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Topic: Five Card Magic Trick (Read 17490 times) |
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luke's new shoes
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Re: Five Card Magic Trick
« Reply #25 on: Nov 12th, 2002, 8:00pm » |
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person B can tell the suit of the card by looking at the suits he is given, and can tell the number of the card by the arrangement of the cards he is given (there is 24 ways to arrange them)
the possible combintaion of suits that person B will be given are:
four hearts (or four clubs etc). he then concludes that the suit person A took out was hearts. and uses the arrangement of the four cards to work out the number of the card.
three hearts and a club (or diamond or spade). the suit of the card is either hearts or a club. so there is 26 combinations of what the cards number and suit could be, but he can eliminate the cards he has, reducing the number of combinations to 22 (below the 24 possible ways to arrange them).
two hearts, a club and spade. the cards suit is either hearts or diamonds (the suit person B doesnt have). 26 combinations, can eliminate the 2 hearts he has, reducing number of combinations to 24.
2 hearts and 2 diamonds. the cards suit is hearts or diamonds. 26 combinations, can eliminate the 4 cards he has, reducing number of combinations to 22.
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luke's new shoes
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Re: Five Card Magic Trick
« Reply #26 on: Nov 12th, 2002, 9:23pm » |
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alright i was close, but doesnt work.
if person B is given 2 hearts and 2 diamonds then the cards suit will be clubs or spades (not hearts or diamonds as i stated). 26 possible combinations, only 24 ways to arrange the cards!
this method will fail 1/26th of the time.
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luke's new shoes
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Re: Five Card Magic Trick
« Reply #27 on: Nov 12th, 2002, 10:00pm » |
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ok i'll try again... this time suit as well as colour (sometimes) come into play.
if person A is given (i'll use 5 hearts as an example, but that could also mean 5 diamonds etc):
*5 hearts: it doesnt matter what card they give back.
*4 hearts and a club: they give back a heart
*3 hearts, a club and diamond: give back club (or diamond)
*3 hearts and 2 clubs: give back a heart
*2 hearts, club, spade, diamond: give back diamond (now it is important to give back the diamond because it is the same colour as the hearts)
*2 hearts, 2 diamonds and clubs: put back a card so u are left with a pair of suits and two cards of differnt colours.
Person B can tell what suit was given back by:
*4 diamonds: suit is diamonds
(13 possible cards)
*3 hearts and a club: suit is hearts or clubs
(26-4 possible cards = 22)
*2 hearts and 2 clubs: suit is hearts or clubs
(26-4 possible cards = 22)
*2 hearts, club and spade: suit is diamond
(13 possible cards)
*2 hearts, diamond and club: suit is clubs or diamonds
(26-2 possible cards = 24)
the number of the card can be determined by the way person A arranges the cards (24 ways to arrange them).
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Rajan
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Re: Five Card Magic Trick
« Reply #28 on: Dec 15th, 2002, 10:20pm » |
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I guess i have a solution to this problem... read on
Spoilers warning!!
The 4 cards can/will be used to index to 24 cards. So we can start from the lowest of the 4 cards and count from it the number of cards that are indicated by the combination number formed by the 4 cards (ofcourse, skipping the cards that are already in the hand).
All this has been amply clear to everyone on this list.... now to the stuff which is not clear.
The magician A has to keep to himself such a card that can be indexed by the other 4 cards... that means the 5 card should not be farther than ~ 24 cards from the lowest of the 4 cards that he is giving to B. It all boils down to.... either A will start counting from the lowest of the 5 cards or the second lowest of the 5 cards. So if the difference between the lowest and the second lowest card is < 24, then B can keep the second lowest and give A the rest of the cards ordered in such a way that they point to the second lowest card that he has kept to himself. If the difference betweeen lowest and second lowest is > 24, then the distance detween second lowest and the lowest will be <24 if we loopback from the other end(count to King and loopback to Ace). Now B will keep the lowest to himself and give A the rest with the same criteria... A will now loopback from King of spades to Ace of diamonds whicle counting... and the lowest card can be indexed successfully.
Hence proved.
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mike1102
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Re: Five Card Magic Trick
« Reply #29 on: May 23rd, 2003, 9:23am » |
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I think everyone is thinking way too hard on this one....
Of the five cards selected by the audience member, two must be the same suit. The second magician selects two cards of the same suit and gives one of them back to the audience member and places the second on top of the other three in a neat pile. When the second magician sets the pile of four cards down, or hands it back to the first magician, he can use either his right or left hand and places his thumb at a particular location on the stack - like the numbers on the face of an analog clock. Using any pre-arranged coding scheme, all the first magician need do is watch the second magician's hands and view the top card. Its magic - It all comes down to slight of hand.
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Leo Broukhis
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Re: Five Card Magic Trick
« Reply #30 on: May 23rd, 2003, 12:50pm » |
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Mike1102,
you're right about the suit but wrong about the rest. This trick can be shown on a computer, using any spectator to enter data - no set-up, cooperation or sleight of hand required.
