Skip to main content
Made it clearer that this is a math problem and not some way to communicate with card positions or verbal cues.
Source Link
Trenin
  • 9k
  • 22
  • 53

A magician (Yes, that's you) has 10 cards, say labelled 0 to 9. Now, the magician's assistant and the magician perform a magic trick:

  1. The magician turns around (so he can't see the cards)
  2. A volunteer from the crowd comes up and arranges the ten cards in any order they want
  3. The assistant now flips overpicks four cards and asks the volunteer to flip them over without changing the order, so they are now facedown (now we have four facedown and six faceup cards). The assistant leaves the stage (i.e. does no further communication with the magician, verbal or otherwise)
  4. Now the magician turns around and works out what each of the facedown cards is

The audience applauds!

So the question is how did the magician (you) do it?

Of course, you talked to your assistant beforehand and decided on a strategy - what is the strategy?


I'm not 100% sure it's possible, although I am pretty confident (90%) it is. The best I've got has a >50% (not sure of the exact amount) chance of succeeding, which may even be 100% although I haven't checked.

A magician (Yes, that's you) has 10 cards, say labelled 0 to 9. Now, the magician's assistant and the magician perform a magic trick:

  1. The magician turns around (so he can't see the cards)
  2. A volunteer from the crowd comes up and arranges the ten cards in any order they want
  3. The assistant now flips over four cards without changing the order, so they are now facedown (now we have four facedown and six faceup cards)
  4. Now the magician turns around and works out what each of the facedown cards is

The audience applauds!

So the question is how did the magician (you) do it?

Of course, you talked to your assistant beforehand and decided on a strategy - what is the strategy?


I'm not 100% sure it's possible, although I am pretty confident (90%) it is. The best I've got has a >50% (not sure of the exact amount) chance of succeeding, which may even be 100% although I haven't checked.

A magician (Yes, that's you) has 10 cards, say labelled 0 to 9. Now, the magician's assistant and the magician perform a magic trick:

  1. The magician turns around (so he can't see the cards)
  2. A volunteer from the crowd comes up and arranges the ten cards in any order they want
  3. The assistant picks four cards and asks the volunteer to flip them over without changing the order, so they are now facedown (now we have four facedown and six faceup cards). The assistant leaves the stage (i.e. does no further communication with the magician, verbal or otherwise)
  4. Now the magician turns around and works out what each of the facedown cards is

The audience applauds!

So the question is how did the magician (you) do it?

Of course, you talked to your assistant beforehand and decided on a strategy - what is the strategy?


I'm not 100% sure it's possible, although I am pretty confident (90%) it is. The best I've got has a >50% (not sure of the exact amount) chance of succeeding, which may even be 100% although I haven't checked.

Source Link
Wen1now
  • 9.3k
  • 4
  • 35
  • 87

Is this 10-card magic trick possible?

A magician (Yes, that's you) has 10 cards, say labelled 0 to 9. Now, the magician's assistant and the magician perform a magic trick:

  1. The magician turns around (so he can't see the cards)
  2. A volunteer from the crowd comes up and arranges the ten cards in any order they want
  3. The assistant now flips over four cards without changing the order, so they are now facedown (now we have four facedown and six faceup cards)
  4. Now the magician turns around and works out what each of the facedown cards is

The audience applauds!

So the question is how did the magician (you) do it?

Of course, you talked to your assistant beforehand and decided on a strategy - what is the strategy?


I'm not 100% sure it's possible, although I am pretty confident (90%) it is. The best I've got has a >50% (not sure of the exact amount) chance of succeeding, which may even be 100% although I haven't checked.