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August 20, 2011 8:07 AM   Subscribe

The Berglas Effect aka The Holy Grail of Card Magic or Any Card at Any Number (ACAAN) and named after its inventor David Berglas is a very simple magic card trick that Berglas claims only two people know.

  1. David or Marc Paul, his apprentice, display a regular shuffled deck of cards to the audience and three participants. Sometimes the deck is borrowed from the audience.
  2. The deck is then handed to the first participant for safe keeping.
  3. The second participant and third participant are asked to think of a playing card and a number between one and fifty-two respectively.
  4. The second participant is asked for their card and the third participant is asked for their number.
  5. The first participant is then asked to deal out the third participant's number of cards.
The final card is revealed to be that picked by the second participant. Berglas hasn't touched the deck since he handed it over for safe keeping and claims that the deck is random and the participants are not working for him.
posted by Mitheral (107 comments total) 30 users marked this as a favorite

 
There is some kind of trick to this!
posted by tomswift at 8:22 AM on August 20, 2011 [4 favorites]


He's a sorcerer. Burn him.
posted by koeselitz at 8:23 AM on August 20, 2011 [3 favorites]


Yes, there are stooges involved. Every card trick is called a card trick because there is a trick involved. Look up Ricky Jay on YouTube to see "REAL" card magic. No plants or stooges involved.. just really good hands, decades of practice, and pure skill.
posted by ReeMonster at 8:24 AM on August 20, 2011 [3 favorites]


My instinct is also to say that there are stooges, but this trick is old, notorious, and Berglas and Paul have done it all over the place for decades. I think if it were actually stooges word would have gotten out long before now.
posted by Navelgazer at 8:31 AM on August 20, 2011 [1 favorite]


On one of the Fool Us Penn & Teller acts, the magician had a volunteer smack a beach ball into the audience, then had the person that caught it throw it again. I thought that was a really nice way to really establish that there weren't any stooges, since those things don't fly predictably enough that you could do that even if you wanted.
posted by Dr.Enormous at 8:31 AM on August 20, 2011 [2 favorites]


Any magic, sufficiently advanced, is indistinguishable from I can't believe I fell for that.
posted by ShutterBun at 8:32 AM on August 20, 2011 [8 favorites]


I would rather not know the secret of how he does it. Knowing would ruin the fun.
posted by Daddy-O at 8:32 AM on August 20, 2011 [1 favorite]


Stooges, Ericsonian Hypnosis, reading the audience to get the person with the number you want. Pick a solution any solution. Also we only see the clips of the times it works.
posted by humanfont at 8:34 AM on August 20, 2011 [1 favorite]


The claim that only two people know how this trick is performed is false. I also know the secret. For $50 I will share that secret with you.
posted by (Arsenio) Hall and (Warren) Oates at 8:35 AM on August 20, 2011 [4 favorites]


Spoilers?

---

The trick is a decision tree that ends in different individual tricks, using a memorized deck and audience manipulation to end up in the most impressive end of the decision tree as often as possible.

----


End Spoilers
posted by michaelh at 8:40 AM on August 20, 2011 [15 favorites]


Here's the trick -- you ask one person to name a card, then you ask them to name a number. You've got the entire deck memorized. If they say a number that matches the card (a one in 52 chance) then you do this particular version of the trick. If they don't then you do a different trick with the card and number they asked for, probably some kind of cut, since the deck is memorized. Note that he doesn't tell him how he's going to use the number until after he's been given it.

It's kind of a brilliant method for doing a trick, because the only way to prove that that is what he's doing is to watch a bunch of his performances and see if A) he does the trick in every performance or B) he always leaves the deck untouched.
posted by empath at 8:41 AM on August 20, 2011 [9 favorites]


You know a trick is good when the only way laymen can imagine it being accomplished is through the use of stooges or camera tricks. Good magicians use neither, and neither does this trick.
posted by Decani at 8:41 AM on August 20, 2011 [2 favorites]


Daddy-O: See, I'm the opposite. Knowing how the trick is done makes it more interesting to me because you can then see all the work that went into it, and you can appreciate a novel trick and why it's novel.
posted by Grimgrin at 8:42 AM on August 20, 2011 [3 favorites]


Yes, there are stooges involved. Every card trick is called a card trick because there is a trick involved.
Absolutely false. Berglas has done this trick in many many setting with three famous participants. Whatever the trick is, it can't possibly involve teaching hundreds of random famous people sleight of hand.
The trick is a decision tree that ends in different individual tricks, using a memorized deck and audience manipulation to end up in the most impressive end of the decision tree as often as possible.
Although this is often cited as an explanation, and it may be right, I've seen him do it three times in a row with the same guessers. It would be extraordinarily difficult to do this with a force. If he is capable of that, he is capable of some serious mind games.

Also, the last guy who claims that the odds of guessing it correctly are 1 in 52*52 is not a skilled mathematician. The odds remain 1 in 52 of guessing correctly.
posted by Lame_username at 8:46 AM on August 20, 2011 [5 favorites]


If you've memorised the deck, it's a 1 in 13 chance, not 1 in 52 - are you counting from the top or the bottom, and are you counting to the eighth card or are you discarding eight cards and then revealing?
posted by kithrater at 8:49 AM on August 20, 2011 [9 favorites]


Although this is often cited as an explanation, and it may be right, I've seen him do it three times in a row with the same guessers. It would be extraordinarily difficult to do this with a force. If he is capable of that, he is capable of some serious mind games.

I think he is indeed capable, and I don't think he would do it three times in a row with just anybody, let alone promise it three times before meeting a person/audience.
posted by michaelh at 8:52 AM on August 20, 2011


It's kind of a brilliant method for doing a trick, because the only way to prove that that is what he's doing is to watch a bunch of his performances and see if A) he does the trick in every performance or B) he always leaves the deck untouched.
He does it every time I've seen him, because its pretty much his signature trick. He doesn't always leave the deck untouched, but he almost always does. He sometimes offers the option to count from the top or bottom. He never has the audience member who holds the deck cut or shuffle, which does suggest some sort of memorized deck. I've seen him let people select the deck from a group of available decks
If you've memorised the deck, it's a 1 in 13 chance, not 1 in 52 - are you counting from the top or the bottom, and are you counting to the eighth card or are you discarding eight cards and then revealing?
Probably 1 in 26, because I have seen the top or bottom thing. I've never seen him discard the extra card -- Its always the number chosen.

