No, we can't call him/her a stooge, as the spectator doesn't know how the trick works or how their choice may affect the outcome of the trick...Originally posted by Carlo Morpurgo:
New question: can we call a stooge someone who's been told before the show: "Pick a number, don't tell me. Do this and that with the deck. Later, pick the same number."
Berglas Effect .... WOW...
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Re: Berglas Effect .... WOW...

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Re: Berglas Effect .... WOW...
We're getting into framing or scope issues here.
The effect is what the audience sees at the performance.
Conditions... troublesome as it's a vendor term used to excuse deletions of pertinent aspects of the effect.
"The Beglas Effect (X)" ... would that be a trademark, a servicemark or if he's used the term himself  a copyright symbol. Anyway it's supposed to be ACAAN where the audience gets to take the deck and count down according to RK's reports.
On the realistic side, if you can have fifty two decks in reasonable proximity and a formula to recall which is required then you can get the effect as described. On the more portable and practical side Bob Farmer's Deckronomicon offers a reasonable set of compromises which leave the performer with the most options when it comes time to have the dealing done and still reliably get the effect to complete correctly.
The effect is what the audience sees at the performance.
Conditions... troublesome as it's a vendor term used to excuse deletions of pertinent aspects of the effect.
"The Beglas Effect (X)" ... would that be a trademark, a servicemark or if he's used the term himself  a copyright symbol. Anyway it's supposed to be ACAAN where the audience gets to take the deck and count down according to RK's reports.
On the realistic side, if you can have fifty two decks in reasonable proximity and a formula to recall which is required then you can get the effect as described. On the more portable and practical side Bob Farmer's Deckronomicon offers a reasonable set of compromises which leave the performer with the most options when it comes time to have the dealing done and still reliably get the effect to complete correctly.
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Re: Berglas Effect .... WOW...
what you call someone is one thing.
But there are two scenarios here. One is where you ask the person to name a card, then ask them to say it again later. the second is where you get them to think of a card before the show, but not to reveal it to you or anyone else.
These two scenarios will have a hugely different impact on that spectator's understanding of the effect.
The trouble with all types of preshow work is that, once the idea becomes known to the general public, every type of mindreading act is dead, whether it uses preshow or not.
But there are two scenarios here. One is where you ask the person to name a card, then ask them to say it again later. the second is where you get them to think of a card before the show, but not to reveal it to you or anyone else.
These two scenarios will have a hugely different impact on that spectator's understanding of the effect.
The trouble with all types of preshow work is that, once the idea becomes known to the general public, every type of mindreading act is dead, whether it uses preshow or not.

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Re: Berglas Effect .... WOW...
Stooge:Originally posted by Marco Pusterla:
No, we can't call him/her a stooge, as the spectator doesn't know how the trick works or how their choice may affect the outcome of the trick...
1. one who plays a subordinate or compliant role to a principal
2. any underling, assistant, or accomplice.
Now it really does depend on who the primary target is. If the trick is addressed to the two guests then they are not stooges (even though after the initial puzzlement they will do 1+1 quickly).
If the trick is addressed to the folks watching the show then the two guests ARE stooges, since they have prearranged something with the performer in order to fool other people (regardless whether they are aware of it or not). If this is not the definition magicians use (for their own peace of mind), it's what most people would associate with the word.
For example imagine the following.
As seen live on National TV: Magician asks guest to name a card, (without touching the deck in full view); he says King of clubs. Now he asks him to look at the top card of the deck. It's the King of clubs. Astonishing.
Prework: Magician asks the guest to merely think of a card and to place it on top of the deck. And then asks him to name the same card later.
Now, who on earth would not think that the guest is a stooge? And thinking of it, condition 2 that the spectator *freely names* a card is certainly not met.
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Re: Berglas Effect .... WOW...
On the other hand  if working for a TV studio audeince you could borrow a deck (so all the cards are in fact theirs) and then get any number  do a long shot for the "name a card" when they speak it (so can dub in voicetrack later for that second) and then let the effect fly.
just a bit of direction, use of an applause sign and that blip of post production if you want to get the name right.
