Jimmy Alan - I think that is your name... thanks you so much for this super ACAAN gift. It is so magical once you learn to perform it with confidence... So cool... Thank you for sharing this wonderful trick procedure.
I've got a Jamie Allen DVD that my parents got me from a Cruise Ship show of Jamie's and it's a great DVD. I've used and modified a few of his card tricks. Really nice. I do like how he does big illusions and smaller tricks. Lovely guy!
Not the genuine Berglas, but still BRILLIANT! Very well thought, prepared and performed. Warmest congrats! P.S. I bet all the bitter comments are coming from colleagues of yours with severe bile infection, hahahahaha.
Excellent explanation...gotta hand it to you mate your card control is tops! Btw I'm a guitarist who loves watching magic..and loads of it..and you would also make an excellent guitarist with that dexterity! I've been researching the Berglas effect most of the day, and you have come closest to it
In my case, after they choose the colour of the card, to force the card's suit I use to say "which one do you dislike? Hearts or diamonds?" if they say they dislike hearts (which is the suit that I want to force) I simply proceed. If they say they dislike diamonds, I'd say "so, okay let's eliminate diamonds because you dislike it." And then proceed to the next phase.
Would it not make sense to have the J, Q, K in the x3, x4,x5 positions respectively? You then have multiple outs and you can direct them to land on the magicians force regardless of the number picked. Hell it doesn't even have to be all the court cards just the same three.
Bonjour je suis.... français et j'aimerais juste ....comprendre s'il vous plaît ce n'est qu'un caan et non acaan j'ai une petite chaîne de magie je suis pas avec mais j'en fait et j'ai pas compris le tour si vous pourriez m'aiguiller s'il vous plaît
Bonjour je suis.... français et j'aimerais juste ....comprendre s'il vous plaît ce n'est qu'un caan et non acaan j'ai une petite chaîne de magie je suis pas avec mais j'en fait et j'ai pas compris le tour si vous pourriez m'aiguiller s'il vous plaît
Kleqy they won’t pick 5 because they need fingers up on both hands, but let’s say they put 5 fingers up on one hand and 2 on the other, then you say ok so 25 or 52? If they say 25 then great just do the old card stuck in the box thing when pulling them out and just throw it on top and force the queen of hearts. If they say 52 you could explain that you need a number between 1 and 52 so you steer them towards 25. Hope that helps 👍
The odds are not in the thousands though. They are simply 1 out of 52. Any card has 1/52 chance to be at any position. Once a card is named, there is 1/52 chances that the named number will correspond. The odds are not 52 times 52.
Are you that stupid and arrogant? Not everyone will (for example) name the Ace of Spades at (for example) position 13, so your reply is nonsensical and rubbish Different cards and different nu bers are named EVERY TIME, and whilst some cards and numbers chosen may be the same but not often mentioned by different spectators, the odds are NOT 1 in 52 FACT !!!!!!
the odds are not 1/52 if the question is to choose a specific card and a specific number, then it would be 1/52 multiplied by 1/52. If the question is what is the chance that a queen is at position 10, for example, then it is 1/52 like you said.
@@Bidoof504 Actualy no, That is a common error to make. Choose a random number, any number. Go. Now, Whatever the number you choose, you are now going to have one single position in the deck. That step is done. There is no probability of anything involved yet, you just chose a number. Say you choose 23. Once you have choosen that number, there are 52 cards you can name after that. One of them will necessarily be the right one, so now there is a probability, you have 1/52 chance that the card will match your number. That is true whatever number you choose. "If the question is what is the chance that a queen is at position 10, for example, then it is 1/52 like you said." Exactly. Every scenario of ACAAN is exactly that. Whatever the card, whatever the position. But the illusion that the chaces are 1/52 times 1/52 is very good, so you can work with that, very few people will figure out that the chances are really 1/52. Unless it is a mathematician, then he will know instantly and will make himself a pleasure of correcting you. I personnaly don't stress that the odds are astronomical, because some people will figure it out and then your trick gets dimished. Whereas if you just don't say a thing, the feeling of impossibility and astronomical coincidence remains. The event itself is a mindfuck, without mentionning odds. But if you have never been called out on it and it feels right doing it like you do, go ahead, that is whats important. Anyways very few people will get it, and maybe even mathematicaly clever people will get fooled, cause the probability illusion is quite good and people dont tend to calculate things they just experience the magic.