I'll stop dominating the conversation and return to my smoker. Pulled pork makes me so happy.
posted by Lame_username at 8:54 AM on August 20, 2011 [1 favorite]


Good magicians entertan their audience. The code of secrets, the denial if various tricks like stooges and camera tricks are all simply part of the aura of bullshit.
posted by humanfont at 8:54 AM on August 20, 2011


Berglas claims only two people know.

But he probably pays Ricky Jay to keep quiet.
posted by Smart Dalek at 8:54 AM on August 20, 2011 [6 favorites]


If you've memorised the deck, it's a 1 in 13 chance, not 1 in 52 - are you counting from the top or the bottom, and are you counting to the eighth card or are you discarding eight cards and then revealing?

You could buy more flexibility still by opting to count off the letters of the number—if you've gotten "eight" but you need 5 or 6, count E I G H T [BAM], etc. That said, if there's no footage or reportage of Berglas or Paul using any of these variations, that doesn't get us very far on the scooby front.
posted by cortex at 8:55 AM on August 20, 2011


I recall Penn Jillette himself once saying "Magic tricks can be summed up by the sentence "Here's a quarter, now it's gone, you're an idiot."
posted by longsleeves at 9:00 AM on August 20, 2011 [4 favorites]


Well, one more thing. Steve Martin once wrote of this trick "An effect as "pure" as the one I have described above (and that some people think there is some special method for) simply does not have a method other than "being lucky". . . It is entirely possible that David Berglas "got lucky" on several occasions, and these are recounted as MIRACLES." Berglas has done this trick to 100s of professional magicians, who universally report stunned amazement.
posted by Lame_username at 9:01 AM on August 20, 2011 [1 favorite]


If you've memorised the deck, it's a 1 in 13 chance, not 1 in 52 - are you counting from the top or the bottom, and are you counting to the eighth card or are you discarding eight cards and then revealing?

In the versions I've seen, he doesn't count the cards himself - the "participant" holding the deck does.
posted by Navelgazer at 9:01 AM on August 20, 2011


I would be seriously disappointed if I were to find out how he is doing this. I like magic in my magic.
posted by Xoebe at 9:02 AM on August 20, 2011


Oh yeah, he does this trick one on one frequently too, which eliminates the stooge theory completely. He'll have his guest pick the card and the number and the top or bottom and then they count out the cards themselves. He never touches the deck in between.
posted by Lame_username at 9:04 AM on August 20, 2011 [1 favorite]


Via, I'm guessing.

In the versions I've seen, he doesn't count the cards himself - the "participant" holding the deck does

The performer does the count out loud, and could certainly direct the person holding the cards as to which way to hold them, etc.
posted by ShutterBun at 9:04 AM on August 20, 2011


Mitheral: “The Berglas Effect aka The Holy Grail of Card Magic or Any Card at Any Number (ACAAN) and named after its inventor David Berglas is a very simple magic card trick that Berglas claims only two people know.”

Well, given that there appear to be at least three dozen youtube videos of various people performing it, I would guess that the secret is out.
posted by koeselitz at 9:06 AM on August 20, 2011 [1 favorite]


Well, as I said, I can't do magic at all and I can upload a perfect version of this trick without using any stooges. I just film myself trying it 13 times or so (using cortex's method).
posted by empath at 9:08 AM on August 20, 2011 [2 favorites]


Having done magic my whole life, I lean toward stooges here- specifically the final counter.

They all hold the deck like they've been around cards before, and the magician always pauses before the reveal. Forcing the card to the bottom would have worked in the first video, and palming the card and dropping it on top would work in the second.

I'm not positive that's how it's done, but that's how I'd approach it.
posted by Benny Andajetz at 9:09 AM on August 20, 2011 [1 favorite]


He's also published a book earlier this year which (apparently) explains it.
posted by ShutterBun at 9:10 AM on August 20, 2011 [1 favorite]


ShutterBun writes "Via, I'm guessing."

Nope, via a private gaming mailing list.
posted by Mitheral at 9:18 AM on August 20, 2011


It stands to reason that Berglas is either able to manipulate the deck on the sly or force the card in a subtle way. That or stooges.

I mean, if it wasn't one of those options then presumably three random people from the audience
could do the trick among themselves without any help from the magician.

If you ever want to figure out how a magic trick is done, just look for extraneous persons or objects in the trick that don't "need" to be there if the trick were "real" (that is, if the magician really had the power to teleport a pretty lady from place to place, why does he need to put her in a box first?, etc.)

In this trick the extraneous items are the deck, the magician, the counting of the cards, and the number of audience participants. The secret lies somewhere in there and it's probably both more and less clever than you think.
posted by Avenger at 9:22 AM on August 20, 2011 [1 favorite]


I once dated a girl who said she did lots of tricks.
posted by Postroad at 9:28 AM on August 20, 2011 [6 favorites]


In this trick the extraneous items are the deck, the magician, the counting of the cards, and the number of audience participants.

Uh... you can... how can you do the trick without the deck of cards? It's a card trick!
posted by LogicalDash at 9:32 AM on August 20, 2011


Uh... you can... how can you do the trick without the deck of cards? It's a card trick!

If the trick could be done with an arbitrary list of things (e.g. index cards with letters of the alphabet in random order) then it's not really about the cards, but about creating the illusion that the "random" selection from the audience matches the "random" order of the cards.

I would guess that he has the order of the cards memorized. When he asks the first participant for the first card or number, he then knows which number or card to ask for. Then he merely has to "force" the other value.

For example, I'm performing the trick and I first ask participant #1 for a card. He says "King of Diamonds" - I've memorized the deck and I know that the KoD is at number 14. Now all I have to do is force participant #2 to say "14"

Forcing someone to say 14 on demand is the trick.
posted by device55 at 9:42 AM on August 20, 2011 [1 favorite]


This doesn't seem that difficult to me.

1. The deck is never shuffled.
2. The card is called before the number.

Conclusion: the number called is calculated from the card. The "number caller" is in on the trick.