So many ways to get the job done  just have to find the right one for the venue. ;)
just a bit of direction, use of an applause sign and that blip of post production if you want to get the name right.
So many ways to get the job done  just have to find the right one for the venue. ;)
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Re: Berglas Effect .... WOW...
The second scenario is not compliant with condition 2, since the spectator (i.e. the guest) is instructed to name the card later, and that would not be "free choice".Originally posted by El Mystico:
One is where you ask the person to name a card, then ask them to say it again later. the second is where you get them to think of a card before the show, but not to reveal it to you or anyone else.
These two scenarios will have a hugely different impact on that spectator's understanding of the effect.
In the first scenario there could still be a loophole in that the card is freely chosen and named, the first time, prior to the show. This is of course insulting my intelligence as a spectator (=whoever watches the show), since I am assuming that the show starts when I see it, not two hours ago, or last year.
But even then, it seems that if the first scenario occurs then condition 1 will be violated in that the deck is not in full view before the effect starts, which happened before I see it on TV. Unless of course they think of full view to the guests, not us TV viewers.
Carlo
Re: Berglas Effect .... WOW...
The effect as shown is impossible (actually, it has a chance of happening 1 in 2704 tries).
Therefore, the posts above are correct  either a stooge was used, or preshow work. Tom Stone has a very well thought out possibility.
The first video Pete linked to stated the conditions of the Berglas effect, which I do not believe were accurate. The condition stating that the deck is in view the entire time does not match up with what I read about the Berglas effect. In one instance, I remember reading that Berglas did the effect to someone in his car. After the card and number were arrived at, he had the participant open the glove box, pull out a deck, and conclude  the deck was not in sight before the effect began.
Richard  when Berglas did the effect to you, was the deck in sight before the questioning started? If it was, then Berglas had to touch the deck before you did your counting. If Berglas doesn't touch the deck, then the only solution is that he has multiple outs. Or, he is a master of hypnosis, and is causing his spectators to forget that they told him the card and number, or something to that effect.
Therefore, the posts above are correct  either a stooge was used, or preshow work. Tom Stone has a very well thought out possibility.
The first video Pete linked to stated the conditions of the Berglas effect, which I do not believe were accurate. The condition stating that the deck is in view the entire time does not match up with what I read about the Berglas effect. In one instance, I remember reading that Berglas did the effect to someone in his car. After the card and number were arrived at, he had the participant open the glove box, pull out a deck, and conclude  the deck was not in sight before the effect began.
Richard  when Berglas did the effect to you, was the deck in sight before the questioning started? If it was, then Berglas had to touch the deck before you did your counting. If Berglas doesn't touch the deck, then the only solution is that he has multiple outs. Or, he is a master of hypnosis, and is causing his spectators to forget that they told him the card and number, or something to that effect.

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Re: Berglas Effect .... WOW...
Since when is what you talk about in your patter the same as what you actually do?
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Re: Berglas Effect .... WOW...
Cohiba wrote:The effect as shown is impossible (actually, it has a chance of happening 1 in 2704 tries).
Actually, there's a 1 in 52 chance of the named card being at the named number. And by allowing oneoff variations and counting from either end of the deck, it goes down to 1 in 13. One way to improve the odds further would be to let the spectator(s) cut the deck, each cut giving roughly a new 1/13 chance of getting the card into position.
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Re: Berglas Effect .... WOW...
The deck was out in the open when he did it for me, though I had not been told where to look at the time. On subsequent visits to David's hoome, I noted that the deck was still in the same position, indicating he has been doing it this way to others.
There is no preshow work, no stooges are used, and there is no hypnosis. He asks you to name a card, then asks you to name a number, then tells you to pick up the deck and count down to the named number and there's your card!
It's like the Hooker Rising Cards: when someone explains the effect to you, you think the person has been unobservant in some way and that YOU could see something they don't. Not true. It's impossible.
