The berglas effect- How is it done?

[dilletante]

I would think that the magician is the only one who would have to memorize the deck. The second man (the stooge) would only have to learn to recognize prearranged vocal or gesture signals from the magician as to what number to pick. So as soon as the magician hears "Queen of hearts", he could signal using feet, hands, or whatever, the number 8 to the second man. The trick would be inventing signals that are easy yet subtle.

 

[ThomasT]

Assuming it is a shill with an excellent memory, you're right - it's not a particularly magician-type trick.

But that is an assumption, and it is an assumption based on no evidence except process of elimination.

He doesn't shuffle the deck!

What would be cool is if he could get a proper celebrity in position #2 - someone who was unlikely to memorize 52 cards in a deck. (Actually, there's probably a simple pattern, making memorization super easy. For example, the simplest is that card 7 is the 7 of diamonds and card 8 is the 8 of diamonds - but that's impractical.)

I don't think there's any memorization going on either. Why take that chance when it's so easy for everybody involved to know that the card is, say, the ten of spades, and it's the 12th card down from the top of the deck?

 

[Oldfart]

It would be easy to "stooge proof" this by challenging the performer to spin a fair wheel to select a number after the card has been chosen. Refusal to do this would prove a stooge is involved.

 

[I_am_learning]

It would be easy to "stooge proof" this by challenging the performer to spin a fair wheel to select a number after the card has been chosen. Refusal to do this would prove a stooge is involved.

How would you know if the fair wheel wan't gimmicked?

 

[DaveC426913]

How would you know if the fair wheel wan't gimmicked?

We have to draw the line somewhere reasonable.

1] It is virtually impossible to prove a human is gimmicked; it is much easier to prove a device is.
2] It is also much harder to gimmick a wheel than a person.

 

[Oldfart]

How would you know if the fair wheel wan't gimmicked?

It would need to be gimmicked (set) within seconds after the card is chosen, pretty hard...

Or use one of those ping pong ball do-hickies like they use to select lottery numbers.

 

[DaveC426913]

The trouble is that this is not fair. Of course totally deconstructing magic trick will make it break, or reveal its secrets.

We don't want to destroy it, all we want to do is eliminate the one variable we consider a "cheat".

And the variable is not even the players, (the trick is still just as good magic if he's supposedly beaming a suggestion into the player's head, or somehow otherwise causing the player to pick a card). The only variable we really want to eliminate is the cheat itself: the pre-knowledge of a given player of the order of the cards.

We want to disrupt the trick as minimally as possible while still giving the magician the freedom to perform the trick and preserve the effect.

 

[TheStatutoryApe]

I've known a couple of professional magicians. They have this 'code' that they take very seriously, and they really love fooling people. As for the other people involved, as I speculated to Dave, maybe they got paid or gave their word or both. And then they are among the few who get to be part of this continuing practical joke.
From what I've read, the money that's been offered for the solution isn't all that much.

I would love to be proven wrong, but it doesn't really appear to be much of a mystery.
He doesn't shuffle the cards!

Haven't you ever kept a secret? It's really easy to do provided you're not being tortured.

There always turns out to be some sort of collusion wrt stuff like the Berglas Effect.

At some point the stooge theory has to break down under the sheer number of people involved in the secret. I think thirty years of practice has probably taken us well beyond that mark. Of course we could run an experiment. Pay some PFers today to keep a secret for you and 30 years from now we will see just how faithful they have been in keeping that secret shall we? edit: Perhaps we could make it a bit more similar by having you pay a couple new people each year for thirty years to keep the same secret and let some various friends in on it as well along the way.

 

[Oldfart]

OK, now what? Dave doesn't want to force the trick by spinning the wheel, so what would you do, Dave, if you could minimally disrupt the trick to try to eliminate the cheat? And StatApe, do you think there is a "cheat" involved here, or is it just, well, magic?

 

[DaveC426913]

OK, now what? Dave doesn't want to force the trick by spinning the wheel, so what would you do, Dave, if you could minimally disrupt the trick to try to eliminate the cheat? And StatApe, do you think there is a "cheat" involved here, or is it just, well, magic?

