

Jack Veteran user 370 Posts 
The first card trick that I learned was from a friend in elementary school. It works by mathematics.
You need exactly 52 cards. 1. Lay the top card face up onto the table. Look at its value. 2. You will now lay cards, face up, oneatatime, on top of that card, counting up from the value of it until you reach 13 (ex. the first card laid down has a value of 7. You will end up adding 6 cards face up on top of it, totaling 13). Face cards have a value of 10. 3. Once you get to 13, deal another card face up onto the table beside that pile and count up to 13 from its value. 4. Continue this until you have a four or more piles on the table. 5. Turn each pile face down. 6. Have the spectator point to any piles, removing each until 3 piles are left. Drop each removed pile on top of the remaining deck portion. 7. Have the spectator turn the top card over on top of any two piles. The remaining pile is still face down. 8. Add the two face up cards together, then add 10 to that total. 9. Remove that many cards from the deck. Count how many cards are left in the deck proper. That number will match the value of the top card of the pile that wasn't turned over. No giant mystery, but I always thought it was interesting. Do any of you know where this selfworking math card trick originated from? Magically, Jack 
Scott Cram Inner circle 2677 Posts 
If the trick requires 52 cards, how come your picture only shows 3 Jacks? <G>
Seriously, I've seen people perform this trick many times (usually without much more than explanatory patter), but I, too, have often wondered where it originated. 
Slim King Eternal Order Orlando 17672 Posts 
Took me a little bit but I see said the blind man. It would be good with some cool patter.......AND more impotantly...Can be done "Over The Phone"!!!
THE MAN THE SKEPTICS REFUSE TO TEST FOR ONE MILLION DOLLARS.. The Worlds Foremost Authority on Houdini's Life after Death.....

drkptrs1975 Elite user North Eastern PA 452 Posts 
You Lost me. Do the Cards have to be in a certain order.

Foucault Elite user New Jersey, USA 424 Posts 
Not at all  it works automatically, if you've followed the rules correctly. And there's the potential problem with this effect. It works well if you can describe the moves in a way that the subject can follow along correctly.
My peasized brain can't get around why this works. I can see that you are setting up a relationship between the values of cards and the number of cards at step #2, but that's where my head begins to hurt! 
rgranville Elite user Boston area 462 Posts 
A similar (but not exact) effect can be found in Hugard's More Card Manipulations (originally published in 1938) as "Infallible Prediction". The underlying principle is the same.
To understand Jack's effect, first we must realize that we only care about the three remianing dealt out stacks after the spectator has eliminated the other(s). The stack(s) eliminated by the spectator become indifferent cards at that point. The fact that cards were originally dealt to "bring them up" to 13 is never used. So in effect, we only need to deal out three cards, bring them up to 13, turn the stacks over, and go from there. So we originally had three cards, each of random value. Let's call them x, y, and z. (And note that there's nothing to prevent x being the same as y or z or any other combination  we don't care.) Starting with x, we deal enough cards to bring it up to 13 and turn the stack over (leaving x on top). That means we dealt 13  x cards to that stack. Counting our original x (which is there, too, don't forget!), that means our first stack has 14  x cards. Similarly, our second stack will have 14  y cards, and our third will have 14  z cards. Now for two of the stacks, turn the top card over. Without loss of generality (mathematicians love phrases like that ), it can be assumed that we're turning over the top card in the x stack and the top card in the y stack. But that's just x and y. Now add x and y, add 10, and remove that many cards from the remainder of the deck. That's x + y + 10. So how many cards have we removed from the deck after all these steps, and how many remain? With our original x stack we removed 14  x cards, and we removed another x cards (from the x + y + 10 step). That's 14  x + x, or just plain old 14. In the same way, we removed 14  y cards with the y stack, and another y cards (x + y + 10 again), so 14  y + y is just 14 again. That's 28 cards. And then 10 more (as the last bit of x + y + 10) makes 38 cards. Since we started with 52, that leaves just 14 cards. But wait! We also removed the z stack of 14  z cards. Take that away from the remaining 14, and we have 14  (14  z) or 14  14 + z or z cards left in the remaining deck. Turn over the top card of the z stack, which is just z itself, and it magically matches the number of remaining cards in the deck. :banana: 
Foucault Elite user New Jersey, USA 424 Posts 
Amazing! Thanks for the explanation, rgranville. I'm still not sure I entirely understand how this works, but I am a little closer, and further examination will make things a lot clearer, I'm sure.

Foucault Elite user New Jersey, USA 424 Posts 
Quote:
On 20051204 21:34, rgranville wrote: I just discovered that another effect using this principle is "Ultra Card Divination" in "The Royal Road to Card Magic" (in the chapter on the Riffle Shuffle, although it has nothing to do with said shuffle). 
Patrick Differ Inner circle 1540 Posts 
Yeah, this was one of my first card tricks, too.
I would lay them out and watch the suits. That way I could name the card.
Will you walk into my parlour? said the Spider to the Fly,
Tis the prettiest little parlour that ever you did spy; The way into my parlour is up a winding stair, And I've a many curious things to show when you are there. Oh no, no, said the little Fly, to ask me is in vain, For who goes up your winding stair can ne'er come down again. 
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