The real trick to this is that the "number caller" doesn't need to "memorize the deck." It only requires a simple calculation, based on the suit and number of the card called. So if Berglas goes on Oprah, let's say, he can use Oprah as his "stooge", because everyone will assume that there's no way that Oprah could memorize a deck of cards.

Well, OK, there is one other trick. Only select stooges that won't talk. And, really, why would anyone want to spoil the fun for others, given how simple the trick really is?
posted by SPrintF at 9:57 AM on August 20, 2011 [4 favorites]


device55 writes "Forcing someone to say 14 on demand is the trick."

However the magician asks both participants to choose their card/number before asking for the card. So he'd not only have to somehow force the number he'd have to force it strongly enough to make them change their original choice.

It would be fascinating to see the distribution of the card/number choices over the thousands of times he has performed this trick.
posted by Mitheral at 10:02 AM on August 20, 2011 [1 favorite]


The "number caller" is in on the trick

Am I really expected to believe that Parky is a stooge? I find it hard, I'm sorry.
posted by howfar at 10:02 AM on August 20, 2011 [3 favorites]


As a thought experiment for how to construct this trick, it would be trivial to do it with a stack of n sorted decks, where each successive deck starts one later than the previous. So the first deck starts with the Ace of Spades. The second with the 2 of Spades, and so on.

Now you have two inputs: card and num, and you want an algorithm of num iterations that gets you to a position in this stack of cards that contains card. The magician gets to announce the algorithm to the person holding the deck.

1 ≤ num ≤ 52

So, there are two memorization steps. 1) the magician must know the contents of the deck, 2) the magician must know, or be able to compute, an iterative algorithm that yields card in num steps.

Let's say the number is 3 and the card is the Ace of Spades and there are 52 cards containing each card rather than duplicates (these last two are assumptions that, if violated, give the magician more options for trickery)

If the Ace of Spades is in position m, what series of n steps starting at card 1 or card 52 will get us to m?


If n = m then it's trivial. Count n - 1 cards from the top, revealing the nth card.

If n = m - 1 then count n cards from the top, revealing the nth + 1 card.

If n = (i * m) mod 52 then count n cards, i cards at a time. (I think the notion here is right, but the actual math is almost certainly off)

These three options give the magician a lot of choices. I think the third, with the modulo operator, may allow the magician to cover most or all of the space of choices.
posted by zippy at 10:05 AM on August 20, 2011


An important thing to remember is never trust the rules set up by the person who invented the effect.
posted by Benny Andajetz at 10:24 AM on August 20, 2011 [2 favorites]


An important thing to remember is never trust the rules set up by the person who invented the effect.

Absolutely. Tricks are lies.
posted by inturnaround at 10:51 AM on August 20, 2011 [1 favorite]


Uh... you can... how can you do the trick without the deck of cards? It's a card trick!
posted by LogicalDash at 9:32 AM on August 20 [+] [!]


As device55 points out above, if the magician "really" had the power to psychically determine a random card draw (or telepathically force someone to choose a certain card) then he really wouldn't need a deck of cards to do the feat. He could just ask a audience member to choose a number between one and a trillion. Or between one and a quadrillion plus a random string of letters.

But he doesn't do that, because psychic powers and telepathy aren't real. Instead, he asks them to choose from a set of 52 playing cards -- which means that the cards are integral to the trick somehow.

That's what I mean when I say to look for things that don't "need to be there" if the trick were "real". If I have the "magic powers" to make a coin disappear with a wave of my hands, why couldn't I just give YOU the coin and make it dematerialize in your own hand? It must be because the wave of my hands (or rather, my hands in particular) are integral to the trick somehow, and so on.
posted by Avenger at 10:57 AM on August 20, 2011 [1 favorite]


I should also say "which means that the cards [and the physical arrangement thereof] are integral to the trick somehow."
posted by Avenger at 10:58 AM on August 20, 2011


Or why should I give you a coin at all? Why couldn't you just pull one of your own coins out of your own pocket? If my powers were "real" I could make a coin disappear no matter where it came from or who was holding it at the moment -- presumably. But my powers aren't real, which is why I'm going to make this coin disappear like this -- *quick wave of hands* SHAZAAM! -- and look, it's gone! Amazing, no?
posted by Avenger at 11:02 AM on August 20, 2011



Absolutely. Tricks are lies.

I thought that a Trick was something that a whore does for money?
posted by mikelieman at 11:04 AM on August 20, 2011 [2 favorites]


device55: Forcing someone to say 14 on demand is the trick.

I'm sort of leaning towards this right now. When I was watching the video and the first part was over I decided to pick a number myself for the 2nd part but to pick the number only after intently listening to him speak and instruct the participants. As soon as he instructed the number picker person to pick up I did so too: and, whaddaya know... I picked 23. So did the participant. That weirded me out.
posted by Hairy Lobster at 11:08 AM on August 20, 2011 [2 favorites]


michaelh and empath both say the "decision tree" hypothesis. But then this is easily falsified. We should see the magician in multiple locations starting the speech and ending in different tricks. Anyone ever follow this guy on the road to confirm this occurs?
posted by scunning at 11:08 AM on August 20, 2011


to pick up to pick one

derp
posted by Hairy Lobster at 11:09 AM on August 20, 2011


However the magician asks both participants to choose their card/number before asking for the card. So he'd not only have to somehow force the number he'd have to force it strongly enough to make them change their original choice.

I see what you're saying, but he only asks them to "think" of a number and to "think" of a card - he then asks for a card, and then asks for a number, in both clips in the same order.

I've seen (here on the blue someplace) some very subtle and amazing "slight of brain" tricks where people are made to say all sorts of things via suggestion.

In the first clip, he gets 8 from the older gentleman - which is 1 + 5 + 2 (one and fifty-two is repeated a ton)

In the second clip he selects a woman from the audience to get "23" - he appears to be selecting a person...is he selecting someone who is signaling "23" inadvertently?
posted by device55 at 11:11 AM on August 20, 2011


Okay should have read down thread as several more said they see him perform this one.

