Andi Gladwin FOOLS Penn & Teller

Some thank yous from Andi! It started with an idea shared by Mark Elsdon, with key method elements from a Tenyo trick (invented by So Sato), Karl Fulves’ work, and a packet trick (with lightbulbs of all things!) and then encouragement from Eli Bosnick, Harrison Greenbaum, Josh Jay and Brent M Braun to personalise it more.
Add in some help with the method from Matt Baker plus George, Brian Watson, and Mark Woodsford from the Vanishing Inc. team and Roger Nicot who built the stand and we’ve got a cocktail made from ideas from many encouraging and patient friends.
Thank you to all of them. And as always, thank you to Josh Jay and George Luck who continuously encouraged me to push the trick further along.
And obviously, thanks to Penn and Teller, Michael Close, and the whole Fool Us team who were beyond encouraging and accommodating of a crazy Brit who thought it was a good idea to fly across the world in a pandemic to perform a quick trick!
----
Vanishing Inc. Cofounder Andi Gladwin appeared on Fool Us, Jan 8th 2022 on the CW Channel. (Uploaded with permission).

[@ebadkhan7649]

For people wondering about the backstory of this performance: Basically everything after 7:59 was re-filmed. Originally, Penn and Teller said they were NOT fooled by this trick. However, Andi recalls after his performance, he got a call from a producer saying that they lost the footage of the performance and need him to come back to do it. He agreed and once they got him on stage, they didn't make him perform again, but instead P&T did their little talkback that we see at 8:00. They had read the explanation packet that Andi gave the producers that detail how his trick is done, and P&T decided they were WAY OFF from what the true method is. They brought Andi back on stage just so they can tell him that he fooled them, and awarded him the trophy. Andi recounts all of this in an interview on the Vanishing Inc. Magic channel.

[@mutatednut]

That’s sick!

[@billyeveryteen7328]

​@@mutatednut How does that happen, where P&T get it wrong, but they're not corrected until later? Because a similar thing happened to Simon Coronel and his poker chips. From what I understand, the way the show actually works in real time, P&T chat not only with each other, but also with the producers backstage via headset first before going into the coded "you fooled us" or "you didn't fool us" spiel they give the performer. Wouldn't the producers be able to step in before that point and let P&T know they got the methodology wrong?

[@mutatednut]

@@billyeveryteen7328 I have no idea man, doesn’t make too much sense.

[@wrenboy2726]

@@billyeveryteen7328 Because they’re talking in code and yes, although they do have direct contact with the producers, it may sound like to them that P&T are on point. It’s simply miscommunication. I don’t see any mystery behind it.

[@billyeveryteen7328]

@@wrenboy2726 They only talk in code to the magician on stage, and that's only done to not give away the secrets to their trick to the audience. According to a few magicians who have been on the show, as well as the backstage producer to whom every magician has to explain how the trick is done, Penn & Teller and the producers backstage already know whether or not a magician is getting a trophy even before Alison says "let's go to the boys" or whatever. There was a notable incident with a magician whose name I don't remember, but he did a trick that involved a ring that jumped back and forth between his fingers. Penn and Teller guessed that his ring came apart or was in some other way gimmicked, the magician said it wasn't, and they gave him a trophy. The KZbin comments section accused him of cheating and lying to P&T, and the producer himself went on various podcasts to defend the magician, saying that not only did he not lie, but the way the show works, they wouldn't be able to lie, because by the time P&T talk to the magician on stage, the producer has already decided whether or not they were fooled. If that's the case and that's how the show works (and it would have to work that way, otherwise it would be very easy for every magician to just lie), then there shouldn't be any incidents where the magician was incorrectly not awarded a trophy.

[@re-de]

Penn and teller buy tricks from him, yet he’s surprised that he fooled them 😂 such a humble guy

[@jacobniedermayer8924]

Penn and Teller likely buy any publicly sold trick to keep current with the magic scene in all fairness.

[@edward9643]

Maybe he didn't fool them - maybe it was an act of generosity for all the help they've received from him.

[@timhitt9541]

@@edward9643 no they were fooled

[@enamelpin628]

I think part of it is that they slipped in some code, you could see his face wince at some key phrases

[@harshharsh2124]

@@20catsRPG oh my god ...............youre so right man i noticed that the order didnt matter because every thing lead to the same thing .................i think penn and teller are rocking this show with one covalent brain cell ....................and youre so right about the dollar magic too

[@nicholashudson5020]

Looking at some of the cards they seem to be a combination of different shifts or a direct translation. All of the starting points are in the same order as the ending points. They just get shifted everything up one, or down one, or stay the same. So the trick isn't so much the odds, it's seeing where Penn and Teller place themselves, computing the up/down total, and then selecting those shift cards from his stash. Since they chose the same level, he just had to even out the up shift and down shift totals. +1,+1,0,-1-1. If Teller had ended up 2 spots above Penn then he would have chosen different cards, +1,+1,0,0,0. The card spin is clever, but pointless. Because as long as all the cards move the tracks in unison up and down, then the card is the same backwards and forwards. Also the order doesn't matter at all so they can be in any arrangement, up or down. Asking them to change position and rotate is just a distraction. Perhaps the real smarts here is that by designing it the way he did, he only needs 7 cards in his folder to make any combination. +1,+1,0,0,0,-1-1, (or +1+1+1,0,-1-1-1) Because no starting positions are ever more than 2 moves away. Even if they pick top and bottom, it's only one move away because when the track cards shift up the top track drops to the bottom. So it's not like he needs a whole binder of 15-20 cards, and has to fumble through them all. Probably has them in order of value, and has memorized the order, so he can give the prompts without looking at the card, which is meant to convince mark that the cards are in a predetermined order and he is not selecting them based on the positions. Maybe there is more to it, but that seems like the simplest plausible explanation to me.

[@stephenledford3808]

Seems pretty accurate but I wouldn't of posted this out of respect..... I caught one on his channel and just left a suggestion....

[@voidremoved]

thats great but he probably just predicted they would put miami vice at 1 and upside down

[@briankarcher8338]

That's a way to do it. Andi mentioned that his trick has no extra items in the envelope though. Wondering how to do it without any extra cards....

[@rogerskitt1542]

@@briankarcher8338 he picked 5 very easy to discerne how they would rank it. Would prob take 5 minutes of research to learn Miami vice was a terrible experience and hated teaching Latin/going to clown school. Broadway show would easily be the best.

