[bonanova]

It was talent night at Morty's and the theme was amateur magicians.

Not surprisingly, Alex and his buddies filled the first row.

A week of drinks says I can guess how the first trick is done, me lads,

boasted Alex. I'm somewhat of a prestidigitator, myself, ya know. It

was a few years ago, mind you, but the mind is still sharp, and it's an

exceptional trickster that will outsmart me. Anyone care to differ?

Poor Davey. Even after stroking his beard he couldn't resist the challenge.

You're on, he said.

Then, with a deafening trumpet fanfare The Great Waldo appeared on the stage.

He shuffled an ordinary deck of cards and then handed the deck to Alex!

Pick any five cards and hand them to Amber, my beautiful assistant, he commanded.

As Amber leaned forward to receive the five cards that Alex selected, it became

apparent that what she lacked in beauty she more than made up for in cleavage.

Thank you, she smiled, and Ian swore later that he saw Alex actually blush at her

words.

Amber looked over the cards, stuffed one of them into the front of what might be

called a dress, and handed the others to Waldo. The rattle of a cheap set of snare

drums pierced the silence as Waldo pored over the cards. Finally Waldo spoke.

Ladies and gentlemen, he intoned, the card now resting safely within the bosom

of my beautiful assistant is .... the Seven of Diamonds! To prove that I'm correct,

I'll ask that gentleman [pointing at Jamie] to come and inspect the card. In two

leaps Jamie was on the stage and, none too quickly, he retrieved the card.

Please show it to the audience, ordered Waldo.

Jamie held up the card. It's ... the Seven of Diamonds!

Two minutes later, Waldo was gone, and Alex was deep in thought.

Finally Alex spoke.

Yes, he said, I think I have it!

Was Alex correct? Is there a legitimate way to perform this trick?

[Guest]

No time to look at this in too much detail, but my starting point:

Assuming Amber gets to choose the card she keeps and orders the cards she gives Waldo, she can describe the card she has:

For example, taking the four cards she gives to Waldo as ordered cards (by size/suit or any other reliable ordering), they can be labelled A, B, C, D. There are 24 (4x3x2x1) permutations of those cards which she needs to describe a unique card from the remaining 48 cards. Clearly, there is a factor of two she needs to deal with, perhaps by handing them to Waldo upside down, or the right way up, or better, by the choice of card she keeps (maybe she only ever takes red cards).

Impressive mental calulations *and* cleavage - clearly there is no legitimate way that Waldo and Amber can have played the trick!

[Guest]

I think he knew Waldo. Prestidigitator not only refers to magic and sleight of hand, but also decietful cleverness. <_< The fact that i cant figure it out has nothing to do with my opinion.

[Guest]

First, assign every card a unique number from 1 to 52. It doesn't really matter how, but I would go Ace of Spades =1, 2 of spades =2, etc.

Then, when the assistant gets the 5 cards, she selects one at random to keep. The rest get renumbered (I'll use roman numerals for clarity) I, II, III, and IV, with I being the lowest of the 1-52 numbers, II the second-lowest, etc.

She can then send a binary-type code as follows: the cards can be either facing forwards or backwards to give a 0/1 sequence, so [face backwards][face forwards][face forwards][face forwards] would be 0111, which means the card is a 7. Then, whichever of the roman-numerals is first could signify the suit - ie, card I being first might mean diamonds.

I'm sure there's a simpler way to do it, partly because I switched plans halfway through, but I think that would do it.

[unreality]

In two leaps Jamie was on the stage and, none too quickly, he retrieved the card.

my answer:

Each card is indexed by a suit (4) and a rank (13). The assistant uses the order of the cards to send the suit value... ie, let's say you rank the other 4 cards from A to B to C to D where A is the highest ranked. ABCD would mean clubs, BACD would mean spades, ABDC would mean hearts, BACD would mean diamonds. If there is a tie of two cards, clubs is higher than spades is higher than hearts is higher than diamonds. So Waldo knows the suit. Now he just needs to know the rank. The four cards are then either put face-up or upside-down when handed to Waldo. This can account for 2⁴ possibilities or 16 different ranks. Only 13 different ranks exist, so we're good ;D

[Guest]

my answer:

