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Impossible Card Magic Trick Math Question

Posted on 2005-04-10
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Don't think this question has a solution but was asked this:

With two decks of cards, one with red and one with orange backs.  Audience picks five cards. The Magician's Helper shows both sides of four of the five cards in some order to the Magician.  The Magician needs to determine the value, suit and back color of the fifth card.  How would you go about doing this using Counting/Pigeonhole methods?
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Question by:bcsmess
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Expert Comment

ID: 13749165
The 5 cards given to the first player must contain two cards of the same
suit. Of these, one must be six or less greater than the other, wrapping
around from K to A to 2. For example, suppose that there are two spades, the
5 and the J. The spades which are six or less greater than the J are the Q,
K, A, 2, 3 and 4, and the 5 is not among these; but the spades which are six
or less greater than the 5 are the 6, 7, 8, 9, 10 and J, and the J is among
these. Thus of the two spades, the J is the one which is six or less greater
than the 5.

The first player then keeps the card which is six or less greater than the
other (in the example, the J of spades) and shows the other (in the example,
the 5 of spades) as the first (leftmost) of the 4 exposed cards. The
remaining cards are considered ordered by rank, with A highest and 2 lowest,
and among cards of equal rank in the order SHDC, with spades(S) highest and
clubs(C) lowest. The first player shows the remaining three cards in one of
six permutations according to this ordering, corresponding to the numbers 1
to 6, whichever of these is the difference between the hidden card and the
one already shown. This means, if the three cards are called L (for low), M
(for middle) and H (for high) according to the ordering, then they should be
placed in one of the following orders:

1   LMH
2   MLH
3   LHM
4   HLM
5   MHL
6   HML

thus in the example, he would place them in the order HML, since the J of
spades is 6 greater than the 5.

When the second player returns to the room, he knows that the hidden card is
of the same suit as the leftmost card, and he adds to the rank of this card
the number 1 to 6 corresponding to the permutation of the remaining three
cards, thus identifying the hidden card uniquely. In the example, the second
player sees

5S  H  M  L

and since HML corresponds to 6, he adds 6 to the 5 of spades to get the J of

Of course, the first player may have more than one choice, since there might
be three cards of the same suit, or two cards each of two different suits,
but he simply chooses one pair of cards of the same suit, and the fact that
there are other choices is irrelevant.

http://www.anamorph.com/docs/ct/cards.html
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Expert Comment

ID: 13749279
The above method does not give enough information with a deck of 104 cards, but if the Helper can choose which side of each of the four cards to show first, that's 4 additional bits of information.
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LVL 31

Expert Comment

ID: 13749417
There are numerous ways of doing nothing to do with Maths at all, suggesting it is Mathematical to the audeince just sends them off in the wrong direction

1st Card:-  If back shown first then then suit is clubs or spade if the  first finger of hand holding card is close to half way up the card then spades else clubs. Same idea for hearts but front shown first.

2nd Card and 3rd Card: You represent numbers from 1 to 6 with a card. Showing colored back first means 1, 2 or 3. If first finger of hand is near the bottom then 1, center then 2 and top then 3. Similar idea for 4,5,6. By splliting the value into tthe sum of 2 numbers from 1 to 6 then then the number can represented. (for 13 see below)

last Card:- If back shown first then the back of 5th card is that color else it is the other color. Add one to the number obtained  for cards 2 and 3 if held close to center.

There are multiple ways of doing this, use your imagination
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Expert Comment

ID: 13749481
this internet based trick is wonderfull.   http://urbanlegends.about.com/library/bl_card_trick.htm
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Author Comment

ID: 13749524
Is there any method without using fingers and all...as in just purely from revealing the value, suit, and backing of the first 4 cards?
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Expert Comment

ID: 13749536
Due to the multitiude ways of doing this ( eg using left and right hands or the order they are shown as the cards form form an ordered set) this sounds like a good job exam/interview question
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Expert Comment

ID: 13749568
You have 4 cards to use and can be thought of as from a set of values from 1 to 104 ( ie each card can be said to be higher or lower than another.) You now after 5 have been removed got 97 left.  The best you can do with 4 cards is map them onto 4*3*2*1=24  values by ordering them onto the table but you have 97 possible value to identify, (eg there is not a lot you an do with 2 aces of hearts and 2 aces of clubs)  ie it is not possible.
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Expert Comment

ID: 13749671
..you can mutliple this by 8 if you use the 3 gaps between the cards, ie a gap means 1, touching means 0, so 3 bit binary number of information.
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Expert Comment

ID: 13749727
oops arithmetic:-  "You now after 5 have been removed got 99 left."
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Expert Comment

ID: 13750183
> The best you can do with 4 cards is map them onto 4*3*2*1=24  values by ordering them onto the table but you have 100 possible value to identify,
It's not quite that bad.  With 52 cards, there are potentially 48 possible values to identify,
but you can still do it because the Magician's Helper can choose one card out of the 5 given to be within a more easily identifed subset of the 48
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Expert Comment

ID: 13750199
I thought there were 2 decks of cards?