[@bender8109]

@@rogerskitt1542 The order doesn't matter. Nicholas was exactly right. Each card is rotationally symmetrical (that is, it doesn't matter if it's upside down) and it amounts to a simple shift of the order. In the first card position 1 went to 3, 2-4, 3-5, 4-1, 5-2, so it shifted everything down by 2. The second card was 1-4, 2-5, 3-1, 4-2, 5-3, and so on for +3. The 5 cards were +2, +3, +4, +0, and +1. If you add these all together, you get +10 which is a multiple of 5 (which is to say that if you add all the cards modulo 5, the result is 0). This means that regardless of what order they arrange the cards and regardless of any flipping, each line will match up with the same position when you read across. The core of the trick is just basic math (and cleverly designed cards and routine that make it difficult to notice in real time how simple it is).

[@sethsheber9088]

“Pen Andi Teller” not going to lie, that is a great name.

[@trippyripty8310]

That what I was thinking Teller but he never speaks🤣 Penn cause he rights the show🤷?

[@johncoops6897]

@@trippyripty8310 - Is this an illiteracy conference? A meeting of the 'bad at spelling' club?

[@LukeHatchet]

“Rick Andy Morty’s!” “Dumb.”

[@Uprising771]

@@LukeHatchet what is that bullshyt

[@SyzygyNoon]

Thank you for not lying. It’s a relief to know you speak the truth.

[@Ellenebert9999]

I love this trick even more now that I know the backstory of how Penn and Teller reversed their original decision and called Andi back to surprise him with a Fool Us trophy. This made my day

[@craig.tanner]

I didn't know this. What's the backstory?

[@Edwardmd7]

@@craig.tanner kzbin.info/www/bejne/h5iXfGysjZ53bac Timestamp: 19:50

[@cregkly5444]

Podcast where Andi talks about this appearance. kzbin.info/www/bejne/h5iXfGysjZ53bac

[@JadesOfGrey]

What?

[@feepy]

@@craig.tanner kzbin.info/www/bejne/h5iXfGysjZ53bac

[@zacharygarza1]

This is perfect. The premise, the effect, the presentation... Andi crushed it!

[@PeanutwormZ]

I can always tell how great a magic trick is by rating how much rage i feel after seeing it performed. Maximum rage.

[@DavidSmith-pg1ob]

The trick is quite simple really. He's using ACME disappearing/reappearing ink on those cards!

[@LasseLundster]

@@DavidSmith-pg1ob No lol, that's not the trick at all

[@tryingtolearnthis]

One of the main reasons I love watching fool us is how genuine the reaction foolers get when they find out they stumped Penn and teller. The hard work that goes into this preparation is astounding.

[@harshharsh2124]

sheep

[@Darkjonny79]

It's even better when it's one of the rare times they rerecord the talk and change their decision. He is genuinely surprised there because he originally didn't fool them. Or rather, he did, but they said enough that he kinda agreed. Then they heard about his method in talks to the producers and apparently were like "Oh, that's not how we thought it was at all." And so called him back to refilm things.

[@Uprising771]

you do realize its a show

[@james_fisch]

@@harshharsh2124 sheep

[@MrAndyStenz]

Great trick, Andi! And your face when they said you fooled them was priceless. Bravo and congrats!!

[@oz_jones]

He was glad to win. :p

[@StevenMcFlyJunior]

@@oz_jones you're fired. oops, wrong show.

[@davidchan9485]

@@StevenMcFlyJuniorx7uxxz6

[@jwvandegronden]

what a lovely routine! Elegant, so well put together; Even without knowing what you were aiming for, you took us by the hand and led us through the maze of the trick. I loved every second!

[@VanishingIncMagic]

Glad you enjoyed it!

[@bobnazrul5278]

Finally owner Vanishing inc Joshua and Andi got the trophy!!!

[@glenntamblyn3271]

Great trick. So clean!!! Don't know if I have worked it out but... Andi in another interview said there were no cards left in the folder at the end. Look at each card as he turns them over and map from left to right to see how each 'input' maps to each 'output'. Number positions from the top as positions 1 to 5. On any one card, for all the convolutions, each 'input' on that card maps to the same shift on the 'output'. So one card shifts by 1. 1-2, 2-3, 3-4, 4-5, 5-1 Another shifts by 2. 1-3, 2-4, 3-5, 4-1, 5-2. And if you flip the cards, they still produce the same net shift!!! There are five cards, and they have shifts of 0, 1, 2, 3, 4. But the total shift across all 5 cards is the sum of their shifts divided by 5 (mod 5 for the mathematical). so 0 + 1 + 2 +3 + 4 = 10. mod 5 equals zero. No shift. And Penn & Teller picked the same positions, position 4. No shift. But, if we take just the cards with shifts from 1 to 4, we add up to a 10 - no shift. So these 4 cards, together, do absolutely nothing!! After traversing these 4 cards, nothing has changed!!! They are dummies. Now follow the order in which Andi puts the cards out, and map that to what we see when they are flipped. The first 4 cards are the 1, 2, 3 & 4 cards. That collectively do nothing! All that is just patter, showmanship etc. But importantly, he gets them out of the way early - empties the folder of stuff that doesn't matter. It is the last card that matters. And it is a zero shift card. It is also the one put up when the patter/misdirection is at it's maximum. The 'I want to be on Penn & Teller' card. Look VERY closely at 4:14. He shortly pulls out the 5th card. What is his right hand doing, hidden behind the folder, for just a few seconds? What if? What if there are 4 standard cards, the first 4, that do not matter at all. Then he needs to select from one of 5 possible 5th cards, depending on whether the final shift he needs to produce is 0, 1, 2, 3 or 4? He has got the dummy cards out of the way, making it easier to work, he hits the high-point of his patter - misdirection. What if his 'folder' has an open, visible part, that contained the 4 dummies. So if confronted later, he can show that it is empty. But a hidden section contains the 5 control cards that he needs to pick just 1 from. Those few seconds with his right hand hidden doing what? Then he is done with the folder and can put it aside, rather than having the money-moment in the middle. In fact, knowing from the beginning what 'shift' he needed to produce, he could have possibly used each step with the dummy cards to prepare for selecting the control card. So this is one conjecture. It doesn't need 5 sets of 5 cards, just 4 dummies and one set of 5 different control cards. And really great/subtle sleight of hand and stage craft. And a simple trick prop that is immediately set aside. That it is enough of a trick that the prop can be 'examined' afterwards! But Andi has said that the folder was empty afterwards - simpler versions of this trick have a hidden, unused card. Does he mean the 'visible' part of the folder was empty - the 4 cards were gone? Or truly empty? If this was done with skilled, hidden, control cards, a brilliant, clean, sweet trick. If there truly were no hidden cards, then I am totally flummoxed! And this sort of analysis is only possible on KZbin with repeated rewind/replay/stop. Live you would have a snowflake's chances in 'that other place' of spotting how this might work. Bravo Andi!

