card trick
Michael Jørgensen
I recently learnt of a card trick involving three people: The magician, his
assistant, and the victim. The victim draws (or chooses) five cards from an
ordinary deck of 52 playing cards and gives them to the assistant. The
assistant removes one of the five cards, and then hands the remaining four
cards to the magician. Upon inspecting the four cards, the magician
announces what the fifth card was.
The method is based on the order of the four cards, as well as the selection
of the fifth card. In other words, this involves cooperation between the
magician and the assistant.
Are there any references to this trick, e.g. the name or origin, or perhaps
variations?
-Michael.
Dgates
Ahh, yes. I remember this. The assistant finds two cards of the same
suit -- say Spades.
She then removes the spade that's the least number of cards higher
than the other spade (even if it means counting around in circles).
For example, the 3 of spades is four higher than the queen of spades.
In this case, the assistant removes the 3 of spades and places the
queen of spades at the top of the stack.
She then arranges the three remaining cards to indicate what number
the magician should add to the queen. She uses a code based on
(L)ow-(M)edium-(H)igh values of the three cards.
Any 7 is higher than any 6, and the 7 of diamonds is higher than the 7
of clubs.
1 LMH
2 LHM
3 MLH
4 MHL
5 HLM
6 HML
So, since the magician has to add 4 to the queen of spades, she
arranges "MHL" -- Medium-High-Low -- or possibly "7 of clubs," "7 of
diamonds," "6 of clubs."
If the magician's stack contains:
queen of spades
7 of clubs
7 of diamonds
6 of clubs
then he knows the missing card is the 3 of spades.
--
dga...@spamfreelinkline.com
John Bailey
<Joe...@SoftHome.net> wrote:
History of this Card Trick by Robert Orenstein
(quoting)
This trick was originally posted as a puzzle in the rec.puzzles
newsgroup. Someone in the newsgroup had been applying for a
programming job, and as part of the interview, they gave him half an
hour to figure out how this trick might be done. He didn't have to
solve the puzzle to be hired, but progress on it would have helped
him. (I don't remember whether or not he got the job).
I tried to solve the puzzle, and came close, but my solution wasn't
perfect. A few different solutions were posted, the most elegant being
by Bob Vesterman (does anyone know where to reach him?). I implemented
Bob's solution in HyperCard, and later adapted it for the Web.
(quoted from: http://www.anamorph.com/docs/ct/cards.html)
John Bailey
http://home.rochester.rr.com/jbxroads/mailto.html
Runyn
"John Bailey" <john_...@rochester.rr.com> schrieb im Newsbeitrag
news:4061e285....@news-server.rochester.rr.com...
> [...]
> (quoted from: http://www.anamorph.com/docs/ct/cards.html)
>
I laughed about the following quote from the same page:
"The rec.puzzles newsgroup is a lot of fun (well, it's fun if you like
puzzles).
But please don't go there and start posting puzzles madly; they've seen most
of
them before and these people do bite! ..."
:)
\/ `/ `|´ | `´ \´ ``´
Mark J. Tilford
> On Wed, 24 Mar 2004 19:35:43 +0100, "Michael Jørgensen"
><Joe...@SoftHome.net> wrote:
>
>>Hi,
>>
>>I recently learnt of a card trick involving three people: The magician, his
>>assistant, and the victim. The victim draws (or chooses) five cards from an
>>ordinary deck of 52 playing cards and gives them to the assistant. The
>>assistant removes one of the five cards, and then hands the remaining four
>>cards to the magician. Upon inspecting the four cards, the magician
>>announces what the fifth card was.
>>
>>The method is based on the order of the four cards, as well as the selection
>>of the fifth card. In other words, this involves cooperation between the
>>magician and the assistant.
>>
>>Are there any references to this trick, e.g. the name or origin, or perhaps
>>variations?
>
> History of this Card Trick by Robert Orenstein
> (quoting)
> This trick was originally posted as a puzzle in the rec.puzzles
> newsgroup. Someone in the newsgroup had been applying for a
When was that? In _The Unexpected Hanging_, by Martin Gardner, he says it
was mentioned in chapter 14 of _Math Miracles_ by Wallace Lee (1950).
