The continuity department at Fool Us are the true magicians, I couldn't spot the seam between the takes. Not a hair or wrinkle out of place. Card 4 was even slightly offset to the right so there was a gap between Card 3 and 4. Thanks for stating the folder is empty at the end. I am shocked there weren't 4 extra cards in there.
This makes me so happy!!! Congrats to VIM crew, Now I know why you looked so sad and then shocked. How cool that they “ tricked “ you back on the stage and surprised you with a trophy!!!!
I watched the performance a few days ago and loved it. This interview was fascinating-- Truly. Thank you!! and thank you Vanishing Inc for all you do!!!!
25:13 "Every now and then you read that 'Fool Us' is a little bit rigged, but that to me shows "That's not rigged."." I can see how people think that. Because the show isn't live the acts are edited for TV, and have to be, to make a decent length TV show .. but as one who tries to work out the method, I can see sometimes vital moves are cut out (not in a sinister way).
Fascinating behind the scenes! Thanks for sharing that! Loved the performance a whole lot!! The problem with a puzzle or maze type trick is that it invites busts. I’d love to figure this one out, just for myself and because it seems a satisfying endeavor.
Yup. The history of maze tricks like that go way back earlier than the Tenyo release. Main difference is you end clean with this version. The folder is empty at the end.
ATTENTION: The maths in the video is simply wrong. Well, if it's true that there was not even one extra card in the envelope, then the "maths" actually is very easy: Because of the nature of the linking inside the cards, the result of any order and orientation of the cards the result would always be that the input position equals the output position. So my guess is that he simply forced the two "bad boys" of magic select the same location at the beginning and that's it. What do you think about? Card linking is as given here: Card at position 4: direct linking left to the right: so this card doesn't change anything Card at position 1: brings everything 2 down (1->3, 2->4, 3->5, 4->1, 5->2) Card at position 2: brings everything 2 up (3->1, 4->2, 5->3, 1->4, 2->5) so this card is simply the undo for the card at position 1 Card at position 3: brings everything 1 up (2->1, 3->2, 4->3, 5->4, 1->5) Card at position 5: brings everything 1 down (1->2, 2->3, 3->4, 4->5, 5->1) so again this card is again the undo for the card at position 3 To rotate a card changes nothing at all. To unintuitive surprise also the order of the cards doesn't change anything. So there maybe comes the only "math" part into the game, to consider that a+b+c+d = a+c+b+d = .... any order you like. However: Andis calculations of the probability of positions in the original presentation are definitely wrong. With 5 inputs linking to 5 output positions he says that there would be only 25 possibilities. The fact that this might be "simple" maths that nearly every viewer could calculate (oh yeah I got it, 5 x 5) doesn't make it correct. Cause When it's true that the first input position has 5 output positions to to got there remain only 4 additional output positions then for the 2nd card to go for each of the 5 positions for the first card. That would sum up to 4x5 =20 positions what the first two inputs could do. However taking into account the additional remaining 3 positions for the 3rd card and the 2 positions for the 4th card for each of the selected combination before that results in the amount of 5x4x3x2 (x1) = 120 combinations if, and only if, the cards would really mixup the starting incoming with the outgoing side. Which they don't. It's also unclear, why the possible number of input-output combinations should go into the calculations of the setup in between, which he does, when calculating 25 x 24 x 16 = 9.600. Indeed, the number of input-output combinations is only the limit of possible results the complete independent way in between could result in. For instance think of a different setup: 5 inputs, 5 outputs, but only one card to place. Then simply the number of cards to be selected results in the possible versions the way may go. Which means, if you only have one card to select (actually no selection), then there will be only one possibility the way may look like. However if you have for example 10,000 cards to select. You would have 10,000 possibilities. Of course then a lot of cards would maybe look different concerning the winding of the path, but still result in one of the maximum 120 combinations of mapping the input into the output. If you have more then one slots to place cards and a limited set of cards the calculation works out as follows. e.g. think of 50 cards to select for 3 slots: 1. card: 1 out of 50 possible selections, 2. card: 1 out of the remaining 49 3. card: 1 out of the remaining 48 resulting in a total product of 50 x 49 x 48 possibilities = 117,600. (That's the way the number increases so fast in the lottery.) Back to the situation in the magic show: 5 slots for 5 cards to select from results again in 5 x 4 x 3 x 2 (x1) = 120 = (5! ... spoken 5 factorial) again. Although it seams to be linked because of the same result as with the input-output combinations, it is not. Would there have been 6 slots with 6 cards to select from it would calculate as 6! = 720 for the different combinations of cards to go to, however still limiting the number of the input-output result to 120. So at least 