There is no preshow work, no stooges are used, and there is no hypnosis. He asks you to name a card, then asks you to name a number, then tells you to pick up the deck and count down to the named number and there's your card!
It's like the Hooker Rising Cards: when someone explains the effect to you, you think the person has been unobservant in some way and that YOU could see something they don't. Not true. It's impossible.
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Re: Berglas Effect .... WOW...
Jonathan Townsend wrote:Since when is what you talk about in your patter the same as what you actually do?
Eeeeeexactly...
Carlo Morpurgo said "The use of stooges is not fair to the public." Really? Why not? What's the difference between using sleight of hand or using a stooge? The effect is everything isn't it? Who cares what the method is?
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Re: Berglas Effect .... WOW...
There are no stooges used if one performs the Berglas Effect as Berglas does.
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Re: Berglas Effect .... WOW...
Jonathan Townsend wrote:Since when is what you talk about in your patter the same as what you actually do?
I've heard of several people explaining the steps in overhand stacking a deck as they went through them. To a layperson, the technical language has very little meaning, so it doesn't really explain anything.
Re: Berglas Effect .... WOW...
Jonathan Townsend wrote:Since when is what you talk about in your patter the same as what you actually do?
Jonathan:
When I first read your reply, I didn't understand what you meant. After some of the comments that followed, I think I now understand what you mean.
I have no problems with how the effect was accomplished. I was just pointing out that what was shown on the clip was not the whole story. Others above believed this to be the Berglas effect. If the Berglas effect does not use stooges or preshow work (as Richard pointed out), then the above clip is not the Berglas effect.
Obviously, what you do and what you say are always at odds. That's how you create the illusion of magic.
Re: Berglas Effect .... WOW...
Richard Kaufman wrote:The deck was out in the open when he did it for me, though I had not been told where to look at the time. On subsequent visits to David's hoome, I noted that the deck was still in the same position, indicating he has been doing it this way to others.
There is no preshow work, no stooges are used, and there is no hypnosis. He asks you to name a card, then asks you to name a number, then tells you to pick up the deck and count down to the named number and there's your card!
It's like the Hooker Rising Cards: when someone explains the effect to you, you think the person has been unobservant in some way and that YOU could see something they don't. Not true. It's impossible.
Richard:
My guess is that Berglas has some clever methods of reducing the number of possibilities that are available. As mentioned above, dealing from back, face, next card, etc. greatly reduce the amount of decks you would need. From what I understand, he probably has a lot of psychological things going on as well.
He could have 13 (likely fewer needed) decks around his house / car / etc. The fact that the deck was in the same place the next time you went to his house would lead me to believe that this is part of the work. If only one deck was needed, what are the odds that it would be sitting in the same place? He'd have it on his person, or it would be laying anywhere in the house.
Re: Berglas Effect .... WOW...
Bob Coyne wrote:Cohiba wrote:The effect as shown is impossible (actually, it has a chance of happening 1 in 2704 tries).
Actually, there's a 1 in 52 chance of the named card being at the named number. And by allowing oneoff variations and counting from either end of the deck, it goes down to 1 in 13. One way to improve the odds further would be to let the spectator(s) cut the deck, each cut giving roughly a new 1/13 chance of getting the card into position.
I don't believe the 1 in 52 is correct. I understand that 52 decks would give you all the possibilities necessary, but that's because each deck can be used for 52 different options. The JS in one stack might be number 1, but that same deck would be used for the KC at position 2, etc.
If you were only tracking one individual card, then the odds would be 1 in 52  shuffle the deck and check where the AS is, for example. Statistically, you would have to shuffle around 52 times to have it show up in any particular spot. To throw in the odds of any named card being in that spot makes the odds go up by 52 again.
Therefore, I would agree with you that as shown in the video clip, the odds would be 1 in 52 that they had the right deck (which is again, not much different from impossible), but the overall odds of ACAAN is 1 in 2704.
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Re: Berglas Effect .... WOW...