--
------------------------
Mark Jeffrey Tilford
til...@ugcs.caltech.edu
Risto Lankinen
"Michael Jørgensen" <Joe...@SoftHome.net> wrote in message
news:c3sko8$1sr8$1...@news.cybercity.dk...
>
> Are there any references to this trick, e.g. the name or origin, or
perhaps
> variations?
Here's one variation called Devil's Poker:
This game is played using one complete suit of cards (total 13 cards).
The devil chooses five cards and hands the remaining eight cards to
you. You choose another five cards and pass the remaining three
cards, stacked so that all three card face the same way, to your
guardian angel. Your guardian angel must then guess the hands
[of five cards] each of you [the Devil and the player] possesses.
An exacting comment: The only information you can pass to your
guardian angel is the choice of face values, and the ordering, of the
three cards you pass her.
Cheers!
- Risto -
Dgates
I think I like this one.
I've got two ideas:
1. You'd like to take each of the devil's cards and increment them by
some number, then encode that number with the remaining three cards.
(Alas, this one isn't working for me, since the devil can prevent you
from incrementing by 1,2,3,4,5 and 6 pretty easily. Maybe the six
possible increment numbers should be 2,3,5,7,11 and 13 instead?)
2. No matter what the devil takes, you grab the five lowest possible
cards you can, then somehow communicate something with the three cards
remaining. (But this one doesn't work either, since there are so many
possible ways that you could hand the angel a J,Q and K.)
Eric Farmer
Rather than a variation, the generalization of this trick is
interesting. That is, given m cards chosen from a deck of n cards,
the magician hands k of them to the spectator to be announced, after
which the partner announces the remaining m-k chosen cards. The
original trick corresponds to n=52, m=5, and k=4.
1. Determine conditions on (n,m,k) for which the trick can be done.
This is relatively straightforward; see
http://home.comcast.net/~erfarmer/puzzles/HTMLLinks/index_10.html.
2. Describe a general procedure (i.e., for general (n,m,k)) for
performing the trick. Note that the usual solution for (52,5,4)
depends on the "13 ranks by 4 suits" labeling of cards unique to this
instance of the problem. Is there a method which is more easily
extended to, say, a 137-card deck with labels 1 through 137? I have
asked this question here before, but have been unable to find a
satisfactory solution.
Eric Farmer
Patrick Hamlyn
>...Note that the usual solution for (52,5,4)
>depends on the "13 ranks by 4 suits" labeling of cards unique to this
>instance of the problem. Is there a method which is more easily
>extended to, say, a 137-card deck with labels 1 through 137? I have
>asked this question here before, but have been unable to find a
>satisfactory solution.
Since this is an 'encode in advance' puzzle, the actual face markings of the
cards are effectively irrelevant, and you can make up your own 'categorisation
scheme' and map card numbers to your scheme accordingly. So your question
reduces to 'what scheme for mapping cads into categories can you come up with
for arbitrary numbers of cards?'
In fact mathematically you don't need any scheme at all, it only helps as a
memory aid. So make yourself any convenient category which is a power of two
near the square root of the number of cards.
If the number of cards gets large, use three categories etc.
--
Patrick Hamlyn posting from Perth, Western Australia
Windsurfing capital of the Southern Hemisphere
Moderator: polyforms group (polyforms...@egroups.com)
Michael Jørgensen
news:dsv8c.12407$k4.2...@news1.nokia.com...
> Here's one variation called Devil's Poker:
>
> This game is played using one complete suit of cards (total 13 cards).
>
> The devil chooses five cards and hands the remaining eight cards to
> you. You choose another five cards and pass the remaining three
> cards, stacked so that all three card face the same way, to your
> guardian angel. Your guardian angel must then guess the hands
> [of five cards] each of you [the Devil and the player] possesses.
