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A long time ago, I saw on a website (with a grec sounding name i can't remember), a card trick I've passed a long time trying to solve. I hope it has never been posted there and i'm sure it will interest you.

The trick is done by a magician, two assistants and a 53 card pack (with one unique Joker). The trick needs the participation of a spectator.

(1) The spectator draw one card from the 53 card pack
(2) the card pack (52 cards) is separated in two packs of 26 cards
(3) each pack is given to each assistant
(4) assistants choose two cards from their pack
(5) the card from the spectator and the 4 cards of assistants are shuffled together and given to the magician
(6) the magician guess the spectator card.

I have to say i've never seen this trick in action, i'm not sure it exists a perfect solution, I was able to find some partial solution.

I assume there is no form of communication (oral or scratch on cards) between assistants and magician after the start of the trick.

Do you see what strategy can be used by assistants and magician to choose cards and guess the spectator card?

EDIT: Partial solution found
Sneftel found a partial solution, it is ok for most of the possibilities but there is still some cases were they magician won't be able to guess which of the card is the one.
Any idea for a perfect solution (all-time winning strategy)

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  • $\begingroup$ They can just mark their cards with a minor scratch that only they and the magician knows.. $\endgroup$ – prog_SAHIL Jan 16 '18 at 10:45
  • $\begingroup$ Can the assistants communicate in the slightest? If they can settle on a set of unnoticeable, minor cues (4 is enough, 1 for each suit of card) then it makes the problem much easier. $\endgroup$ – votbear Jan 16 '18 at 10:46
  • $\begingroup$ I will edit the question, but I assume there is no scratch or mark on card, they can't communicate after the start of the operation but they can set up a strategy before. $\endgroup$ – Untitpoi Jan 16 '18 at 10:49
  • $\begingroup$ is the card pack of 52 cards randomly separated in two packs or not? $\endgroup$ – athin Jan 16 '18 at 11:08
  • $\begingroup$ @athin Yes it is $\endgroup$ – Untitpoi Jan 16 '18 at 12:21
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Based on Jaap's answer (which seems to be addressing a slightly different problem):

The cards are implicitly numbered 0-52, based on a combination of suit and rank. Each assistant chooses two cards whose numbers, added together, produce a sum of 26 mod 53. Once the magician has the cards, he simply adds their numbers and 1 together mod 53, and converts the resulting number back to suit/rank.

(EDIT:)

Changed 53 to 26, since it's not guaranteed that both assistants can come up with a sum of exactly 53 (such as if one has all even-numbered cards).

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  • $\begingroup$ I Think you've got the answer but why not using a modulo 52 instead of 53 and adding 1? $\endgroup$ – Untitpoi Jan 16 '18 at 13:58
  • $\begingroup$ The math requires modulo 53. With modulo 52, there'd be no way to select any of all 53 cards. $\endgroup$ – Sneftel Jan 16 '18 at 14:01
  • $\begingroup$ There are combinations where that is not possible. For example, one assistant A gets 1-13 and 27-39. However, the odds are very low. And they can compromise by both picking cards that add up to 13 or 39 modulo 53. That will often work. Also, never pick the joker. $\endgroup$ – user3294068 Jan 16 '18 at 15:50
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    $\begingroup$ Also, given the "never pick the joker" rule, modulo 52 might work well, and is much easier to calculate. $\endgroup$ – user3294068 Jan 16 '18 at 15:51
  • $\begingroup$ Hmm, if you do modulo 52 (never select joker), then there are 2^26 ways for it to go wrong, out of 52 choose 26, for the odds of a problem occurring are about 1 in 7 million. $\endgroup$ – user3294068 Jan 16 '18 at 15:54
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NOTE: I misread the problem, and thought the spectator kept their card, and that you had to find out what it was by looking at the four cards the assistants give. I'll leave this answer here however, since it describes a technique that can be useful for solving the actual problem.


It is easier to do without the joker, i.e. with a 52 card pack.

Let's give the cards values, with A=1, J=11, Q=12, K=13, and number cards their normal number value. The total value of the 52 card deck is then 4*(1+2+..+12+13), which is a multiple of 13. After the spectator's card is chosen, if you knew the total value of the remaining cards modulo 13, then you would also know the value of the missing card. For example, if the total were 4 modulo 13, then you would know that the missing card is 9 so as to make the total 13.

So to find out the value of the missing card:

Each assistent looks at the cards they have and removes one card that would make their remaining hand a multiple of 13.
When you see the first card each assistant produces, you add them together, and know that this number together with the spectators card will make a multiple of 13 (i.e. 13 or 26).

