This is Simon Aronson’s “Prior Commitment”, based on his “Undo Influence” principle. It can be found in his book “Try the Impossible”. He was an awesome creator.
Simon Aranson is a genius. If you can memorize a stack, there are impossible tricks that you can pull of which can absolutely floor people, including magicians!!
TLDR: There are two independent variables -- the number of cards before the first joker (call it X) and the number of cards between the jokers (call it Y). After the procedure, Spectator 1's card will be Y cards from the top and Spectator 2's card will be (52 - X) cards from the top. In the video, X = 9 and Y = 18, so Spectator 1's card is 18th from the top while Spectator 2's card is (52 - 9) = 43 from the top. Why it works: I will try to show the state of the playing cards from top to bottom... The SETUP: 9 cards Joker43 18 cards Joker18 25 cards LOCATIONS OF THE SPECTATOR'S CARDS: Spectator 1's card is the Nth card after the first joker. Spectator 2's card is the Mth card after the second joker. 9 cards Joker43 N cards (Spectator 1's card is the Nth card) -------------- Spectator 1 cut ------------- 18 - N cards Joker18 M cards (Spectator 2's card is the Mth card) -------------- Spectator 2 cut ------------- 25 - M cards Note in the above, the dashed lines indicate the locations of the spectator cuts. Checking consistency, there are still (N + 18 - N) = 18 cards between the two jokers and there are (M + 25 - M) = 25 cards after the second joker. FIRST REASSEMBLY: As in the video, keep the bottom packet and transpose the positions of the spectator's packets... 18 - N cards Joker18 M cards (Spectator 2's card is the Mth card) 9 cards Joker43 N cards (Spectator 1's card is the Nth card) 25 - M cards JOKER SPLIT: 18 - N cards -------------------------------------------- M cards (Spectator 2's card is the Mth card) 9 cards -------------------------------------------- N cards (Spectator 1's card is the Nth card) 25 - M cards SECOND REASSEMBLY: As in the video, keep the top packet and transpose the positions of the bottom two packets. 18 - N cards N cards (Spectator 1's card is the Nth card) 25 - M cards M cards (Spectator 2's card is the Mth card) 9 cards You are done! Spectator 1's card is (18 - N + N) = 18 cards from the top and Spectator 2's card is 18 + (25 - M + M) = 43 cards from the top.
I recognized the 18 from the setup, and 52-9=43. I knew there was some switching to get the cards between the jokers. Thanks for getting the HOW of that.
If you want to change the numbers (17,44) it’s easy. This works for the second tutorial (color changing selection) may be different for the other version. Idk. 44 is 52 minus the amount of cards you count off prior to placing the first prediction card. In this case it was 8. 52-8=44 17 is one more than the number of cards you count between the prediction cards. So if you want 39, you’d put 13 cards on top (52-13=39). If you wanted 15 you’d count 14 cards between the two prediction cards during setup. Now please understand this will change the spaces between selections and prediction cards. So play with it first. But I think it’s nice to have options. Maybe their age, or sport number.
I love this! I feel like using the breather card in Red Hot Mamma makes that trick stronger as well. Gonna play with this for a bit to see what I can get out of it. Thanks for sharing it!
It's actually a Simon Aronson routine from a book called *Try The Imposible* if i'm not wrong. You can find different routines based on the same principle. Very clever method. Thank to refresh my mind on that trick. It's a good one. Very deceptiv.
GREAT TRICK!!! I like very much those self working tricks, your explanations are absolutly clear and plesent. Thank you so much for the fun you bring into our lives!!!
Great trick (especially your version). However I'm struggling with the "kicker." Since the magician is the one who generates the numbers 18 and 43, will it feel like a reveal to see them on the jokers? It seems less like a kicker and more like confirmation of two arbitrary numbers that you brought into the trick.
I know what you mean, but when he said the numbers i though "well, they could be calculated depending on what happened until there" as you would do in some tricks with stacked decks. When here turned the jokers I was like "huh, never mind". But I kinda agree, it wouldn't take much away if the numbers where shown before counting in my opinion...
Hola Steven, me parece un truco espectacular y muy desconcertante para el espectador. ¿Puedes por favor, activar los subtitulos en este video de youtube, para que pueda leer correctamente la explicación del tutorial en español? Muchas gracias por el canal, soy un suscriptor encantado de tu trabajo. Un saludo desde España Steven.
Im interested.....noce self worker! Sorry if you mention it later in the lesson but can these be done together as a two- phaser?.Or you can pnly do one or thr other because of the setup?
NICE self-working trick and my older grandkids will love it. You explained it well and earned yourself a new subscriber. One question...how do you go to your next card trick without ditching the deck or adding the cards that the blue back replaced? Gonna have to do some binging of your channel this weekend. 😁
I guess you’d just have to do tricks that don’t require exactly 52 cards. Remove the odd backed cards and say, maybe if I do this 25 more times I’ll have a complete blue deck.