>
> An exacting comment: The only information you can pass to your
> guardian angel is the choice of face values, and the ordering, of the
> three cards you pass her.
Well, the angle must determine the distribution of five cards out of the
remaining ten, i.e. C(10,5) = 252 possibilities.
Given eight cards, you have C(8,3) * 3! possible ways to choose 3 cards.
This amounts to 336 possibilities.
These values are so close, that there is not much room for redundancy. You
need to encode almost three bits of information in each of the three cards.
A total of eight bits should suffice.
It appears to be solvable, and a solution could perhaps be found using brute
force. However, I have no ideas how to devise a scheme that is easily
memorized.
Any clues?
-Michael.
Risto Lankinen
"Michael Jørgensen" <in...@ukendt.dk> wrote in message
news:4063e945$0$184$edfa...@dread11.news.tele.dk...
>
> Well, the angle must determine the distribution of five cards out of the
> remaining ten, i.e. C(10,5) = 252 possibilities.
>
> Given eight cards, you have C(8,3) * 3! possible ways to choose 3 cards.
> This amounts to 336 possibilities.
>
> These values are so close, that there is not much room for redundancy. You
> need to encode almost three bits of information in each of the three
cards.
> A total of eight bits should suffice.
>
> It appears to be solvable, and a solution could perhaps be found using
brute
> force. However, I have no ideas how to devise a scheme that is easily
> memorized.
>
> Any clues?
Here's a clue to one possible solution.
Call the three cards that the player passes to the angel
as A, B, and C. There's a function F that takes the face
values of two cards in specific order, and returns a face
value of an unknown card. The following six invocations
of the function will then return six cards:
F(A,B) = x1, F(B,A) = x2, F(A,C) = x3, F(C,A) = c4,
F(B,C) = x5, and F(C,B) = x6.
Finally, use the ordering of the cards A,B,C to indicate
which xN must be discarded so that the remaining results
correctly indicate either hand [player's or devil's].
An example (deliberately wrong so as to just act as a clue)
of what kind of a function F should be is the following:
F(A,B) = 2*A+3*B mod 13;
The tricky part then becomes finding the simplest possible
F that can cover all possible permutations of 8-out-of-13
cards.
Cheers!
- Risto -
P.S. Another solution would involve a similar but different
function G(A,B,C) using all permutations of A,B,C except,
say, the one in which the cards were actually handed. In
fact, any two variable F-function solution can be easily
transformed into an instance of a three variable G-function
solution that uses logical operators.
moriman
news:57n360hejfrc3jmh2...@4ax.com...
> On Wed, 24 Mar 2004 19:35:43 +0100, "Michael Jørgensen"
> <Joe...@SoftHome.net> wrote:
>
> >Hi,
> >
> >I recently learnt of a card trick involving three people: The magician,
his
> >assistant, and the victim. The victim draws (or chooses) five cards from
an
> >ordinary deck of 52 playing cards and gives them to the assistant. The
> >assistant removes one of the five cards, and then hands the remaining
four
> >cards to the magician. Upon inspecting the four cards, the magician
> >announces what the fifth card was.
> >
> >The method is based on the order of the four cards, as well as the
selection
> >of the fifth card. In other words, this involves cooperation between the
> >magician and the assistant.
> >
> >Are there any references to this trick, e.g. the name or origin, or
perhaps
> >variations?
>
> Ahh, yes. I remember this. The assistant finds two cards of the same
> suit -- say Spades.
>
What if there aren't two cards of the same suit?
mori
moriman
news:Vml9c.366$N2.200@newsfe1-win...
Apologies, hadn't read the OP correctly, i.e. the *assistant* withdraws a
card, so it is (if relevant) for there to be guaranteed two same-suit cards
mori
> mori
>
>
>
>
Jeffrey Johnson
>"Dgates" <dga...@nospamlinkline.com> wrote:
>> On Wed, 24 Mar 2004 19:35:43 +0100, "Michael Jørgensen"
>> <Joe...@SoftHome.net> wrote:
>>
>> >Hi,
>> >
>> >I recently learnt of a card trick involving three people: The
>> >magician, his assistant, and the victim. The victim draws (or
>> >chooses) five cards from an ordinary deck of 52 playing cards and
>> >gives them to the assistant. The assistant removes one of the five
>> >cards, and then hands the remaining four cards to the magician. Upon
>> >inspecting the four cards, the magician announces what the fifth card
>> >was.