The same system can be used for the suit of the missing card:

Give the suits values (e.g. according to CHaSeD order, Clubs=1, Hearts=2, Spades=3, Diamonds=4). The full pack total is a multiple of 4. So the assistants again remove a card that makes their remaining cards (including the card removed for signalling the value) are a multiple of 4. You again determine the suit of the missing card by knowing that it will make a multiple of 4 when added to the suits of the two assistant's cards.

The method above has a slight problem, in that it is possible for an assistant to not have the right card available amongst the ones in his hands. An improved method is as follows:

The assistants have to do value and suit at the same time. I.e. They remove two cards such that the remaining cards have a total value that is a multiple of 13 and a total suit that is a multiple of 4. This gives many more possibilities for signalling the right totals. You determine the missing card by looking at the sum of all four offered cards instead of two for the value and two for the suit.

I'm not sure what would be the easiest way to tweak this to accommodate a joker, but you could

consider the joker to have value 14 and suit 5. Then use the same method except modulo 14 and 5 instead of 13 and 4.

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Here's a solution that is effective (guaranteed to work), but may be too simple to be impressive unless the assistant's cards aren't shown to the audience.

By the pigeon-hold principle, each deck of 26 cards must have at least 1 pair of duplicate numbers (e.g. 2 of clubs and 2 of spades; or 3 of hearts and 3 of diamonds). You can only have 14 unique 'numbers', counting picture cards and the joker, so a deck with more than 14 cards must have duplicates. Further, with 26 cards, the deck must have multiple duplicate numbers.

Each assistant picks a pair of duplicate numbers different from the spectator's number.

The magician simply discards all duplicates. The remaining card was chosen by the spectator.

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  • $\begingroup$ What if the spectator picks the Joker, Assistant 1 has A-K clubs and A-K hearts, Assistant 2 has A-K spades and A-K diamonds? $\endgroup$ – Jay Jan 17 '18 at 5:00
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    $\begingroup$ I like this thought, but note that this approach does require the spectator to show their card to the magician's assistants, which maybe weakens the effect. It changes the magic from "How did the magician know which card was the spectators'?" into "How did the assistants communicate the card's identity to the magician?" And no magician wants their assistants to be the ones the audience thinks did something amazing (even though the assistants are almost always the ones doing the really impressive work). $\endgroup$ – Trevor Powell Jan 17 '18 at 5:00
  • $\begingroup$ @Jay I believe in that case, each assistant has thirteen duplicate numbers? One set each of A-K? $\endgroup$ – Trevor Powell Jan 17 '18 at 5:11
  • $\begingroup$ @TrevorPowell I see, but it definitely requires that the assistants see the spectator's card $\endgroup$ – Jay Jan 17 '18 at 5:24
  • $\begingroup$ @Jay Then Asst 1 picks (say) 2C + 2H, Asst 2 picks (say) 3S + 3D. M discards both 2s and both 3s, retaining the Joker. $\endgroup$ – Lawrence Jan 17 '18 at 5:47
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I think it's important to remember that this is a magic trick, and most card magicians (like most folks in the general public) aren't entirely fantastic with maths. So... relying on a maths-based method is probably not the best plan. (particularly one that involves doing a modulus by 53 in your head while also giving an entertaining performance; I mean, it's flattering that you think I could do that, but it does seem just a wee bit optimistic.)

Personally, if I was going to perform this trick such that a spectator would describe it as was done in the OP, then I would use one of two methods:

(That's a lie; I'd use Method Two every time and sleep like a baby. But Method One would work and is provided here for completeness)

Method One (aka: the complicated way):

Before the trick began, I would remove the 9 of hearts, the 5 of diamonds, the 7 of spades, and the 4 of clubs. (Really, any cards will do, as long as they're unremarkable cards; don't use aces or face cards). Then, I'd re-seal the deck, complete with both jokers, the advertisement cards, etc. When presenting the deck, I'd make a big deal about removing one of the jokers and the advertisement cards as I introduced what was about to happen, and explaining that it was now a deck of 53 cards, including the single remaining joker.

Caveat: If you would feel bad about lying to the audience about there being 53 cards in the deck when there are actually only 49 cards (what sort of magician are you??), then you can include a second copy of the 9 of diamonds, the 5 of hearts, the 7 of clubs, and the 4 of spades (the alternate same-color-suits of the missing cards), to bring the deck's actual size back up to 53 cards. This trick will work the same, either way, but you're slightly cleaner at the end of the trick if you don't add these duplicate cards.

Anyhow. At the start of the trick, the two assistants would each have two cards; one would have the 9 of hearts and the 7 of spades, the other would have the 5 of diamonds and the 4 of clubs; the cards that we removed from the deck, earlier. These could be tucked away in a coat pocket or any other standard card-stashing positions. The assistants palm their cards while audience attention is on the spectator, who is selecting their card.

While being handed their half-decks after the spectator has made their selection, the assistants would then covertly add their palmed cards to the tops of their half-decks, and then do some false shuffles or whatever, and eventually would each "select" their two cards.