He does not have 13 decks when he does the trick.
Like I said, only a person who hasn't seen it would be forced to think of something silly like that in intellectually thrashing around for some sort of explanation.
One deck.
Like I said, only a person who hasn't seen it would be forced to think of something silly like that in intellectually thrashing around for some sort of explanation.
One deck.
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Re: Berglas Effect .... WOW...
So, Berglas does not use stooges, he does not use preshow work, he only uses one deck, the deck is in view from the getgo.
Based on this information, he either has to touch the deck at some point, or psychologically gets you to name a particular card / number. Or it's real magic. Those are the options.
Based on this information, he either has to touch the deck at some point, or psychologically gets you to name a particular card / number. Or it's real magic. Those are the options.
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Re: Berglas Effect .... WOW...
He did not touch the deck either time I did the effect.
He did not use any psychology to get me to think of a particular card.
It's as close to real magic as I've seen with a deck of cards short of the Hooker Card Rise and some other wellknown things.
He did not use any psychology to get me to think of a particular card.
It's as close to real magic as I've seen with a deck of cards short of the Hooker Card Rise and some other wellknown things.
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Re: Berglas Effect .... WOW...
Cohiba wrote:If you were only tracking one individual card, then the odds would be 1 in 52  shuffle the deck and check where the AS is, for example. Statistically, you would have to shuffle around 52 times to have it show up in any particular spot. To throw in the odds of any named card being in that spot makes the odds go up by 52 again.
There's no difference between the named card and the AS in your example. Every card in the deck has a 1/52 chance of being at any given position. So it doesn't matter what card (or number) the spectator chooses.
Re: Berglas Effect .... WOW...
Bob Coyne wrote:Cohiba wrote:If you were only tracking one individual card, then the odds would be 1 in 52  shuffle the deck and check where the AS is, for example. Statistically, you would have to shuffle around 52 times to have it show up in any particular spot. To throw in the odds of any named card being in that spot makes the odds go up by 52 again.
There's no difference between the named card and the AS in your example. Every card in the deck has a 1/52 chance of being at any given position. So it doesn't matter what card (or number) the spectator chooses.
Bob:
The fact that you say "EVERY CARD in the deck" is what changes the odds from just 1 in 52. You can't say the odds of a specific card are 1 in 52, and that by opening it up to EVERY CARD it won't make a difference in the odds.
If you knew what card they were going to choose, the odds would be 1 in 52 that they would get the right number. So let's say for sake of argument they are going to choose the AS. You place it in the deck at number 20. Your odds are 1 in 52 of them choosing 20. However, let's just say you defy the odds and they choose 20. Now, they forget what card they were supposed to pick. They now have 52 more options for a card. The odds are no longer 1 in 52.
I think that this is proven by the well known example of how to accomplish this effect  with 52 stacked decks, each of which has 52 possibilities.
Re: Berglas Effect .... WOW...
Richard Kaufman wrote:He did not touch the deck either time I did the effect.
He did not use any psychology to get me to think of a particular card.
It's as close to real magic as I've seen with a deck of cards short of the Hooker Card Rise and some other wellknown things.
Richard  thanks for playing along. I don't really know what to say at this point, since any potential solution is negated. However, a solution does exist, and if it takes balls of steel, and no sleight of hand, then I can only assume there must be some psychology at work. If it's good psychology, you wouldn't know if it's happening or not.
Anyway, I hope some day to learn the secret. Hopefully he doesn't take it to his grave with him.

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Re: Berglas Effect .... WOW...
Cohiba wrote:I think that this is proven by the well known example of how to accomplish this effect  with 52 stacked decks, each of which has 52 possibilities.
This is actually proof that the odds ARE 1 in 52. When a card and a number are named, only 1 of those 52 decks will match.
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It's no different than seeing the Hooker Card Rise. All possible solutions are negated by what you see. These are perfect tricks and also perfect examples of why the Too Perfect Theory is probably baloney if you're a great magician. If you're not a great magician, running afoul of the Too Perfect Theory will get you in trouble.