>> >
>> >The method is based on the order of the four cards, as well as the
>> >selection of the fifth card. In other words, this involves
>> >cooperation between the magician and the assistant.
>>
>> suit -- say Spades.
>
>What if there aren't two cards of the same suit?
Considering the restrictions above (standard 52-card deck, draw five
cards) this outcome is very unlikely. How unlikely I'll leave you to work
out yourself. Clue: a standard 52-card deck has four suits. :)
Jeffrey Johnson
--
Replace the rent-a-car with 'avis' to reply
moriman
> <Joe...@SoftHome.net> wrote:
>
> >Hi,
> >
> >I recently learnt of a card trick involving three people: The magician,
his
> >assistant, and the victim. The victim draws (or chooses) five cards from
an
> >ordinary deck of 52 playing cards and gives them to the assistant. The
> >assistant removes one of the five cards, and then hands the remaining
four
> >cards to the magician. Upon inspecting the four cards, the magician
> >announces what the fifth card was.
> >
"Upon inspecting the four cards"
Does this include the backs of the cards? i.e. are the backs of the cards
symmetrical?
mori
moriman
"Jeffrey Johnson" <jsjo...@ucdhertz.edu.INVALID> wrote in message
news:Pine.GSO.4.58.04...@vici.ucdavis.edu...
Yup, realised and posted apology for mistake ;-)
mori
Gerard Michon
as "Fitch the Magician", 1894-1974) who earned the first math Ph.D. ever
awarded by MIT (in 1927).
A nice history of the trick appears in a 2002 article by Michael Kleber,
published in the "Mathematical Intelligencer" and available online (in PDF
format) at the following URL:
http://people.brandeis.edu/~kleber/Papers/card.pdf
In that paper, Kleber also deals with a generalized case where all cards but
one are shown (namely k=m-1 with the notation used by Eric Farmer below, or
k=n-1 using the notation of Kleber, who calls d the size of the deck).
Admittedly, his approach may not be very practical...
Also, I've just uploaded a NEW webpage on this subject (with other comments
and links) which I intend to update as time and/or new discoveries permit...
Here's the address:
http://www.numericana.com/answer/magic.htm#fitch
Enjoy...
Gerard P. Michon, Ph.D.
www.numericana.com
"Eric Farmer" <erfar...@comcast.net> wrote in message
news:a1babb92.04032...@posting.google.com...
Chris Thompson
Risto Lankinen <rlan...@hotmail.com> wrote:
[...]
>Here's one variation called Devil's Poker:
>
>This game is played using one complete suit of cards (total 13 cards).
>
>The devil chooses five cards and hands the remaining eight cards to
>you. You choose another five cards and pass the remaining three
>cards, stacked so that all three card face the same way, to your
>guardian angel. Your guardian angel must then guess the hands
>[of five cards] each of you [the Devil and the player] possesses.
>
>An exacting comment: The only information you can pass to your
>guardian angel is the choice of face values, and the ordering, of the
>three cards you pass her.
Others have pointed out that there is enough information (by a 4/3
ratio) to do this, although I haven't seen a neat human-memorable
algorithm posted yet. [Although I suppose only the encoding needs
to be human-memorable and not the decoding: the angel is, presumably,
super-human...]
But suppose the devil, who is after all a greedy fellow, grabs six
cards, leaving you with seven. You are still allowed to give three
cards to the angel, as before. Now there's _exactly_ enough information
available: 13*12*11 = {13 choose 6} (or 10! = 7!*6!) so there's no
leeway for error.