@@metatrix4251 your right and thanks for the explanation. the chance of a specific card being at a specific number is always 1/52. but what I was trying to say was that the chance that a spectator names the number you want and a card you want is 1/52 * 1/52. (assuming no forces used)
@@Bidoof504 OK, I see. Then for that probability to exist, it would need to be a prediction. Like write a card and a number on a peice of paper, then tell a spectator to choose a card and a number between 1 and 52. Then if their choice matches your prediction, that would be a probability of 1/2704. But as soon as there is no prediction on a peice of paper or something similar, meaning as soon as the reveal is just to show the card in the deck at the mentioned position, then the chances are 1/52. Another thing you could do to make it 1/2704, is have the named card match the named number AND be the only face down card. That would work.
I didn't watch the whole thing honestly because I had the sense it wasn't for me. But you say something early on which I think is wrong. The chances of them picking the card at the number is 1 in 52. To see this, just imagine any card. Once it's chosen, there are only 52 possibilities for where it is.
I doubt this is right but I would say hold up 'any fingers on each hand behind your back'. They hold up 5. I would say 'I did say fingers, not thumbs so we will class that as 4!!!!'
@@stonewallbaron09 I'm not a math expert but I fail to see how the odds are not 1 in 52. If two cards were involved and both had to be in specific positions, then I get that exponents will be involved, making the odds far less. But if it's just 1 card, then that card has got to be in one of the 52 positions, thereby making the probability 1 in 52.
@@gauravtee You are right, it's obviously 1/52. However, with simple variations, a TRUE (unforced) any card any number can be reduced to 1:13 by counting from either the top or bottom as needed, and stopping AT, or ONE AFTER the selected number, as needed. Anyway, the odds are far from "thousands" as this video pretends.
it's not a thousands of possibilities. the chance of being correct is 1 out of 52. i don't say that the trick is easy... it's magnificent, but if you take a card random from the deck, it doesn't matter where you take the card, it is still 1 chance out of 52. maybe it makes it more difficult and more possibilities for the magician, but not as the spectator sees it
You are very wrong. Guessing the card is a 1 in 52. 1 in 52 for the proper card. 1 in 52 for guessing what position you were going to say. But to get both, you use 6th grade math. 1 in 52 times 1 in 52. Thus, you have a 1 in 2,704 chance of the exact card ending up in the guessed position.
@@RandomBJJGuy If the magician before the show should write down on a paper which position and which card the audience would pick, then it's 1/2704. but if the public picks a card and he get a position from the audience, then there is 1/52 chance it would be right. You can try to explain which cards are the 2703 that would fail the trick. It's 51 out of those 2703 which is also being correct. That is 52/2704=1/52. Maybe you must prepare as an magician for the 2704. possibilities that the audience chooses. But I still think that chosing 1 card out of 52 is a random chance of 1/52 being correct. The difficulty in the trick is how the magician makes 100% being correct
@@stefanjohansson6316 the probability of it working is 100% if you do it right. The 1/2704 represents the PERCEPTION of what is happening. The person picking the card is a separate 1/52 from the position. Im reality it's a force so its a 1/1 chance. In other words, for a person to pick an exact card that's a 1/52 chance. For them to then pick it's exact location is a separate 1/52.
@@RandomBJJGuy but if you have 1/52 to get the card correct and 1/52 to get the position correct. then it must be 51/52 to get the card wrong and 51/52 to get the position wrong. that is, out of those who are both wrong card and wrong position there will be a chance of getting the right card in the right position. there are 2704 cases with the same probability out of them there will be 52 cases where the card will be correct. 52/2704 =1/52. else you must be able to pick out 2703 cases that doesn't contain the correct card. If I understand you correct, then you say that the perception is that you have to guess card correct (1/52) and position correct (1/52). then you can like 6th graders multiply those probabilities, but that is iff they're independent of each others... but they are not indepedent and then you can't multiply them I don't know what the Perception ought to be, but getting a card at a random position, and guessing it would to me always be 1/52
@@stefanjohansson6316 you're right, and I'm glad to see I'm not the only one here who thinks that the odds are 1 in 52. The selected card has to be in one of the 52 positions, and if it's not in one, it's in another, but there are only 52 possibilities. That is to say, if you were to do a random ACAAN for a card, say the 4ofC, you'd expect to guess the position of the card correctly 1 out of every 52 times. At least that's what I think
Hello friend, I would like to ask you as a great favor, if you could upload the videos with subtitles in Spanish, it would be of great help, thanks for your understanding, it is difficult because many of us do not fully master the English language and perhaps it would help you to add more subscribers , A greeting and thanks again
It never works with my native language. Language is a mindset, and the magician skillfully plays with English speakers' habits. In Polish it simply does not work. As soon as you say pick numbers or pitch they say numbers you say so that leaves me pitches and you are IMMEDIATELLY called out. We are very sensitive for word tricks.
