Berglas Effect .... WOW...

[Cohiba]

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.

In this example, you are singling out the AS. I agree that the odds of the AS being at the chosen position are 1/52. The problem is, you don't know that the AS is going to be chosen. Once a card is chosen, you have 1/52 odds. Until that point, there are still 2704 possible combinations. The magician "predicting" is the same as the spectator randomly calling out a card and number.

[Cohiba]

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.

-Jim

Jim, each deck has 52 combinations, so 52 decks of 52 combinations = 2704 total combinations.

Even if a magician knew which card was going to be named, he'd still need 52 decks to cover all 52 positions that could be named. Then, the order of all the indifferent cards wouldn't matter, so long as the named card rotated through all 52 positions.
As I mentioned above in the 2 card packet example, by stacking all of the decks, each deck can cover 52 of the total combinations, so that 52 decks can cover all 2704 total combinations.

[Denis Behr]

It is certainly right that there are 52*52 combinations.

Still, in performance one has a full deck of 52 cards that covers 52 possibilities. Hence the odds of succeeding with a single deck are 1/52.
(If you would predict the card and number on a piece of paper, it's 1/2704, but you are not predicting. The deck of cards holds 52 different predictions.)

Denis

[Carlo Morpurgo]

Jim, each deck has 52 combinations, so 52 decks of 52 combinations = 2704 total combinations.

Even if a magician knew which card was going to be named, he'd still need 52 decks to cover all 52 positions that could be named. Then, the order of all the indifferent cards wouldn't matter, so long as the named card rotated through all 52 positions.

Look, in probability you have to be careful in choosing the "sample space", which basically represents mathematically what experiment you are doing. In this case the experiment is very simple: a position is given and a card is spotted in THAT position. The possible outcomes of this experiment are 52 not 52*52. For example if instead of 1 position 10 positions are randomly chosen by ten spectators and then the highest one is chosen (or the average is computed or what have you) you still get 1/52, since the moment you are TOLD which position to search, that part of the uncertainty gets killed.

The fact that you know the card in advance is completely irrelevant. Indeed, the experiment is equivalent to this one:

- one spectator picks a position randomly and looks at the card
in that position, wihtout showing.

- another spectator writes a card on a piece of paper.

What are the chances that the cards match?

Carlo

[Carlo Morpurgo]

Jim, each deck has 52 combinations, so 52 decks of 52 combinations = 2704 total combinations.

Another way to convince yourself is to actually perform several times the effect with a deck of 2 cards. According to your theory success will occur approx. 1/4 of the times. In reality you will instead get approx. 1/2 of the times, provided you set it up right.
All you need is one coin: first toss determines the position and second toss the card. Your punishment is to perform this experiment 1000 times today.

Carlo

[Carlo Morpurgo]

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.

I just saw this post after I wrote my 2-card deck experiment above. As I said, just try it out 1000 times and compute the ratio of successes. I think the apparent confusion is due to you not clarifying HOW this experiment is performed. Can you spell this out for us? To me is:

- position picked
- card picked
- position revealed (say X)
- card revealed (say Y)
- check that at position X there is card Y

Do we agree on this?

[greg manwaring]

Maybe he's discovered that certain cards when named get a common number response from people? Believe me, I'm as perplexed as anyone about this! Just letting my mind wander here.

[naquada]

I've just re-read the section in the david berglas book regarding thought of cards, and ACAAN, and it's kinda clear from that what the basics are... the rest is just pure berglas-ness.. and kind of impossible to describe.. probably hence the in-ability for anyone to put there finger on the full method...

[Cohiba]

Jim, each deck has 52 combinations, so 52 decks of 52 combinations = 2704 total combinations.

Another way to convince yourself is to actually perform several times the effect with a deck of 2 cards. According to your theory success will occur approx. 1/4 of the times. In reality you will instead get approx. 1/2 of the times, provided you set it up right.
All you need is one coin: first toss determines the position and second toss the card. Your punishment is to perform this experiment 1000 times today.

Carlo

Carlo:

If you take a deck of two cards, name one of the cards, and then name a position, you obviously will get a 1/2 chance of success. I agree that shuffling and trying this 1000 times will yield 1/2. The problem is, for this effect to be completed successfully by a magician, you don't know which card will be called, and therefore you need TWO decks of two cards to ensure success.