Chris Thompson
Email: cet1 [at] cam.ac.uk
Jugoslav Dujic
| 30 spammers agree that moriman wrote:
|| "Dgates" <dga...@nospamlinkline.com> wrote:
||| On Wed, 24 Mar 2004 19:35:43 +0100, "Michael Jørgensen"
||| <Joe...@SoftHome.net> wrote:
||| Ahh, yes. I remember this. The assistant finds two cards of the same
||| suit -- say Spades.
||
|| What if there aren't two cards of the same suit?
|
| Considering the restrictions above (standard 52-card deck, draw five
| cards) this outcome is very unlikely. How unlikely I'll leave you to work
| out yourself. Clue: a standard 52-card deck has four suits. :)
Easy: a spade, a heart, a diamond, a club, and a hippogriff :-))
The origin of the story is a humoresque (I forgot the author)
issued in Bridge World. Briefly, a bridge player dies and goes
to Hell. In one game against the devil and its assistants, he
holds S AKQJ H AKQ D AKQ C AKQ as dealer, and opens 7NT
(all 13 tricks, no trumps). His opponent doubles and leads
a card with a strange-looking symbol. His partner asks amazedly
"No hippogriffs partner?" :-).
--
Jugoslav
Jeffrey Johnson
That's pretty funny. I suppose that in this case, a standard deck is
whatever the devil says it is.
-Jeffrey Johnson
Boggle
I've spent some time thinking about this puzzle and haven't seen any
posts in the last week. Risto's follow-up about the two- or
three-variable function is a neat idea, but I can't figure out how to
generate a function such that (a) it's obvious which cards to pass to
impart the information and (b) it is guaranteed that the cards I need
will be among the eight I have.
I've come up with a different approach that is based on the
distribution of cards in the Devil's hand. The difference between the
first and second cards I pass to my angel indicates the distribution.
The third card (and some other info from the first two) nails down the
ranks.
The methodology isn't exactly *simple*, but it wouldn't be too bad to
memorize. You could probably learn all the tricks in fifteen minutes,
after which you'd want to practice with a cheat sheet for several
hands before going from pure memory. Tedious, sure, but less so than,
say, learning how to play bridge. :-)
The whole thing is too long to post. Even just the basic idea is a
little lengthy. I apologize. Here it is, and anyone who wants more
(via post or email) can let me know.
-
-
-
-
-
-
For symmetry, I assume the cards to be circular in rank. In other
words, Q, K, A, 2, 3 is a consecutive five-card sequence. (It's not
that bad, once you get used to it.)
The Devil has 1,287 different possible hands. These have the
following distributions:
5 card run: 13 hands (Ex: Q K A 2 3)
4/1 91 hands (Ex: Q K A 2 7)
3/2 91 hands (Ex: 3 4 7 8 9)
3/1/1 273 hands (Ex: 3 7 8 9 Q)
2/2/1 273 hands (Ex: A 2 4 5 7)
2/1/1/1 455 hands (Ex: 2 5 7 8 K)
1/1/1/1/1 91 hands (Ex: 3 5 8 J K)
Note: Remember, rank is circular; there is no highest card. Make
sure to include "wrap-around" runs that span an ace.
Now, for any first card I throw, the second card can be 1 to 12 more
than that card (modulo 13). If the first card I pass is a 6, then I
can increase by anything from +1 (pass a 7 next) to +12 (pass a 5
next). I assign these increases across the distributions. More
common distributions get more assignments. Each of them -- +1 to +12
-- has 143 different hands it could signify. [143 is 13 x 1 x 11.
The first pass could be anything. The second is determined by the
first. The third could be any from the remaining 11.]
Here's how I choose to distribute them. Yes, there was a method to my
madness:
Distrib. Devil's Add
5 or 4/1: 104 hands 1
3/2 91 hands 12 [Same as minus 1]
3/1/1 273 hands 4 or 9 [Plus 4 or minus 4]
2/2/1 273 hands 2 or 11 [Plus 2 or minus 2]
2/1/1/1 455 hands 5 6 7 8 or 10
1/1/1/1/1 91 hands 3
So if the first two cards you see from me are 5 9, you'll know it's a
3/1/1 distribution. (You'll also know other things since I chose +4
not +9, but I'll get to that.) If I pass you Q 5, then I've added 6,
and it's a 2/1/1/1.