Each card is indexed by a suit (4) and a rank (13). The assistant uses the order of the cards to send the suit value... ie, let's say you rank the other 4 cards from A to B to C to D where A is the highest ranked. ABCD would mean clubs, BACD would mean spades, ABDC would mean hearts, BACD would mean diamonds. If there is a tie of two cards, clubs is higher than spades is higher than hearts is higher than diamonds. So Waldo knows the suit. Now he just needs to know the rank. The four cards are then either put face-up or upside-down when handed to Waldo. This can account for 2⁴ possibilities or 16 different ranks. Only 13 different ranks exist, so we're good ;D

Since there are 5 cards and only 4 suits, then at least 2 of the cards have the same suit. The assistant can pick a card from the 5 that has the same suit as another card. Then she can place that suit on top of the stack so that Waldo knows the suit immediately. She can do the face up - face down algorithm to reveal the number like you said.

[Guest]

The 7 of diamonds could already have been secreted about her person. Waldo did not ask Alex to look at the cards so he doe not know whether the 7 of Diamonds was among cards selected. Amber could have then returned 5 cards back to the pack, or her cleavage may have been big enough to hide another card in.

[unreality]

enlightened: I thought about that already, but it doesn't work. There's no saying that "the suit" will be one of the ones that appears more than once. It could appear only once, or never

[Guest]

enlightened: I thought about that already, but it doesn't work. There's no saying that "the suit" will be one of the ones that appears more than once. It could appear only once, or never

There's no need to create a code for what suit is pictured on the cleavaged card. Since the assistant can pick any card to stow there, she will pick a card of a suit that is represented more than once. She then presents the "sister" card (one of the same suit) to Waldo first of the four cards she gives him, so he knows the suit. The number must be communicated through the binary code already proposed.

Edited August 5, 2008 by Cherry Lane

[Guest]

enlightened: I thought about that already, but it doesn't work. There's no saying that "the suit" will be one of the ones that appears more than once. It could appear only once, or never

Enlightened method could work. Amber looked through the cards and chose one. Therefore there must have been at least 2 the same suit (at worst one of each of the 4 suits plus a second one of any of the suits). Of course it could be any combination of suits from that just mentioned to all the same suit, but enlightened's 's point is that, because she chose the card, not picked at ramdon, she could employ the method suggested.

[unreality]

yeah, sorry- I forgot she got to choose the card herself (mine would still work if the cleavaged card was picked randomly). So yeah, that does work

[Guest]

yeah, sorry- I forgot she got to choose the card herself (mine would still work if the cleavaged card was picked randomly). So yeah, that does work

ha! you used my made-up word - see, I can start things, too!

[bonanova]

Is there a solution if Amber returns the cards to Waldo all face down?

[Guest]

You could do it this way:

OK, so instead of using forwards / backwards to signify binary digits, you can use the overlap - ie, you fan them out, and if the first card is over the second, the second is under the third, and the third is under the fourth, that could be 100. The only problem is that you only get 3 digits with 4 cards, and you need 4 digits to specify the correct card.

To get a fourth digit, you could have the roman numeral of the first card (from my other post, you number the four cards you send I, II, III and IV from lowest to highest) signify the suit of the card, and the next numeral could be the last binary digit - eg lowest roman for 0, highest for 1

[Yoruichi-san]

Is there a solution if Amber returns the cards to Waldo all face down?

If you use the first card to signify the suit, then by ordering the suits you can order the remaining 3 cards from highest to lowest, which gives 6 possible permutations in the order they are handed to the magician...give each of these permutations a value from 1-6, then the card that is kept by Amber can be represented by the first card +/- a number from 1-6 using these permutations...

Now just need a way to signify whether it's + or -...maybe whether she hands the cards over with her left hand or her right hand?

Edit: This value system "wraps around", i.e. 13 (king)+1=1 (ace)

Edited August 5, 2008 by Yoruichi-san

[Yoruichi-san]

If you use the first card to signify the suit, then by ordering the suits you can order the remaining 3 cards from highest to lowest, which gives 6 possible permutations in the order they are handed to the magician...give each of these permutations a value from 1-6, then the card that is kept by Amber can be represented by the first card +/- a number from 1-6 using these permutations...

Now just need a way to signify whether it's + or -...maybe whether she hands the cards over with her left hand or her right hand?

Edit: This value system "wraps around", i.e. 13 (king)+1=1 (ace)

You can make it always + by choosing to keep the card that is the other card + a value from 1-6. I.e., you choose to keep the ace and give the king to the magician and the other three cards in the permutation that corresponds to +1...