The issue about the memorization of the deck. It does sound like he has a memorized deck if you never see him or anyone shuffle it. But what would memorizing the deck do exactly - the other two people are independently determining the position of the chosen card, and if they aren't stooges, then they can't coordinate. So how would magician memorization benefit the trick in this instance (real question - I honestly don't understand)?
posted by scunning at 11:12 AM on August 20, 2011


I read a lot of magic books for my job, and if there's one thing I've learned, it's that most magic tricks are explained by them lying about the one thing that, if you knew they were lying, would make the trick unbelievably lame. That's why magicians protect their secrets so fiercely. The really smart ones build up some other explanation. For instance, if you think Ricky Jay is pure skill and never resorts to gimmicked decks or the like, I'd ask you: Do you do your job in the most difficult way possible for no reason? Well, neither does Ricky Jay. This isn't to say that Ricky Jay isn't very, very good at slight of hand. It's just that he's also very good at branding himself as a pure old-school technician, as if old school people like his mentor Dai Vernon didn't use trick decks.
posted by Bookhouse at 11:13 AM on August 20, 2011 [4 favorites]


Everyone with the theory should try out their theory and post how it goes. I'd be interested in whether you're able to reproduce his results. Or even with a program.
posted by scunning at 11:19 AM on August 20, 2011


It's up his sleeve.
posted by Legomancer at 11:22 AM on August 20, 2011


This version is more subtle, in that the performer asks someone (off camera) to think of the card and number, but the participant never says it out loud until the end.

Which maybe blow my theory out of the water.
posted by device55 at 11:25 AM on August 20, 2011


I used to do magic.

1. You can never reliably force a person to pick a given number .
2. You can almost always reliably force a person to pick a certain card from a deck (amazing but true once the skill is learned)
3. You can always reliably position a card within a given place in the deck within a shuffle or two. Sometimes you can not even see if the person holding the deck is shuffling (you can cut and shuffle with one hand without anyone seeing)
4. The easiest way to do the trick above is if the first person is a shill and trained at card manipulation.

FWIW.
posted by Poet_Lariat at 11:26 AM on August 20, 2011


I thought that a Trick was something that a whore does for money candy?

FTFY - We have kids around here!
posted by a_green_man at 11:32 AM on August 20, 2011 [4 favorites]


A thought: you only need one stooge, and it can be any one of the three:

1. The card picker can choose the right one if they go after the number picker. Especially with some sort of system.
2. Vice-versa for the number picker.
3. A good sleight-of-hand accomplice can pull up the right card.

If it were me, I would just do it each way some of the time. That way, it's harder to catch on and more honest volunteers can be included.
posted by Dr.Enormous at 11:34 AM on August 20, 2011 [1 favorite]


FTFY - We have kids around here!

And you just taught them how to get candy.
posted by prefpara at 11:34 AM on August 20, 2011 [7 favorites]


This guy suggests it's a combination of the several theories offered (or maybe just one of them). It's a decision tree theory, in a sense. But mainly, the trick works only if the two participants call out the same two numbers that precisely identify the location of a given card.

If they don't, then the magician just skips the trick and goes on. He only says it's a miraculous trick after he performs it; ex ante, he keeps it quiet.

This apparently is drawn primarily from Berglas though - the Paul videos, he said, may be doing a different version.

I'm skeptical of this theory if only because the Paul videos do this trick apparently on demand. I've not yet seen Berglas perform it, so I'm only going off the Paul videos. But if it was a probabilistic trick in the way that blogger suggested, you could not become so famous for it and identified with it that on live camera you do it on demand.
posted by scunning at 11:40 AM on August 20, 2011 [1 favorite]


More than just perform it I will up the stakes using mass of MeFites who self select in. Go to the store, and buy a fresh pack or ordinary playing cards. Remove the joker and any miscellaneous cards (instructions, poker hands, etc). Now shuffle the deck as much as you like. Pull a card from the deck at random. Now burn the card (outdoors is probably a good idea, you may need to light it repeatedly, or place it on a fire). Now once you've done that. Post to the thread to let me demonstrate my power. Are you afraid of this magician's all seeing eye?
posted by humanfont at 11:42 AM on August 20, 2011 [1 favorite]


So how would magician memorization benefit the trick in this instance (real question - I honestly don't understand)?

You would have to know the order of the cards very thoroughly. For example you'd have to know that the King of Diamonds is the 4th card and that the 4th card is the King of Diamonds.

To highlight what I mean further - you can probably recite the alphabet in order, and maybe even backwards, but what's the 14th letter?

So. If you have this deck cold memorized and you can immediately come up with the card given a number, or the number given a card - you've got a pretty good trick right there right?

"Think of a card, got it? good." [count some cards out] "Is this your card?"

When you ask some random guy to give you a number, that is give you the number you want, it looks to be completely improbable and therefore awesome.

I'm not saying this is how the trick works....but it seems like a reasonable approach to it.
posted by device55 at 11:44 AM on August 20, 2011


Absolutely false. Berglas has done this trick in many many setting with three famous participants. Whatever the trick is, it can't possibly involve teaching hundreds of random famous people sleight of hand.

He wouldn't have to go to such ridiculous lengths; all he'd have to do is tell them beforehand (or "subtly influence" them, or whatever the weasel words he used in the clip were) which card and number to pick.

And if anyone really thinks that violates some sort of magician's code, well, mission accomplished. Only the presumed existence of strict rules could make something interesting out of such a COMPLETELY OBVIOUS so-called illusion.
posted by Sys Rq at 11:57 AM on August 20, 2011 [1 favorite]


(Which is to say: If a magician says there's nothing up his sleeves, there's probably something up his sleeves.)
posted by Sys Rq at 11:59 AM on August 20, 2011


Sys Rq: Unless he wants you to watch his sleeves.
posted by Grimgrin at 1:12 PM on August 20, 2011 [4 favorites]


I was a stooge at an event where a magician was putting on a show for a bunch of journalists, PRs, marketing types, etc, a lot of whom knew me. While we were milling about prior to the show, one of his assistants came up and asked me quietly if I'd be prepared to help out - and why not. I forget the details, but I had to be picked as a volunteer and then come up with two 'random' choices.