At the end, when the magician is given the shuffled set of five cards, the magician can easily tell that the spectator's card is the only one that isn't one of those four predetermined cards.

Further commentary on Method One:

In summary: the overall approach this method takes is to have four cards not in the deck at all, so that they can't be picked by the spectator. Then these previously-missing cards get added into the half-decks by the assistants, and then shuffled into the final set of five cards, so the deck invisibly becomes complete by the end of the trick.

As a result, you're entirely clean and everything can be examined at the end. Exception: if you added duplicate cards to the deck in order to keep the "53 cards in the deck" claim honest, then there are four extra cards in the deck at the end of the trick. In that case, maybe don't try to squeeze the cards back into the box while cleaning up; might be a bit of a giveaway if they don't fit back in. :)

It's also worth noting that as is common in magic acts involving assistants, the assistants here are the ones who have done all the difficult or skillful work; the magician does literally nothing, except to be engaging and to keep the audience's attention focused elsewhere while the assistants do their dirty deeds.

Method Two (aka: the easy way) (aka: untying the Gordian knot):

Just force the card.

Since you know what card they chose, everything else is just theater; you will trivially be able to identify that known card at the end of the trick.

Further commentary on Method Two:

Forcing a spectator to choose a specific card that you want them to choose, without them being aware that you've done it is an ancient trick. There are hundreds if not thousands of ways to do it. And in this case, it solves the entire trick; nothing else needs to be done.

Personally, I'm fond of using the Classic Force whenever practical. It requires a lot of practice to pull off with any sort of consistency, but if you can do it cleanly, it feels entirely fair and natural and nobody will ever suspect you've done anything untoward. Folks with less card manipulation skill (or audience-member-manipulation skill) may want to use the Hindu Force, or any other ordinary card force with which they are comfortable.

Under this approach, you're completely clean and everything can be examined both before and after (or even during!) the trick; there are no extra cards or missing cards or doctored cards involved. (unless you used them as part of your force, of course!)

It's worth noting that a lot of card tricks – maybe even the majority of them – follow this same "force the selection of a card, then provide an elaborate piece of theater to reveal the card" pattern. The theater is there to distract the audience; give them other things to be suspicious of, when the only real "trick" happened right at the start, with the selection of the already-known card.

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  • $\begingroup$ Note: Where I've discussed slight-of-hand in the above, I have described what each move accomplishes, and the standard names for the moves, but I haven't given instructions about how to actually perform them. Those who are interested can search Google for the names to find teaching resources. :) (It's so much easier to learn this stuff now than it was back when I was learning it!) $\endgroup$ – Trevor Powell Jan 17 '18 at 9:15
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There is a way to make this work mathematically. Some answers have hinted at this, or parts or it already, but here's how I think it can be done.

First, ignore the face value of the cards. We are going to assign each card a sequential value from 0..52. This trick works because the sum of numbers from 0..52 mod 53 is 0. Any numbering method can be used and agreed upon beforehand.

The spectator removes a card and each assistant gets a pile of cards.

Each assistant sums the value of the cards mod 53 and remembers the value. Say there are 3 values, Assistant A's count $= a$, Assistant B's count $=b$, and the spectator's value $=c$. We know that $a + b + c = 53$ so it's a matter of deduction that the spectator's card is $c = 53 - a - b$.

But how do the assistant's communicate their count to the magician? The magician needs to read it out of a shuffled pile so there must be a code that can be deciphered after mixing. Here's my suggestion which isn't perfect. The assistants agree beforehand that the one will give a heart and a club, while the other gives a diamond and a spade. Adding the heart and club should give value $a$ and the diamond and spade should give $b$. It may additionally be decided that whoever holds the most hearts in their pile choses that suit and it's pair.

However,

It isn't actually required that the assistants count the cards. If both assistants give pairs that sum to 0 (mod 53) then the spectator's card will be self evident even after a shuffle. This pairing isn't always possible though.

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  • $\begingroup$ But.. the spectator's card gets shuffled into the pack received by the magician, right? If the spectator's card is any card except the joker, then the magician is going to get a hand with two hearts, or two clubs, or two diamonds, or two spades, and therefore can't tell which of the four cards came from the assistants, and can't actually do this math, right? $\endgroup$ – Trevor Powell Jan 17 '18 at 6:14
  • $\begingroup$ @TrevorPowell Yep, you would have to take two guesses, and only one of them should give you the card that you now see in your hand. $\endgroup$ – Jay Jan 17 '18 at 6:16
  • $\begingroup$ interesting, the other problem is if one of the assistant can't give a spade/ diamond or can't adjust the modulo 53 number he gives to the magician with what he has in spade/diamond. $\endgroup$ – Untitpoi Jan 17 '18 at 8:33

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