[@l_szabi]

If you watch closely when he flips the cards, the "Penn Andi Teller" card looks like multiple sheets of paper behind each other, while the others are just simple two sided papers.

[@crybirb]

Just saw about the reversing of the decision story, and man I gotta say Pen and Teller seems the most genuine people ever. Like, I'm legit impressed that almost everyone who comes in contact with them have good things to say, it looks, from an outsider view, they truly respect the craft and their colleagues at that. It's so wholesome.

[@VanishingIncMagic]

Yeah, the story of the decision reversal is crazy.

[@aminzahedim.7548]

When P&T easily realize they haven’t figured a trick all the way down you know it’s genuine. Kudos👌🏻👏🏻 I was also able to crack parts of the underlying logic (math enthusiast, physics student, etc…) but couldn’t find all the missing bits. Curiously, it’s ever more intriguing this way!

[@OldCountrySeeds]

teller had a shine in his eye at the end! damn right. that was great

[@ConstantlyDamaged]

I love the bit where they shifted things around. Great act-really loved it.

[@jseance]

I saw this last night and what a brilliant act. Congratulations Andi

[@mercedeswilkins5566]

What a blessing

[@Miruj]

"Großartig! Bravo!" Best wishes from Berlin/Germany

[@Ylyrra]

Love this trick, and love the story behind it even more. Like P&T I've perhaps figured out more of this by myself than any other trick I've seen on the show... but I don't know HOW you did it. I know what needs to have happened, and I can see that it has happened, I can see how it works, but I don't know when it happened. So frustrating... congratulations!

[@PhilippBoettcher]

Congratulations Andi, great performance! Loved everything about it.

[@matthewao]

Wow, incredible act! I love the premise and the enormous amount of thinking that must’ve went into creating the trick.

[@michaelwidjaja1502]

Andi almost cried there. Penn's words were that deep.

[@philiphall2733]

Congrats Andi! Great effect and so well performed!

[@Scoupe400]

They’re always so gracious. True gents. And congratulations.

[@electrichellion5946]

Fixing a bad call publicly like they did adds to the respect a fan can have for performers in both the professional business dealings and as humans doing right by fellow humans and as performers to the judges of their performance of their chosen craft.

[@davidlashaway]

Congratulations Andi!!! You definitely earned that. What a Wonderful performance and trick.

[@SeanDevine]

Watching this again after watching Andis interview. Such a cool thing! So happy for ya.

[@sefhapita]

Beautiful magic! Congratulations, Andi!

[@pianotubeleonbricht9635]

Well done Andi! You are an amazing magician! I loved the trick and I love vanishing inc. Keep up the incredible work!

[@johnbmx4christ]

Even if you weren't gonna fool them, it was really awesome incorporating their stories into it making it personal. What a nice touch.

[@X320riginal]

You say it's awesome but what about my feelings? :-(

[@HiyaEverybody.]

Absolutely amazing, very well done and congratulations 👍💯

[@ronb4115]

One of my favorite tricks and presentations.

[@julievanderleest]

I just love magic. After watching different tricks, I started to gain an interest in trying to do a little myself. Awesome trick Andi! Beautiful performance too.😃

[@VanishingIncMagic]

That is awesome! We made a page with five easy tricks to get you started for free: www.vanishingincmagic.com/learn-card-tricks/five-easy-card-tricks/

[@X1Y0Z0]

Great presentation! Excellent act!

[@alexhardao2742]

Great trick and flattering performance. Much respect

[@stefank4959]

What a kind guy! Such a nice way to perform it.

[@timmack2415]

Wow 😳 I was totally baffled about this one. But....I'm an engineer (we solve things) and I have the ability to slow this down and I have the ability to freeze the screen. So I did just that and was able to figure out that you fooled me too.

[@hieronymusnervig8712]

Just think of the maze as a set of relations (the mathematical ones).

[@sup-games]

i'm also an engineer and i don't understand how did you get fooled, you can see that no matter the direction all input fit the same output. for example if a card have the second line as input and then third line as output, if you flip the card it give the same result. As for the card choices, he knew that both choose homes, there is most likely every combination possible of the whole maze in his black folder, in that case, the current combination is every output are the same as the input if you look the complete maze, in 1 give out 1 , in 2 give out 2 , the next cards combo could be for example each output = input +1 , so if in =1 , then out = 2 , then you would have output = input +2 , output = input +3 and finally output = input +4, so that make only 5 possibilities to cover which he had a set ready inside his folder. I could be wrong though, but that would be an easy way to do it. Of course when you do for example +3 when you reach 5 you go back to 1.

[@luxxxurymercedes]

@@sup-games agreed

[@SimKieu]

@@sup-games you're right, it's brilliant, and it's not only the complete maze, it's also for every block too. Each block with have each output = each input + constant n (where n could be 0, 1, 2, 3, 4). And by doing this, after going through one block, all the outputs will be shifted by n. And for each block, we can randomly assign a number n for it. And if we do this, regardless the order of the blocks, the final output will be the same (because the total shifted amount is the same, it's the sum of all the n's of all the blocks). And he doesn't even need 5 sets to cover 5 possibilities. He only need 6 cards (each will have different values of shifted amount n: 0, 1, 2, 3, 4, 5). And depending on the value of n = final output - final input, he can choose to discard 1 card. For example, in the case in the video, since 0+1+2+3+4+5 = 15 which mod 5 = 0, he can just discard the card with n = 0, or n=5 (same thing). If, however, the output is 1 line above the input, he could just discard the card with n = 1.

[@OccupyMusk]

@@SimKieu Andi said in an interview that there was nothing left in the envelop. If it's true we need more research to figure this out.

[@enilenis]

The cards that are picked determine which input hooks up to which output, regardless of the shuffle and orientation. That's where the math portion comes in. The binder contains 4 unused cards in addition to what's shown, and options are picked based on knowing ahead of time, which gates are designed to hook up. Cleverly hidden symmetry. Something only a mathematician/programmer would come up with. If you want to see it more clearly, copy the cards, but draw straight lines, as oppose to maze squiggles. You'll see it.

[@G.Aaron.Fisher]

Well said. I'm a mathematician and programmer (split career) and how this trick works was blatantly obvious to me. Honestly, the math background helped a lot more.