Which I did. The way everything was phrased during the performance, I didn't actually have to lie or be dishonest, but there were true statements made that seemed to say the opposite of what they did say. In front of hacks, I thought that was fair enough...

Afterwards, I was asked by a LOT of my pals, and others, how on earth the trick had been done. I said I couldn't say, but that I'd let people have one guess and if they were right, then I'd make it clear.

Nobody guessed I'd simply been told beforehand what to say. Not one. Made me realise quite how much magic is a matter of relying on preconceptions.
posted by Devonian at 1:18 PM on August 20, 2011 [10 favorites]


But what would memorizing the deck do exactly - the other two people are independently determining the position of the chosen card, and if they aren't stooges, then they can't coordinate. So how would magician memorization benefit the trick in this instance (real question - I honestly don't understand)?

First, magicians don't publish a set list before their set. You never know which tricks they're going to do in which order and how exactly they're going to do them.

So imagine you have two options:

If the first two people pick the right card and number, then you have the third person count from the top of the deck and you look like the greatest magician who ever lived.

If they don't, then you pick up the deck and do a cut or use a stooge for the third person -- in the first video, the camera pans away when the third person opens his hands with the cards in them for the first performance, and he doesn't pick the 3rd person until after the card and number are picked in the second performance. It's not as impressive as the first version, but it's still a decent trick, either way.

You can increase the odds of the first situation happening a lot by changing up the way you ask people to count cards and by front loading the deck with face cards, I imagine.
posted by empath at 1:24 PM on August 20, 2011


It's really important that he doesn't say.

"You will name me a card, you will name me a number, and you will draw that number of cards from the top of the deck, before he asks them to give him their choices."

He asks for their choices first and then decides what to do with them.
posted by empath at 1:27 PM on August 20, 2011 [2 favorites]


empath, Paul, at least, has done this a number of times for skeptical audiences and prefacing his performance by saying that he's going to do this trick. The decision tree would be cool, but it looks like this is done another way.
posted by Navelgazer at 1:59 PM on August 20, 2011


You are all overlooking one simple way to do it. You just select a card. The universe splits off into multiple parallel universes with all the various outcomes, and you enjoy thundering applause in the one where you picked correctly. Who cares about the universes where you were wrong?

Come on, guys, it's simple physics.
posted by Legomancer at 2:10 PM on August 20, 2011 [15 favorites]


I think Legomancer has it. You simply create a tiny computer with voice recognition and store the deck order in it. To that, you attack a tiny syringe full of ricin. If the computer hears anything but a match from the volunteers, it injects you with the deadly poison.

Since you can only, obviously continue to exist in a universe in which you are alive, this guarantees that you would be successful every time.

It's so obvious in retrospect.
posted by empath at 2:19 PM on August 20, 2011 [13 favorites]


*quick wave of hands* SHAZAAM! -- and look, it's gone! Amazing, no?

Yes! Holy crap. How'd you do that?!?!
posted by yeti at 2:44 PM on August 20, 2011


Brief pseudo-algorithmic analysis, possibly not useful:

It takes six bits of information to identify a playing card (52<26, or 4 for the rank, 2 for the suit), and thus to write down a random ordering of playing cards takes 6 × 52 = 312 bits. If there's a way to order the deck so that it looks random when fanning it but allows the number-picking stooge to memorize an order of magnitude fewer bits then we have a plausible hypothesis for one of the three participants having become a stooge just a few minutes before. One possibility:

The magician stacks the deck as follows:
0   1   2   3   4   5   6   7
A♣ 3♦ 5♥ 7♠ 3♣ 5♦ 7♥ 9♠…
The stooge is taught the following:

0. Number the suits like so:

♣ = 0
♦ = 1
♥ = 2
♠ = 3

(This is the bridge suit ranking, suits in alphabetical order.)

1. When the mark picks a card, if it's an odd-numbered card (or J=11, K=13), add one, divide by two, then subtract one. If it's an even-numbered card, divide by two and add 6. Add 13 to any negative numbers.

2. Finally, subtract the suit number from that number, multiply that result by four, then add the suit number back in. Add one to that and say the number.

Now the mark names a card, say 5♦. Our stooge-algorithm springs into action:
(1) 5+1=6, 6/2=3, 3-1=2
(2) 2-♦=2-1=1, 1×4=4, 4+♦=4+1=5, 5+1=6. Say "6".

A mere six operations to memorize and complete in just a few seconds before the audience notices something is wrong!

So this is a fast-ish hashing algorithm for a somewhat random-looking deck, though anyone inspecting the ordering for very long would notice something screwy. This might not be a viable approach to this trick outside of a math or computer science department.
posted by silby at 2:47 PM on August 20, 2011 [1 favorite]


I watched Penn and Teller's UK show Fool Us on YouTube and loved every minute. I particularly love this trick that even though they saw a part of the trick exposed, Penn couldn't for the life of him figure out how he did the last part. So great. My favorite part of the series has been Teller's glee in watching others perform magic for them... he's like a kid.
posted by yeti at 2:51 PM on August 20, 2011 [4 favorites]


Devonian -- that was something I was going to post earlier, but couldn't explain it clearly. Often they can use "instant stooges" which is basically you were. People will usually go along.

The other part you described, which can sometimes be done by itself without the stooge really knowing, is called "dual realities". That's where a phrasing means something different to the person on stage than it does to the audience. Like saying "imagine a book, pick a word you see" to the audience means "a word you imagine" but if the magician is secretly showing a list of words to the guy on stage, to him it means "pick a word from the list".

Another cool trick, I can't remember the name, is when you have everybody think of something, like a number. Then you bring one person up, and while everybody is clapping you quietly ask him his number. Do a lesser routine, and at the end "guess" his number. To him, you're just reinforcing the trick, but to the audience that's the big reveal.

