The berglas effect- How is it done?

[Galteeth]

This is referred to as the "holy grail" of card tricks. I don't think people could figure it out, but it might be fun to try. Here's a youtube video of the trick.



I was reading some magician's forums, and there were various speculations about it possibly involving trick decks and memorizations. Even so, the trick, as performed, doesn't seem like any of that would help. The most obvious solution would be that there are in fact stooges. But the first example involves British tv personalities. From what I've read, Berglas perfomed the trick for other magicians who were stunned. The man performing the trick is Berglas's best friend, supposedly one of the only other people who knows how its done.

It's interesting, because by all means, it looks completely impossible. Any thoughts?

 

[1MileCrash]

I'd think the person who picks the number is in on it and has the deck memorized. You say "british television personality" as if it is some sort of seal of credibility. :P

 

[Galteeth]

I'd think the person who picks the number is in on it and has the deck memorized. You say "british television personality" as if it is some sort of seal of credibility. :P

It's possible, but in this case that would require a British tv show host to do a lot of work to help out a magician. Also, since this trick is considered so valuable, you'd think one of the stooges (over the many times it was done) would have offered to sell the secret.

Anyway, here is an interesting discussion on the trick, recounting several incidents where the trick was performed by Berglas solo for magicians.

http://www.geniimagazine.com/forums/ubbthreads.php?ubb=showflat&Number=31790

 

[Galteeth]

My basic point as to why the stooge explanation seems unlikely is that it requires a fairly large conspiracy. You need several magicians and others, including celebrities to testify falsely that he has done it for them, you need multiple stooges at mulitple shows, all of whom would have to keep the secret, and who have a lot of incentive to come forward (since there are people offering money to learn the secret.) You also need a lot of interested people, including magicians, to not to have been able to sniff it out. It just seems unlikely, although i suppose it is the simplest explanation.

 

[FlexGunship]

James Randi could learn it in his sleep and then wake up and do it better before he had his morning coffee.

 

[TheStatutoryApe]

The first thing that I noticed is that he apparently does not shuffle the deck, at least not during his trick. It gives the impression that the order of the cards is somehow important though I am not sure that means that the deck is actually stacked. He maybe hints at the way he does the trick when he tells them in the first example that he got them to make the choices that they did. In both versions he holds up and fans out the cards before setting them down and not touching them again. I am considering the possibility that something he does while fanning the cards suggests the choices which these people make. For instance if one were in such a situation, was just shown some cards, and then suddenly put on the spot and asked to name a card one may simply blurt out one of the cards which they observed only a second ago. He may be able to suggest the number in a similar fashion.

 

[FlexGunship]

Okay, these are usually "party tricks." I could do a similar one in college. There is always an accomplice. The videos provided are NOT good enough evidence for the card trick itself because there are camera cuts.

Read the book "Moonwalking with Einstein." Card tricks are a common party trick among memory gurus. They will memorize a deck of cards and have people say a number and recite which card it is.

My guess is that Berglas has very little to do with the trick. One person indicates a card (or number), then the plant says the corresponding number (or card). The trick is done looooong before any cards are counted out. Throw in a false shuffle and you've got yourself a magic show!

 

[Galteeth]

Okay, these are usually "party tricks." I could do a similar one in college. There is always an accomplice. The videos provided are NOT good enough evidence for the card trick itself because there are camera cuts.

Read the book "Moonwalking with Einstein." Card tricks are a common party trick among memory gurus. They will memorize a deck of cards and have people say a number and recite which card it is.

My guess is that Berglas has very little to do with the trick. One person indicates a card (or number), then the plant says the corresponding number (or card). The trick is done looooong before any cards are counted out. Throw in a false shuffle and you've got yourself a magic show!

If you read the link I provided, there were numerous examples of him performing the trick for magicians with no accomplices at all, i.e., the person the trick is being done for both picks the number and the card. Granted, there's no video of that, but you can find other instances of magicians and celebrities having seen the same thing.

Nevertheless, ok, we've got the stooge theory. Anything BESIDES that which could produce this effect?

 

[Galteeth]

The first thing that I noticed is that he apparently does not shuffle the deck, at least not during his trick. It gives the impression that the order of the cards is somehow important though I am not sure that means that the deck is actually stacked. He maybe hints at the way he does the trick when he tells them in the first example that he got them to make the choices that they did. In both versions he holds up and fans out the cards before setting them down and not touching them again. I am considering the possibility that something he does while fanning the cards suggests the choices which these people make. For instance if one were in such a situation, was just shown some cards, and then suddenly put on the spot and asked to name a card one may simply blurt out one of the cards which they observed only a second ago. He may be able to suggest the number in a similar fashion.