All you need do is make sure that the one who thinks of a card is not a shill.

That is sufficient to eliminate any coaching or memorizing or otherwise leaves the trick untouched. By pulling from the audience, the magician is implicitly claiming that he's picked an uncoached participant. Thus, any participant is equivalent.

It should not be too hard to ensure that a given participant is not one specifically coached. A random participant picker (say, by seat location) should be perfectly acceptable to the magician. The only possible reason why he might object to a random participant picker is if he's using a shill.

 

[I_am_learning]

All you need do is make sure that the one who thinks of a card is not a shill.

That is sufficient to eliminate any coaching or memorizing or otherwise leaves the trick untouched. By pulling from the audience, the magician is implicitly claiming that he's picked an uncoached participant. Thus, any participant is equivalent.

It should not be too hard to ensure that a given participant is not one specifically coached. A random participant picker (say, by seat location) should be perfectly acceptable to the magician. The only possible reason why he might object to a random participant picker is if he's using a shill.

I like the idea of minimally disturbing the effect. Its called a Magic Show.. Its not a presentation of a new technology, where you can ask the presenter to do every sort of demonstration. All you can do is watch, and do what you are asked. What I mean is, if you are asked to select a random card, then there you have flexibility, choose any card. If you are asked to come any people from the audience, then there you have chance, you can send a person, like Dave suggested.
The maximum we can do is make sure that, what he is saying is actually what he is doing. i.e. If he is saying he is calling a random person, make sure he is random indeed. If you feel its not the case, There You can object him. But you certainly can't ask him to do the trick in completely different way, like using the wheel.

And for Dave's suggestion, who would select the Seat no.? He can be stooge too :)
Edit: I just realized, you could use a random number generator (like the wheel) for this. Difficult part is to ensure, the random no. generator is random indeed.

 

[DaveC426913]

And for Dave's suggestion, who would select the Seat no.? He can be stooge too :)
Edit: I just realized, you could use a random number generator (like the wheel) for this. Difficult part is to ensure, the random no. generator is random indeed.

Have people call out row numbers and seat numbers.

For this trick, the magician can only afford to train a single person to be the shill. It would be very hard for him to ensure that person got picked if he were taking suggestions from the audience.

 

[Loren Booda]

It's done by the power of suggestion, like

"Give me a million dollars!"

 

[TheStatutoryApe]

OK, now what? Dave doesn't want to force the trick by spinning the wheel, so what would you do, Dave, if you could minimally disrupt the trick to try to eliminate the cheat? And StatApe, do you think there is a "cheat" involved here, or is it just, well, magic?

I do not believe it is "magic", I believe it is a card trick. There are no "cheats" per se but the thing about this particular trick is that there are supposedly no "stooges" involved and considering the circumstances for the trick, if this is true, there must be some rather clever ploy in use that no one seems to have quite been able to figure out.

I like the idea of minimally disturbing the effect. Its called a Magic Show.. Its not a presentation of a new technology, where you can ask the presenter to do every sort of demonstration. All you can do is watch, and do what you are asked. What I mean is, if you are asked to select a random card, then there you have flexibility, choose any card. If you are asked to come any people from the audience, then there you have chance, you can send a person, like Dave suggested.
The maximum we can do is make sure that, what he is saying is actually what he is doing. i.e. If he is saying he is calling a random person, make sure he is random indeed. If you feel its not the case, There You can object him. But you certainly can't ask him to do the trick in completely different way, like using the wheel.

And for Dave's suggestion, who would select the Seat no.? He can be stooge too :)
Edit: I just realized, you could use a random number generator (like the wheel) for this. Difficult part is to ensure, the random no. generator is random indeed.