From there, you need conventions for how to assign the ranks. I came
up with some that I think aren't too onerous.
FOR EXAMPLE:
For the 5 or 4/1 distributions, it's +1, and I start by passing two
ascending, consecutive cards, like 2 3. GENERALLY I will start by
passing the two cards immediately below the Devil's four-card run.
The third card I pass will GENERALLY be the card below the Devil's
off-card, i.e., the one outside the run.
I pass 5 6 Q means Devil has 7 8 9 10 K.
I pass 5 6 3 means Devil has 4 7 8 9 10.
I pass J Q 6 means Devel has K A 2 3 7.
Special case 1: Three cards in ascending order will mean a five-card
run:
I pass 5 6 7. This will mean the Devil has 8 9 10 J Q.
Special case 2: If the Devil's off-card is two below the bottom of
the four-card sequence (like 5 7 8 9 10), then I can't follow the
general plan. Instead I pass the two immediately below the off-card,
then the card in between the off-card and the four-card run:
I pass 3 4 6. (Note: Clearly the Devil does not hold 5 6 7 8.)
This will mean the Devil has 5 7 8 9 10.
[For this distribution, the third card I pass will never be four,
five, or twelve higher than the first. I will never pass 5 6 9, 5 6
10, or 5 6 4. These have no meaning. There are "leftover" mappings
because there are 143 passing sequences in the +1 category, and I am
only representing 104 hands. Naturally there are 39 (or 13 x 3)
combinations not in use.]
And that covers the 5 or 4/1 distributions!
The other cases get a little bit more complicated, but they're not so
bad that they can't be memorized. One thing I will point out.
Generally I paid attention to which cards you are guaranteed to be
holding. For the 3/1/1 distributions, I chose +4 and +9 (or +4 and
-4). In most cases, the first two cards I pass will be the cards
above and below the Devil's three-card run. (There are special cases,
of course.) After that it becomes a matter of placing the two
off-cards.
Finally, I will admit there are a few areas in which I'm not thrilled
with my mnemonics for remembering which card to throw third. Like
maybe I'll have nine different passing options for the final card
assigned to nine possible pairs of cards in the Devil's hand and can't
find a really great way to match them up beyond visualizing them in a
3x3 table and walking through in a certain pattern. But even if it had
to be done by rote, 9-to-9 is manageable, I think.
Comments welcome. Interesting puzzle, Risto. Thanks!
P.S. Chris, your suggestion to have the Devil take six cards is,
well, diabolic. I'm not touching that one. :-)
Addison
newhe...@yahoo.com
Risto Lankinen
Spoiler follows...
A
2
3
4
5
6
7
8
9
T
J
Q
K
Here is the description of the simplest method that I have found.
First, the decoder's algorithm:
For each of the three passed cards, the devil's hand will include
the cyclically next card, and exclude the cyclically previous card.
For instance, if the encoder passes 2,5,8 then the devil's hand
must contain all of 3,6,9 and must not contain any of A,4,7.
If a card is included that is one of the passed cards, then the next
available card is included. For instance A,2,J indicates that the
devil's hand must include 3,4,Q and 4,5,7 indicates 6,8,9 . Any
three consecutive passed cards (for instance A,2,3) indicate the
next three consecutive cards (in this case 4,5,6) as included.
Similar adjustment applies to the excluded cards, except that also
the included cards are skipped in the exclusion. For instance, if
the passed cards are 2,4,6 then the included cards are 3,5,7; now
the run of known cards is 2,3,4,5,6,7 so the three excluded cards
are Q,K,A .
This will always leave four unknown cards, of which two must be
in the devil's hand. There are six permutations, any one of which
can be communicated using the the ordering of the three passed
cards.
The encoder's algorithm left as an exercise for the reader :-)
- Risto -