Today while at school, I did Josh Jay's Her Card, Her Name. Basically, you deal from the bottom and tell her to stop you anywhere. When you turn over the stopped card, it has her name on it. I put a little mental twist on it. I "send" her the card mentally, and she "subconciously" knows where to stop you. This time something weird happened. After "sending" her the card, she blurted out Ace of Hearts, before the deck was even out of the box. This happened to be the card with her name on it!!!

Just a coincidence, and no exposure on here buddy. Sometimes it just works out that way.

I wasn't exposing, they always know you're dealing from the bottom, I should have cleared that up in the original post.

He never said it was anything more than a coincidence. And I believe he was describing the effect of the trick, not the method.

Yeah he explained that quite well before you posted. But look at the title of the thread. The question in it made me think he might not think it was a coincidence. That's why I responded as I did.

I just thought the title sounded cool, I know it was a coincidence, albeit a cool one!

I agree, it was most definitely just a coincidence. Assuming that every spectator randomly selects one card,* then the probability of them selecting the force card is 1/52, which is about 2%. Thus, you could expect this to happen about once every 50 times you perform this trick.

~ Kyle

* As people are more prone to picking more recognizable cards, such as the ace of spades or hearts, as opposed to say the 4 of clubs, I doubt the distribution is truly random. Consequently, if your force card is an ace or king, chances are that the probability would be slightly higher than 1/52.

Indeed, I've thought about all of that. From now on, when I do the effect, I'm going to use an "off" card and see if it ever happens again. I love it when things just go your way though!!

The first section works out, of course, only if you keep using the same card

Strangely enough, although counter-intuitive at first, this is actually not the case -- you could choose a different force card every time and, assuming the spectators' choices are truly random, the probability remains the same. This is because the probabilities of each event are independent (i.e. each individual outcome has no influence on past/future events). Thus, regardless of the choice of the force card, one could expect a random spectator to choose it about 2% of the time.

Another way of looking at the problem is to consider the following related problem. Suppose we randomly shuffle a 52 card deck and predict that the top card will be the ace of spades. Because the shuffle is random, the probability that the top card will be the ace of spades is 1/52... but there is nothing special about the ace of spades -- the probability that any given card will be on top is 1/52. So the probability that any card we choose will end up on top is 1/52, and hence we should expect this to happen about 2% of the time.

~ Kyle

P.S. Curiously enough, in the above exercise, there is nothing special about the position of the top card either... we could change our chosen position and the outcome would remain the same!

Would it not be that you must randomly choose a card out of 52 and they as well meaning they have to choose the same one you chose (both random)? That would make it so that you'd have to multiply those two probabilities together to get something like a 0.03% chance of it working instead of 2%? I could be wrong I've been away from combinatorics and probabilities for a while.

I was thinking about that too for awhile before I posted... however, I do not think this is the case because the force card (whatever it's position or value be) is known.

For example, suppose I have a (fair) quarter in my pocket, of which I have drawn an "X" on one side. The probability that you will guess which side has the "X" on it is 1/2, not 1/4, regardless of whether or not I randomly choose which side to mark.

Regards,
Kyle

Very true, I agree with that logic.