The problem is that while the person may not be a plant the card, number, and persons being chosen are not likely random. He has to have some sort of control over the situation or we would literally be asking him to perform magic. The most likely theory, in my opinion anyway, is that he is somehow influencing the persons involved without them realizing it.
In the card trick forum that Galteeth linked one of the magicians, I assume from his comments that he is a magician, notes that good illusions usually rely upon a freeform style and the ability of the performer to influence and direct their audience (also "having balls" apparently helps). He suggests that Berglas begins going through the motions of the routine and if he finds that he is not receiving the desired effect he simply changes the trick. This would follow with at least one of the two examples in the youtube video where he displays the deck and asks the woman next to him to name a card. He did not announce what trick he was doing. Theoretically if she does not name the card he intended for her to name he could just work it seemlessly into another trick as he has yet to ask for the number. This doesn't seem to follow with the second example though where he definitely appears to have announced his "any card at any number" trick before starting. According to the magician I mentioned this would apparently be where "having balls" comes into play.

 

[I_am_learning]

He uses a gimmicked cards. That has tiny hi-tech electronics embedded and special pappery screen, and he controls what card appear on the face of each card wireleslly through a remote control embedded in his shoe, operated by his toes.

 

[DaveC426913]

I do not believe it is "magic", I believe it is a card trick. There are no "cheats" per se but the thing about this particular trick is that there are supposedly no "stooges" involved and considering the circumstances for the trick, if this is true, there must be some rather clever ploy in use that no one seems to have quite been able to figure out.

If there are no stooges then it is a pretty impressive trick indeed. That much I will grant. If there are stooges, then it is not an impressive trick at all.

I think it is fair to verify a magician's explicit claim. If he says "this is a real knife" then we can expect that it's a real knife.

The claim is that there are no stooges, but I/we are dubious. We challenge the claim, and are exploring ways of eliminating a stooge while allowing the magician otherwise complete freedom to perform.

 

[TheStatutoryApe]

If there are no stooges then it is a pretty impressive trick indeed. That much I will grant. If there are stooges, then it is not an impressive trick at all.

I think it is fair to verify a magician's explicit claim. If he says "this is a real knife" then we can expect that it's a real knife.

The claim is that there are no stooges, but I/we are dubious. We challenge the claim, and are exploring ways of eliminating a stooge while allowing the magician otherwise complete freedom to perform.

I am skeptical of the stooge theory but obviously I can not honestly rule it out completely. It is even possible that he occasionally has stooges when he wants to make sure he pulls off the trick and on other occasions does not.

I am uncertain if he has actually claimed anything at all regarding the trick. Everything I read in the link Galteeth posted is based on the claims of observers both amateur and professional. Theoretically their observations should carry some added weight though one of the posters there considered this to be a poor assumption.

 

[Galteeth]

I do not believe it is "magic", I believe it is a card trick. There are no "cheats" per se but the thing about this particular trick is that there are supposedly no "stooges" involved and considering the circumstances for the trick, if this is true, there must be some rather clever ploy in use that no one seems to have quite been able to figure out.The problem is that while the person may not be a plant the card, number, and persons being chosen are not likely random. He has to have some sort of control over the situation or we would literally be asking him to perform magic. The most likely theory, in my opinion anyway, is that he is somehow influencing the persons involved without them realizing it.
In the card trick forum that Galteeth linked one of the magicians, I assume from his comments that he is a magician, notes that good illusions usually rely upon a freeform style and the ability of the performer to influence and direct their audience (also "having balls" apparently helps). He suggests that Berglas begins going through the motions of the routine and if he finds that he is not receiving the desired effect he simply changes the trick. This would follow with at least one of the two examples in the youtube video where he displays the deck and asks the woman next to him to name a card. He did not announce what trick he was doing. Theoretically if she does not name the card he intended for her to name he could just work it seemlessly into another trick as he has yet to ask for the number. This doesn't seem to follow with the second example though where he definitely appears to have announced his "any card at any number" trick before starting. According to the magician I mentioned this would apparently be where "having balls" comes into play.

On the forums I read, this was a common speculation by magicians. Essentially they were saying, there were several methods, including "outs" that involved sleight of hand or switching to a different trick. People then remember the most spectacular times where everything works perfectly. It is true that Berglas did not always perform the trick exactly the same way. However, this all seemed like speculation. No one had any evidence of this, and it's possible that Berglas did it slightly differently at times to keep people from figuring it out (misdirection) or simply to vary up the routine.Edit:
Here's an Irish guy doing it:

Not the berglas effect, but a good version.