This is probably the worst ACAAN I've ever seen and the audio makes my ears bleed. It's a 32 minute video telling you to use the magician's choice WAY more than you should ever use it in one trick.
Jimmy Alan - I think that is your name... thanks you so much for this super ACAAN gift. It is so magical once you learn to perform it with confidence... So cool... Thank you for sharing this wonderful trick procedure.
I've got a Jamie Allen DVD that my parents got me from a Cruise Ship show of Jamie's and it's a great DVD. I've used and modified a few of his card tricks. Really nice. I do like how he does big illusions and smaller tricks. Lovely guy!
Not the genuine Berglas, but still BRILLIANT! Very well thought, prepared and performed. Warmest congrats! P.S. I bet all the bitter comments are coming from colleagues of yours with severe bile infection, hahahahaha.
হাহাহা...hahaha
Excellent explanation...gotta hand it to you mate your card control is tops! Btw I'm a guitarist who loves watching magic..and loads of it..and you would also make an excellent guitarist with that dexterity! I've been researching the Berglas effect most of the day, and you have come closest to it
Congratulations brother, Thank you so much. Request to teach us .we are very interested to learn mind reading card tricks. Suresh and family. Odisha.
I do it a little different, but I use the concept you teach. Thanks!
Why is the sound cut off at 25.04 ? After which the video is soundless.
Hi Tanjeeb the video stopped the voice over after 25 minutes I was watching with no sound. Thanks
Tout le monde s'en bas les...'🤷🏼♂️🤣c'est dommage
This is awesome, but the audio shuts off about 3/4 of the way through the video. What's wrong?
In my case, after they choose the colour of the card, to force the card's suit I use to say "which one do you dislike? Hearts or diamonds?" if they say they dislike hearts (which is the suit that I want to force) I simply proceed. If they say they dislike diamonds, I'd say "so, okay let's eliminate diamonds because you dislike it." And then proceed to the next phase.
Bonjour je suis français et j'aimerais connaître svp Merci
Oui c'est ce qu'il faut fair Sinon 😉
Sa s'appelle un forçage patéo 😊
Donc ce n'est pas un acaan mais un caan ??!!
@@Hagstone. Oui. cest caan. Cest nes pas acaan
Would it not make sense to have the J, Q, K in the x3, x4,x5 positions respectively? You then have multiple outs and you can direct them to land on the magicians force regardless of the number picked. Hell it doesn't even have to be all the court cards just the same three.
I think it would create a pattern of JQK and it would diminish the result
LEGEND 🖤🔥👍
Bonjour je suis français j'aurais bien aimé le comprendre
Berglas Effect is mentioned only. ("To explain would not be correct") Title is misleading. Very good explanation though.
Bonjour je suis.... français et j'aimerais juste ....comprendre s'il vous plaît ce n'est qu'un caan et non acaan j'ai une petite chaîne de magie je suis pas avec mais j'en fait et j'ai pas compris le tour si vous pourriez m'aiguiller s'il vous plaît
Bonjour je suis.... français et j'aimerais juste ....comprendre s'il vous plaît ce n'est qu'un caan et non acaan j'ai une petite chaîne de magie je suis pas avec mais j'en fait et j'ai pas compris le tour si vous pourriez m'aiguiller s'il vous plaît
This trick is amazing I fooled like 20 people with this trick just doing it once
when someone picked 5 what did u say? please help
Kleqy they won’t pick 5 because they need fingers up on both hands, but let’s say they put 5 fingers up on one hand and 2 on the other, then you say ok so 25 or 52? If they say 25 then great just do the old card stuck in the box thing when pulling them out and just throw it on top and force the queen of hearts. If they say 52 you could explain that you need a number between 1 and 52 so you steer them towards 25. Hope that helps 👍
So they don't pick 5 but I performed it with many people and I asked them all then I toofk the fingers of persons I needed
Or just say it's 2 and 5 that's 25 and Don't give them the chance to choose
Wait so what happened to the 50s again ?
Wow its magic♦️♠️♥️♣️😎
The odds are not in the thousands though.
They are simply 1 out of 52.
Any card has 1/52 chance to be at any position.
Once a card is named, there is 1/52 chances that the named number will correspond.
The odds are not 52 times 52.
Are you that stupid and arrogant?