As you've mentioned, out of 5 cards at least 2 will be of same suit (let's say exactly 2 WLOG). The magician will use the first card out of the other 4 to indicate the suit, and the other 3 can only encode 6 combinations, but the magician has the liberty to choose which card out of 2 to return and which to use for encoding: Need I say that the circular distance between two cards of the same suit is never greater than 6?
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ManjeetBothra
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Re: 132-card Variation on 5-card Magic Trick
« Reply #31 on: May 10th, 2004, 11:01pm » |
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but how do u decide that after transferring the cards tht the magician who has to tell the card has to move in clockwise or anticlockwise direction ??
on Aug 7th, 2002, 4:07pm, william wu wrote:Awesome! I'm interested in the history behind all these riddles. Thanks for the info.
Have you heard of a 132 card variation of this same riddle? I heard that apparently you can pull off this same trick on a 132 card deck, using only 4 cards to encode a fifth. However, I don't know enough about this riddle to attack it. Particularly, what kind of cards does the 132 card set consist of? If we double a regular 52 card deck, we get 104 cards; what are the remaining 28 cards? Let me know if you know the details about such a variation ... maybe it's just a rumor. |
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Nigel_Parsons
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Re: Five Card Magic Trick
« Reply #32 on: Jul 11th, 2004, 12:20pm » |
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An additional method of encoding one bit of information is the inversion of a card. In a standard deck, roughly half (A 3 5 6 7 8 9) of the cards have a right-way-up. If the four cards are all of the same orientation it would have one meaning, if any showed differing orientations this would have another.
Clearly if you receive 5 cards which cannot be oriented differently you must pass on 4 with the message that the fifth is also undifferentiable. (hence a 2 4 10 J Q or K)
If you receive 5 cards in which only one card is orientable you pass this back with three others. (message is 5th card was not orientable) The orientable card must be in the 4 otherwise you would pass the message in the previous paragraph.
If you receive 2 or more orientable cards you pass back at least 2, and whether their orientation matches or opposes gives your message.
I don't know how much this helps, but it avoids the need for secret handshakes, thumbs in certain positions etc.,
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sushma
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Re: Five Card Magic Trick
« Reply #33 on: Dec 4th, 2004, 11:28am » |
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how abt this?
i guess it works
the cards in the deck would be numbered 1 to 52.
first magician would choose the card in the middle of the sorting order, eg: 1,3, 34, 51, 52 are the five cards say then he would choose 34 and then arrange the reamining in a particular order.
As someone rightly said, we would arrange the remaining four cards in such a way that each permutation corresponds to a number which is to be added to the 2nd highest of the cards or subtracted from 3rd highest.
Now i'll talk abt how to assign the number to each combination and how to decide whether to add or subtract.
the four cards can be arranged in 24 ways and say
(least, second least, third least, fourth least) corresponds to the muber 1.
similarly (least, third least, second least, fourth least) correspobd to the number 2.
similarly 24 numbers can be assigned for 24 permutations.
now we can give the set of 4 cards with all the faces up or all the faces down.
faces up --> add the number to the 2nd highest
faces down --> subtract the number from 3rd highest
ne suggestions on how to make this better?
acc to me if the number of cards is more than 52, this sol doesn't work!!! ne bright ideas r welcome
~Sushma
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Icarus
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Re: Five Card Magic Trick
« Reply #34 on: Dec 4th, 2004, 12:45pm » |
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If you read the replies on the first page, you will see a link that goes to a description on how to do this puzzle with 132 cards.
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Dostoevsky's Mouse
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Re: Five Card Magic Trick
« Reply #35 on: Apr 24th, 2005, 4:07am » |
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I came up with this yesterday, and while I see that it isn't technically the "correct" solution, I'm pretty sure it's equally accurate. Let me know if you guys see any problems with it.
To Denote the Cards:
"Mystery Card" -- The card Magician #2 must guess.
"Suit Card" -- The top card in the stack of four. Shares the suit of the Mystery Card.
"Value Cards" -- The three bottom cards in the stack of four, used to encode the numerical value of the Mystery Card.
The Strategy:
Mentally label your three Value Cards "1," "2," and "3," in order from low to high (Ace high, and Clubs < Diamonds < Hearts < Spades). For instance, if the cards were an Ace, a Jack, and a 4, they would have the following relative values:
4 = 1 (low card)
J = 2 (middle card)
A = 3 (high card).
There are six possible ways to arrange these cards: 123, 132, 213, 231, 312, and 321. Note that if you discount the Aces, there are twelve possible values for the Mystery Card (2-10, J, Q, and K). Assign each of the six possible Value Card combinations to two of the twelve possible Mystery Card values, as follows:
Combination 123 = Mystery Card is either 2 or 3.
Combination 132 = Mystery Card is either 4 or 5.
Combination 213 = Mystery Card is either 6 or 7.
Combination 231 = Mystery Card is either 8 or 9.