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Re: Berglas Effect .... WOW...
Cohiba wrote: If you were only tracking one individual card, then the odds would be 1 in 52  shuffle the deck and check where the AS is, for example. Statistically, you would have to shuffle around 52 times to have it show up in any particular spot.
Actually, you have to shuffle > 26 times before the odds are greater than even that you've put the AS into the wanted position. If you want to guarantee that it will get into a wanted position, and the shuffles are random, you have to shuffle it an infinite number of times.
Coyne is right, you are wrong  the odds of ACAAN working right just from chance is 1 in 52.

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Re: Berglas Effect .... WOW...
Cohiba wrote:If you knew what card they were going to choose, the odds would be 1 in 52 that they would get the right number. So let's say for sake of argument they are going to choose the AS. You place it in the deck at number 20. Your odds are 1 in 52 of them choosing 20. However, let's just say you defy the odds and they choose 20. Now, they forget what card they were supposed to pick. They now have 52 more options for a card. The odds are no longer 1 in 52.
This is the same sort of problem as the odds of rolling doubles with dice. It doesn't matter what the first number is  you only care that the second number is the same as the first. Hence the chance is 1 in 6. It's not 1 in 36. Same with ACAAN, it's irrelevant what card they pick. It only matters that it's at the given position.
If you don't believe it, then try ACAAN with a packet of 4 aces and you'll see that a randomly selected ace will be at a randomly selected position one time in 4.

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Re: Berglas Effect .... WOW...
Bob Coyne wrote:Cohiba wrote:The effect as shown is impossible (actually, it has a chance of happening 1 in 2704 tries).
Actually, there's a 1 in 52 chance of the named card being at the named number. And by allowing oneoff variations and counting from either end of the deck, it goes down to 1 in 13.
There are 52*52=2704 ways to pick randomly a position and a card. However, for a given position each card has exactly the same probability of being there, so 1/52 is the probability that the effect will succeed, given that everything is random.
I am not sure about "counting from either end of the deck, it goes down to 1 in 13." What do you mean by this?
It would be 1/2704 if for example the magician marks in advance one card in a given position, and then the spectators happen to choose precisely that card and that position.
A simple but often baffling related example is: I remove one card from a regular deck of 52 (without showing it). What is the probability that a randomly picked card from the rest of the deck is the Ace of Hearts?
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Re: Berglas Effect .... WOW...
Carlo Morpurgo wrote: I am not sure about "counting from either end of the deck, it goes down to 1 in 13." What do you mean by this?
The chance is 1/52 for a chosen card to be at a chosen number. If the performer is allowed to count from either end of the deck, then it's 1/26 (since the chosen number can count to two different positions, depending on which end of the deck we start at). The odds could be cut in half again, to 1/13, by allowing the card to be one off (e.g. for the number 8, you'd count off 8 cards and the chosen card would then be revealed at the top of the deck...really the 9th card).

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Re: Berglas Effect .... WOW...
Bob Coyne wrote:The chance is 1/52 for a chosen card to be at a chosen number. If the performer is allowed to count from either end of the deck, then it's 1/26 (since the chosen number can count to two different positions, depending on which end of the deck we start at).
I agree with this, IF the magician performs both counts. If the magician is allowed one count only, then it's still 1/52, again by the same principle that once you choose where to count from
each card has the same chance of being in a given spot.
I doubt that allowing two ways of counting is reasonable for any effect, however.
Carlo
Re: Berglas Effect .... WOW...
Originally posted by Bob Coyne: "The chance is 1/52 for a chosen card to be at a chosen number".
Correct me if I'm wrong but I believe that for a chosen card to be at a chosen number, the odds are MUCH higher than that. 52 cards to chose from located at 52 positions to chose from.......I'm not a statistical guy, but it's much more that a 1 in 52 shot. Right???