Here is how I figured it out. We can think that there are five stacks: S1 = 9, S2 + S3 = 18 and S4 + S5 = 25. We do not know what S2 is, because it depends on how the spectator cuts. The same with S4. As a result of all, the stacks are in the order S3 S2 S5 S4 S1. That’s the solution.
I learnt the first one just a few days ago it's quite mind blowing how it's self working. Your version is ofcourse great but relying on spectators to cut to the right card worries me 😂
This would be more impressive if those 17 and 44 were somehow forced to the spectators and named by them. Otherwise it’s kinda pointless to have it written on the cards.
But in the beginning of the trick the 6 of diamonds on the left and the 8 of diamonds on the right”I noticed when the cards are revealed the 6 of diamonds come out for the spectator on the left side, and the 8 of diamonds come out on the right. But in the beginning of the trick the 6 of diamonds on the left and the 8 of diamonds on the right
Not a fan of the magician cutting the deck to show the card. It's too obvious. And I'm not sure I can trust the spectator to cut accurately. What would you do if the spectator cuts it wrong?
Joe Diamond says a good magician will take risks. I'm in the same boat what if they don't do this or that and that's true for most tricks. I guess the best tricks have to have some element of possible failure to them
This is Simon Aronson’s “Prior Commitment”, based on his “Undo Influence” principle. It can be found in his book “Try the Impossible”. He was an awesome creator.
Time to re visit and read my copy I am glad so many know what can be found in what books👍👍👍👍
I'm not sure why Steven rarely credits the authors of these tricks (or even seeks permission in some cases).
Simon Aranson is a genius. If you can memorize a stack, there are impossible tricks that you can pull of which can absolutely floor people, including magicians!!
@@alanbrownmusic I am very disappointed with the lack of crediting.
@@alanbrownmusic It's because he doesn't care, he just want to win money with the views, sad.
TLDR:
There are two independent variables -- the number of cards before the first joker (call it X) and the number of cards between the jokers (call it Y). After the procedure, Spectator 1's card will be Y cards from the top and Spectator 2's card will be (52 - X) cards from the top. In the video, X
= 9 and Y = 18, so Spectator 1's card is 18th from the top while Spectator 2's card is (52 - 9) = 43 from the top.
Why it works:
I will try to show the state of the playing cards from top to bottom...
The SETUP:
9 cards
Joker43
18 cards
Joker18
25 cards
LOCATIONS OF THE SPECTATOR'S CARDS:
Spectator 1's card is the Nth card after the first joker. Spectator 2's card is the Mth card after the second joker.
9 cards
Joker43
N cards (Spectator 1's card is the Nth card)
-------------- Spectator 1 cut -------------
18 - N cards
Joker18
M cards (Spectator 2's card is the Mth card)
-------------- Spectator 2 cut -------------
25 - M cards
Note in the above, the dashed lines indicate the locations of the spectator cuts. Checking consistency, there are still (N + 18 - N) = 18 cards between the two jokers and there are (M + 25 - M) = 25 cards after the second joker.
FIRST REASSEMBLY:
As in the video, keep the bottom packet and transpose the positions of the spectator's packets...
18 - N cards
Joker18
M cards (Spectator 2's card is the Mth card)
9 cards
Joker43
N cards (Spectator 1's card is the Nth card)
25 - M cards
JOKER SPLIT:
18 - N cards
--------------------------------------------
M cards (Spectator 2's card is the Mth card)
9 cards
--------------------------------------------
N cards (Spectator 1's card is the Nth card)
25 - M cards
SECOND REASSEMBLY:
As in the video, keep the top packet and transpose the positions of the bottom two packets.
18 - N cards
N cards (Spectator 1's card is the Nth card)
25 - M cards
M cards (Spectator 2's card is the Mth card)
9 cards
You are done! Spectator 1's card is (18 - N + N) = 18 cards from the top and Spectator 2's card is 18 + (25 - M + M) = 43 cards from the top.
I recognized the 18 from the setup, and 52-9=43. I knew there was some switching to get the cards between the jokers. Thanks for getting the HOW of that.
OMG you suk
Love a self-working trick ,Stephen. This one is cool, and I do like your twist on the end.
If you want to change the numbers (17,44) it’s easy.
This works for the second tutorial (color changing selection) may be different for the other version. Idk.
44 is 52 minus the amount of cards you count off prior to placing the first prediction card. In this case it was 8. 52-8=44
17 is one more than the number of cards you count between the prediction cards.
So if you want 39, you’d put 13 cards on top (52-13=39). If you wanted 15 you’d count 14 cards between the two prediction cards during setup.
Now please understand this will change the spaces between selections and prediction cards. So play with it first.
But I think it’s nice to have options. Maybe their age, or sport number.