[Guest]

This is going to be complicated...but I think there's a solution which doesn't rely on any reversed/folder/fanned/marked -type fiddling.

Firstly, I'm assuming that, with four ordered cards, they can show 24 different permutations (I think others have shown this sufficiently already). Any ordering method is fine as long as both know. This is Method A and will uniquely identify one card amongst a set of 24 possible solutions.

The key extra bit of information is that Amber chooses her card and, as long as she follows a pre-agreed method, I think she can impart enough information to Waldo to uniquely identify her card. There are several ways, I expect, and this is just one strategy this solution that works by partitioning the cards. I'm using suits but others categories would be possible (maybe even using 5 categories instead?...)

So, there are 6 permutations of the suits of 5 cards:

0 0 0 5 -> 0 0 0 4

0 0 1 4 -> 0 0 1 3

0 1 1 3 -> 0 0 1 3

0 0 2 3 -> 0 0 2 2

0 1 2 2 -> 0 1 1 2

1 1 1 2 -> 0 1 1 2

Against each I have identified the permutation that Amber will then choose to give to Waldo.

(There is also another permutation of four cards (1 1 1 1) which can tell you nothing useful, yet!)

Based on what Waldo recieves, we consider a number of scenarios:

Waldo recieves

(0 0 0 4) - obviously we know the suit, so the 24 permutations of Method A is easily sufficient

(0 0 2 2) - we know the suit matches one of the suits we already have, and 2 cards from each suit are already identified. Method A is just sufficient

(0 0 1 3) & (0 1 1 2) - these require some more thought....

The first point to note is that the pattern (A B C C) has twelve permutations (if the Cs are indistinguishable). Therefore, there are 12 permutations of (0 1 1 3), (0 1 2 2), (0 0 1 3) & (0 1 1 2). Fortunately, this means we can define a one-to-one relationship between the necessary patterns. For example, one possible pre-determined method might be (considering suits ordered as Clubs, Diamonds, Hearts, Spades, Clubs, ...and so on):

(0 1 1 3) -> remove the card which is in the suit after the empty suit -> (0 0 1 3)

(0 1 2 2) -> remove a card in the suit which is after the single suited card -> (0 1 1 2)

As this is one-to-one, it is reversible and I call this Method B:

(0 0 1 3) -> the card is in the suit before the single suit) -> (0 1 1 3)

(0 1 1 2) -> the card is in the suit before the empty suit) -> (0 1 2 2)

So,

if Waldo receives (0 0 1 3), he either knows the card was removed from the 'tripled' suit, which, assuming the highest card is always removed, uses up 10 of the Method A permutations, or is from an empty suit which we can deduce (by method B), and requires another 13 Method A permutations. Done!

if Waldo receives (0 1 1 2), he either knows the card was removed from the singled suits (and we can deduce which one by method B) which requires 12 Method A permutations, or from the empty suit, which requires 13 Method A permutations. WE ARE 1 SHORT! BUT Amber can help by careful choice of removed card. She now never removes an Ace from (1 1 1 2), so reduces the empty suit permutations to 12. Hooray! Except.....Oh dear.....what if all the single suits are Aces? Then, Amber just takes either card (except an Ace) from the paired suit, leaving (1 1 1 1). This is the only time Waldo sees it, and it contains three Aces and a non-Ace. The other card is therefore in the non-Ace suit and is identified by Method A.

Wow. I made it.

This isn't quite complete, as I haven't detailed the order that Amber/Waldo use to make complete use of Method A, but I think it demonstrates that it works.

I bet there's a mistake in it....

edit: correction to the (0 0 1 3) solution.

Edited August 6, 2008 by foolonthehill

[Prime]

There are many simple ways to make a code:

Assuming there are 52 cards in the deck, If the Great Waldo sees the four cards, he must find one out of 48. Assign an order to all cards in the deck. For example order suits: spades, clubs, diamonds, hearts, and the cards within each suite according to their rank. Thus 2 of spades is 1, 3 of spades is 2, Ace of spades is 13, 2 of clubs is 14, and so on.

Then the four cards handed to the magician may be arranged according to their order in 4! = 24 ways. To make it 48 combinations, Amber can choose her left or right bosom to hide the card. The resulting number from 1 to 48 may indicate in what order the hidden card fits into the remainder of the deck.