That said, I don't see how any of those tricks could be in play here.
posted by mad bomber what bombs at midnight at 2:55 PM on August 20, 2011


silby, that is so far beyond Martine McCutcheon you wouldn't believe.
posted by cromagnon at 2:57 PM on August 20, 2011


Ah see, there's the good part, it's the other guy that's the stooge who has to memorize the hashing algorithm. Martine McCutcheon can say whatever card she wants if the other guy can hash it.
posted by silby at 3:00 PM on August 20, 2011


I still think that might be pushing it, to be honest.
posted by cromagnon at 3:08 PM on August 20, 2011


Oh, probably. SPrintF just said The real trick to this is that the "number caller" doesn't need to "memorize the deck." It only requires a simple calculation, based on the suit and number of the card called. and I felt like…cooking up an implementation I guess.
posted by silby at 3:18 PM on August 20, 2011


I have to say that the comments on this thread which take the tone of "It simply has to be done this way or that way; simple logic dictates it" are cracking me up. The people making these comments are clearly - oh so very clearly - non-magicians.
posted by Decani at 3:55 PM on August 20, 2011 [1 favorite]


Well, obviously, it's all RFID and electronic paper -- the real trick being having everyone believe that some 'Berglas' has been doing this for decades.
posted by Anything at 4:43 PM on August 20, 2011 [1 favorite]


This looks like an example of silent sound presentation, or presenting sound subliminally. The idea is that if you can't hear the sound consciously, then you process it subliminally. Asking someone for a number while "silent-sounding" that number will probably produce that number in most people, and it is easy to spot those open to suggestion in the audience beforehand to further guarantee the results. He pretty much admits to doing this somehow. Furthermore, with a parabolic speaker setup pointed at the target, it can probably be done with very low-tech methods, no broadcast required.
posted by Brian B. at 4:53 PM on August 20, 2011


I'm sorry, Brian B., the fact that your link is to a post by a conspiracy theorist and 9/11 Truther makes me suspicious of its evidential worth. Mainstream psychology doesn't take subliminal messaging of this type particularly seriously, certainly. The other problem is that it doesn't eliminate the stooge. In the first clip Michael Parkinson and his production team would have had to be in on it. Given Parky's previous experience of possession, I can't see him being up for it. Joking aside, Occam's Razor surely applies in this instance.
posted by howfar at 5:11 PM on August 20, 2011 [1 favorite]


The whole deck is made out of psychic paper, obviously.
posted by jeather at 5:14 PM on August 20, 2011 [2 favorites]


I buy the teaching a quick hash function to the number picker theory, as well.
posted by empath at 5:34 PM on August 20, 2011


I figured out that trick from pen and teller. The 7 of diamonds was a force, there was a clear piece of plastic on the top of it that penn signed that he peeled off. The other deck was premade with the 7 of diamonds reversed, all he did when he said, 'let me pull it out lightly' was stick penn's signature on the face of the card.
posted by empath at 5:41 PM on August 20, 2011


I was able to figure out how this works rather easily, however I'm still completely stumped by magnets.
posted by milnak at 6:54 PM on August 20, 2011 [3 favorites]


The multiple claims that Berglas has done this one-on-one sort of undermine the theory that the number picker is in on it.
posted by kenko at 7:53 PM on August 20, 2011


Well then the answer is obviously that it's magic.
posted by empath at 7:55 PM on August 20, 2011


there was a clear piece of plastic on the top of it that penn signed that he peeled off.

Surely someone who's handled as many cards as Penn has would notice that.
posted by kenko at 7:55 PM on August 20, 2011


It seems likely that there are a variety of ways to approach it, and depending on the situation, the illusionist chooses his strategy based on various elements of the setting and individuals involved. For example, in the first situation in the posted video, if I had to guess which card the woman celebrity would choose, I would have guessed "Queen of Hearts."

So... I imagine that in many cases, working with "known quantities" in terms of interests, personality, and avocation is going to be a lot easier. Some types of personalities are going to be more likely to choose certain types of cards; performers are likely to have big egos and identify with the card they select (I'm just saying this as a simplistic example of the sort of thing the magician-psychologist might determine about somebody)... likewise, he might know quite a bit about the sorts of numbers/cards amateur magicians might choose. There is additionally the stooge factor, of course. In most cases my guess would be that there is some level of stoogery, and, again in the first part of the video, the stooge seems like the guy counting out the cards. It seemed to me that he was getting some instructions as Marc Paul was doing quite a bit of pointing and hand gestures, and his dealing seemed a rather strained and covert.

Why wouldn't an actor/performer be willing to participate? Their work is all about illusion and creating belief. If the magician is a master psychologist, he will know who is his best bet for collaborating, to what level, without spilling beans all over the place. And also, there is probably still a lot that appears amazing and magical even to the stooge, because even if s/he cooperates, it is probably apparent that the magician has predetermined a number of results.

For the most amazing examples of the trick (when the illusionist does the trick with a single person who holds the deck, chooses the card, chooses the number, and counts off the cards), my guess would be that this could never be an on-demand trick... that the magician picks his subjects opportunely by observing that they are of a certain type or temperament, and determining furthermore that they are easily suggestible or susceptible to light hypnosis. The conversational patter of the magician and other possible prompts like certain colors and patterns on clothing or certain accessories or jewelry may affect the subject (diamond ring, diamond tie pattern, arranged in groups of four to create yet another diamond shape, the color red, may reliably get the right sort of person to choose the four of diamonds... deftly exposing the subject to a number before s/he is even selected to do the trick, may predispose this sort of person to choose that number, etc.).

This would be the kind of trick that the magician chooses solely on his own terms when he feels like he has the perfect subject, while standard performances would be more likely to rely on stooge backups, and celebrity appearances would give him the chance to learn quite a bit about the other people and make a lot of decisions about his approach pre-appearance.

So, I'm guessing that it's not one trick, but many tricks of many components, employed as needed – or when opportunity presents itself, and requiring skills in psychology, mathematics, memorization, statistics, social engineering, persuasion and sales... all of which is still amazing and magical.
posted by taz at 10:02 PM on August 20, 2011




And no one could fool Dai Vernon.

... except Slydini.
posted by RavinDave at 11:49 PM on August 20, 2011 [3 favorites]


More than just perform it I will up the stakes using mass of MeFites who self select in. Go to the store, and buy a fresh pack or ordinary playing cards. Remove the joker and any miscellaneous cards (instructions, poker hands, etc). Now shuffle the deck as much as you like. Pull a card from the deck at random. Now burn the card (outdoors is probably a good idea, you may need to light it repeatedly, or place it on a fire). Now once you've done that. Post to the thread to let me demonstrate my power. Are you afraid of this magician's all seeing eye?
posted by humanfont at 11:42 AM on August 20 [+] [!]