[@enilenis]

@@G.Aaron.Fisher I'm a programmer mathematician myself and an electrical engineer. This thing is like a shift register designed by Rube Goldberg. Each card represents a base 5 addition. The cards that are picked in this particular setup represent shifting of the input bit by +3, +2, +1, 0, -1, for a total sum of +5, or a full wraparound, where each input ends up back in the initial position. If you trace all the lines, you'll see that all the matching gates line up, and not just the ones that had Penn and Teller. The binder contains 4 additional cards that produce other sums in the 1-5 range. Either he has 5 sets of Miami Vice card from -1 to +3), or he has 2 of each card with values off by 1 with respect to eachother. This would make a great magic kit for kids.

[@es8450]

the order of the cards doesnt make any difference thats true but if one of them didnt pick the library the trick wouldnt have worked?

[@williamjansen1]

@@es8450 If they didn't pick library, then another five cards from the binder would have given the same result. We are working on the presumption that he had multiple events in that binder (first SNL-appearance, being on West Wing, the debut of Fool Us etc.)

[@enilenis]

@@es8450 What the cards say doesn't matter. What matters is whether the lines on them go diagonally up or diagonally down. Each card represents by how much a line travels up or down. All lines shift in parallel, which is what makes the trick possible. Magician looks at line arriving where it started, meaning line can either go straight, travel 5 paces up, or 5 paces down. If you want to understand, draw lines on 5 paper cylinders representing gates and start stacking and flipping them. You'll see that the path is always parallel on all drums. It's a math thing.

[@NeuralNetProcessor]

The mental gymnastics required to make the maze alone is staggering to me.

[@Mr-vp8kw]

Wow!!! Thanks for putting this in youtube absolutely blown away hot chills from this!

[@Pathfinder-Doc]

Absolutely beautiful. That was a perfect routine.

[@simonproctor9187]

Amazing, completely folded. Wonderful performance, well deserved a masterpiece in all elements 👌👍

[@YYYValentine]

It isn't a typo, right?😀

[@samuell.hodgesjr.1577]

Oh, glad to meet sha'.Andi. I buy Vanishing Inc. Products. That maze effect is beasty!

[@mattesrocket]

A really N I C E trick, nice to watch, more tricky than you think at the beginning, great stuff for the brain, nice story. Magic!

[@LEGITBOSSS]

Yeees Penn Andi Teller 🥳 Couldn’t stop smiling for 1 sec watching you finally perform on FU . You made that classic so relevant , loved your presentation , loved the jokes , loved everything about it . The story is as good as if it was from Joshua 😬 Congrats , and Yes Vanishing Inc is the Best indeed . Thank You so much for your Magic Andi 🥳👏🏆👏😍👏🎉👏

[@RayAtchley]

This is Andis second time in FU

[@ke6gwf]

In Theory, this Group of cards is one of the most enjoyable tricks to figure out! Especially after seeing the back story that p&t thought they had you busted, and then had to call you back in when the Producer found out they had been fooled after all! Another brilliant trick and great show, you must have had fun coming up with this one. Lol

[@floating_iceberg]

amazing perfomance! bravo! bravissimo! superbtly done!

[@sbak1972]

Brilliant. Congratulations!

[@lucmercatoris8290]

From a mathematical point of view, the trick is quite straightforward. Watching closely, one can see that the permutations used in the maze are the potencies of the cyclic permutation π =(1, 2, 3, 4, 5), with π^5 = id (the identical permutation). Especially, {π, π^2, π^3, π^4, π^5} is forming a subgroup of the symmetric group S_5, with the nice property that π^x o π^y = π^(x+y). Turning the page actually has no effect on the permutation (It does not invert the permutation as i thought at first!!!). Thus, turning pages can be done arbitrarily without changing the maze. All Andi has to make sure is that the sum ∑ of the exponents modulo 5 is equal to the „distance“ between Penn and Tellers choices. (Here: Penn chooses 4, so does Teller, meaning the distance is 0). As there are 5 possible values for the distance, Andi only has to differentiate 5 cases. My guess how he realizes this is as follows: In the envelope there are 5 copies of the last piece of paper (the one with himself as 3rd member of the show), all with a different permutation on the back (one for π, one for π^2, …). The mazes on the other 4 pieces of paper sum up to the permutation π^10 = id. So Andi has to choose the copy of the last piece for which the potency of π corresponds with the distance. Leaving aside the whole math stuff, Andi does a great performance. Congratulations for fooling them! Edit: From other comments i saw only now that the envelope is empty at the end. I‘m a very beginner in magic but my math is right:) Definitely, the maze doesn’t have another ending whit another ordering (in general cyclic permutations are not commutative, but in this special case they are). So, one way or another, a corresponding 5th card must appear somewhere. A force can be excluded in my eyes

[@yviruss1]

Good one.

[@adellp8515]

He can turn the paper in 4 different ways. That‘s the trick, I guess. He needs only five ways because itˋs really unlikely that Teller would say stop at the House.

[@MrKockabilly]

The 9,600 combinations he mentioned is just an illusion. Even if there are million ways of placing the cards, the odds of making a match - without employing any trick - is simply 1 in 5. The five starting points each has only 5 final destinations to land into. Anyway, he actually had 6 cards in the envelope. He simply used the 5 that would be needed to make the match. Note how he only took them out after the start and end points (home, coffeeshop, etc) were already established and Andi never asked P&T if they want to change any of the starting and end positions after he has taken 5 cards.

[@richie_is_boss4977]

This routine was incredible. I’m curious how many times Andi went through and practiced it. His word choice and everything was just great.

[@VanishingIncMagic]

Check out yesterday's episode of our podcast - Andi talks about rehearsing it.

[@gernotg8480]

@@VanishingIncMagic Hello. Can I please work for you? I need a job and would love to be Part of the magic Industry

[@robinvids2628]

And everything?

[@gernotg8480]

@@robinvids2628 ?

[@amcconnell6730]

kzbin.info/www/bejne/h5iXfGysjZ53bac&ab_channel=VanishingInc.Magic

[@mattmushrush8494]

Been watching a bunch of these lately and this was one of the first on that I figured out

[@VanishingIncMagic]

Well done! You've worked out what the best minds in magic got fooled by! Unless, you've not worked it out, of course :) - clue: the folder is empty at the end.

[@LobsterPuncher]

Your presentation is great. I really enjoyed this one.

[@SoundVoltage]

Rotations, permutations, commutivity. Four of the five cards just mapped 0->0, 1->1, 2->2, etc. The remaining card was the heart of the trick, and Andi just had to pull out the one that represented the distance (mod 5) between P&Ts initial choices.