Correct me if I'm wrong but I believe that for a chosen card to be at a chosen number, the odds are MUCH higher than that. 52 cards to chose from located at 52 positions to chose from.......I'm not a statistical guy, but it's much more that a 1 in 52 shot. Right???

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Re: Berglas Effect .... WOW...
Carlo Morpurgo wrote:Bob Coyne wrote:The chance is 1/52 for a chosen card to be at a chosen number. If the performer is allowed to count from either end of the deck, then it's 1/26 (since the chosen number can count to two different positions, depending on which end of the deck we start at).
I agree with this, IF the magician performs both counts. If the magician is allowed one count only, then it's still 1/52, again by the same principle that once you choose where to count from
each card has the same chance of being in a given spot.
I doubt that allowing two ways of counting is reasonable for any effect, however.
Carlo
The spectator wouldn't be told until after the card and number are named how to pick up the deck (face up or down). So there would only be one count, but it would be magician's choice, cutting the odds in half. Others on this thread have suggested the same earlier.

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Re: Berglas Effect .... WOW...
Randy wrote:Originally posted by Bob Coyne: "The chance is 1/52 for a chosen card to be at a chosen number".
Correct me if I'm wrong but I believe that for a chosen card to be at a chosen number, the odds are MUCH higher than that. 52 cards to chose from located at 52 positions to chose from.......I'm not a statistical guy, but it's much more that a 1 in 52 shot. Right???
If the magician were to *predict* the card and position, then the probability of the spectator choosing those same values would be 1/2704 (1/52 times 1/52). So maybe that's what's confusing you.
In this case, however, the magician isn't predicting either the card or position, so all that matters is that the chosen card be at the chosen position. Since there are 52 positions, the chance that the chosen card is at any given position is 1/52. For example, let's say the spectator chooses AS and then randomly picks a position. One out of those 52 positions will give a match, so the chances of the AS of being at the chosen position is 1/52. The same is true for any chosen card. So no matter what card they choose, the probability is 1/52.

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Re: Berglas Effect .... WOW...
Bob Coyne wrote:The spectator wouldn't be told until after the card and number are named how to pick up the deck (face up or down). So there would only be one count, but it would be magician's choice, cutting the odds in half. Others on this thread have suggested the same earlier.
Maybe I am missing something. I am assuming that the magician
does not know anything about the cards in the deck. In this case, it does not matter one bit which side he chooses, or how he chooses it (tossing a fake coin even): the probability is still 1/52.
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Re: Berglas Effect .... WOW...
Carlo Morpurgo wrote:Maybe I am missing something. I am assuming that the magician
does not know anything about the cards in the deck. In this case, it does not matter one bit which side he chooses, or how he chooses it (tossing a fake coin even): the probability is still 1/52.
Carlo
Ah yes, I'm making the implicit assumption that the magician is using a memorized deck and knows the position of every card. So after hearing the card/number the magician can give the appropriate instructions to make it work out (and thereby improve the odds). If the deck is in unknown order, you're right, then none of that will work.

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Re: Berglas Effect .... WOW...
Bob Coyne wrote:
Ah yes, I'm making the implicit assumption that the magician is using a memorized deck and knows the position of every card. So after hearing the card/number the magician can give the appropriate instructions to make it work out (and thereby improve the odds). If the deck is in unknown order, you're right, then none of that will work.
Ah! in that case I agree :)
Carlo
Re: Berglas Effect .... WOW...
The odds could be cut in half again, to 1/13, by allowing the card to be one off (e.g. for the number 8, you'd count off 8 cards and the chosen card would then be revealed at the top of the deck...really the 9th card).
Bob, the odds would be cut to 1/13 by allowing the card to be one off. This is for a count in one direction.
Now, wouldn't the odds again be cut in half (1 in 6.5) if you allow for another count in the opposite direction?
I'm no expert on statistics, nor a mathematician so if this is incorrect please excuse my ignorance.
Re: Berglas Effect .... WOW...
Interesting topic.

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 Location: New York, NY
Re: Berglas Effect .... WOW...
