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The Magic Cafe Forum Index » » Latest and Greatest? » » A.A.C.A.A.N by Asi Wind (48 Likes) Printer Friendly Version

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theothermentalist
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Quote:
On May 4, 2016, Mikey7938 wrote:
Has anyone else encountered that when the 9D is selected the card ends up one position short? It seems I need to add one card to the stack to get it into the correct position... anyone else have this happen? Confused in Pasadena.


Mikey,

I do think this is a problem with the teaching on the "Three Card Routines" DVD. If the sum of the stack number of the named card and the number of cards you have to shift is over 52 then you can't do that extra special retention of the 9D Asi teaches.

**ninja edit**
The above is still true, but doesn't answer Mikey's question. Sorry.
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Steven Conner
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Quote:
On May 16, 2017, theothermentalist wrote:
Quote:
On May 4, 2016, Mikey7938 wrote:
Has anyone else encountered that when the 9D is selected the card ends up one position short? It seems I need to add one card to the stack to get it into the correct position... anyone else have this happen? Confused in Pasadena.


Mikey,

I do think this is a problem with the teaching on the "Three Card Routines" DVD. If the sum of the stack number of the named card and the number of cards you have to shift is over 52 then you can't do that extra special retention of the 9D Asi teaches.

**ninja edit**
The above is still true, but doesn't answer Mikey's question. Sorry.


Here is Asi's answer: To be honest if somebody says nine of diamonds I skip the whole Acaan plot because nothing would be stronger than saying I have placed the card first and show it. However if you still want to perform any card at any number just don't slide the card in and perform the cut without sliding one card in

Best

Steve
"The New York Papers," Mark Twain once said,"have long known that no large question is ever really settled until I have been consulted; it is the way they feel about it, and they show it by always sending to me when they get uneasy. "
wandering star
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A great trick - but it definitely needs practice in performing it fluently due to various simultaneous actions you need to master. Still there are ocassions when the best thing happens and when spectators do all the work and everything simply fits. I cannot name the exact mathematics behind it in terms of propability calculation but this happened to me not only once - funnily always with the ace of spades with the mnemonica.
Last Laugh
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The chances of the spectator hitting is 1 in 52.
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wandering star
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Quote:
On Jun 23, 2017, Last Laugh wrote:
The chances of the spectator hitting is 1 in 52.


... spontaneously I would say it's not - because you do not just say "select a card and I tell you what card you are going to choose" (this would mean 1 in 52, doesn't it?) - but in addition (to keep in terms of maths ;-) you predict at what position the card will be put back in the fan - so this would mean 1/52 x 1/52 if I hit this right. And to make it even more complicated: it should even be less possible to have - let's assume a number of hundred times me perfroming Asi's routine - to have someone name the ace of spades plus fitting number... I am curious to know the solution of the stochastic problem knowing for sure that I am not a mathematical genius
wandering star
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Quote:
On Jun 23, 2017, Last Laugh wrote:
The chances of the spectator hitting is 1 in 52.


..just realising your MC-name - funny though - maybe the last laugh is on you?
Dave the Knave
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Quote:
On Jun 24, 2017, wandering star wrote:
Quote:
On Jun 23, 2017, Last Laugh wrote:
The chances of the spectator hitting is 1 in 52.


... spontaneously I would say it's not - because you do not just say "select a card and I tell you what card you are going to choose" (this would mean 1 in 52, doesn't it?) - but in addition (to keep in terms of maths ;-) you predict at what position the card will be put back in the fan - so this would mean 1/52 x 1/52 if I hit this right. And to make it even more complicated: it should even be less possible to have - let's assume a number of hundred times me perfroming Asi's routine - to have someone name the ace of spades plus fitting number... I am curious to know the solution of the stochastic problem knowing for sure that I am not a mathematical genius

Last Laugh is right. The math isn't complicated; it's a 1/52 proposition. The trick is Any Card At Any Number, not I Predicted A Number You Will Choose and I Also Predicted The Card You Will Choose.
tomd
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Not this again...: it's 1/52, nothing more.

If they can state any card they wish, then that card can only be at any one of 52 positions.. meaning 1/52 times it will be a miracle.
wandering star
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At least it is a great trick and seems impossible to do for the audience (and I prefer possibilities of being 100% flexible in many ways in comparison to lots of other ACAAN plots) - ... don't want to have the last word on the maths-issue, but I guess you are right (..last word)
Mr. Mindbender
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Of course it's 1/52. Simple proof is that all I need is 52 decks of cards to have every single outcome - thus 1 in 52!
Last Laugh
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While it is indeed 1 in 52, the fact that it seems like it would be much more can definitely be used in performance. I love starting my ACAAN by talking about 52! (factorial) and talking about how a truly shuffled deck is likely in an order that has never existed before in the history of the universe.
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Mr. Mindbender
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Yes, I've always been enamored with that fact(orial!).
Last Laugh
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It makes something seemingly simple - a shuffled deck of cards - into a truly unique cosmic occurrence. And that's before the effect even starts!
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Charles Adams
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>>Of course it's 1/52. Simple proof is that all I need is 52 decks of cards to have every single outcome - thus 1 in 52!

This assumes you know which card the spectator was going to choose in advance. You would need 52 decks for each of the 52 cards in order to cover all possible cards at all possible locations.
EndersGame
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Another performance video to share with this thread.

This is David Blaine performing the routine at a private birthday party.



Link: https://www.youtube.com/watch?v=4JjVnaRHcjk
ZakRPW
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I’m having a problem with this trick.

What if the spectator asks for a low-stack card, let’s say 3 clubs, and asks for it to be in a high-stack position, let’s say position 48?
8 cards need shifting.
The problem here is that using the 9d in the way explained in the video results in the selected card being off by 1 (in position 47, as the 9d stays in its original place). NOT using the 9d in the way explained makes the handling look suspicious because you have no guide for the secret move.

Any solutions?
langston3711
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ZakRPW,

Watch or read the explanation again. When the number called for is higher than the number of the card there is a totally different way to produce the deck without using the 9 like you're thinking. The flap is then cover for the slights. Hope this helps, trying to keep it vague, here.
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Trentonmatthew
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The other day I felt lazy and didn’t feel like doing the math in my head. I hauled out my iPhone calculator and did the calculations on there and stated that I was “re-programming the deck” the reactions I received were the same or better then when I preform without a calculator. Just a thought...
niva
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Quote:
On Apr 28, 2018, EndersGame wrote:
Another performance video to share with this thread.

This is David Blaine performing the routine at a private birthday party.



Link: https://www.youtube.com/watch?v=4JjVnaRHcjk


Did I hear correctly? He told him to turn over the 39th card. HEHE!!
Yours,

Ivan
Magic1
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Does anyone have any tips/resources/apps/courses/drills for getting comfortable with speed arithmetic?
Thanks!
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