[@JonathanBartlesSWBGaming]

exactly! I didn't see your comment until after I posted mine, but you said it much better than me

[@ke6gwf]

The problem with this is, after all the cards were up there, they had the option to flip or move them. And the Miami vice card that got flipped was NOT a 1-1 card. Can you explain that? I can't lol

[@SoundVoltage]

@@ke6gwf Yeah, it's a bit weird. Not sure how well I can explain it without diagrams and stuff. Look at the first card, the one that got flipped. Treat all the entry paths on the left as 'inputs' numbered 1 to 5 down the card. Treat all the exit paths on the right as 'outputs' also labeled 1 to 5. If you start with input one, you can follow it to output three. Now go to input 3, follow it, and it goes to output 5. Go to input 5, follow it to output 2, and go to input 2 and follow it to output 4. Output 4 goes back to input 1. You can write that as (1 -> 3 -> 5 -> 2 -> 4). Now do a screen cap, flip that upside down and do the same thing. You'll see the exact same pattern. Check the second card, same deal, it's (1 -> 4 -> 2 -> 5 -> 3) and if you flip it upside down, it does the same thing. So that handles the 'flipping' situation. Now, each of those 'paths' like (1->3->5->2->4) is a permutation of the numbers from 1-5. Getting into something called Group Theory here, but 'permutations can commute' -- which is to say that it doesn't matter what order you read them in, you'll always get the same thing. That commutativeness is what lets them swap the cards around.

[@ke6gwf]

@@SoundVoltage ok, when I read that explanation, about 3 brain cells nod and understand the concept, and then I go look at the trick and can't see how it's possible! Lol And THAT'S the beauty of the trick! I don't feel like recreating the trick to prove it out, but your explanation makes sense so I am going to accept it, because it makes it a really awesome trick, since no one knows about this group theory... Sheesh According to other comments and the magician's pod cast, p&t originally told him they had figured it out, so he gave them the trick and left, but then the producer, who was calling in remotely, was still on the line as p&t were discussing the trick in their dressing room, and when they said something about how it was done without the code, the Producer said Hey! That's not how it's done! So they actually had to call the guy to come back, and reshoot the end where they get fooled! So I am guessing they probably followed a couple of lines and figured he had to get the cards in the right order or something, and came up with a simple solution, that was wrong. But if you are right that any combination will arrive at the same result, flipped or not, which does make sense, then he just needed to pull out the right offset card or set of cards to match the 4-4 pair, and the rest is just part of the show. I love figuring out tricks more than I enjoy being awed by the unknown, so thank you for being the only one here who seems to understand it! Lol

[@SomeRandomName999999]

"just had to pull out the one that represented.." apparently, there was only that 1 single card left and folder is empty at the end of the trick

[@shinfaux7717]

I love this trick!!! This is such an amazing and creative idea!! So talented Andi, thanks for inspiring me:)

[@jj-fk9ez]

bro this guy has not only written acts that have fooled them before, BUT HE FOOLED THEM HIMSELF. ultimate magician.

[@tolgaozbalci]

Wow. Such a good trick. Bravo 🎊👍👏

[@koszegimatyas]

That is certainly a very-very clever trick! I too like programming, math and magic, so this one was a special treat. Thanks and congrats!

[@gernotg8480]

Szia. Tudsz nekem segiteni?

[@koszegimatyas]

@@gernotg8480 Szervusz! Miben tudok segíteni?

[@gernotg8480]

@@koszegimatyas munkat keresek... kerlek adj egy eselyt ha tudsz

[@koszegimatyas]

@@gernotg8480 Nem tudom, hogyan tudnék ebben segíteni. Nincsen semmilyen cégem, alkalmazott vagyok.

[@gernotg8480]

@@koszegimatyas hol? Be tudsz nyomni? Kerlek

[@georgeyounts9391]

Really loved this trick. Wonderful & due to video editing I think Andi was pretty stunned he fooled them.

[@daftsquiddy]

Congratulations, Mr. Gladwin! 🎊🎉🎊👏🏻👏🏻🦑🤙 That trick was Flippin amazing!

[@sharpo]

Congrats! Brilliant routine.

[@MrJonathanwhyte]

Nice trick with an interesting mathematical background (permutation groups and so). The trick would have been more fun if Penn and Teller would not both have selected the Library. Because of that, the “counter” example Home obviously and unfortunately also led to Home, which was a bit of a spoiler.

[@ninjazprojects5246]

This is why there are not only 5 but 25 cards in the envelope to cover every possible outcome. Even that I found out the trick, it was still a great act, what a wonderful guy!

[@user-mo1jq9zo1c]

Good point! The three replies are not worth reading.

[@TorQueMoD]

I really loved that Penn really made him think he'd lost to the point that I think Andi was more shocked than the audience was when he won :P

[@mpr746]

Fun fact. This footage was a reshoot. Originally they thought they weren't fooled. A few hours later they found out how it was done and realized they were indeed fooled. Andi was told the footage was somehow lost and asked him to redo the trick. But actually Penn and Teller wanted to change only the finale, giving Andi the trophy. That is why he was so shocked!

[@kwijung]

@@mpr746 Source for that story? Like how would you even find out about that

[@mpr746]

@@kwijung kzbin.info/www/bejne/h5iXfGysjZ53bac Timestamp: 19:50 I found that reading through the comment section.

[@darkcoolyo]

@@kwijung it was revealed to him in a dream

[@wrenboy2726]

There is no audience.

[@kalvinhepple5975]

Bravo Andi!Brilliant!

[@88Doug]

Soooo cool! This is why I love this stuff!!

[@duran3d]

Great trick. My idea: Andi knew where the penn and teller thumbnails were located before taking the cards out of the folder, so he produced the cards that would join those locations, and those cards would work regardless they were upside down or in any order.

[@VanishingIncMagic]

That would totally work, apart from the fact the portfolio is empty after the cards have been removed... :)

[@Error6503]

If you pause and study the cards you'll see that no matter what order and rotation you'll get the same result; a start location always connects to the same end location. When you rotate a card it still connects the same start to end, and card #1 has the opposite connections to card #2 so effectively cancels out ... the same goes for #3 and #5. Card #4 is the crux; it connects the same in to out and is the only one responsible for deciding the ultimate path. It would only take a magician's sleight-of-hand to select 1 from the necessary 5 different versions of this card, especially as the trick is performed before the audience are aware of what is going on.

[@IvanCherganski]

@@VanishingIncMagic How about the "Pen Andi Teller" card? Isn't that a bit of a fishy one?

