Do you know The Fitch Cheney Five Cards Trick? The following is a much harder twist of it.

Standard 52 cards deck. Magician is outside the room. Assistant asks random guy from the public to shuffle the deck and to select 3 cards from it. Then asks another random guy to select 1 card from these three. Then Assistant lays out the three cards in a row on the table - the last selected face down, the other two face up. Then Assistant calls Magician to enter the room. Magician looks for a while the three cards on the table and names the face down card.

Question: What code is used by Assistant and Magician?

Keep the following in mind:

  1. There is no communication of any sort between Assistant and Magician.
  2. Only the three cards arranged on the table give information to Magician.
  3. Repeating the trick several times is safe.
  4. It is possible to perform this trick over phone, i.e. Assistant to tell arrangement of the three cards (naming only the two face up cards of course) to Magician.
  • $\begingroup$ I could do it over the phone only ;) The assistent could encode information by saying "Ace Heart" or "Heart Ace" for example, something that can't be done on the table. $\endgroup$ Aug 20 '15 at 0:25
  • $\begingroup$ I've heard a version of this that's phone-only; as soon as the magician picks up, they start naming the cards from ace to king, and the assistant says the name of the first card when they're on the right one. They do the same thing with the suits. But that wouldn't translate here. $\endgroup$
    – Patrick N
    Aug 20 '15 at 0:34
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    $\begingroup$ @kamenf If more than the order is able to be used, I don't see how the trick can be used over the phone and in person without changing or making it completely obvious (like saying over the phone "8 of clubs turned at 7 degrees") $\endgroup$ Aug 20 '15 at 1:38
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    $\begingroup$ Hey, what am I doing here ? $\endgroup$ Aug 20 '15 at 6:36
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    $\begingroup$ I would say that when the assistant calls the magician to enter the room, they have plenty of leeway to use one of 52 phrases that could signal the card, but you specify no communication. I'm going to assume the "no communication" doesn't apply to calling the magician into the room, even though my response is completely valid. $\endgroup$
    – Kingrames
    Aug 20 '15 at 16:57

I think this is impossible, but I hope someone points out an error in my reasoning.

There are $\binom{52}{3}\cdot 3=52\cdot 51\cdot 25$ situations the assistant can be handed.

However, there are only $52\cdot 51\cdot 3$ possible arrangements that the magician can be handed by the assistant. This is because what he sees is completely specified by the two face up cards (52 choices for one further to left, 51 for other) and the location of the one face down (3 choices for its location).

Since $52\cdot 51\cdot 25>52\cdot 51\cdot 3$, by the pigeonhole principle, there must exist two situations the assistant is handed which result in the same arrangement the magician sees. This means the magician will make the same guess in two situations where there are different face down cards, so he must be wrong for one of them.

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    $\begingroup$ For a simpler approach, consider the two up cards, They together with the down card can express $3! =6$ different situations. However, there are $50$ possible values for the third card. $\endgroup$
    – Rohcana
    Aug 20 '15 at 1:24
  • $\begingroup$ I suggest using all conditions that are given than only the 3 cards. Then math can show a lot more. Else based only on 3 cards it is impossible for sure. $\endgroup$
    – kamenf
    Aug 20 '15 at 2:26
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    $\begingroup$ @kamenf Can you elaborate on what you mean by "using all conditions that are given"? Specifically, in what ways is the assistant allowed to lay out the cards? $\endgroup$ Aug 20 '15 at 3:23
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    $\begingroup$ @kamenf Since you said this trick can also be done over the phone, the table placement is either irrelevant or there are secret verbal cues for this version instead of visual cues. $\endgroup$
    – Trenin
    Aug 20 '15 at 13:07
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    $\begingroup$ @kamenf I think the issue here is that it is trivial to find a solution that will work based either on the placement of the cards on the table or on some spoken code. However, there is no apparent way to find a solution that satisfies both these requirements without fudging your stated guidelines. I think this isn't a good puzzle because it's extremely broad, but then you give impossibly narrow constraints that you yourself don't seem to expect us to follow. $\endgroup$
    – Patrick N
    Aug 20 '15 at 13:55

Partial answer - only satisfies 1, 2 and 3 and assumes that the magician is permitted the luxury of using a pack of cards that has an assymetric back. I doubt that the audience would be suspicious if this was a otherwise plain back blatantly advertising the name of the show, the magician or the venue.

The magician and assistant have to agree on a suit priority in the event of both face-up cards having the same value. This will allow for one of the cards to be a high card (H) and the other a low card (L).

There are six possible arrangements of face-down card, H and L. Adding the orientation of the face-down card into the mix (^ for upside-down and v for right-way-up), this now gives 12 combinations. If the cards are usually laid close together, the 13th value can be indicated by spacing the cards out a little more. In this last case the order of the cards is unimportant.

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The suit would then be indicated by placing the cards slightly off-centre on the table e.g towards the top for spades, right for hearts, bottom for clubs and left for diamonds

It's not perfect, but will certainly work once and, for most audiences would survive repetition.

