Hi all,

Great suggestions, but perhaps I didn't make my question clear enough. I'm more interested in what presentational idea you have to elevate the effect from "just another card trick" into a more mental effect. I'm working on a precognition script, where the number is already known to me then "transmitted" to the spectator, but I would be happy to hear how others present it.
Cheers.

While that may be a good way to present the routine to the specs (makes it seem even more impossible as 1/52 times 1/52 yields a probability of 1/2704) it is not mathematically true, so beware if you have any mathematicians in your audience. The reason is although the spectator has a "sea" of 2704 combinations to choose from, 52 of them would be correct because there are 52 combinations of card/number in any shuffled deck. Thus the probability is really 52/2704 which simplifies to 1/52.

But if you present it as a piece of mentalism and state something like "Out of 52 different cards I was able to make you think of the <suit/value>, and out of 52 different numbers I was also able to make you think of <number>" then that would make it sound like the probability really was 1/2704. It's just the way you prove it with the deck (which has 52 possible correct answers) that reduces the true probability to 1/52.

I use Barrie's ACAAN in his Theatre of the Mind book on page 257 and think it's brilliant!

** Bill **

Bill is right. While the number of combinations is 2,704, the probability of the spectator naming the correct position of any card is 1/52. The probability of a spectator naming the correct locations of two cards is 1/2,652.

You can certainly get by with the (1/52)*(1/52)=1/2,704 spiel, but as Bill said, beware the mathematicians!

I suppose the confusion is in the presentation. When I present( those of you who have read "Mendacity" will know exactly what I'm talking about here) the ACAAN routine I first have someone think of a playing card and then someone else think of a number. The odds are 1/2704. While I agree if one is simply getting a subject to predict where one card of fifty-two lies in a pack of cards the odds drop considerably. To me, this is not what the effect has been about.

It all depends on the specific question being asked.

There are 52 X 52 unique ways that a card and a number can be specified. So if you were asking someone to predict a specific card and a specific number it would be a 1/2704 chance that they would pick both correctly (assuming the choice was truly random. Some cards and some numbers are more popular choices).

But in ACAAN, either the card or the number are specified first. There is a free choice, i.e. there is no wrong card to choose and no wrong number as in the case mentioned above.

The probability that any given card in a truly random order deck is at a particular specified number from 1 to 52 is 1/52 or the probability that any given number from 1 to 52 is the position of a particular specified card is 1/52.

Thus, once the card or the number are specified, the last person to choose has a 1/52 chance of getting it right. Essentially, the first person who chooses either the card or the number reduces the number of possibilities by a factor of 52.

It makes the trick less amazing, that is, if the audience thought about the real odds.

However, it can be presented that the performer has a specific card and a specific umber in mind that he or she is projecting. Interestingly, the odds that the person would hit a card that matches a number by chances is still 1/52, but if the performer asserts they, in fact, got the card and number he or she had chosen, then the audience would believe the probability is 1/2704, assuming that they bothered to think about probability. (They wouldn't). One could supply proof that the correct choices were made. Because I doubt that people analyze probabilities and would instead be amazed at the performer's success, I don't think this extra proving is necessary, but I also haven't tried it when performing so I might be wrong about that.

Also, using the Magic Café Search Engine, I found the following topics about Any Card At Any Number.

David Berglas

Mentalism slant; any card at any number...

Any Card at Any Number

Any Card At Any Number?

Simon Aronson and Any Card to Any Number

Any card at any number...closest to perfect?

Card at named number

Any Card At Any Number (sort of)

ACAAN Richardson's Impromptu or Stephen Tuckers

Stephen Tucker: Any Card at Any Number

Mendacity - Consecution by George Tait

What is your favorite "Any Card At Any Number" effect?

Any Card At Any Number

Signed card at any number named

Boomerang vs. The Grail vs. Impossible 1......nowledge

Selected card at any number

The only reason I made a point of clarifying the true probability is so folks wouldn't try to correct the spectator if they challenged the 1/2704 probability claim. It's bad etiquette to correct a spectator when he/she is wrong and really bad etiquette to correct them when they are is right! <grin>

Cheers!

** Bill **

Grail is ridiculously clean. Great effect and with a few alterations you can do it without any gimmicks. Probably my favorite method. Richardson's version let me down. Too much work for my taste.

