician will then announce: "Here are 2 cards (fanning the two cards in his right hand, backs visible, and placing them on the table). Here are 23 cards (placing the cards from his left hand on the table) and here (pointing to the case)... 27 cards (picking up the two cards previously placed on the table and turning them face up).

The magician will finish by saying: "You can check..."

27 – 23 – 2

The bookmark

This will allow you to perform two magic effects.

Ask your spectator to choose 4 coins (one of each value) and add up the corresponding years.

Your spectator will arrive at a number corresponding to a multiple of 9.

They will then only need to add the digits of this multiple to reach the number 9.

The spectator will turn over the bookmark and the prediction will be accurate, regardless of the coins chosen.

Place this bookmark on page 81 or 171 of your Book Test and you can present a magical effect before performing your Book Test routine. (See routine 81 / 171 in the book)

THE NUMBER 9 AND THE LUCKY NUMBER

PROPLESS TRICK THAT CAN BE PERFORMED OVER THE PHONE

The magician asks the spectator to type their lucky number into their phone's calculator. Then to type the "x" sign. The magician states that such a number could allow them to win the lottery or any other type of money game.

The magician asks the spectator to imagine a sum between 9000 and 10000 euros, specifying to choose a complicated sum with only different digits. Of course, the first digit typed will be a "9", hence the presence of this trick in this book.

The spectator types this 4-digit number into their calculator, after the "x" sign. Then the spectator types the "=" sign.

The magician asks the spectator for the result of the multiplication, stating that he cannot know either the lucky number or the desired sum.

However, the magician says that he will (after a pseudo-study of the spectator's personality) try to find these two numbers.

EXAMPLE:

Lucky number: 8

Desired sum: 9543

8 x 9543 = 76,344

For the lucky number, just add 1 to the first digit of the result.

7 + 1 = 8

From there, just divide: 76344 by 8 to get: 9543

PLUS:

It is possible to suggest that the spectator choose a number between 90,000 and 100,000; the trick will also work.

Thus, the spectator's choice will be among 10,000 possibilities, instead of 1000.

THE NUMBER 9 AND THE CHRONOFORCE APPLICATION

This presentation follows the TIME IS MONEY trick

The magician says he will use his phone's stopwatch to illustrate the "TIME" part of the trick. In reality, he will open Chronoforce's stopwatch which will allow him to force the number 45 (or any other two-digit number that is a multiple of 9).

Prepare the application to force the number 45 (number of hundredths of a second)

The magician announces the name of the trick: TIME IS MONEY and specifies that he will start with the TIME part. For this, he opens his phone's stopwatch and explains how it works. The spectator presses "Start" then "Stop" to display a first random 2-digit number.

These first two presses are just to show how the stopwatch works. It is only with the next two presses that the number 45 will be forced.

The spectator complies, notes the result obtained, and places the phone screen down on the table.

The magician moves on to the MONEY part to also obtain a multiple of 9.

The magician will address a second spectator.

Several possibilities may arise, and the magician will take them into account (if they do) to strengthen the effect.

FIRST POSSIBILITY

The spectator takes 4 ten-cent coins and 5 one-cent coins.

The magician will stop there to show that he obtained the number 45 (corresponding to the stopwatch number).

SECOND POSSIBILITY

The spectator takes 2 ten-cent coins and 7 one-cent coins.

The magician will have them add 2 + 7 to get 9.

He will ask the first spectator to do the same: 4 + 5 = 9

Both spectators will get the same result, thus confirming that time,

is indeed money.

Same for 1 + 8, 3 + 6

THIRD POSSIBILITY (the most frequent)

This is the one described in the book.

You will be asked to subtract the number of coins taken from the sum they represent. The result will be a multiple of 9, which will be reduced to 9.

For the 3-pile version.

You can secure the trick by having two cards moved from the leftmost pile to the middle pile, before counting the number of cards in each pile.

Likewise, before revealing your prediction, it's a good idea to have one card moved from the middle pile to the left pile, to get: 10, 17, and 25 cards.

This way, the principle of the number 9 is even more hidden.

CHESTNUT DAY

It's the 258th day of the year. September 15th or 29 Fructidor.

The magician places a box on the table. Then he takes out a deck of cards and has a spectator shuffle it (dovetail shuffle). The spectator is asked to take the top three cards of the deck and form the smallest 3-digit number.

This number will necessarily be: 258.

The magician says that this morning he had an intuition and that he had to put an object in the box. The magician opens the box and takes out a chestnut.