That would be pretty impressive, and from what I was reading on various magician forums, this was a popular suggestion as to how its done, and then there are "outs" if the people don't state the right cards that involve sleight of hand. Still, it seems incredible to me that this method would produce consistent enough results and that it would work on professional magicians.

Also to clarify for anyone who misunderstands me, I am not suggesting this is "genuine magic." It is of course an illusion, it's just a darn good one, and having some familiarity with card tricks and sleight of hand, I find it a fascinating puzzle.

 

[FlexGunship]

That would be pretty impressive, and from what I was reading on various magician forums, this was a popular suggestion as to how its done, and then there are "outs" if the people don't state the right cards that involve sleight of hand.

David Blane does a good version of a trick like that by offsetting a single (known) card in the deck and riffling at the right speed. The result is that he tells you to think of only one card, but only one seems to be visible during that time. In the TV version of the trick it is a queen of spades. Also one of the most common cards guessed as "random."

Try this, I'm giving away one of my great party tricks; the cardless card trick.

Tell someone (a bit inebriated, perhaps), to think of a card. Any card. Give him or her a second to think. Then say: "...and make it random, don't make it a 2 or an ace or a king or something stupid. Pick something really random." (You've just eliminated 23% of the deck, by the way.)

They will either immedaitely say "okay, got it." Or think of somethign new. If they think of something new, they are the type to have a "preferred random number." They think they're outsmarting you by picking something obvious (if they think of a new card, it's probably because they were thinking of the ace of spades).

When they're done, stare at them, and tell them to think of the card really hard. You basically have a 1 in 40 chance of just guessing. Double your chances by saying: "hmm, you're not thinking of it very clearly... you're really thinking of two. I might get this mixed up... hmm, it seems to be red." Judge there response. You might get a hint or you might not.

In the worst case scenario, tell them you've got two cards, say "it's either the seven of hearts, or the queen of spades." (Don't pick two random cards, actually use THOSE two cards.)

What you've done is you've made quite a few options available:

• 7 of hearts
• 7 of spades
• queen of hearts
• queen of spades

So you've actually quadrupled your chances of guessing. You're ROUGHLY at a 1 in 10 chance at this point. However, most people are so eager for you to be right, then will say something like: "No, but it was the 6 of hearts, holy crap man! That's crazy!"

And your respond, "well, I knew I was getting a crossed signal there. Have you been drinking?" And everyone laughs it off. If you're totally wrong, just shrug... "oh, well."

If you're spot on... you are the king of the party. But you MUST refuse to do the trick again. YOU WILL FAIL.

 

[Galteeth]

David Blane does a good version of a trick like that by offsetting a single (known) card in the deck and riffling at the right speed. The result is that he tells you to think of only one card, but only one seems to be visible during that time. In the TV version of the trick it is a queen of spades. Also one of the most common cards guessed as "random."

Try this, I'm giving away one of my great party tricks; the cardless card trick.

Tell someone (a bit inebriated, perhaps), to think of a card. Any card. Give him or her a second to think. Then say: "...and make it random, don't make it a 2 or an ace or a king or something stupid. Pick something really random." (You've just eliminated 23% of the deck, by the way.)

They will either immedaitely say "okay, got it." Or think of somethign new. If they think of something new, they are the type to have a "preferred random number." They think they're outsmarting you by picking something obvious (if they think of a new card, it's probably because they were thinking of the ace of spades).

When they're done, stare at them, and tell them to think of the card really hard. You basically have a 1 in 40 chance of just guessing. Double your chances by saying: "hmm, you're not thinking of it very clearly... you're really thinking of two. I might get this mixed up... hmm, it seems to be red." Judge there response. You might get a hint or you might not.

In the worst case scenario, tell them you've got two cards, say "it's either the seven of hearts, or the queen of spades." (Don't pick two random cards, actually use THOSE two cards.)

What you've done is you've made quite a few options available:

• 7 of hearts
• 7 of spades
• queen of hearts
• queen of spades

So you've actually quadrupled your chances of guessing. You're ROUGHLY at a 1 in 10 chance at this point. However, most people are so eager for you to be right, then will say something like: "No, but it was the 6 of hearts, holy crap man! That's crazy!"

And your respond, "well, I knew I was getting a crossed signal there. Have you been drinking?" And everyone laughs it off. If you're totally wrong, just shrug... "oh, well."

If you're spot on... you are the king of the party. But you MUST refuse to do the trick again. YOU WILL FAIL.