[@final1037]

@@Error6503 still, if teller choose different place other than library...

[@Error6503]

@@final1037 If you take the difference between start and end location there are only 5 possibilities. Cards # 1,2,3 & 5 have no effect whatsoever on the outcome, only card #4 needs to change to account for the difference. The start and end point are chosen at the very beginning so all the performer needs to do is select the correct card #4 once he knows this. I can't see when it's done but the trick boils down to that; hiding the fact that he has pre-selected one from five cards before the audience is aware of what they are watching out for.

[@stephennetu]

I feel as if I could develop something that would at least pretend to be something such as this, if it were fewer rows and cards... But the fact he let them rearrange *and then* flip *any* of them over, too... That...just did me in.

[@wrenboy2726]

Flip any of the cards and the maze stays exactly the same.

[@kevboard]

the trick isn't that card placement, any beginning track leads to the exact same position on the other side. the question is what would have happened if they didn't pick the same starting place. he must have had a method to make them meet anyway.

[@trentvlak]

@@wrenboy2726 not the first card. maybe others but i didn't look at them.

[@tommarquette9706]

That was fun. Got four out of six moves. Then lost. Excellent. Hope he releases a mini version. Oh, big finish.

[@OwenStaton1]

Great work Andi so glad your fooled them massive well done

[@blakebelladonna9075]

I love how this trick is so simple, but effective.

[@CalTek]

Love how no matter whether the card is rotated or not the path remains the same (yes I screenshot it and rotated them :) )

[@CainSuzuko]

Same thoughts. Was looking for someone actually trying it and here you are.

[@tk20channel]

Thank you, because I was wondering the same thing. Appreciate you spending the time for us.

[@phiupan]

Oh, that is nice. I thought I had it figured out except for rotating the pieces. Now it is even cooler.

[@johnsmithe4656]

This was so classy. Such a massive compliment to P&T.

[@funfever8484]

i was not expecting this kind of trick it was amazing fresh and new keep going :)

[@Martin_Vail_Esq.]

The 5th card is the key...that is the piece that “connects” them. To be clear, the 5th card refers to the final card placed, not the numbered location.

[@Sam_on_YouTube]

There are some clever ways to not have that be the case, involving quick mental math and double lifts. You could get away with as few as 6 cards and do it with no envelope. But it would be harder. To do it with 6 cards, the unseen one would have to look like 5th card placed (the one that doesn't change the order). Then he'd have to swap that for one of the other 5, so that it alters the order as needed. I think that is what he did, having 6 events. The last one was the punchline, which is why it was the one that mapped things to themselves. He didn't want to skip that. But one of the 6 events was going to get skipped. To eliminate the envelope, you'd need to do a double lift to skip one event. I don't think he did that, but it might have improved the trick by removing the envelop that hid his extra card

[@alvinkoeswanto8622]

This is a magic trick I've never seen before. Good job Andi!

This was awesome! Great job!

[@NandyKudry]

Congratulations, that was amazing ! :)

[@karlsultana8]

I think trick is something like this (but involves more maths): he has many "cards" to show Penn & Teller not just those 5. Since they choose the library on both sides he shows them just those "cards" that when placed will create a route from library to library. And you can place the cards in any order or upside down the routes will be the same. The 5 cards he placed create a route from library to library always. Another set of cards creates a route from library to home, and so on.

[@ser_igel]

you're right, there some explanation with really simple math language: it's a pretty easy trick when you know what permutation is he has at least 9 different cards, four with (1, 2, 3, 4, 5) => (1, 2, 3, 4, 5) permutations, and another five for each (1, 2, 3, 4, 5) => (1, 2, 3, 4, 5); (2, 3, 4, 5, 1); (3, 4, 5, 1, 2) and so on order of permutation doesn't change the composition result but you don't need to know that when you have four identity permutations (f(a) = a) you can draw this permutations and see for yourself that turning it upside down doesn't change the permutation (i.imgur.com/ni5MlOt.png) so all he needs is take first four cards, see the difference between p&s choices and pick the card with corresponsive permutation (only the difference, doesn't matter which places they choose, for example coffee p-coffee and t-library is the same as p-movies and t-theatre) it's an easy to perform and easy to understand trick, but you need to know some math to understand that quickly enough to say "you didn't fool us"

[@Jibbitz6019]

The envelope was empty at the end after the five cards were taken out.

[@thiagorfpinheiro]

@@Jibbitz6019 How can you know?

[@theofficialczex1708]

@@thiagorfpinheiro He said so in another comment.

[@michaelmcintyre3871]

First time I've figured a trick out and Penn and Teller didn't. I was baffled that they didn't get it. Reminds me a lot of a trick for the home viewers David Copperfield did one time that was easy to figure out. If the result is based on math, taking just a little time to go over a trick reveals where the force is very quickly.

[@Ryan-pv7nw]

oh cool! very interesting. thank for explaining how the trick actually worked!

[@q-tuber7034]

The math is very clever, but the math alone isn’t enough to explain the trick. The podcast reveals that the envelope is empty at the end of the trick.

[@wrenboy2726]

@@q-tuber7034 the main envelope is empty, but that doesn’t mean there aren’t more cards on that stage (the joke card is noticeably different from the other 4)

[@mrcolley6334]

I do love the small flash he has with the coin during his intro! Overall amazing routine!! Way to go!

[@mbrooker79]

Small flash? It was blinding.

[@MydasNeomagie]

Yeah and why would you love it :D

[@davidshareefChTPhD]

That was an amazing trick! Also being on fool us was an awesome experience I loved it and I want to go back

[@davidevens8486]

Very cool, sort of reminds me of bit rotation and in this case, 4 cards combined rotate through all of the 5 positions twice (4 of the cards cause 10 shifts total, 1+2+3+4=10 shifts) which keeps the original order. Then the last card keeps the order as-is.

[@thedeviator]

Very astute of you to notice. I really love the premise of this. This would have been really difficult to work out on the spot. I had to rotate some things around visually to just see how it fits in. Very well performed. Wonderful trick.

[@Qwentar]

I suspect there are four more panels in the envelope.

[@VanishingIncMagic]

you're wrong.

[@mirpcatalan1578]

Neither order nor orientation matters. And you need just one more card with a zero shift to replace one of the first 4 card to make the trick work. Mathematically it's not that hard - but the idea and the performance were both great. Congrats for a well earned trophy!