Hint/SPOILERS:2:27

 

[Galteeth]

Haha Here is a "lesser" version of the trick by some dude on youtube. No stooges here!

 

[Bhokara]

I'm coming a couple years late to this thread, but I think I know how this trick is done. No memorization of the deck is necessary.

Participants 1 and 2 indeed are not stooges, and they can be randomly chosen from the audience. They are unimportant.

Participant 3 is the magician's permanent paid assistant, a card sharp with the kind of sleight-of-hand skills in palming cards that make most card tricks work.

Participant 3 is always chosen in an offhand, "he doesn't matter" sort of way by Berglas, sort of an afterthought which focuses the audience's attention on the two "important" participants. He is seemingly unimportant because all he does is just count out the cards.

Participant 3 has an identical deck of cards on his person. After exciting Participant 1 chooses the random card, and while everyone's attention is now turned to exciting Participant 2, Participant 3 simply locates that card in the identical deck he has in his pocket (or wherever) and palms it.

Then when it is time to count the cards out, he simply makes it seem like the palmed card is the 10th, or 17th, or whatever number card it is supposed to be.

Participant 3 is an ordinary, forgettable sort of person of average height capable of many disguises with wigs, beards, glasses, fake extra weight, and so on. Anyone good with make up can make the same person look very different at different venues.

Participant 3 is paid a good annual salary to travel with Berglas, much more than is offered to reveal how it is done, and he will inherit the profitable trick when Berglas dies, so he has zero incentive to reveal the secret, even if the magician's code of not revealing secrets wasn't important to him.

That seems to me to be the simplest way to do it. I'll bet if somebody compiled photos of Participant 3 at various performances of this trick, they would always be men of roughly the same height.

 

[zoobyshoe]

If you watch the video, Bohkara, they demonstrate the trick twice with two separate audiences. Participant #3 is obviously not the same man in both performances.

Also, there seems to be no opportunity for #3 to find and palm a card at all in either performance. Both hands are visible at all times.

 

[Bhokara]

I disagree. Note especially that the four criteria state only that Participants 1 and 2 are not stooges. Participant 3 is identified as a "spectator." That's a loophole you can drive a truck through.

In both videos, there is extensive palming of the cards going on.

Ask yourself, "Why do Participant 3 in both performances hold the card deck in two hands with one hand over the top of the deck"? Why can't the card deck stay on a flat surface? Why does it have to be held in two hands by Participant 3?

Watch the YouTube video again:



Now stop the video at 4:37. Participant 3 moves his left hand over the deck, while at 4:38, the magician makes a jerky motion with his upper body and both hands of the exact type which evolution has trained our eyes to move to automatically.

Participant 3's hands remain over the deck until 4:41.

If the deck were already in the correct order somehow for the trick, wouldn't it be much more of a show to leave all the cards on the table, and have Participant 3 reach down with his thumb and forefinger of one hand and daintily turn them over one-by-one?

There are thousands of professional card sharps who can palm a matching card onto the deck at 4:37. And there are thousands of people who can professionally disguise themselves.

http://tvtropes.org/pmwiki/pmwiki.php/Main/ManOfAThousandFaces

The simplest explanation for how this trick is done within the confines of the four criteria (Participants 1 and 2 are not stooges) during the second performance on the YouTube video is this:

The magician calls out a random person to be Participant 1 and name a card. The card is named. The magician's assistant has an organized deck secreted on his person, and pulls out the matching card. The magician gives him plenty of time to do this while selecting Participant 2 and bantering with her. Then and only then is Participant 3 identified and anybody notices him in his boring white shirt and boring white hair. He comes up on the stage, picks up the deck, and at 4:37 palms the correct card onto the top of the deck when the magician pauses in the counting for the Big Reveal and makes a classic audience distraction motion at 4:38.

Presto. No paranormal activity, no mental powers of suggestion, no memorization of card order, and Participants 1 and 2 can indeed be anyone on earth.