I was friends with David Blaine's brother in high school. He showed me some pretty interesting stuff. I've met Blaine a few times, and he is a funny character.

 

[DaveC426913]

This is easy!

The chances of a given card coming up at a given location are 1 in 52. So, before the show, he simply thinks of a number and a card and tests it - 51 times in a row. They will all fail. But now the deck is loaded for the next one to be a hit.

Easy!

 

[DaveC426913]

 

[DoggerDan]

It said the first person wasn't a stooge. How about the second and third folks?

 

[DoggerDan]

This is easy!

The chances of a given card coming up at a given location are 1 in 52.

The chance of a given card coming up at any given location is 1 in 52. The chance of a given card coming up at a specific location is 1 in 2,704.

 

[DaveC426913]

OK, for the record, before anyone gets the wrong idea, I was kidding around, however...... he spoke incorrectly. He gave the impression that the chances of a hit are 1-in-52 times 1-in-52.

That is untrue.

The chance of a given card coming up at any given location is 1 in 52. The chance of a given card coming up at a specific location is 1 in 2,704.

Yes, but there are 52 ways that could happen.

See, it's not like participant A must pick card 8 (a 1-in-52 chance) and participant B must pick a Queen of Spades (a 1-in-52 chance).

No, any spot in the deck works - as long as they get the card right.
So:
pA picks card 1 - 1-in-52 that it's the Ace of spades (a 1-in-2,704 chance).
plus
pA picks card 2 - 1-in-52 that it's the 4 of hearts (a 1-in-2,704 chance).
plus
pA picks card 3 - 1-in-52 that it's the 10 of clubs (a 1-in-2,704 chance).
...etc.

i.e there are 52 of 1-in-2,704 chances of getting the right card in the right place. i.e. one in 52.

 

[Pythagorean]

Easy one:

Get a deck that has assymetry on the back pattern. Align all the cards. After they randomly pick a card, turn the deck 180 (don't flip it, rotate!) so that the back pattern will be opposite from their card. Let them shuffle (90% of the time they keep the cards aligned).

Let them look at the cards while you sift through them (hild between you so you can be looking at the backart while they anxiously wait for you to pas their card)

Pass the card, act like you caught something in their eye, scroll back and say "aha" and give them their card.

They'll say your trick is reading their eyesmind.

 

[DaveC426913]

Easy one:

Get a deck that has assymetry on the back pattern. Align all the cards. After they randomly pick a card, turn the deck 180 (don't flip it, rotate!) so that the back pattern will be opposite from their card. Let them shuffle (90% of the time they keep the cards aligned).

Let them look at the cards while you sift through them (hild between you so you can be looking at the backart while they anxiously wait for you to pas their card)

Pass the card, act like you caught something in their eye, scroll back and say "aha" and give them their card.

They'll say your trick is reading their eyesmind.

Does this have anything to do with the trick being discussed, or is this a distinct trick?

 

[turbo]

Does this have anything to do with the trick being discussed, or is this a distinct trick?

This is not a trick. It's something anybody can do with a doped deck. I can do the same thing with a shaved deck, without even looking at the deck. These are "tricks" that you can buy anywhere and require zero skill.

 

[Thetom]

This is easy!

The chances of a given card coming up at a given location are 1 in 52. So, before the show, he simply thinks of a number and a card and tests it - 51 times in a row. They will all fail. But now the deck is loaded for the next one to be a hit.

Easy!

Ha, I'm going to try that next time we flip a coin to see who breaks at pool

 

[DaveC426913]

This is not a trick. It's something anybody can do with a doped deck. I can do the same thing with a shaved deck, without even looking at the deck. These are "tricks" that you can buy anywhere and require zero skill.

Uh, that's still a card trick, no matter how you dress it up.

The question was: are you describing the berglas trick in the OP, or are you striking out on your own? I just didn't know which.

It sounds like the latter.

 

[Pythagorean]

Dave, no this isn't the berglas effect, just one I remember form high school.

turbo, just like with all "magic", the 'trick' is the psychological part:

Pass the card, act like you caught something in their eye, scroll back and say "aha" and give them their card.

They'll say your trick is reading their eyes.

 

[ThomasT]

The first thing that I noticed is that he apparently does not shuffle the deck, at least not during his trick.

