ACAAN ROUTINE | ANY CARD AT ANY NUMBER | DAVID BERGLAS EFFECT | BEST Any card at any number magic 

হাহাহা...hahaha

I didn’t understand how he force the card can you explain me how he do it pls?

You're welcome

I think it would create a pattern of JQK and it would diminish the result

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

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.

You could get them to deal Into your hand and double lift to show the queen

traduree per conto proprio - fai da ti - fallo da solo - capisci?

TAROTAI grazie ma già ho capito 3 mesi fa ☺️

Sorry, I don't no

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

true

That's why he says its not the perfect answer right at the beginning.

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.

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

No, this is a hack solution.

Because there are 4 suits C H S D

CHaSeD

The first 10 minutes was good not gonna lie Then