People also fail to remember the extremely clean tabled d***** version. You make a prediction. Remove it from the deck and keep it in full view. Spectator shuffles deck as much as he'd like. You take deck back with clean empty hands and count down to any number called out by spectator. Or spectator can just say stop at any time as you deal of cards.

You lost me Bill. Math has never been my strongest field. I am not able to understand the logic. How does it come down to 1/52 again?
The statement is: The probability of ocurence of any card at any number is 1/52 times 1/52. How is this again multiplied by 52?
Let's take a specific case: Whats the probability that spectator names 24? 1/52, since he has only 52 choices.
Whats the probability that the card at 24th position is King of Spades? I believe this will be 1/52 again. So the probability that the spectator names 24 AND the card at that position is King of Spades is (1/52 times 1/52). Am I missing any other possibilities?

You're missing the basic premise. Look at it like this: The spectator names a card. That can be any card because you haven't predicted anything yet. Now is when the mathmatics come in. Now that the card is named, there is a 1/52 chance they'll name the number at which that card (any card named) lies. Make sense?

Here's a little math thing that messes with people: what are the odds of flipping a coin the same way two times in a row? A lot of people would say one in four: 1/2 X 1/2. It's actually one in two or 50/50. Because you made no prediction on the first flip. Likewise, you made no separate prediciton on the card called for. You only predicted where it would be in the deck. If you first revealed that you predicted the named card. And secondly revealed that it's located at the number they called, that would be 1/52 X 1/52. Subtle but big difference.

I guess its the way I look at it!
You say the probability is after the number is named. But I look at the whole problem starting from any number, proceeding with any card.

I am lost again here.

Examples for independent events. For independent events like coin tosses, let P(A) be the probability of getting heads on the first toss, equal to 1/2. Let P(B) be the probability of getting heads on the second toss, also 1/2. The probability of getting two heads in a row (P(AB)) is 1/2 times 1/2 = 1/4. That is, the joint probability of two independent events is the product of their individual probabilities. The probability of getting heads on the first toss or getting heads on the second toss is the sum of their individual probabilities minus their joint probability: P(A)+P(B) - P(AB) = 1/2 + 1/2 -.1/4 = 3/4. The conditional probability is the same as the probability. Thus, the probability of getting heads on the second toss, given that the first toss is heads, is unchanged: P(B|A) = P(B) = 1/2.


"Taken from http://www2.chass.ncsu.edu/garson/PA765/probability.htm

Based on the independent events probability theorem, the probability of ACAAN should be 1/52 times 1/52, since the two are independent events.

When two events, A and B, are independent, the probability of both occurring is:
P(A and B) = P(A) · P(B)

Harishjose,

That example isn't the same since the choices are not totally independant. Each card is already at a specific position that will not change (at least according to the audience the positions won't change!). The choices are free choices, but that is not the same as being statistically independant. Each position is tightly coupled to a specific card and each card to a position.

The probability would be 1 in (52 X 52) if the audience had to pick, for example, the three of spades and the number 17. That trick would be named Specific Card At Specific Number. For that trick to be effective, the mentalist or magician has to (ostensibly) write down the prediction in advance.

But, they get to pick any card and any number.

So while there are (52 X 52) possibilities total they can choose from, there are only (52 X 52)/52, or 52, possibilities remaining after their first free choice. Only the second choice really matters after that in determining the actual probability of the card being at the specified number.

For a concrete example, I've chosen the three of spades. You choose the number now. How many choices do you have?

How many choices of a card would you have if I chose the number 17 first?

In each case, you have 52 choices and only 1 is correct. So the probability of you, the second chooser, getting the right choice is 1 in 52.

Now, if after the chosen card is shown to be at the chosen number, the mentalist or magician happen to predict the spectator chose the predicted card and the predicted number, then the probability will appear to be 1 / (52 X 52) to the audience. However, since such a prediction will be rigged (for me anyway, I can't really read minds) the actual apparent probability for generic ACAAN is 1 in 52. (This is all ignoring that some choices are more popular than others).

Of course, as I wrote before, the real probability is close to 1, since the performer will almost certainly succeed.