600 combinations would result in similar input-output combinations. But let's remain with the 120 possible positions the cards could be go to. + adding the option to rotate some (or all of them), then the possible options increases as follows: 1 option: all 5 cards no rotation 5 options: 1 out of 5 cards rotated 10 options: 2 out of 5 rotated (1+2, 1+3, 1+4, 1+5, 2+3, 2+4, 2+5, 3+4, 3+5, 4+5) 10 options: 3 out of 5 rotated (1+2+3, 1+2+4, 1+2+5, 1+3+4, 1+3+5, 1+4+5, 2+3+4, 2+3+5, 2+4+5, 3+4+5) 5 option: 1 out of 5 NOT rotated 1 option: all 5 cards rotated _______________ or simply (2 ^ 5) resulting in 32 possible rotation/no-rotation combinations So the maximum of possibles ways the cards could have gone to turns out to be 120 (possible places) x 32 (possible rotations) = 3,840, with 3,720 replications as the limit of 120 possible outcomes where the 5 inputs might go to remains still valid.' However Andi calculates this taking the number 25 input->output possibilities (which indeed is still wrong) x 24 (where I have no clue, where this number comes from) x 16 (again no clue). While indeed the correct calculated number of combinations turns out to be 3,720 (or incorrectly 9,600) and still goes into the 1000 the limit to 120 possible outcomes remains. But with the carefully selected layout of the cards as described at the very beginning of this explanation that number goes down to 1. (input = output) Final conclusion: Taking for granted, that there was no remaining card in the envelope, the trick was to have Penn and Teller selected the same spot. And while I do not know, how he has done that, I know for sure, that his math is completely incorrect.
I'm a little bit confused about one thing: If they reshot the corrected reveal before the supposed repeat performance, why is the completed maze still in the background?
There wasn't a repeat performance. The producer told Andi on the phone that they had to reshoot it not to give away that Andi was getting a trophy, so it'd be a surprise ☺️
at 12:02, he says that having an assistant who's doing some of the method is against the show's rules (if I understand right). Is that true? I can remember a few times where I thought the on-stage assistant (or an off-stage one) is manipulating things to make a trick work.
From the comment I left on the fooler video: I'm gonna guess that he has multiple sets of the same card questions, each one of which corresponds to a location pair. So one card for home to library. One card for home to theater, etc etc. That's just 25 cards he has to hold. Additionally, I am going to guess that the maze pieces are designed such that they follow the same path even when flipped upside down. In fact, I cut one of the cards out of the video into photoshop, and determined that I am right about the orientation. The input output paths are identical no matter if the card is right side up, or upside down. So, I imagine all five cards are the same way. That means that no matter how the cards are arranged all the way across the board, and no matter their orientation, there is only ONE actual maze. All the fooler has to do is pull the right cards from his stack: Home to library, stack one. Home to Coffee shop, stack two. Library to theater, stack three. And so on. If Penn would have asked, _"Can we look at your card stack folder?"_ it would have been all over.
But I thought this trick was CLEVER. But, like Andi said, it's a one time trick. You couldn't do it again later because your audience could alway do like I did, and AFTER the fact, examine it closer. This one works only because folks do not have to to critically analyse it. But it's wonderful in real time.
5:48 - See? Exactly. It's all about the experience. You, taking advantage of limited time, gave us a wonderful experience. That's a GREAT way of putting it.
Same thought here. I thought I was so clever, having figured out the mathematics of it, and knowing that in addition to the “0” card we saw there must be four more versions. But knowing that the folder is empty, I am blown away! There must be a gimmick…. Ok, here is a theory: magnetic sheets. The four missing maze sections are duplicates normally stuck on top of their originals/equivalents, but if necessary one can be held back and stuck on the keystone card instead. (If that’s the real answer, Andi / Vanishing should feel free to respond and then delete this comment, or tell me to delete it!) EDIT: simpler, seems like it could be done with six cards rather than nine? The last is not the key, it is just a gag; the other five are all unique, and Andi can choose which to “hide” by leaving it flat on top of another.
I’m absolutely astonished that Penn and Teller didn’t already know about this age-old trick, and the secret to it. Andi didn’t show that the envelope was empty after five cards were removed from it.
Integrity, Penn and Teller exude it and are willing to admit they messed up.
The continuity department at Fool Us are the true magicians, I couldn't spot the seam between the takes. Not a hair or wrinkle out of place. Card 4 was even slightly offset to the right so there was a gap between Card 3 and 4.
Thanks for stating the folder is empty at the end. I am shocked there weren't 4 extra cards in there.
19:47 If you came for the story about “Fool Us” then here you are.
This makes me so happy!!! Congrats to VIM crew, Now I know why you looked so sad and then shocked. How cool that they “ tricked “ you back on the stage and surprised you with a trophy!!!!