Not everyone will (for example) name the Ace of Spades at (for example) position 13, so your reply is nonsensical and rubbish
Different cards and different nu bers are named EVERY TIME, and whilst some cards and numbers chosen may be the same but not often mentioned by different spectators, the odds are NOT 1 in 52
FACT !!!!!!
the odds are not 1/52 if the question is to choose a specific card and a specific number, then it would be 1/52 multiplied by 1/52. If the question is what is the chance that a queen is at position 10, for example, then it is 1/52 like you said.
@@Bidoof504 Actualy no, That is a common error to make.
Choose a random number, any number. Go.
Now, Whatever the number you choose, you are now going to have one single position in the deck. That step is done. There is no probability of anything involved yet, you just chose a number. Say you choose 23. Once you have choosen that number, there are 52 cards you can name after that. One of them will necessarily be the right one, so now there is a probability, you have 1/52 chance that the card will match your number. That is true whatever number you choose.
"If the question is what is the chance that a queen is at position 10, for example, then it is 1/52 like you said." Exactly. Every scenario of ACAAN is exactly that. Whatever the card, whatever the position. But the illusion that the chaces are 1/52 times 1/52 is very good, so you can work with that, very few people will figure out that the chances are really 1/52. Unless it is a mathematician, then he will know instantly and will make himself a pleasure of correcting you.
I personnaly don't stress that the odds are astronomical, because some people will figure it out and then your trick gets dimished. Whereas if you just don't say a thing, the feeling of impossibility and astronomical coincidence remains. The event itself is a mindfuck, without mentionning odds.
But if you have never been called out on it and it feels right doing it like you do, go ahead, that is whats important. Anyways very few people will get it, and maybe even mathematicaly clever people will get fooled, cause the probability illusion is quite good and people dont tend to calculate things they just experience the magic.
@@metatrix4251 your right and thanks for the explanation. the chance of a specific card being at a specific number is always 1/52. but what I was trying to say was that the chance that a spectator names the number you want and a card you want is 1/52 * 1/52. (assuming no forces used)
@@Bidoof504 OK, I see. Then for that probability to exist, it would need to be a prediction. Like write a card and a number on a peice of paper, then tell a spectator to choose a card and a number between 1 and 52. Then if their choice matches your prediction, that would be a probability of 1/2704.
But as soon as there is no prediction on a peice of paper or something similar, meaning as soon as the reveal is just to show the card in the deck at the mentioned position, then the chances are 1/52.
Another thing you could do to make it 1/2704, is have the named card match the named number AND be the only face down card. That would work.
Wonderful trick amazing
you need to reload the video the end of it has no sound -which is more than a little annoying. Cheers
Don't really need the sound at the end
We get the picture and I'm going to use that technic. Thanks man!
You're welcome
@@TanjeebKhan Who's the magician talking in this video?
You're a genius
Please upload more magic vedio bro
Haha brilliant about removing the jokers! But wait, I thought you're not supposed to give away magician's secrets?
Come posso fare per tradurre il video in italiano?
traduree per conto proprio - fai da ti - fallo da solo - capisci?
TAROTAI grazie ma già ho capito 3 mesi fa ☺️
This is really the best ACAAN
THANK U SO MUCH
Card forcing is amazing
I didn’t understand how he force the card can you explain me how he do it pls?
cool cool music tv theme song jason born that trickbusters copperfield youtube videos use also
Can you made a if/or key for that? (like maps have a key)
I didn't watch the whole thing honestly because I had the sense it wasn't for me. But you say something early on which I think is wrong. The chances of them picking the card at the number is 1 in 52. To see this, just imagine any card. Once it's chosen, there are only 52 possibilities for where it is.
NO AUDIO starting @25:44
Sounds disappears at 25:03.
I only have 1 joker in my new deck :( And I don't have any other deck with a card box can anyone help?
Sorry, I don't no
How can we get rid of 5? How do I control the card to 34 if the spectator shows 35?
You could get them to deal Into your hand and double lift to show the queen
The audio cuts out at 25:04. What did you say about 32 and what do you do if someone holds up 5 fingers?
I doubt this is right but I would say hold up 'any fingers on each hand behind your back'. They hold up 5. I would say 'I did say fingers, not thumbs so we will class that as 4!!!!'
Andrew Bickham Thanks
For five fingers and one card to top from the bottom.
And for 32 remove those jokers right
Bonjour hello pdf français 🇫🇷? Please
30:00 👂🏼pas de son ?
Good trick
But it’s not the genuine ACAAN.
true
That's why he says its not the perfect answer right at the beginning.
Me going on a cruise and seeing this guy is the magician: ಠ_ಠ
The chances are definitely not higher than a royal flush.. it’s 1 in 52 to name a card and the number it will be at.