I love this! I feel like using the breather card in Red Hot Mamma makes that trick stronger as well. Gonna play with this for a bit to see what I can get out of it. Thanks for sharing it!
It's actually a Simon Aronson routine from a book called *Try The Imposible* if i'm not wrong. You can find different routines based on the same principle. Very clever method. Thank to refresh my mind on that trick. It's a good one. Very deceptiv.
GREAT TRICK!!! I like very much those self working tricks, your explanations are absolutly clear and plesent. Thank you so much for the fun you bring into our lives!!!
A brilliant trick! Please may I ask where you got the lovely green board you place the cards on? Thank you
I used to have one It was a cut down foldable card table just reduced playing size area and removed legs
Wonderful trick, and tutorial. Graham Jolley fooled Penn & Teller several years back with this. 😊👍😊👍😊
Amazing. Great tutorial
The first prediction I have always set up the second stage is new to me so thank you so much for taking it to a new level👍👍👍👍👍👍👍👍👍
Great trick (especially your version). However I'm struggling with the "kicker." Since the magician is the one who generates the numbers 18 and 43, will it feel like a reveal to see them on the jokers? It seems less like a kicker and more like confirmation of two arbitrary numbers that you brought into the trick.
I know what you mean, but when he said the numbers i though "well, they could be calculated depending on what happened until there" as you would do in some tricks with stacked decks. When here turned the jokers I was like "huh, never mind".
But I kinda agree, it wouldn't take much away if the numbers where shown before counting in my opinion...
Agree totally. Love both versions of the trick, but don't think I like the numbers written on the cards 👍
Just ask "And do you have any idea how they told me where to look? (pause) Simple!" and turn over the jokers.
@@MartyHirsch It still feels like a double beat. 1) These jokers made a prediction. 2) These jokers made the same prediction I already told you about.
Wonderful trick.. killer trick indeed
Hola Steven, me parece un truco espectacular y muy desconcertante para el espectador. ¿Puedes por favor, activar los subtitulos en este video de youtube, para que pueda leer correctamente la explicación del tutorial en español? Muchas gracias por el canal, soy un suscriptor encantado de tu trabajo. Un saludo desde España Steven.
yes it's Simon's trick with the wispering queens!! not surprised P&T about they are not mathematicians! Latin? IV? MM?
'Whispering Jokers' by Graham Jolley, it's been around a while.
Really cool trick
Im interested.....noce self worker! Sorry if you mention it later in the lesson but can these be done together as a two- phaser?.Or you can pnly do one or thr other because of the setup?
Excellent
NICE self-working trick and my older grandkids will love it. You explained it well and earned yourself a new subscriber. One question...how do you go to your next card trick without ditching the deck or adding the cards that the blue back replaced? Gonna have to do some binging of your channel this weekend. 😁
I guess you’d just have to do tricks that don’t require exactly 52 cards.
Remove the odd backed cards and say, maybe if I do this 25 more times I’ll have a complete blue deck.
Very nice Sir thank you❣️♠️♣️♦️
As soon as I saw the ending to the first performance I said: hey this is the trick that tricked Penn and Teller! Great stuff mate!
@@Zxhir_10 I haven't got to the episode where Penn and teller get fooled by this trick so that's exciting
Mr Aronson was a beast.
😱😱😱 not another one
Here is how I figured it out. We can think that there are five stacks: S1 = 9, S2 + S3 = 18 and S4 + S5 = 25. We do not know what S2 is, because it depends on how the spectator cuts. The same with S4.
As a result of all, the stacks are in the order S3 S2 S5 S4 S1. That’s the solution.
I learnt the first one just a few days ago it's quite mind blowing how it's self working. Your version is ofcourse great but relying on spectators to cut to the right card worries me 😂
Just put a nice breather into them
R u sure if we ask the spectators to cut, they will get this position? Or for safety, we cut for them?
If you want to know how it works Steven, you should buy Simon's book.
This would be more impressive if those 17 and 44 were somehow forced to the spectators and named by them. Otherwise it’s kinda pointless to have it written on the cards.
That would certainly be ideal. I was thinking the exact same thing.
Do you have any ideas?
But in the beginning of the trick the 6 of diamonds on the left and the 8 of diamonds on the right”I noticed when the cards are revealed the 6 of diamonds come out for the spectator on the left side, and the 8 of diamonds come out on the right.
But in the beginning of the trick the 6 of diamonds on the left and the 8 of diamonds on the right
First❤❤
Not a fan of the magician cutting the deck to show the card. It's too obvious. And I'm not sure I can trust the spectator to cut accurately. What would you do if the spectator cuts it wrong?
I believe he address that situation. I’m more concerned if they expose one of the blue backs. That would take a lot away from the trick.
Joe Diamond says a good magician will take risks. I'm in the same boat what if they don't do this or that and that's true for most tricks. I guess the best tricks have to have some element of possible failure to them