Hell, I'll take you up on this, since I actually happen to have a sealed deck of cards that was recently bestowed upon me, and no strong urge to play poker. What's the card?
posted by FatherDagon at 12:13 AM on August 21, 2011


If it was the 7 of Diamonds, you're gonna think I'm a genius.
posted by ShutterBun at 11:46 AM on August 21, 2011


Hell, I'll take you up on this, since I actually happen to have a sealed deck of cards that was recently bestowed upon me, and no strong urge to play poker...

Using my special powers, I will make a random mefite burn a playing card in his backyard.
posted by humanfont at 5:47 PM on August 20 [+] [!]
posted by TedW at 3:05 PM on August 21, 2011


The Secret revealed - MASSIVE SPOILER ALERT

Ok, I did some digging and found out how this can be done. Yes, I can say that now there are at least three people in the world who know the secret. And after you read this, the number will be certainly greater :)

This explains how the Berglas effect can be achieved as performed in the FPP. It can be done without stooges, and even without memorizing a deck. The secret lies in something called "pre-show" work. All four conditions of the pure Berglas effect are respected.

Before the show, Marc would have approached the woman and said something like this: "We're going to be doing an illusion on the show and i'll be asking you to pick a card. To save time, can you pick a card now and tell me what it is?"

She says "Queen of Hearts". Fine. Marc takes a deck and finds out where in the deck the Queen is. Suppose it is the 8th card as in the video.

He then leaves the woman and approaches the other fellow (Parky). He says something like: "Parky, I'm going to be doing a neat trick on the show and I'll be asking you to pick a number. To save hassle, can you pick a number now? Oh, here I've got a deck of cards with a set of numbers written on it. Maybe you can pick one of these cards and use the number written on it"

Marc takes out a deck of numbered cards (not playing cards, just cards with numbers from 1 to 52 written on them). From here, it is a cinch to force Parky to take the 8 card using any Magic 101 technique.

Now it's show time. Marc does a short intro then he hands the deck to the third guy. He asks Parky to think of a number. He asks the woman to name a card. She's confused because her card isn't a number card but that doesn't matter, says, Marc. The woman says "Queen of Hearts". He goes back to Parky for his number. Parky says '8'. The third guy counts 8 down into the deck and holy crap, it's the Queen of Hearts!

Notice how Marc cunningly doesn't ask for Parky to give him a number right away. He asks him to *think* of a number then he asks the woman to think of a card. Only after the woman says QofH does he go back to Parky. This makes it look like Parky has just thought of his number because he's been given a stretch of time and we assume he's been using that time to come up with something. Very cool.

And since neither the woman or Parky know that Marc has spoken to the other pre-show, the trick remains a secret.

It's true that Marc could have used any number of different techniques to get Parky to choose a number. There may be some subtler techniques than using a deck of numbers.

Finally a word from Berglas himself - he said that there is no single Berglas effect. There are Berglas *effects*. So in other words the same illusion can be done in a multitude of different ways, depending on circumstance, audience, opportunity and luck.

I have to admit The Holy Grail of mentalism is somewhat disappointing with its pants down.
posted by storybored at 8:45 PM on August 21, 2011 [1 favorite]


While that may be how he does it in the majority of cases it doesn't account for how he does it one-on-one, three times in a row, or when the participants are chosen with a great degree of randomness like with the twice tossed beach ball noted up thread.

Also it'd be pretty risky. Considering how long they've been doing it it there would be a significant number of people who would spoil the trick by changing their mind (or just plain forgetting) between pre-show and and giving their answer on stage. If the trick failed 1 in 50 times it wouldn't be well known.
posted by Mitheral at 12:39 AM on August 22, 2011


While that may be how he does it in the majority of cases it doesn't account for how he does it one-on-one, three times in a row, or when the participants are chosen with a great degree of randomness like with the twice tossed beach ball noted up thread.

People are terrible at recalling what happened during magic tricks they watched. I'm not going to accept second hand accounts of what people saw him do at face value.
posted by empath at 1:25 AM on August 22, 2011 [2 favorites]


Here's a nice first-hand account (not "100% pure" Berglas effect). I was actually looking for another one that I read earlier on a forum posting, but this one is a lot more interesting. There are at least a few first-person anecdotes around.
posted by taz at 2:41 AM on August 22, 2011


You notice that in every case, he asks him to think of a card or mention a card, but doesn't tell him what he's going to do with it?
posted by empath at 5:45 AM on August 22, 2011


Before the show, Marc would have approached the woman and said something like this: "We're going to be doing an illusion on the show and i'll be asking you to pick a card. To save time, can you pick a card now and tell me what it is?"

But it doesn't save any time, at all. All this does is make the person who you pre-asked for a card think "Oh, that's right, he asked me for the card I was going to say before the show. He probably just asked the other guy for the number before the show and pre-arranged the deck." The guy who picks the number is going to be thinking the exact same thing. If your two accomplices are going to think "This is the dumbest trick ever" when you do it right in front of them, then you might as well just make them shills.
posted by 23skidoo at 11:07 AM on August 22, 2011 [1 favorite]


Well I've been watching this thread develop but keeping my mouth shut. But since the (almost certainly) correct answer has finally hit the board I'll throw my two cents in.

The method given above by storybored is essentially correct apart from this:
"To save time, can you pick a card now and tell me what it is?"

As 23skidoo says, doing it this way isn't going to fool the people involved though it may fool the rest of the audience. A more likely scenario would be something along these lines:

"We'll be doing an illusion later and I wonder if you would help me out? Sometimes when I ask people to help me on stage they freeze up and don't know what to do. To avoid that I'd like you to think of a card right now. You can think of any card and I don't want to know what it is. Have you got one? Good. Now when I ask for your help tonight you just need to call out that same card and everything will go smoothly. Oh and one more thing. Just in case something goes wrong would you write down the card your thinking of? Don't let me see it. Tear of the page and put it in your pocket. If something goes wrong I may ask to see that paper just to verify, okay? Great."