The magician seems surprised and doesn't understand the connection to 258. Then he corrects himself (as if the idea just came to him) and asks to check on the internet (the spectator's mobile phone) what the 258th day of the year is.

We learn that it's September 15th and it's Chestnut Day!

DECK SETUP

From top to bottom (backs visible): a TWO, a FIVE, an EIGHT, twenty-three indifferent cards, the joker, an EIGHT, a FIVE, a TWO, and twenty-three indifferent cards.

PRESENTATION

The magician removes the joker from the deck and entrusts the two halves to the spectator, so that they can dovetail shuffle them. That's all, on top of the deck, you will have the three cards necessary for the routine.

The ultimate version of Estimations

There are magic routines that, like good wine, improve with age. This is the reason for this QR-Code.

I had a feeling that this "Estimations" routine was one of them. So, here is what I consider to be the ULTIMATE version of Estimations.

The 52-card deck is borrowed and will not be touched by the magician at any point. The magician asks the spectator to lift an unknown number of cards and count them, secretly, one by one.

Example: 21 cards

The magician asks the spectator if they indeed have a two-digit number. The spectator answers in the affirmative, and the magician asks them to add these two digits and remove from the lifted packet the number of cards corresponding to this result.

The removed cards are placed on the bottom of the deck.

Example: 2 + 1 = 3

Three cards are removed and placed on the bottom of the deck. The magician then asks the spectator to slide the remaining cards in their hands into the card case.

This number of cards will then be 9, 18, 27 or (very rarely) 36.

The closed case is placed on the table. The magician asks the spectator to take the bottom of the deck in their hands and hide it (for example, under the table). The magician asks the spectator if they prefer him to guess the number of cards the spectator has in their hands or the number of cards in the case... As the magician says the word "case", the magician lifts the case and estimates (by weight) if there are 9, 18 or 27 cards inside it.

It seems impossible to do, and yet it is extremely simple. From then on, the magician is able to say how many cards are in the case and how many cards are in the spectator's hands.

And this with 100% success, even though the magician has seen nothing and the deck is borrowed.

He just lifted, for a very brief moment, the case with the cards it contains.

The magician's justification to support his estimation.

Very simple, the magician will tell the truth.

As he lifts the case, he will say: "Since there are 52 cards in the deck, I will guess the number of cards you have placed in the case, and from that, I will determine the number of cards you have in your hands. And to do that, I will estimate this number of cards by weight by lifting the case."

And it's true, that's exactly what he does. Try this routine, and as others have written, you will realize that: "This routine is a real atomic bomb!"

The very latest version

The magician says he will perform a trick with two dice and a 52-card deck.

To do this, he takes two dice from his pocket and places them aside: "We'll deal with them later."

He continues by asking for a deck of cards. The magician, with his back turned, asks the spectator to lift a little less than half the deck and secretly count the number of cards lifted.

The magician says: "Do you have a two-digit number? Yes? Well, to make things a little more complicated, I'm going to ask you to add these two digits and remove that many cards from your packet to place them on the bottom of the deck." The magician continues by asking the spectator to put the cards remaining in their hands into the card case, close it, and leave the case on the table. So, on the table, there is the case with an unknown number of cards to the magician and the bottom of the deck. The magician turns around and asks the spectator to count, one by one, the number of cards making up the bottom of the deck.

The spectator complies and announces (for example): "34 cards"

The magician says: "We agree that I couldn't have known in advance how many cards you were going to lift and how many cards would be in the bottom of the deck." The spectator replies in the affirmative.

The magician continues by saying: "Do you want to take a look at the dice that were rolled there, earlier...?"

One die shows 3 and the other 4... 34, the number of cards in the talon.

NOTE

The magician has his back turned, it's not his deck, the magician sees nothing and touches nothing.

EXPLANATIONS

Of course, after the slight calculation and the removal of a few cards, the lifted packet

can only contain 9 or 18 cards.

This is one of the principles of the number 9.

Example 1: 24 - 6 (2 + 4) = 18

Example 2: 17 - 8 (1 + 7) = 9

PS: If the spectator has lifted a little less than half the deck, the cases of 27 or 36 cards are impossible.

The dice are rigged to show 3 on one and 4 on the other. The dice are casually rolled at the beginning of the routine and set aside without paying attention. If there are 9 cards in the case, there will be 43 left in the talon. If there are 18 cards in the case, there will be 34 left in the talon. We are lucky: 3 and 4 cover both possibilities. Once the number of cards in the talon is announced, the spectator themselves will notice that the dice had indeed predicted this result.