  • $\begingroup$ "Standard 52 card deck" $\endgroup$ Aug 20 '15 at 17:07
  • $\begingroup$ Yes, "Sdandard deck" is the problem here, else has some nice ideas :) $\endgroup$
    – kamenf
    Aug 20 '15 at 20:03
  • $\begingroup$ An alternative approach to give the same number of combinations is to lay the cards out in a row for the first six values and in a column for the next six and again space them out a little for the ace. The problem here is that this change in layout would be more obvious to the audience on repetition. $\endgroup$
    – Gordon K
    Aug 20 '15 at 20:12

So assuming the trick is different over the phone (as the asker has sort of implied?), this would work:

Label the deck in any arbitrary way giving the cards each a value 1-52. Let's say it is Ace->King, Spades->Clubs->Hearts->Diamonds. So ace of spades is 1, 2 of spades is 2, ace of clubs is 14, king of diamonds is 52. Now imagine the table as a 3x3 grid with squares labeled 1-9. When the assistant places the cards down, they can be closest to one of these sections of the table, without being obviously placed there. So now there is a value 1-9 that can be conveyed. There are six ways the cards can be arranged in a row. Label them arbitrarily 1-6. The three cards can be given unique identities because one is flipped, and of the other two, one is higher, one is lower (based on the numbering above). Now there are $9*6=54$ possible combinations of order/location on table. Arbitrarily label those and you have one for each card, plus 2 extra if you use the jokers in your trick.

Then for phone:

Just use any vocal queue, like the magician listing off all suits/values

  • $\begingroup$ I like the table version. I would use the middle card placing it in the similar nine spots in the rectangle formed between the other two - the offsets are much much smaller compared to placing the whole row, easily detectable by the magician but not so for the public, keeps the same place of the row on the table in repeating tricks. $\endgroup$
    – kamenf
    Aug 20 '15 at 20:59

It could all depend how the cards

are placed.

Number the cards 0 to 51
The assistant then lays down each of the three cards in one of four positions.
Card Positions
Turn this into a three digit Quaternary Number
convert it back to decimal, and match it up with the numbered cards.

This system can actually encode for 64 different cards if you had some weird deck. :)

  • $\begingroup$ This works for sure but magical part is wasted ;) $\endgroup$
    – kamenf
    Aug 20 '15 at 3:04
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    $\begingroup$ And this works over the phone? $\endgroup$
    – warspyking
    Aug 20 '15 at 3:11
  • $\begingroup$ The audience can be dumb, but not that dumb $\endgroup$
    – Rohcana
    Aug 20 '15 at 5:08

The table top is specially designed so that it is translucent/opaque from an angle and is transparent from straight on. Below the table is designed a small mirror that is hidden unless viewed from a certain angle.

When the magician approaches the table, he is able to see the face of the face down card. He can then easily say what the card is.

This solution relies on having the audience members do their selecting away from the table, or having a special bottom on the table that can drop away when the magician approaches, obscuring the trick.

As for over the phone, there have been many possible ways given how this could be done.


I think this is a viable solution for over the phone and on the table although kind of complicated.

The spacing between the cards can be used as Morse code, which can be used over the phone. A space between the cards is a dash and side by side is a dot so you will have 4 combinations that equal 1-4.

-- = 1

.- = 2

-. = 3

.. = 4

From here we can use the face down card so determine a 4 bit binary number that has the ones spot as always zero so you can get

xxd = 110 = 12

xdx = 101 = 10

dxx = 011 = 6

Now comes the part that feels kind of incomplete or needs tweaking. The suit of the card can specify whether to add or subtract. You will have 16 combinations of Spades, Clubs, Hearts, and Diamonds.



For the suit you can place the cards slightly up or slightly down and change your voice inflections over the phone. To make it less obvious you could get the suit based off the card touching the face down card. In all the examples the first arrow is the face down card and you can use your imagination for the case when the face down card is last

v^ = S

^^ = C

vv = H

^v = D

Also think of the deck as circular so if you subtract 4 from 2 you will get Jack. I'm not sure how to justify this reasoning but if someone finds something wrong with it please let me know.

You should be able to get all combinations with that.

12 + 1 = A

12 + 2 = 2

12 + 3 = 3

12 + 4 = 4

12-1 = Q

12-2 = J

12-3 = 9

12-4 = 8

10 + 1 = Q

10 + 2 = K

10 + 3 = A

10 + 4 = 2

10 - 1 = 9

10 - 2 = 8

10 - 3 = 7

10 - 4 = 6

6 + 1 = 7

6 + 2 = 8

6 + 3 = 9

6 + 4 = J

6 - 1 = 5

6 - 2 = 4

6 - 3 = 3

6 - 4 = 2

  • $\begingroup$ Morse is good idea here - allows encoding of 4 states and can be used over phone too. $\endgroup$
    – kamenf
    Aug 20 '15 at 21:09

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