A spectator freely selects a card, and it is returned to the deck and shuffled. The spectator then cuts a small packet, and the deck is dealt into three piles. From each pile, a number of cards - freely chosen by the spectator - is cut from the top. The three resulting cards are not the selection. However, the total of their values locates the selection, and the bottom cards of the three piles turn out to be the other three cards of the same value.
Performance transcript:
- First, shuffle the deck.
- Then I riffle like this.
- Have the spectator call stop anywhere to select a card.
- Have the spectator remember the card they stopped at.
- Return the selection to the deck.
- Give the deck a cut.
- Now I'll involve the spectator in locating the card.
- Have the spectator cut off a small packet.
- No more than 10 cards.
- Assume this is the packet they cut.
- Now I deal three piles, 10 cards each.
- Have the spectator count how many cards they cut.
- Let's say they cut 5 cards.
- Why not more than 10?
- Because I'll use that number to adjust the packets.
- For example, if they cut 5 cards.
- I'll transfer the top 5 cards to the bottom.
- The number determines the position of the selection
- Do the same with each pile—cut 5 from top to bottom.
- We end on the Two of Spades, Four of Clubs, and Seven of Diamonds.
- The spectator says it's not their card.
- But stopping at these three cards must mean something.
- 2+7+4=13
- This may indicate the final position of the selection.
- This is the spectator's card—the Jack of Diamonds.
- But here's the question.
- If I'm only finding the Jack of Diamonds, why use three cards?
- Because these three cards not only locate the Jack of Diamonds.
- They also lead me to the other three Jacks.
1st edition 2026, video 6:02.