Additional ideas by Sébastien Thill

Upon finishing reading Gérard Bakner's excellent "Estimations," I came across version 3 of the principle, which is based on forming three packets and allows for a triple prediction under laboratory conditions (page 42)

I found the principle so brilliant that I tried to find alternative uses for it, which in reality keep the predictions secret (knowing in advance the number of cards making up each packet), to make it a secret weapon to add to our magical arsenal.

Hoping you will enjoy this approach!

Preparation: none.

Procedure:

As a reminder, have the deck shuffled and ask the spectator to create three roughly equal piles of cards.

Have them count the number of cards in the first pile, add the two digits (e.g., 1+7 for 17 cards), and remove that number of cards (8 in our example) and add them to the central pile.

Repeat the same sequence by having them count the number of cards in the modified central pile, and remove the corresponding number of cards, adding these cards to the pile on the right.

At the end of these procedures, the left pile will consist of 9 cards, the middle one of 18 cards, and the right one of 25 cards!

Yes: it's brilliant!

Here's what I propose now that the three piles are formed!

Version 1 - Control a card.

Have one of the three piles chosen, which is then shuffled again.

Its bottom card is looked at and remembered by the spectator.

Have this pile placed on one of the two remaining piles, left on the table, and finally place the last pile on top of the whole.

By observing the order in which the spectator handles the deck, it's very easy to know the position of the card, by adding the number of cards above the chosen card, since we know the number of cards in each pile!

Version 2 - Hamman style.

The three piles have just been formed.

Ask the spectator to pick up the smallest pile and place it in the card case (you can have your back turned).

Then, they pick up the largest pile, shuffle it, look at the top card, remember it, and place this pile on the pile remaining on the table.

Turn back to face the spectator and perform a cut (or double cut) of 9 cards to the bottom, then a false shuffle.

Have them count the cards placed in the card case.

There are 9.

Have them count 9 cards in the rest of the deck, and reveal the chosen card.

Version 3. Hamman 2.0 style

It's pretty much the same, except the spectator puts whichever pile they want into the card case (you can even turn your back).

When you turn back to face the spectator, it's easy to spot which pile was put into the card case.

They choose their card from a second pile they select, and replace the remaining pile on top of everything.

It's then very easy to adjust with a cut, double cut, etc., to achieve the result of version 2, since you know the number of cards in the pile placed in the card case, but in reality, in each of the three piles!

Version 4 - Any card at number of coins.

This is probably my favorite version!

You bet the contents (currently unknown) of a purse placed on the table.

Have the deck shuffled and ask the spectator to form three roughly equal piles of cards, making the modifications of the original trick.

Once all that is done, ask the spectator to pick up the largest pile (the 3rd), shuffle it, and look at the top card.

Have them place one of the other two piles on top of this pile, and the last remaining pile under this pile on the table.

In a small purse placed on the table from the beginning are 9 two-euro coins.

Two possibilities are open to you:

-If the smallest pile was placed on the deck, say that there are 9 coins and that if you remove 9 cards from the top of the deck, you will find the card (the next one).

-If the largest pile was placed on the deck, say that there are 18 euros and that if you remove 18 cards from the top of the deck, you will find the card (the next one).

Hoping that these versions will amuse you as much as I enjoyed playing with this principle.

Thank you, Gérard Bakner, for passing it on to us: a revelation!

The fact that there are three piles in Version 3 of Estimation immediately made me think of a shell game, and by association, of Bob Hummer's principle, brilliantly reinterpreted in Gabriel Werlen's Green Eck System.

Here is the result!

Version 5 : By Hummer.

Preparation: none.

Procedure :

Perform the usual procedure.

Pile 1 consists of 9 cards, pile 2 of 18, and pile 3 of 25.

While the magician has his back turned, the spectator chooses one of the three piles, looks at the face card, and replaces the pile in its spot.

"To avoid revealing which pile has been moved, the spectator is asked to reverse the positions of the other two piles, and finally to reverse the positions of the two thinnest piles."

The chosen card will then be at the bottom of the pile in the position originally occupied by pile 3 (of 25 cards).

All that remains is to reveal the location of the card and the number of cards that make up the pile!

Version 6 : Bob Hummer again.

Preparation: none.

Procedure :

Perform the usual procedure.

Pile 1 consists of 9 cards, pile 2 of 18, and pile 3 of 25.

While the magician has his back turned, the spectator chooses one of the three piles, looks at the face card, and replaces the pile in its spot.