[@davidevens8486]

@@denisott5557 No it doesn't, actually. Flipping it upside down flips both the positions as well as the shifts - after the rotation it remains the same because they cancel each other out. Consider the shift 1->2, 2->3, 3->4, 4->5, 5->1. First if we ONLY consider the new positions after a flip, 1s become the 5 spot, 2 becomes the 4s spot, 3 stays in the same spot, 4s become the 2s spot and 5s become the 1s spot. Taking that shift i mentioned earlier and changing positions only, we get 5->4, 4->3, 3->2, 2->1, 1->5, now if we reverse the shifting direction which happens after a flip, we get 5

[@bezalelgeretz]

Each card has an effect on the outcome regardless of its orientation. The cards shown are [+2,-2,-1,0,+1]. One more card combined with four existing cards can get you to any outcome you want. Andi just had to select 5 cards (from a deck of 6) with a net outcome of zero.

[@vsh555]

And nothing depends of orientation.

[@shaunmeredith6437]

That's one of my suppliers in the UK , lovely to see him on fool us

[@poloprankster]

Lovely! Bravo! 👏🏼👏🏼👏🏼

[@CJtheThird]

Anyone thinking there are multiple pages in the envelope arent correct. He said in a podcast that it was empty and its a different method

[@tobyfitzpatrick3914]

Then why not have them already layed out in the open at the beginning of the trick? I'm not sure I completely believe him on that. An "empty envelope" can still mean that he (magnetically?) stuck pages together, effectively giving him multiple outs.

[@CJtheThird]

@@tobyfitzpatrick3914 It could have been a red herring but I think has some form of algorithm or something where depending on which location P&T choose he can use the maze to force them to a specific location. Each maze piece moves the person 1 down, 2 down, 3 down, 4 down, and 5 down (no change). So i think that is part of the trick

[@RaveyDavey]

@@tobyfitzpatrick3914 That's what I'm thinking. I cannot think of another way of doing this without one of the career events having extra cards.The envelope is the only thing he has full control of. So if the envelope is empty at the end, then the extras must be on the board.

[@doodles990]

What a great one! The only thing I recognized about this trick is that the ending maze is guaranteed success from the start. No matter what order those cards are in, even if you turn some upside down, every single one of those lines traces from a symbol on one side to the same symbol on the other. Somehow, and if I thought long enough I could probably work out how, those cards are designed such that as long as Penn and Teller both chose the same symbol, no matter what that symbol was, and no matter what order and orientation they put the cards in, the path would always work out. I expect that's what Penn and Teller worked out too; it was an easy guess after you demonstrated how House leads to House. What I can't figure out is what the strategy would have been if Penn and Teller had chosen different starting locations. I know, intuitively, the math should work out that the same properties could be created in any set of cards such that every start leads to an end of x offset (i.e. every symbol leads to the next symbol down, or to two symbols down, or to one above, etc). If you had five sets of cards, cases 0 through +4 (or -2 through +2 if you prefer), you could just pick the set corresponding to the relationship between the two chosen start positions, and the trick is finished as soon as it begins. Some cards might be able to be shared between sets, if the math works out, so you could have less than 25 total cards required. Not sure exactly how small, maybe as few as 9 total. That being said, and according to what you give away in a couple replies below, you apparently didn't have any other cards. Those are, allegedly, the only five. So I've hit a brick wall in my reasoning.

[@carlostower4807]

Probably one card is the one that decided the final relationship between the 5 icons left and 5 icons right, so probably he picks the correct one from the envelope and there are still more there, since the way of getting the cards out is a little bit fishy...

[@rjdverbeek]

With six cards you can probable generate all the necessary combinations. They should have checked the envelope at the end.

[@unclejeffmagicnsong7397]

Amazing, awesome, nice presentation 👍👍👍

[@customerservices4u]

Andy seems so happy, good for you. You deserved it.

[@gavinsong7637]

"I personally take offense that I had to make the commute across the board, and not Teller! But the commute aside, either way you flip it, it's a great trick. I think it really added to it that we could modify the order of the cards freely. After all, a single card could have completely shifted the outcome, AND it was completely dependent on the audience, who you have no control over. Now, we think we know how you did it, but that doesn't matter, because in the end, it all adds up to a fantastic trick. Thank you." -- Penn, if he had figured out the trick

[@f52_yeevy]

This must have taken ages to create, amazing performance!!

[@koszegimatyas]

@@danquaylesitsspeltpotatoe8307 There is always a hidden person connected via an earpiece to Penn and Teller. This guy knows how the tricks are done, so that he can tell Penn and Teller, whether they are correct or not. It is possible that they asked him about the number of the cards in the envelope after the trick was done. I am guessing the two were thinking a lot more than shown to us, but those parts were cut not to bore the watchers at home.

[@TheOldMan-75]

@@tangente00 I highly doubt that there were only 5 cards because if there were you wouldn't bring that super suspicious folder. Most stuff you see in a magic performance has a purpose and that folder definitely isn't there because it makes it easier to hold 5 cards.

[@gaganbajwa2707]

This is the excellent invention by this man because it doesn't matter how many times you turn the cards up side down or anyway you want the result will be the same.

[@scottbelcher9026]

Wow! That was awesome!!!

[@fep_ptcp883]

If you turn Miami Vice upside down the result would be completely different. Also Penn and Teller had free choice of starting points in the beginning. This trick is lovely 👌

[@sairatisbona]

it is lovely, but it still would be from first position to third postition on the paper.

[@sairatisbona]

you have to rotate it not mirror it

[@wrenboy2726]

No, it wouldn’t lol

[@trevorlambert4226]

Just finished it, and going to make a guess here. All the combinations and orientations of the cards create the same map. He has 5 different sets of these that he can choose from after Penn and Teller decide there starting positions, to account for the five different possibilities.

[@albertlwj]

there's more than 5 possibilities

[@VanishingIncMagic]

Nope, the folder is empty at the end.

[@trevorlambert4226]

@@albertlwj There are more than 5 possible combinations of arrangements of lines between starting and ending points, but only 5 of them are needed to account for any choice Penn and Teller make. Remember that you only care about one starting and one ending point, the other four branches can go anywhere. If you replace the location descriptions with numbers 1-5, then imagine the first pattern is 1-1, 2-2, 3-3, 4-4, 5-5. (Incidentally, I don't think it's a coincidence that the house went to the house when the library went to the library in the trick). The next pattern you just increment the second number: 1-2, 2-3, 3-4, 4-5, 5-1. Third one is 1-3, 2-4, 3-5, 4-1, 5-2, fourth one is 1-4, 2-5, 3-1, 4-2, 5-3, fifth one is 1-5, 2-1, 3-2, 4-3, 5-4. You should be able to see that every possible starting point and ending point are covered within those five patterns. All he has to do is subtract the end number from the start number after Penn and Teller make their choices, and that tells him which of the 5 patterns yields the required result.