Combination 312 = Mystery Card is either 10 or J(11).
Combination 321 = Mystery Card is either Q(12) or K(13).
This narrows the Mystery down to two possible cards. All that remains to be encoded is whether the the answer is even or odd.
To do this, choose your Mystery Card in the following manner. Check to see whether your two same-suit cards are both odd, both even, or one of each. If both cards are odd, or both cards are even, designate the higher card as the Mystery Card. If one is odd and one is even, designate the lower card as the Mystery Card.
How does this work? Well, say we arrange our cards 10d, Jc, 4s, Ac. The Suit Card, 10d, tells our partner he's looking for Diamonds. The order of the Value Cards, 213, tells him he's looking for a 6 or a 7. And the Suit Card is a 10, an even-numbered card, which tells him immediately that if the Mystery Card is higher, it must be even; if lower, it must be odd. Since both 6 and 7 are lower than 10, the answer must be 7. We would only have made the Mystery Card lower than the Suit Card if one was odd and one was even. Had both cards been even, the 10 -- the higher card -- would have been designated the Mystery Card instead.
This rule works for every card except the Ace, for which there is a separate rule:
Say your two matching-suit cards are the Ace and the 4 of Clubs. Because 4 is even, make it your Suit Card, and make the Ace your Mystery Card. However, arrange the three Value Cards in the order 132, to indicate that the Mystery Card is a 4 or a 5. Your partner will know that the Mystery Card is not really a 4 or a 5. How? Well, the 4 of Clubs is already in his hand; it's the Suit Card. The 5 of Clubs is higher than the 4, but it's odd while 4 is even, which breaks the rule for selecting a Mystery Card. The only answer is the Ace.
Now say your two matching-suit cards are the Ace and the 5 of Clubs. Because 5 is odd, make it your Mystery Card, and make the Ace your Suit Card. Arrange the Value Cards 132, to indicate a value of 4 or 5. Your partner will know, by seeing the Ace used as a Suit Card, that the Mystery Card is odd, and hence, the 5. If it had been even, you would have followed the rules in the previous paragraph, and the Ace would have been the Mystery Card.
Strategy Summary:
Arrangement of Value Cards:
123 = 2 or 3
132 = 4 or 5
213 = 6 or 7
231 = 8 or 9
312 = 10 or J
321 = Q or K
Choosing the Mystery Card:
If the two same-suit cards are both odd or both even, the higher becomes the Mystery Card.
If the two same-suit cards are mismatched, one odd and one even, the lower becomes the Mystery Card.
If the two same-suit cards comprise an Ace and an odd-numbered card, the odd card becomes the Mystery Card.
If the two same-suit cards comprise an Ace and an even-numbered card, the Ace becomes the Mystery Card -- but encode the value of the even card.
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scepe
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Re: Five Card Magic Trick
« Reply #36 on: Aug 24th, 2006, 5:47pm » |
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The question goes on to state how *large* of a deck of cards can you encode. So far, it looks like people have only come up with a way to encode the 52 cards. (As soon as I said this, I read a bit more and found the .pdf on the first page. oops.)
One thing I haven't seen mentioned is the Orientation of the cards. A spade upside down could represent a bit turned on, as opposed to a spade facing right side up. This follows the requirement that the cards are neatly aligned and face down when passed. The passer and passee would have to have a system to know how to handle the cards once they've been passed, but yeah.
Seems like this would give 4! * 2^4 possible card combos. The way I had been doing 'that extra bit' of info commented on for the 52 card deck was just whether or not the cards were all oriented the same way or not. But, I do realize that some decks of cards are non orientable. My Marlboro deck, however, is. :)
Any of that aside, when considering a 'larger' number of cards, we'd have to assume no duplicates, no? There would have to be some discernable difference whether it be a new suit, or a different look to the cards in order to give each card an absolute number.
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| « Last Edit: Aug 24th, 2006, 6:08pm by scepe » |
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Bishamon
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Re: Five Card Magic Trick
« Reply #37 on: Jun 8th, 2007, 3:52am » |
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For the second magician to identify the card, he would need the suit and the value.
Like someone said earlier in this thread, there is bound to be at least 2 cards that have the same suit amongst the five. Choose one of them to be returned to the spectator and put the second card of the same suit on top of the pile to be returned.
For figuring out the value of the card(2,3,4,5,6,7,8,9,J,Q,K,A) assign each type of card a value between 1 and 13.
The four cards that are being returned to the magician B can be arranged to represent a 16 different values by doing the following.
Each card can facing up or down.(i.e. its colorful side(dunno, wat u call the side) facing up could be 1 and other side facing up 0.)
we can get a sequence of 4 binary digits from which the value can be determined.
P.S. : I am sorry if this solution has been proposed before. I read through all those above and although most were suggesting some similar, I do not recollect coming across the exact same solution.
-Edited grammatical mistakes.
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| « Last Edit: Jun 8th, 2007, 3:55am by Bishamon » |
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