Film, Media & TV2 mins ago
Another Card Trick
4 Answers
Apologies for putting this in Chatterbank - my previous post was in Riddles which doesn't show up on the main page.
Like that post, what I'm trying to do is understand the mathematics which makes this trick work. I'll try to explain the procedure as best I can.
You ask someone to select any THREE cards from the deck. The only requirement is that they're able to remember them all - so offer them pen and paper if needs be, but they could just choose J, Q & K from one suit; it makes no difference.
The next bit is important to get right. You deal four piles of cards, 10 in the first, then 15, 15 again, and finally 9 cards. It makes the performance better if you can hide the fact that you're counting.
You tell your volunteer to place the first of his/her three cards face-down on top of the first pile of (10) cards.
They then split the next pile (15) wherever they wish, putting those cards on the first pile, and their second card on the remainder. Stack this on the first pile.
Now they split the next pile (15) and put this on the second one, adding their final card to the remainder. Again, stack this on the first pile.
You now take the last pile of 9 cards and put them on top so you have a complete deck.
Now deal the cards into two stacks, alternating between face up and face down. Tell your volunteer to call "Stop" if they see one of their cards. They won't.
Take the face down cards (you can remove the rest) and do the same thing again, face up then face down to form two stacks. Again, they won't see one of their cards.
Change things up with the remaining cards; deal face down then face up. Same result.
Go back to face up / face down. Guess what? You are left with three face down cards, the three they chose.
It's definitely impressive, but apparently based on straightforward arithmetic - and I don't have a clue!
Like that post, what I'm trying to do is understand the mathematics which makes this trick work. I'll try to explain the procedure as best I can.
You ask someone to select any THREE cards from the deck. The only requirement is that they're able to remember them all - so offer them pen and paper if needs be, but they could just choose J, Q & K from one suit; it makes no difference.
The next bit is important to get right. You deal four piles of cards, 10 in the first, then 15, 15 again, and finally 9 cards. It makes the performance better if you can hide the fact that you're counting.
You tell your volunteer to place the first of his/her three cards face-down on top of the first pile of (10) cards.
They then split the next pile (15) wherever they wish, putting those cards on the first pile, and their second card on the remainder. Stack this on the first pile.
Now they split the next pile (15) and put this on the second one, adding their final card to the remainder. Again, stack this on the first pile.
You now take the last pile of 9 cards and put them on top so you have a complete deck.
Now deal the cards into two stacks, alternating between face up and face down. Tell your volunteer to call "Stop" if they see one of their cards. They won't.
Take the face down cards (you can remove the rest) and do the same thing again, face up then face down to form two stacks. Again, they won't see one of their cards.
Change things up with the remaining cards; deal face down then face up. Same result.
Go back to face up / face down. Guess what? You are left with three face down cards, the three they chose.
It's definitely impressive, but apparently based on straightforward arithmetic - and I don't have a clue!
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Well you have got them to make irrelevant cuts of the partial decks and still ensured their cards go exactly where you want.
I guess one can work out the maths from that point if bothered, but it seems a reasonable result to me. All the resplitting of the piles do is bring their cards together. Until the end.
I guess one can work out the maths from that point if bothered, but it seems a reasonable result to me. All the resplitting of the piles do is bring their cards together. Until the end.
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