With one pack by arranging the cards horizontally or vertically will give 1 more bit of info and hence the 48 possiblities.
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Expert Comment

ID: 13750253
With one deck, no more bits of info are necessary than the order of the shown cards.  (and the choice of which of the 5 cards to show)
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Expert Comment

ID: 13750389
This would infer that the helper could impart 1 of 2 properties on the set of 4 cards. I can not see immediately how this would be done. I am open to argument on this and will think about it, but bed now.
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Expert Comment

ID: 13750788
i would simplify what maybe written above a little:

2 decks = 26 cards, 4 are known => 22 unknown cards .
4 cards will be ordered from strong to weak, and red stronger then orange.
the order will present the missing card.

cards, has values X1,X2,X3,X4 there are 24 options to order them.

x1,x2,x3,x4 => strongest card missing is the 5th card,
x1,x2,x4,x3 => 2nd strongest card missing is the 5th card,
...
x2,x1,x3,x4 => 7th strongest card is the 5th ,
...
etc...
x4,x1,x2,x3 => 19th card is the 5th
..
x4,x2,x3,x1 => the most weak card is the 5th

nice question, nice trick.

tal
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Expert Comment

ID: 13752065
2 decks = 104 cards  not 26
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Expert Comment

ID: 13752273
GwynforWeb, thanks. here is my correction :

basically, the order of exposuring cards, is a way of coding a number. for 4 cards, prooved below, that the spanned interval of numbers that can be encoded is 384 < 104

each card exposed 2 sides: if it's front exposed before back it is A, else it is B.
exposure vectore = {AAAA} : all cards exposed front first
another example = {ABAB} : 1st and 3rd cards exposed front first.
exposure vector gives a number A='0' , B='1' from 0 to 15

the vector given by the order (see previous commnet) can span values from 1 to 24

=> the Magician's Helper can give HINT to the Magician about any card out of remaining 16*24 cards = 384 cards.
Magician find the number represent by the order vector (OV), and by the exposure vector (EV)
now, card_number = OV*16 + EV

card_number should be agreed between Mag and Helper : Spade_Red = 7*13+card, Heart_Red = 6*13+card ... Club_Orange=card

cards game for home practice :

for 3 cards exposure :
8 * 6 = 48 => if only 1 deck involved , magician take off one card(known), audience take 4, helper expose 3 of them both sides, and Magician has all information to know the missing 1 card out of remaining 48 cards.
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Expert Comment

ID: 13752560
> Don't think this question has a solution but was asked this:

with the solution I suggested, the Magician can find missing card out of remaining 7 decks (2 colors back)
{384 / 52 = 7.38} and exactly from a precised deck.
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LVL 31

Expert Comment

ID: 13753526
I think we are looking at the cards being layed out before the magician arrives back, we have been through various ways of adding more information, please the thread.
I have not eliminated ozo's idea that the choice of cards to show can be used and results in an interesting problem that is purely mathematical. ie given 5 numbers from 1 to N is there a mapping that will map 4 onto a 5th, if so how big can N be. This assumes the helper can choose the 4 to show.
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Expert Comment

ID: 13753740
....just thought, the chosen card can not possibly be part of the 4 as need only be placed in the last position to be identified.
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Expert Comment

ID: 13753790
wrongly typed : 384 > 104 ...
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LVL 31

Expert Comment

ID: 13759266
I misread ozo's comment >> Helper can choose one card out of the 5 given to be within a more easily identifed subset of<<

I agree this can be done, I have thought of a way. (order 5 cards and choose either 2nd or 4th, which ever give the smaller gap in number between the cards either side)

In short I believe this can be done for 1 pack but not for 2 as yet.
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Expert Comment

ID: 13759896
GwynforWeb, read my proof, not only 1 pack, even not only 2 packs, but a full 7 packs.

challenge me if you want...
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LVL 31

Expert Comment

ID: 13761881
Talmash , Please read the thread.  There are multiple ways of communicating more information eg by the order in which they are put down, what side is shown first  etc. and some of these types of ideas have been discussed The trick has the cards placed on the table with out the magician seeing them being put down.
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Expert Comment

ID: 13762190
but he must see both sides, isn't it ?

> The Magician's Helper shows both sides of four of the five cards in some order to the Magician

I defined 2 vectors :
1) exposuring vector (0-15 values = EV )
2) value order vector (1-24 values = OV)

since Mag. sees both sides, Helper choose the exposuring vector value(Helper chooses foreach card, how to show it to the Mag. front to back OR back to front)
and since 4 cards are order somehow on the table (from left to right etc...) this represent the value order vector.

therefor, both Helper and the Mag. can be agreed on the same algorithm to represent each card with a number 0-103 = X

the helper do:
EV = X % 16
OV = int(X/16)

the Mag. do : OV*16+EV = X

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Expert Comment

ID: 13764984
Given 4 additional bits, 14 packs can be done
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Accepted Solution

GwynforWeb earned 2000 total points
ID: 13765866
if the order they are shown can be different from the order they are placed on the table as well (ie 24 ways so 1 more bit per card) then 31 packs can be done.
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Expert Comment

ID: 13770224
Gwyn, as a digital hw eng. I had to give digitized solution.

I'll post a nice question named "3 colors 100 hats", challenge yourself - but this time ,
no finger tricks are allowed.

antway nice question, nice & various solutions.

tal
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