Sorry, that's just wrong. Spectator names a card and a number between 1 and 52, and on average once out of every 52 tries the card will be at that number, it's as simple as that.

Are you talking to me?
How so?
Think of it this way I write two numbers down.
both between 1 and 52.
what is the probability that someone else would write down the exact same numbers? 1/2704
Now say I write down 52 sets of two numbers between 1 and 52.
What is the probability that someone would pick pick two numbers that match one of the
two I have in my list of 52? 1/52

I do not understand any other way around that unless I am completely missing something.
If so let me know. I do not think I am.

Q: How many magicians on the Café with attention deficit disorder does it take to discuss, bash, praise, rip apart, make wild accusations, and do math about a trick?

A: Look..... over there.... its a bird!


:)

Ha craig I am just interested in the problem now.

I am also assuming this is the probability before the spectator makes any choices
after that they are no longer probabilities

Arios,


Let's assume you ask for the card first and then the number.

When the spec names a card it can be any card and the trick will work right? One out of 52 doesn't play into it because any card will be fine.

Now when they name a number it has to be the right number for the trick to work. So the trick will work one out of 52 times.

If you wrote down a prediction ahead of time (for instance "the spectator will name the two of diamonds and it will be found 24 from the top"), then had the spectator shuffle the cards and name one randomly the odds would be 52X52.

Exactly. That is what I said. I just thought Indigo was saying that was incorrect.

Then, Once the spectator starts making choices there are no probabilities involved.

If you wanted a trick where the odds are one out of 2704 here is one plot:

Ask the spec to name a card

Ask them to name a number

They count down and find that card at that number

The rest of the cards are turned face up and shown to be blank!

Yes, or just have the card and number written down before hand. I was just trying to explain that.

I see that I jumped in without reading the previous page. Sorry arios you explained it correctly in a previous post.

Yes, but same idea to get the point across I still was not sure if indigo still thought I was wrong about
probabilities changing. You could just have the back be different then the rest. Either way the probabilities come out the same.

edit: lol the source of the confusion...

No, I was responding to the post directly before mine.

It helps to think of just a two card deck: the AS and the 2S.

Now what is the probability that the AS is in position 1?
What is the probability that the AS is in position 2?

Ask yourselves the same questions for the 2S.

Should be all clear now.

ACAAN is 1/52 period.

Yes, except if they chose a card at any number and then the card they chose was a red back in a blue back deck then it would not be 1/52.
If you want to actually predict what the person will chose it is 'NOT' 1/52. I think this is where most people are getting confused.
It is 1/52 if there is no prediction.

Yes, but we are talking about the "normal" ACAAN, nothing about prediction, red/blue back, blank faces, just the classic and pure ANY CARD AT ANY NUMBER.

I was just talking about ACAAN in general and explained both possibilities in previous posts.

You, are talking about the "normal" ACAAN. There are many ACAANs and it's not always so clear for everybody. Anyway the distinction made between two broad categories of ACAAN may have helped to dispel the confusion about the probability of the outcome.

I think nearly everyone will agree ACAAN is "Name a card, Name a number, Deal deck find card at number". And the original post on math were talking about that trick, the additions were only added later on.

With a prediction, the probability of the prediction itself is 1/2704. However even predicted, the probability of the card actually being in that spot is still 1/52. If you only cared about the math, you would just reveal the prediction and forget the cards.

And if math made a good effect, then sympathetic cards must surely be the best trick ever! 1/52^52

What are the odds on a thought of card???? to any#
my effect "Arlene the physic Elephant" I can do this with out a miss, with a normal deck, that is shuffled by the spectator before the effect.Also no dupes or crimp cards and it is impromptu,
sounds impossible but it is true..
vinny

That would be 1/52 still vinny. But, it does sound pretty awesome though. Do you have it published anywhere?