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Shuffle the deck.

Flip over the first $26$ cards (half the deck), face-up, in a pile (must keep them in order)

As you're doing this remember the $7$th card. Take those first $26$ cards and put them on the bottom of the deck. Now flip over the top three cards (next to each other). Count to $10$ with each one. (If you flip a $3$, $K$, $6$, you place $7$ cards from the deck on the three, $0$ on the king, and $4$ on the six. All face cards count as $10$, and the Ace counts as $1$. Add up the three cards you just flipped face up. (In the example it would equal $19$).

From the deck you are holding, the $7$th card from the initial flipping of $26$ will be the $19$th card (or whatever the sum of the $3$ cards you count to ten with is).

How does this card trick work mathematically?

Link: See this Youtube video

V.G
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Niraj Raut
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1 Answers1

6

This works because in the steps

  • flip over the top three cards
  • count to 10 with each one
  • add up the three cards and flip that many cards

you're always flipping exactly 33 cards. This puts you at the 7th card of the second half of the deck.

(If you call the value of top three cards $x_1$, $x_2$, and $x_3$, you're flipping $3$ cards in the first step, $(10-x_1) + (10-x_2) + (10-x_3)$ in the second step, and $x_1 + x_2 + x_3$ cards in the last step. This gives a total of 33 cards, independent of the actual value of $x_1$, $x_2$, and $x_3$.)

Magdiragdag
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  • "(If you call the value of three cards $x_1, x_2, x_3$, you're flipping $3$, $(10-x_1) + (10-x_2) + (10-x_3)$, $x_1 + x_2 + x_3$ cards in each of the three steps)" — It's inefficacious for me to comprehend. – Niraj Raut Aug 31 '20 at 10:24
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    @NirajRaut "inefficacious"... it has been a long time since I've seen that word. Google translate? I'll try rephrasing this anyways, but if you are relying on google translate to communicate, I don't know how helpful we can be. You might be better off asking someone who speaks your language. So... we flip a card. Suppose the value of that card was $x_1$. We flip $10-x_1$ more cards to get the total of $x_1$ and $10-x_1$ to $10$ (Note that $x_1+(10-x_1)=10$). We do that for the second and third cards too. Later, we additionally flip $(x_1+x_2+x_3)$ more cards. – JMoravitz Aug 31 '20 at 13:21
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    @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 happened to be. As such, we can be certain where in the deck we wind up and can predictably arrive back at our desired card every time. – JMoravitz Aug 31 '20 at 13:22