I watched the performance a few days ago and loved it. This interview was fascinating-- Truly. Thank you!! and thank you Vanishing Inc for all you do!!!!
Our pleasure!
25:13 "Every now and then you read that 'Fool Us' is a little bit rigged, but that to me shows "That's not rigged."."
I can see how people think that. Because the show isn't live the acts are edited for TV, and have to be, to make a decent length TV show .. but as one who tries to work out the method, I can see sometimes vital moves are cut out (not in a sinister way).
congradts Andi - a wonderful performance to watch and you definately found and created a wonderful piece of magic. Thank you !!
Fascinating behind the scenes! Thanks for sharing that! Loved the performance a whole lot!! The problem with a puzzle or maze type trick is that it invites busts. I’d love to figure this one out, just for myself and because it seems a satisfying endeavor.
Congratulations on your fool us win.Fabulous routine.👏🎩
That was a great trick. It did eerily look like a Tenyo effect from a couple of years ago. Magic Maze.
Yup. The history of maze tricks like that go way back earlier than the Tenyo release. Main difference is you end clean with this version. The folder is empty at the end.
ATTENTION: The maths in the video is simply wrong.
Well, if it's true that there was not even one extra card in the envelope, then the "maths" actually is very easy:
Because of the nature of the linking inside the cards, the result of any order and orientation of the cards the result would always be that the input position equals the output position. So my guess is that he simply forced the two "bad boys" of magic select the same location at the beginning and that's it.
What do you think about?
Card linking is as given here:
Card at position 4: direct linking left to the right: so this card doesn't change anything
Card at position 1: brings everything 2 down (1->3, 2->4, 3->5, 4->1, 5->2)
Card at position 2: brings everything 2 up (3->1, 4->2, 5->3, 1->4, 2->5)
so this card is simply the undo for the card at position 1
Card at position 3: brings everything 1 up (2->1, 3->2, 4->3, 5->4, 1->5)
Card at position 5: brings everything 1 down (1->2, 2->3, 3->4, 4->5, 5->1)
so again this card is again the undo for the card at position 3
To rotate a card changes nothing at all.
To unintuitive surprise also the order of the cards doesn't change anything.
So there maybe comes the only "math" part into the game, to consider that a+b+c+d = a+c+b+d = .... any order you like.
However: Andis calculations of the probability of positions in the original presentation are definitely wrong.
With 5 inputs linking to 5 output positions he says that there would be only 25 possibilities.
The fact that this might be "simple" maths that nearly every viewer could calculate (oh yeah I got it, 5 x 5) doesn't make it correct.
Cause When it's true that the first input position has 5 output positions to to got there remain only 4 additional output positions then for the 2nd card to go for each of the 5 positions for the first card.
That would sum up to 4x5 =20 positions what the first two inputs could do.
However taking into account the additional remaining 3 positions for the 3rd card and the 2 positions for the 4th card for each of the selected combination before that results in the
amount of 5x4x3x2 (x1) = 120 combinations if, and only if, the cards would really mixup the starting incoming with the outgoing side. Which they don't.
It's also unclear, why the possible number of input-output combinations should go into the calculations of the setup in between, which he does, when calculating
25 x 24 x 16 = 9.600.
Indeed, the number of input-output combinations is only the limit of possible results the complete independent way in between could result in.
For instance think of a different setup: 5 inputs, 5 outputs, but only one card to place.
Then simply the number of cards to be selected results in the possible versions the way may go.
Which means, if you only have one card to select (actually no selection), then there will be only one possibility the way may look like.
However if you have for example 10,000 cards to select. You would have 10,000 possibilities.
Of course then a lot of cards would maybe look different concerning the winding of the path, but still result in one of the maximum 120 combinations of mapping the input into the output.
If you have more then one slots to place cards and a limited set of cards the calculation works out as follows.
e.g. think of 50 cards to select for 3 slots:
1. card: 1 out of 50 possible selections,
2. card: 1 out of the remaining 49
3. card: 1 out of the remaining 48
resulting in a total product of 50 x 49 x 48 possibilities = 117,600. (That's the way the number increases so fast in the lottery.)
Back to the situation in the magic show:
5 slots for 5 cards to select from results again in 5 x 4 x 3 x 2 (x1) = 120 = (5! ... spoken 5 factorial) again.
Although it seams to be linked because of the same result as with the input-output combinations, it is not.
Would there have been 6 slots with 6 cards to select from it would calculate as 6! = 720 for the different combinations of cards to go to,
however still limiting the number of the input-output result to 120. So at least 600 combinations would result in similar input-output combinations.
But let's remain with the 120 possible positions the cards could be go to.