Essentially the same conversation goes on with the person choosing the number. This presentation solves a lot of problems:
The paper (actually pad of paper) is given to the volunteers by the magician and is an "impression pad." Basically carbon paper but the carbon copy appears a few pages down rather than on the next page. The copy obviously allows the magician to know the card and number.

The pages in the volunteers pockets guarantee that they won't change their mind just to be dicks because they've been told that they might be asked for those pages, "if something goes wrong. "

The framing of the pre-show work is that, "sometimes people panic and I want to make this easy on you." It's not, "hey pick a card and tell me what it is."

Finally, because of the framing and use of an impression pad, neither spectator is aware that the magician knows what they picked. This is important because it means those spectators will still be impressed by the effect when they see it performed because they won't know how the magician figured out what they were thinking even though they've been thinking it for a half hour or however long.

There are infinitely more ways the pre-show could be done, but the above, more or less, is how I would do it.

As to multiple performances, one-on-one performances etc. Most people, especially non-magicians, often don't realize just how utterly obsessed magicians are about methods. Seriously we're all a bit crazy. My personal library of magic books numbers in the dozens and even in that small-ish number of books there are at least tens of methods for doing ACAAN. I've even created a few of my own. And that barely scratches the surface. To a layman, especially one trying to put it all together hours or even days later, all the methods look the same. Worse than that, the strengths of one method often cover up the weaknesses of another. So doing the same effect with varying methods can make the effect seem more impossible than it would with just a single viewing.

storybored:
"I have to admit The Holy Grail of mentalism is somewhat disappointing with its pants down."

Yeah most magic tricks are which is the real reason magicians don't tell their secrets. There's nothing worse than showing people something and getting a, "wow! that was the most amazing thing I've ever seen! How did you do it?" reaction just to tell them and have them say, "oh it was up your sleeve? Well that's stupid!" It's terribly depressing for us magicians. Especially considering how much freaking time we put into learning the stuff in the first place. It's why I always tell people there is a world of difference between knowing how a trick is done and knowing how to do it.

These days I tend to be of the opinion that people knowing how the magic works gives them a chance to appreciate it on a different level. They can still enjoy the performance, if it's good, but they can also appreciate the skill.
posted by Mister_Sleight_of_Hand at 12:31 PM on August 22, 2011 [8 favorites]


While that may be how he does it in the majority of cases it doesn't account for how he does it one-on-one, three times in a row

Yes, that's why I said there was more than just one Berglas Effect. To understand how he would have done it "three times in a row", we would first need a reliable description or video. One of the Berglas Effects can indeed work three times in a row, but it requires that the magician be allowed to touch the deck once (and only once!).

...or when the participants are chosen with a great degree of randomness like with the twice tossed beach ball noted up thread.

If the audience is small enough, less than twenty people say, you could use the same technique I described above. I.e. set up the left side of the audience so they've prepicked a card. And set up the right side of the audience so they've prepicked a number. This is time consuming but it would work. It doesn't matter where the ball lands, the choices are still forced.

Also it'd be pretty risky. Considering how long they've been doing it it there would be a significant number of people who would spoil the trick by changing their mind (or just plain forgetting) between pre-show and and giving their answer on stage. If the trick failed 1 in 50 times it wouldn't be well known.

It would indeed be risky if you're doing this with an audience of skeptics or "troublemakers" :). But 99.9% of the time, when you ask for help, people are quite docile. They will do what you ask them to. Part of the reason is that they want to be part of the magic, and they're curious as to what's going to happen. The best mentalists ask for help in a way that enlists them without making them shills.
posted by storybored at 12:45 PM on August 22, 2011


"We'll be doing an illusion later and I wonder if you would help me out? Sometimes when I ask people to help me on stage they freeze up and don't know what to do. To avoid that I'd like you to think of a card right now. You can think of any card and I don't want to know what it is. Have you got one? Good. Now when I ask for your help tonight you just need to call out that same card and everything will go smoothly. Oh and one more thing. Just in case something goes wrong would you write down the card your thinking of? Don't let me see it. Tear of the page and put it in your pocket. If something goes wrong I may ask to see that paper just to verify, okay? Great."

Yes! That's terrific. I'm a relative beginner at this, so my guess at the pre-show spiel was pretty crude. Totally, an impression pad would be the way to go, particularly since the pads are technically so good these days.
posted by storybored at 12:51 PM on August 22, 2011


Mister_Sleight_of_Hand: There's nothing worse than showing people something and getting a, "wow! that was the most amazing thing I've ever seen! How did you do it?" reaction just to tell them and have them say, "oh it was up your sleeve? Well that's stupid!" It's terribly depressing for us magicians.

Well, personally I'm always more impressed when the solution is something simple, but which I would never have thought of in a million years. That's my personal holy grail.
posted by Kattullus at 5:21 PM on August 22, 2011


I'm still at the stage where I see a trick and part of me wishes that the coin really did get inside the bottle via a magical wormhole.

So far I've come across a couple of tricks that are quite ingenious. And there's a good feeling from seeing how creative the inventor was. The Invisible Deck is one of them. Simple to do, cool and baffling effect, and a smart idea.

But back to ACANN.

How would someone do ACANN if they allowed the spectator to shuffle the deck BEFORE he names his card and his number? The magician doesn't touch the cards. The trick is from Andrew Gerard.

Video starts at 5:50, runs for about 2 mins. I'm stumped and awestruck.
posted by storybored at 7:12 PM on August 22, 2011


Well, personally I'm always more impressed when the solution is something simple, but which I would never have thought of in a million years.

That's the case for a lot of people which eventually led to my belief that telling "secrets" really wasn't the end of the world. Though it took some years of magic snobbery to get there. Still the, "that's stupid" reaction is annoying when you get it.

But really the presentation is more important than the secret. I figure everybody knows it's a trick because, you know, people aren't stupid. So it's not my job to make them think it's "real" magic it's my job to make them, however briefly, not care that it's a trick. And that's all in the presentation. If you figure out how it was done later, and that gives you enjoyment, I don't mind so long as you enjoyed the show. But YMMV and all that.
posted by Mister_Sleight_of_Hand at 3:07 AM on August 23, 2011


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