You're still missing the subtle differences. Remember, I said what are the odds of a coin being flipped to the same side twice in a row. It's 1/2. That's different than asking: what's the odds a coin will be flipped heads twice in a row. In the original queston, the first flip is irrelevant because either heads or tails can land. It's only the second flip that's relevant. There's a 1/2 chance it will match the first flip that could have been either heads or tail.

So... back to ACAAN. It's not two independent 1/52 events. It's one. You're not making a 1/52 prediction on what card they name. In fact you're making no prediction at all on the matter. They're simply naming a card. You've gotten nothing right at this point, as again, you have not uncovered any prediction. No matter what card they name (it does not matter), you have a 1/52 chance at guessing where it is located in the deck.

Just like the coin hitting the same side twice, no matter what comes up on the first flip (it does not matter) you have a 1/2 chance of hitting it again on the second toss.

The title says any card at any number. The two choices are freely selected. At least from my stand point of view or how I would present the effect, it will be like that. The choices are statistically independent. The effect's presentation or the "magic" behind the effect is this: Spectator freely names a card and a number. The card is present at that number.

Here is my question, If one spectator chose 21, and after the card is chosen and just before you start dealing, he says, "I want to change my number. I want to change it to 31". Will you say, it's not possible to do that and stop the trick?

I know Barrie Richardson will say, "Sure, lets do it!"

What Platt is saying is conditional probablity and not independent probability. You say that whats the probability afetr the first event happens (heads). If the first event happen, whats the probability that the second event will happen (which is again another heads). This is what you are saying.
What I am saying is different. I am talking about indpendent events. The difference is in the presentation of the effect.
In Act Two: Barrie Richardson stress on this type of presentation. He asks the spectator or gives him/her a chance to change the number again. that's whats "magical" about ACAAN in my opinion. The difference, from what I understood, is in the effect and in its presentation. In other words, I present this effect as two independant events.

As a side note, I have a similar effect out called Lord of the Envelopes.

Yes, but that means that the choice of 21 will be considered wrong later by the spectator. They still will realized the card was only at one position, unless you present ACAAN as a magical effect instead of a mentalism effect.

Here's another way to look at it: There are 52 X 52 possibilities for the spectator to pick, but there are not that many choices that will produce a card-position match. There are only 52 card-position combinations that can work from the audience's perspective. So there are really only 52 choices for the trick to be successful.

The fact that we know we can make anything they choose work means that we as performers much transform all of this larger set of possibilities down to the specific possibility out of the 52 that the audience will imagine if they analyze this. To them, hopefully we're not moving the cards or switching decks. So to them, there are only the 52 choices, and somehow out of the 2604 possibilities, they happened to choose one of the 52 card-to-number pairings that are correct. So to them, the probability of being correct is (the total number of correct pairing choices)/(the total number of possibilities), which is 52/2604, which equals 1/52.

If you allow for the audience to imagine there are more than 52 correct card-number pairings that will produce a match, then the trick is a magic trick and not mentalism. However, it can still be presented as mentalism if they change their choice midstream!

Bill,
Four words,
Conditional Probability - Independent Probability.

This is interesting. I wonder how a mathematician wil approach this?

Harishjose, I don't see what's interesting about this.

Are you really not realizing that in reality this is a 1/52 effect? Or are you looking to make a tricky worded mathematical argument that it's 1/52 x 1/52?

Here's another way to think of it that may make more sense to you.

If the performer was to write down a number and a card as a prediction ahead of time and then ask the spectator to pick any card and any number, the probability of the performer's prediction being correct would be 1/52x1/52 = 1/2704.

But with ACAAN the performer is making 52 predictions ahead of time. Each card in the shuffled deck represents one of the performer's predictions and they would be correct choices if the spectator chooses any one of them.

So for the performer to be successful there are 52 possible correct choices divided by 2704 total choices which simplifies to 1/52. While the spectators choice of the card and number are independent, the rub comes when you forget to realize you really have 52 predictions or chances for success (thus the probability is 52/2704 and not 1/2704).

Hope that helps clear things up.

** Bill **

Ok.....

So, if I'm reading this correctly the presentation you all use is to confuse the spectators with some really difficult math?

(That was a joke by the way, but I can't even figure out how to add a smiley to my posts, let alone figure out what any of you are talking about!)