For the cards face down, Waldo must guess one out of 52. There may be plenty of variations. Left or right bosom – 2 times 2 (left or right hand) times 7 (between which two of her fingers Amber holds the cards), times 2 (whether she holds cards wide side forward, or narrow). And that makes it 56 combinations – more than enough.

Card cheaters often use arrangement of their fingers on the cards to give signals.

I say where magician's assistant saw cards -- it is not a trick.

[Prime]

Small correction: I said that if Waldo does not see the cards he must find one out 52, whereas he can still narrow down his choice to 48.

[unreality]

how can she have a 'left bosom' and 'right bosom'? Bosom refers to the whole bodily area... ie, what I'm asking is, she only has one space to put the cards in, unless she has three, er, 'breast organs'

Edited August 6, 2008 by unreality

[Prime]

how can she have a 'left bosom' and 'right bosom'? Bosom refers to the whole bodily area... ie, what I'm asking is, she only has one space to put the cards in, unless she has three, er, 'breast organs'

There was so much space devoted to the subject and such a vivid description... I got confused. Sorry. But you must agree, bosom has to play some role there.

[brainz]

It was talent night at Morty's and the theme was amateur magicians.

Not surprisingly, Alex and his buddies filled the first row.

A week of drinks says I can guess how the first trick is done, me lads,

boasted Alex. I'm somewhat of a prestidigitator, myself, ya know. It

was a few years ago, mind you, but the mind is still sharp, and it's an

exceptional trickster that will outsmart me. Anyone care to differ?

Poor Davey. Even after stroking his beard he couldn't resist the challenge.

You're on, he said.

Then, with a deafening trumpet fanfare The Great Waldo appeared on the stage.

He shuffled an ordinary deck of cards and then handed the deck to Alex!

Pick any five cards and hand them to Amber, my beautiful assistant, he commanded.

As Amber leaned forward to receive the five cards that Alex selected, it became

apparent that what she lacked in beauty she more than made up for in cleavage.

Thank you, she smiled, and Ian swore later that he saw Alex actually blush at her

words.

Amber looked over the cards, stuffed one of them into the front of what might be

called a dress, and handed the others to Waldo. The rattle of a cheap set of snare

drums pierced the silence as Waldo pored over the cards. Finally Waldo spoke.

Ladies and gentlemen, he intoned, the card now resting safely within the bosom

of my beautiful assistant is .... the Seven of Diamonds! To prove that I'm correct,

I'll ask that gentleman [pointing at Jamie] to come and inspect the card. In two

leaps Jamie was on the stage and, none too quickly, he retrieved the card.

Please show it to the audience, ordered Waldo.

Jamie held up the card. It's ... the Seven of Diamonds!

Two minutes later, Waldo was gone, and Alex was deep in thought.

Finally Alex spoke.

Yes, he said, I think I have it!

Was Alex correct? Is there a legitimate way to perform this trick?

this totaly possible i do a card trick similer to this step one buy a pack of ordanary playing cards step two (the most important step) go throght the deck and find all the queens well say get all the queens of spades a with a pair of scissors evenly trim the top of the card about a hairs width now insert the cut cards back into the deck now practise since the cut card is smaller it will act like a identifire and makes it easy to get to the top of the deck and youre traind helper will know wich one to choose. (and remember mke sure you dont take more than a hairs width and that its a straight cut or the gig is up it must look as close to normle as you can help)

[bonanova]

there are 4! = 24 card permutations, but 48 cards from which Waldo must determine a single card.

Let's assume Alex got that far, also.

Alex can't expose the trick?

Or can Alex handle that factor of 2?

[without using other signals]

[Prime]

there are 4! = 24 card permutations, but 48 cards from which Waldo must determine a single card.

Let's assume Alex got that far, also.

Alex can't expose the trick?

Or can Alex handle that factor of 2?

[without using other signals]

I thought I gave plenty of examples of additional factors. See my post (#18).

Most notably, my use of Amber's bosom.

[Guest]

there are 4! = 24 card permutations, but 48 cards from which Waldo must determine a single card.

Let's assume Alex got that far, also.

Alex can't expose the trick?

Or can Alex handle that factor of 2?

[without using other signals]

Did my (rather convoluted) method not show that she could always choose a card from a set of maximum size 24, which was well-defined by the suits of the other four cards?

[bonanova]

I thought I gave plenty of examples of additional factors. See my post (#18).

Most notably, my use of Amber's bosom.

Yes, you did.

But read the second spoiler above and see that no other signals are involved.