[@trevorlambert4226]

@@VanishingIncMagic Well, that's at least part of it. I have since looked at the maze pieces, and they are unchanging when you flip them top to bottom. I'm no magician, so I don't know all the possible tricks.

[@nodiced1]

I thought about having different outs on different cards, the easy way to do this would be to have 5 folders, each with 5 different cards, but he leaves a single envelope up front the whole time....took my best guess away right there. Yes the orientation of the cards doesn't matter, but the placement of Penn and teller does....so far, very stumped.

[@koustubh132]

A real masterpiece !!

[@shmookins]

Wow. Bravo. I can't even begin to understand how this is done. To me, this is magic. Amazing job and a very well deserved win.

[@ArthurRainbow]

Love the idea of having elements which are self-inverse; but under rotation and not algebraic inverse :p. That's a new axiomatic concept to find here. Because the pictures looks like braid group, it seemed extremely impressive to the mathematician I'm. Until I realized that in theory it was "only" a permutation subgroup, which in practice is Z/5Z. That's the kind of thing I'd say as code word, and probably most magician would not get it either.

[@tk20channel]

I can't tell if this is gibberish or you're a genius. Either way, you're a genius. I have no idea what you said.

[@ArthurRainbow]

@@tk20channel ​ @tk20 Absolutely not making things up. And I graduated in theoretical computer science. Given the end interview, math+comp science, I had hope the performer my have known the math I mentioned. And, exactly as with Penn's codeword, most of the stuff can be googled. "braid group" has a wikipedia page and its images are similar to the one here. "Permutation group" also is on wikipedia, even if there are no images that look immediately similar. Z/nZ (in this case with n=5) is also searchable online In layman term, I'd just state that this has a single property. As with Penn, I could state the things in layman term instead of using algebraic jargon. Honestly, at least from someone with an undergraduate level of algebra, the reason the trick work is really really easy, and I'm confident it can be explained to a child. I just find it fun to add math jargon because that's not often I can use it and it actually make sens (at least to the three people who liked the comment)

[@aj76257]

I can see visually that rotating any individual piece doesn’t change the outcome - I assume that’s the self-inverse you were referring to. I’m having trouble understanding why the permutation subgroup here is Z/5Z as opposed to S5. I haven’t done a careful check but it looks like swapping the order of the cards doesn’t change the outcome either. Is that where the simplification comes from?

[@ArthurRainbow]

​@@aj76257 ​ You're perfectly right regarding the self-inverse part. It looks like it's S5. But actually, if you look at each card c, there is always some integer i_c such that all inputs of c are shifted by i_c. For the 5th card, the first input goes to 3, the second to 4, the third to 5, the fourth to 1 (which is why you should count modulo 5) and the fifth to 2. So i_5=2. You can check that i i_1+i_2+i_3+i_4+i_5 = 0 mod 5. In practice, it means that all path have the same input and output. Home go to home. Theater to theater, and so on. And more importantly, you can reorder the cards, the output is the same. You can even rotate the card. Imagine you rotate card 5. The first input go to 3. It means that if you rotate it, the third input go to 5 (because the first input became the fifth output), so i_5 is still two! Obviously, this only work because Penn and Teller selected where they are at the start of the trick. If Teller decided to be one down, to the theater, it would require i_1+i_2+i_3+i_4+i_5 to be 1 mod 5. But that could easily be done by replacing one of the card. Let's say, instead of having i_4=0, you'd select a card that ensure that i_4=1. The same joke can be made, it'll have the same front. But the back is slightly different. I'm certain those four other cards are in the big envelope he was selecting cards from. He could have make the trick even harder to grok by using a permutation of {1, 2, 3, 4, 5}. Let's say (1, 3, 5, 4, 2). Relabelling input and output would ensure that you can't just compute this number i_c until you have first understand which reordering is used.

[@aj76257]

@@ArthurRainbow I see, and thanks for the explanation. I was thinking about the raw ordering of five objects. I didn’t notice the inputs were shifted by a consistent amount on each card. Clever setup!

[@MagicScorpio]

The programmer in me figured out most of it before he said he was a programmer. No pausing. Then briefly at end I paused for a few seconds to confirm some things and create a final theory. I feel pretty amazing knowing I figured it out within the given 1-2min limit. But good for him winning! Such a clever little trick within a regular trick.

[@djlowtek]

I thought I knew this immediately, but the more I thought about it, the more I realized I'm completely wrong

[@RandomBoxOfWeird]

So: The maze cards permute the locations 1-5, but they are in fact very special permutations. They are all just multiples of the cycle (12345), i.e. every number gets sent to the next number mod 5. So every card is just the operation "+k (mod 5)" for some k=0,1,2, 3,4. Therefore 1) the order of the cards doesn't matter since addition is commutative and 2) they have the property that when you reverse the orders of both inputs and outputs and then switch inputs with outputs (which is what rotating the card upside-down does), they stay the same. [To see this, imagine first just switching the orders, i.e. reflecting the card horizontally: We start with output = input + k which is to say 3 + b = 3 + a + k for a,b=-2,-1,0,1,2. Reflecting horizontally just changes the signs of a and b, so now 3 - b = 3 - a + k, which by some algebra is just 3 + b = 3 + a - k or output = input - k. Now reflect vertically to switch outputs w/ inputs, and we end up back at the operation +k.] So we see that the sum of all k's used in the line-up is the only thing that matters. In this case, we have sum = 0 (mod 5), matching the fact that Penn and Teller both chose the same starting number. Had they chosen another combination, we would need a different sum, but crucially there are only 5 different possible differences between Teller's start and Penn's start. The cards shown are +2,+3,+4,+0,+1. How many extra cards do we need? Perhaps we have 4 extra cards and swap the +0 card for one of the other possible k's to achieve any difference? We can do better if we think about it. Take e.g. just one extra +0 card. For any m, replacing the +m card in the line-up with the extra +0 card gives us an end result of -m mod 5. If m ranges from 0 to 4, then -m will range from -4 to 0, which mod 5 is just the range from 0 to 4, meaning that we can achieve any difference this way. Since Andi has revealed that the extra card does not stay in the envelope at the end, we can only assume that the extra card must be attached to one of the cards in the line up (e.g. by magnets), and I believe there is evidence of this: The fourth card looks and behaves differently when he flips it. This still doesn't cover how the trick is performed (great performance btw), but I sincerely believe the mechanism is what I described.