That's the first thing I noticed too. So wrt the Berglas Effect, I have to defer to the law of parsimony, which has been a most useful guide for scientists. He doesn't shuffle the deck because it's all been agreed beforehand that they'll be picking, say, the 22nd card in the deck from the top, and that card is, say, the four of hearts. Wrt the magicians, well, they're in on it too. It's just a hoax, imo. And this really doesn't require any sort of massive conspiracy. Magicians love to keep secrets. And the other shills probably just love it that they got to be in on it too. It's just a practical joke on everybody in awe of and trying to figure out the Berglas Effect.

As with Fermat's deceit that he had a simple proof for his most famous conjecture, Berglas isn't doing a magic trick, or any sort of mental gymnastics (he's just staging a hoax with collaborators) -- and they both had a bit of fun in their deceits.

 

[DaveC426913]

That's the first thing I noticed too. So wrt the Berglas Effect, I have to defer to the law of parsimony, which has been a most useful guide for scientists. He doesn't shuffle the deck because it's all been agreed beforehand that they'll be picking, say, the 22nd card in the deck from the top, and that card is, say, the four of hearts. Wrt the magicians, well, they're in on it too. It's just a hoax, imo. And this really doesn't require any sort of massive conspiracy. Magicians love to keep secrets. And the other shills probably just love it that they got to be in on it too. It's just a practical joke on everybody in awe of and trying to figure out the Berglas Effect.

As with Fermat's deceit that he had a simple proof for his most famous conjecture, Berglas isn't doing a magic trick, or any sort of mental gymnastics (he's just staging a hoax with collaborators) -- and they both had a bit of fun in their deceits.

OK, but it is not essential - and in fact quite risky - to have too many collaborators. There need only be one - the second guy - who decides on a card. He has memorized the deck.

 

[TheStatutoryApe]

OK, but it is not essential - and in fact quite risky - to have too many collaborators. There need only be one - the second guy - who decides on a card. He has memorized the deck.

I don't know. He has apparently done this card trick for several years. As much as Magicians like their mystery and tend to have professional courtesy they also like to show each other up. That several magicians and random patsies (who apparently all have eidetic memories) would keep such a secret over such a long period of time and with people offering money for the solution just does not seem probable.

edit: according to wiki he has been doing the trick since the 70's.

 

[ThomasT]

OK, but it is not essential - and in fact quite risky - to have too many collaborators. There need only be one - the second guy - who decides on a card. He has memorized the deck.

Sure, there are people who could memorize a deck with probably less effort than it takes me to memorize my various 12-digit passwords. But how many collaborators has Berglas had? And then there are the other people doing the same 'trick' (are there? I don't know). It just seems more likely to me that everybody involved in the 'trick' is also in on the joke. Maybe they got paid. Maybe they gave their word not to reveal the 'secret'.

 

[ThomasT]

I don't know. He has apparently done this card trick for several years. As much as Magicians like their mystery and tend to have professional courtesy they also like to show each other up. That several magicians and random patsies (who apparently all have eidetic memories) would keep such a secret over such a long period of time and with people offering money for the solution just does not seem probable.

edit: according to wiki he has been doing the trick since the 70's.

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.

 

[Galteeth]

Sure, there are people who could memorize a deck with probably less effort than it takes me to memorize my various 12-digit passwords. But how many collaborators has Berglas had? And then there are the other people doing the same 'trick' (are there? I don't know). It just seems more likely to me that everybody involved in the 'trick' is also in on the joke. Maybe they got paid. Maybe they gave their word not to reveal the 'secret'.

Only two people have can perform the Berglas effect. Berglas himself, and his best friend (who is the man in the youtube video.) There are many other magicians who do their own "version" of the effect, but it's not the same, and not nearly as impressive.

 

[ThomasT]

Only two people have can perform the Berglas effect. Berglas himself, and his best friend (who is the man in the youtube video.)

Really? I think I'll do it tomorrow for (and with ) some friends who are coming over.

There are many other magicians who do their own "version" of the effect, but it's not the same, and not nearly as impressive.

How many 'versions' of "name a card and specify its position in the deck, from the top or from the bottom" (and then that card is found to be in that position), can there be?

Anyway, I'm not impressed by the Berglas Effect as I don't think it involves any particularly magician type skills. If some other magicians are doing something roughly akin (??) to the Berglas Effect that does involve magician type skills, then I'm sure I'd be more impressed by their stuff than by the Berglas Effect.

 

[DaveC426913]

Anyway, I'm not impressed by the Berglas Effect as I don't think it involves any particularly magician type skills. If some other magicians are doing something roughly akin (??) to the Berglas Effect that does involve magician type skills, then I'm sure I'd be more impressed by their stuff than by the Berglas Effect.

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.

It may be right, but you need to recognize that it's on kinda shaky ground.

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.)

 

[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.