Then you take the first card from the top of the deck and put it down on the table face up. Now, if it's a king you put it back in the deck and try again. If it's anything else you read the value on the card. Lets say you picked a 3. Now you have to put down cards from the deck in your hand on top of the 3 until you have counted 13. So 13 - 3 = 10.

Double the face value of the card (aces = 1, jacks = 11, queens = 12, and kings = 13). Add 3 to the result. Multiply this by 5. Add 1 if the card is a club, 2 if it is a diamond, 3 if it is a heart, and 4 if it is a spade. Ask them to tell you their number. To predict the card, subtract 15 from the final total.

2018年10月28日 · From that the magician knows the fifth card. The partner and the magician can’t communicate during the trick. How did they do it? I thought that amoung the five cards there will be two with the same sign (Spades,Hearts,Diamonds or Clubs) and one of these two cards will be the fifth, and the other will be the first card to give to the magician...

2020年8月31日 · $\begingroup$ @NirajRaut so... the three cards we flipped, plus the additional cards we flipped to get the values to add to $10$, plus the additional cards we flip at the end corresponding to the total values... that is a total of $1+1+1+(10-x_1)+(10-x_2)+(10-x_3)+(x_1+x_2+x_3)$ cards flipped overall, which simplifies to $33$ cards flipped every time regardless of what the three shown cards ...

2018年9月20日 · Why does the selected card always end up in the center of the deck? What is the mathematical proof for this trick? And, out of curiosity, would this trick work for any number of cards? and if not, what would the largest deck that can be used for the trick?

2016年12月12日 · $\begingroup$ I first learnt this card trick when ... in the positions $8,9,10,11,12,13,14$ before dealing ...

If you follow carefully, the cuts are an illusion. Before you start dealing the cards up and down you have a deck with 15 cards (including the five moved from the final 9), the spectator's card, 15 more, the spectator's card, 15 more, the spectator's card, and 4.

Since (1) works by pigeonhole, (2) works because $\frac{13-1}{2} = 6$, and (4) works because there are 6 possible permutations of 3 cards, the trick succeeds. My question is whether the same trick can work with more than 52 cards.

2019年8月16日 · Here is how we figure that combination: A. there are four groups of six combinations B. those four groups are 1-6, 7-12, 13-18, 19-24 C. the 16 falls in the 13-18 group which is the third group this means that the first of the four signal cards must be the third highest.

2011年4月16日 · The top card has some number k on it. Take the top k cards off of the top of the deck, reverse their order ...