+ adding the option to rotate some (or all of them), then the possible options increases as follows:
1 option: all 5 cards no rotation
5 options: 1 out of 5 cards rotated
10 options: 2 out of 5 rotated (1+2, 1+3, 1+4, 1+5, 2+3, 2+4, 2+5, 3+4, 3+5, 4+5)
10 options: 3 out of 5 rotated (1+2+3, 1+2+4, 1+2+5, 1+3+4, 1+3+5, 1+4+5, 2+3+4, 2+3+5, 2+4+5, 3+4+5)
5 option: 1 out of 5 NOT rotated
1 option: all 5 cards rotated
_______________
or simply (2 ^ 5)
resulting in 32 possible rotation/no-rotation combinations
So the maximum of possibles ways the cards could have gone to turns out to be 120 (possible places) x 32 (possible rotations) = 3,840,
with 3,720 replications as the limit of 120 possible outcomes where the 5 inputs might go to remains still valid.'
However Andi calculates this taking the number 25 input->output possibilities (which indeed is still wrong) x 24 (where I have no clue, where this number comes from) x 16 (again no clue).
While indeed the correct calculated number of combinations turns out to be 3,720 (or incorrectly 9,600) and still goes into the 1000 the limit to 120 possible outcomes remains.
But with the carefully selected layout of the cards as described at the very beginning of this explanation that number goes down to 1. (input = output)
Final conclusion:
Taking for granted, that there was no remaining card in the envelope, the trick was to have Penn and Teller selected the same spot.
And while I do not know, how he has done that, I know for sure, that his math is completely incorrect.
I'm a little bit confused about one thing: If they reshot the corrected reveal before the supposed repeat performance, why is the completed maze still in the background?
Separate maze I think
There wasn't a repeat performance. The producer told Andi on the phone that they had to reshoot it not to give away that Andi was getting a trophy, so it'd be a surprise ☺️
at 12:02, he says that having an assistant who's doing some of the method is against the show's rules (if I understand right). Is that true? I can remember a few times where I thought the on-stage assistant (or an off-stage one) is manipulating things to make a trick work.
From the comment I left on the fooler video:
I'm gonna guess that he has multiple sets of the same card questions, each one of which corresponds to a location pair. So one card for home to library. One card for home to theater, etc etc. That's just 25 cards he has to hold. Additionally, I am going to guess that the maze pieces are designed such that they follow the same path even when flipped upside down.
In fact, I cut one of the cards out of the video into photoshop, and determined that I am right about the orientation. The input output paths are identical no matter if the card is right side up, or upside down. So, I imagine all five cards are the same way. That means that no matter how the cards are arranged all the way across the board, and no matter their orientation, there is only ONE actual maze. All the fooler has to do is pull the right cards from his stack: Home to library, stack one. Home to Coffee shop, stack two. Library to theater, stack three. And so on.
If Penn would have asked, _"Can we look at your card stack folder?"_ it would have been all over.
But I thought this trick was CLEVER. But, like Andi said, it's a one time trick. You couldn't do it again later because your audience could alway do like I did, and AFTER the fact, examine it closer. This one works only because folks do not have to to critically analyse it. But it's wonderful in real time.
5:48 - See? Exactly. It's all about the experience. You, taking advantage of limited time, gave us a wonderful experience. That's a GREAT way of putting it.
12:19 - Oh dear. I hope I didn't reveal the Tenyo method???? Sorry if I did.
No multiple sets. Folder is empty at the end.
@@VanishingIncMagic Empty Folder doesn't equal no multiple sets
I think the black maze cards have to have a scroll mechanism for it to work
If there aren't extra 4 cards then it's really interesting.
The folder is empty...
Same thought here. I thought I was so clever, having figured out the mathematics of it, and knowing that in addition to the “0” card we saw there must be four more versions. But knowing that the folder is empty, I am blown away! There must be a gimmick…. Ok, here is a theory: magnetic sheets. The four missing maze sections are duplicates normally stuck on top of their originals/equivalents, but if necessary one can be held back and stuck on the keystone card instead. (If that’s the real answer, Andi / Vanishing should feel free to respond and then delete this comment, or tell me to delete it!)
EDIT: simpler, seems like it could be done with six cards rather than nine? The last is not the key, it is just a gag; the other five are all unique, and Andi can choose which to “hide” by leaving it flat on top of another.
I’m absolutely astonished that Penn and Teller didn’t already know about this age-old trick, and the secret to it. Andi didn’t show that the envelope was empty after five cards were removed from it.
Because the envelope _was_ empty. he addressed that in the podcast.
@@RabblesTheBinx The podcast is not part of the performance. If he didn’t show it was empty, how could anyone be sure it was at the time?
He showed it!
But that bit was cut from the final edit
@@Raph333 ok - first time anyone has said that, to my knowledge.
@@superman00001 Its in the podcast that Rabbles mentions and the link is in the comments somwhere