No, it’s not. The odds of picking a card and finding it at a specific number is 1 in 2,703
@@stonewallbaron09 I'm not a math expert but I fail to see how the odds are not 1 in 52. If two cards were involved and both had to be in specific positions, then I get that exponents will be involved, making the odds far less. But if it's just 1 card, then that card has got to be in one of the 52 positions, thereby making the probability 1 in 52.
@@gauravtee You are right, it's obviously 1/52. However, with simple variations, a TRUE (unforced) any card any number can be reduced to 1:13 by counting from either the top or bottom as needed, and stopping AT, or ONE AFTER the selected number, as needed. Anyway, the odds are far from "thousands" as this video pretends.
it's not a thousands of possibilities. the chance of being correct is 1 out of 52. i don't say that the trick is easy... it's magnificent, but if you take a card random from the deck, it doesn't matter where you take the card, it is still 1 chance out of 52. maybe it makes it more difficult and more possibilities for the magician, but not as the spectator sees it
You are very wrong. Guessing the card is a 1 in 52. 1 in 52 for the proper card. 1 in 52 for guessing what position you were going to say. But to get both, you use 6th grade math. 1 in 52 times 1 in 52. Thus, you have a 1 in 2,704 chance of the exact card ending up in the guessed position.
@@RandomBJJGuy If the magician before the show should write down on a paper which position and which card the audience would pick, then it's 1/2704. but if the public picks a card and he get a position from the audience, then there is 1/52 chance it would be right. You can try to explain which cards are the 2703 that would fail the trick. It's 51 out of those 2703 which is also being correct. That is 52/2704=1/52. Maybe you must prepare as an magician for the 2704. possibilities that the audience chooses. But I still think that chosing 1 card out of 52 is a random chance of 1/52 being correct. The difficulty in the trick is how the magician makes 100% being correct
@@stefanjohansson6316 the probability of it working is 100% if you do it right. The 1/2704 represents the PERCEPTION of what is happening. The person picking the card is a separate 1/52 from the position. Im reality it's a force so its a 1/1 chance.
In other words, for a person to pick an exact card that's a 1/52 chance. For them to then pick it's exact location is a separate 1/52.
@@RandomBJJGuy but if you have 1/52 to get the card correct and 1/52 to get the position correct. then it must be 51/52 to get the card wrong and 51/52 to get the position wrong. that is, out of those who are both wrong card and wrong position there will be a chance of getting the right card in the right position. there are 2704 cases with the same probability out of them there will be 52 cases where the card will be correct. 52/2704 =1/52. else you must be able to pick out 2703 cases that doesn't contain the correct card.
If I understand you correct, then you say that the perception is that you have to guess card correct (1/52) and position correct (1/52). then you can like 6th graders multiply those probabilities, but that is iff they're independent of each others... but they are not indepedent and then you can't multiply them
I don't know what the Perception ought to be, but getting a card at a random position, and guessing it would to me always be 1/52
@@stefanjohansson6316 you're right, and I'm glad to see I'm not the only one here who thinks that the odds are 1 in 52. The selected card has to be in one of the 52 positions, and if it's not in one, it's in another, but there are only 52 possibilities. That is to say, if you were to do a random ACAAN for a card, say the 4ofC, you'd expect to guess the position of the card correctly 1 out of every 52 times. At least that's what I think
Alas, the video gets bad at two third. Nobody has noticed it?
its okay. kinda sketchy for you average spectator
Donc ce n'est pas un acaan mais un caan ??!!
Hello friend, I would like to ask you as a great favor, if you could upload the videos with subtitles in Spanish, it would be of great help, thanks for your understanding, it is difficult because many of us do not fully master the English language and perhaps it would help you to add more subscribers , A greeting and thanks again
No sound
2020 anyone
⬇️
Why u uploadet it?
This is a magican secret!
No, this is a hack solution.
In spanish please....
It never works with my native language. Language is a mindset, and the magician skillfully plays with English speakers' habits. In Polish it simply does not work. As soon as you say pick numbers or pitch they say numbers you say so that leaves me pitches and you are IMMEDIATELLY called out. We are very sensitive for word tricks.
Any card ? Any number ? Nah - More like pick the Q at any pf the 4 numbers
Terrible and loud music.
Then "S" appears third in the word "CHASED" 🤣🤣
Because there are 4 suits C H S D
CHaSeD
This is probably the worst ACAAN I've ever seen and the audio makes my ears bleed.
It's a 32 minute video telling you to use the magician's choice WAY more than you should ever use it in one trick.
The first 10 minutes was good not gonna lie
Then