Say for example, in packet one, you put the AS in position 1, and the 2S in position 2. This will cover two of the four choices that the spectator might give:

1 - AS at position 1
2 - 2S at position 2

But since the spectator might call either of these other two options:

3 - AS at position 2
4 - 2S at position 1

...you need a second packet, with the 2S in position 1, and the AS in position 2.

Again, in either packet, you have a 1 in 2 chance of success - but as a magician, you don't want a 1 in 2 chance, you want 100% chance of success. So you need a second packet in a different order, which fulfills the 4 different combinations that can be called.

[Cohiba]

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.

I just saw this post after I wrote my 2-card deck experiment above. As I said, just try it out 1000 times and compute the ratio of successes. I think the apparent confusion is due to you not clarifying HOW this experiment is performed. Can you spell this out for us? To me is:

- position picked
- card picked
- position revealed (say X)
- card revealed (say Y)
- check that at position X there is card Y

Do we agree on this?

We agree on this. I think Denis helped clear this up for me - 52 of the 2704 possibilities are accounted for by the deck, so the odds are 1 in 52.

My point is that with ACAAN, the magician must somehow come up with 2704 different possibilities - that is the number of options (combinations) that a spectator can call out. I was trying to prove the fact that there are 2704 combinations, and I now believe there wasn't a disagreement there.

Thank everyone for the insight! This was a good exercise for me at least! ;o)

[Carlo Morpurgo]

If you take a deck of two cards, name one of the cards, and then name a position, you obviously will get a 1/2 chance of success. I agree that shuffling and trying this 1000 times will yield 1/2. The problem is, for this effect to be completed successfully by a magician, you don't know which card will be called, and therefore you need TWO decks of two cards to ensure success.

Sure, but this has nothing to do with your original claim that "(actually, it has a chance of happening 1 in 2704 tries)."
This whole rest of the thread was based on this claim, not on what the performer needs to do in order to be 100% sure to succeed.

It's not so bad to acknowledge to be wrong, you know? People will look up to you and respect you. What a wonderful world this would be.

Carlo

[Cohiba]

Carlo, you are absolutely right - my original post was incorrect. I apologize to Bob (and all) for the lengthy debate! Again, thanks for the discussion!

[Carlo Morpurgo]

Carlo, you are absolutely right - my original post was incorrect. I apologize to Bob (and all) for the lengthy debate! Again, thanks for the discussion!

I actually had fun! No need to apologize for anything, in my opinion. There are always positives in discussing, even more so if it ends well.... By the way I agree completely on everything else you said in that post. My biggest problem is that the conditions specified in the beginning of the performance are really not met, if there was pre-work.

Carlo

[Danny Archer]

[size:11pt]Coincidences happen last night while performing I always do two memdeck tricks back to back first the Wish Trick ( 4 spectators decide on a card by naming color, suit, high/low and then a value the named card is on top of the deck) this is followed by the Birthday Book (a spectator announces their birthday and touches a card from a face down deck they open the Birthday Book and check the card written down, which matches their selected card)

TWICE last night the card from the Wish Trick was the same card written in the Birthday Book of course I played it like I made it happen but inside I was freaking out

What does this have to do with ACAAN?... nothing I guess, but it does show that amazing things can and do happen more often then we would have believed possible
[/size]

[Richard Kaufman]

I'm sure that experienced workers out there know that if asked to name a card, more people will select a particular card. And if asked to name a number, more people will select a particular number.
So, having THAT card at THAT number in your deck might allow you to take advantage of a mircle on a few occasions.

[SoMd]

I have a dilemma. Since people seem to intuitively think that the odds are 1 in 2704, I like to claim that those are the correct odds when I do ACAAN. However, I'm also a mathematician, and I really hate to reinforce the public's innumeracy problem. I don't mind lying through my teeth to achieve a good magic trick, but lying about math presents an ethical problem for me. My solution is to claim that the odds are 1 in 52 times 52 times another 1000 because the trick is so good. At this point, at least everyone should know I'm kidding about the odds.

[Pete McCabe]

Gordy,

The correct odds are one in 52 for the card times one in 52 for the number times one in 365 that you would be doing the trick this specific day of the year times one in 6 billion that you would be doing the trick for this exact spectator.

[Chas Nigh]

I have used Paul LePaul's method for ACAAN using a stooge from the audience (1) and have completely blown away any audience including magicians.