"To avoid revealing which pile has been moved, the spectator is asked to reverse the positions of the other two piles, and finally to reverse the positions of the two thinnest piles.

The chosen card will then be at the bottom of the pile in the position originally occupied by pile 3 (of 25 cards).

When you turn back to face the audience, and observe how the spectator reassembles the deck, you are able to find the card by its rank in the deck, under strict conditions.

Gérard Bakner also clarified that if you yourself reassemble the deck, you can even have predicted the position of the card!

Version 7 : Bob Hummer forever.

Preparation: none.

Procedure:

Perform the usual procedure.

Pile 1 consists of 9 cards, pile 2 of 18, and pile 3 of 25.

Have each of the three piles reshuffled.

While the magician has his back turned, the spectator is invited to take a card from one of the three piles and place this card on one of the other two piles.

To reverse the position of the other two piles and finally to reverse the position of the two thinnest piles.

The chosen card will then be on the pile in the position originally occupied by pile 3 (of 25 cards).

When you turn back to face the audience, not only are you able to point out the pile containing the chosen card, but also, by adding "1" to 9, 18, or 25, in addition to announcing the number of cards it contains.

Prediction

In the version with three piles, you know that the first pile will contain 9 cards, the second 18 cards, and the third 25 cards. You can therefore force the number 91825 beforehand, either with a rigged calculator or a Forcing Matrix.

1834

It is said (I leave it to magic historians to verify the authenticity of this claim (sic!)) that in 1834, in a Parisian salon, young Jean-Eugène Robert-Houdin presented the following trick. The magician asks for a deck of 52 cards (not one more, not one less). The magician asks two spectators to join him. Each sits on either side of the table. The spectator who brought the deck is asked to cut it roughly in the middle. The second spectator is asked to take one of the two piles present after the cut and count the number of cards in that pile. They will get a two-digit number, which they will reduce to a single digit. They will remove from their pile as many cards as the digit obtained and place the removed cards on the unchosen pile left on the table. The magician then says that the trick is now finished and reminds them of the date announced at the beginning of the story: 1834.

18... 34

The magician pushes the two halves of the deck, with his index fingers, towards the two spectators, saying once again: 18... 34.

The spectators understand that they must count the cards.

One will have 18 cards and the other 34 cards.

EXPLANATION

Asking to cut the deck roughly in the middle will imply that each half will contain a little over twenty cards. We will thus arrive at the desired result: 18 and 34. However, if the spectator were to cut the deck such that one of the piles contained a little over 30 cards, you would entrust that pile to them to count the cards and perform the small calculation. Thus, in the end, both piles would each have a little over twenty cards, and you would then address the other spectator to do the same (with the other pile) to achieve the desired result.

The Sibyl of the Salons

You will need the 52 cards of a "The Sibyl of the Salons" deck.

Deck setup (from the top of the deck, backs visible):

10 random cards (X) - 51 - X - 45 - X - 42 - X - 36 - X - 33 - X - 27 - X - 24 - X - 18 - X - 15 - X - 9 - X - 6 - 21 random cards (X)

The positions of the eleven cards: 51, 45, 42, 36, 33, 27, 24, 18, 15, 9, and 6 can be interchanged.

They will be placed head-to-tail with the random cards.

Since the deck has an asymmetrical back, they will thus be recognizable from the back.

Nine of these eleven cards each total 6 or 9 when the two digits composing their number are added. Cards 6 and 9 will not need to be added.

You will also need a prediction: a piece of paper with the number "6" that can also be read as "9" by turning the paper over.

Presentation:

The magician places his prediction on the table, stating that he intends to present a numerological "reading."

False shuffle of the deck.

The magician deals the first ten cards one by one, explaining that one should say "Stop" whenever they wish.

A spectator must say "Stop" (during the next 21 cards).

This should not pose a problem.

If the spectator says "Stop" on one of the eleven cards (recognizable by their asymmetrical backs).

This card will be placed next to the deck.

The remaining cards will then be placed on top of those already dealt.

The magician asks that the two digits composing its number be added, if it is a two-digit number, or that the digit be kept, if it is a one-digit number.

In all cases, a "6" or a "9" will be obtained.

This digit will correspond to the prediction.

If the spectator says "Stop" on an "X" card, the magician will take the next card and place it perpendicular to the cards already on the table, then place the rest of the deck on top. Thus, the apparently chosen card will be the only one perpendicular to all the others.

The magician will remove this card from the deck and, obviously, after the addition (or not: 6 or 9), it will correspond to the prediction.