 

[zoobyshoe]

The magician calls out a random person to be Participant 1 and name a card. The card is named. The magician's assistant has an organized deck secreted on his person, and pulls out the matching card. The magician gives him plenty of time to do this while selecting Participant 2 and bantering with her. Then and only then is Participant 3 identified and anybody notices him in his boring white shirt and boring white hair. He comes up on the stage, picks up the deck, and at 4:37 palms the correct card onto the top of the deck when the magician pauses in the counting for the Big Reveal and makes a classic audience distraction motion at 4:38.

OK. This makes perfect sense. He locates and secretes the proper card way before anyone is even aware of him. Then with 'conventional' sleight-of-hand, he produces the proper card at the proper point in the count.

As to the disguise though, I'm sure you're wrong, and I say that from personal expertise with theatrical makeup. I was a theater major in college and I became quite good at character makeup. I can do foam latex appliances, and hair ventilation for beards, moustaches, and wigs. I was good enough to get paid professionally to do that here and there. So, in my expert opinion participant #3 in the second video is absolutely not the same man as #3 in the first in makeup. It's not that you couldn't convincingly make the first look like the second, the difference is textural. The second #3 is much older and I can see his face hanging as he bends his head down to look at the cards in a way you can't achieve with foam latex. A foam appliance has no weight like flesh and exhibits no pendular properties. You won't observe the appliance reacting to gravity as the actor moves the same way real flesh does. No matter how good all other considerations are, there are two tell tales of an appliance. I just mentioned one, and the other is that you won't see changes of complexion due to emotion: the face of an actor wearing an appliance won't redden with anger, or pale with shock. You will see those things on a good actor without any makeup, though.

Regardless, I think your explanation is perfectly plausible if we remove the limit on how many assistants he has. It could be he has two, or three, or it could be he can tap into the entire body of professional magicians who are sworn to secrecy about the explanations. I don't know much about that world. I'm really, really confident in saying he must have at least two. (Not that that guarantees I'm right. Just saying my opinion about the difference between the two #3's is well informed.)

A girl here (here in real life, not here at PF) taught me a baffling trick whose explanation I could never, ever have figured out in a million years. It is so simple it's amazing. So, I'm aware of the fact that the way these things are done can easily be counter-intuitively simple. If yours isn't the right explanation, I'd bet the real explanation is even simpler than you suggest.

 

[collinsmark]

Participants 1 and 2 indeed are not stooges, and they can be randomly chosen from the audience. They are unimportant.

That assumes that the person who made the Youtube video actually knows the trick and is telling the truth. I'm not convinced of either.

Previous comments regarding the unlikelihood of conspiracy and the inability to keep secrets ignore the unrelenting motivation of a struggling actor to get a stamp on his or her SAG (Screen Actors Guild) card (or in this case the equivalent actors guild/union in the UK -- I think it's AGGB or "Equity" or something like that).

Participant 3 is the magician's permanent paid assistant, a card sharp with the kind of sleight-of-hand skills in palming cards that make most card tricks work.

That's also a possibility.

But I think the simplest solution is at least one of the participants is an actor who is either (a) a struggling/unknown actor who will bend over backwards for to get that notch of experience recognized by the actors guild/union, or (b) an already famous actor who's a SAG/AGGB/Equity/Union member who's taking one for the team.

Would they keep it a secret? Of course. Future hope of fame is motivation enough (not to mention there's probably a contract involved too). Do you think a struggling actor will sell their hope of fame for a few bucks, knowing that there would be practically no chance of them ever landing a role again [not to mention a lawsuit for breach of contract]? Not a chance.

All it takes is a stacked deck, and a simple algorithm that maps a card to a number. Or to make it simpler, both participants I and II could be actors, and nobody has to do anything in one's head.

That, or participant III could be palming cards. But in the second presentation in the video, that didn't seem to be the case.

Either way, somebody's in on it.

 

[lendav_rott]

I want to do that trick more than once while reshuffling, only then would I believe in magic. Sadly, there is no such thing as magic :<