Edward wrote:Now, wouldn't the odds again be cut in half (1 in 6.5) if you allow for another count in the opposite direction?
The chances would be about 3 in 26 since any of three positions (the actual number and one before and after, when those exist) could be used for the number. It seems to me, however, to be more awkward to stop one early...not sure how that could be done/justified in a natural way, whereas it would more easily pass to deal the named number to the table and then reveal the next card.
Assuming the 1/13 approach (counting from either end and oneoff in the higher number direction) then the odds could be reduced further by optionally asking the spectator to cut the deck once or twice. The chance of the card getting into position somewhere in that process would be (1  12/13 * 12/13 * 12/13) = 22%.
Re: Berglas Effect .... WOW...
Bob Coyne wrote:Cohiba wrote:If you knew what card they were going to choose, the odds would be 1 in 52 that they would get the right number. So let's say for sake of argument they are going to choose the AS. You place it in the deck at number 20. Your odds are 1 in 52 of them choosing 20. However, let's just say you defy the odds and they choose 20. Now, they forget what card they were supposed to pick. They now have 52 more options for a card. The odds are no longer 1 in 52.
This is the same sort of problem as the odds of rolling doubles with dice. It doesn't matter what the first number is  you only care that the second number is the same as the first. Hence the chance is 1 in 6. It's not 1 in 36. Same with ACAAN, it's irrelevant what card they pick. It only matters that it's at the given position.
If you don't believe it, then try ACAAN with a packet of 4 aces and you'll see that a randomly selected ace will be at a randomly selected position one time in 4.
Bob:
I'm formerly a math guy, but am rusty on my statistics. Thanks for this discussion, it's good to clear out the cobwebs.
Play along with me here, because I question your dice proposal as well.
Are you saying that rolling two dice at once yields a different probability (of both getting a six, for example) than throwing a single die twice? If you throw a single die twice, the odds are 1/36 that both throws are a 6. It seems the same to me if you throw two dice at the same time. Logically, this makes sense as well. Do you think the odds of getting a 6 with one die are the same as throwing 10 dice at once where they all turn up 6's? The odds of all 10 dice ending in 6 is astronomically higher than just one die (1/6).
Re: Berglas Effect .... WOW...
Maybe this will help clarify.
Assume you have a two card packet consisting of an ace and a 2. These two cards are shuffled.
If a spectator is told to name a card (ace or 2), then a number/position (1 or 2), there are 4 potential ways this could go:
1  ace at position 1
2  ace at position 2
3  two at position 1
4  two at position 2
Now it is correct that once a card is named, it has a 1 in 2 chance of being at a particular location. However, you don't know which card will be named. So I think both sides of the argument are correct in their own way:
Any card has a 1 in 52 chance of being at a particular location in the deck. However, there are 2704 possible combinations that could be called out by a spectator.
For the above example, when the ace is in position one, the 2 is in position two. This packet covers two of the four combinations, and a second packet would cover the other two combinations. Therefore, you would only need two packets to cover all four combinations.
Similarly, with ACAAN, each deck contains 52 combinations, so 52 decks are what are needed to cover all 2704 combinations.
Assume you have a two card packet consisting of an ace and a 2. These two cards are shuffled.
If a spectator is told to name a card (ace or 2), then a number/position (1 or 2), there are 4 potential ways this could go:
1  ace at position 1
2  ace at position 2
3  two at position 1
4  two at position 2
Now it is correct that once a card is named, it has a 1 in 2 chance of being at a particular location. However, you don't know which card will be named. So I think both sides of the argument are correct in their own way:
Any card has a 1 in 52 chance of being at a particular location in the deck. However, there are 2704 possible combinations that could be called out by a spectator.
For the above example, when the ace is in position one, the 2 is in position two. This packet covers two of the four combinations, and a second packet would cover the other two combinations. Therefore, you would only need two packets to cover all four combinations.
Similarly, with ACAAN, each deck contains 52 combinations, so 52 decks are what are needed to cover all 2704 combinations.