[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?

[bonanova]

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?

It might have been too convoluted for my fading brain ...

I'll go back and check.

---- minutes pass ----

OK, I'm back

I guess that part of your method is not clear to me.

You say, I comment:

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) Are you saying 0 clubs, 1 diamond, 1 heart, 3 spades;

or does "after" mean next higher in suit rank?

(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) ... does "before" mean next lower in suit rank?

(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, ... that part I understand

or is from an [the?] empty suit which we can deduce (by method B), [not clear how] and requires another 13 [26?] Method A permutations. Done!

I think I will understand if you can work this example:

Let's say Amber hands Waldo, in this order: 2 of hearts, 3 of hearts, King of clubs, 4 of hearts. [the 0 0 1 3 scenario]

Then the missing card is one of the following : 5 6 7 8 9 10 J Q K or A of hearts [10 Method A possibilities] or ... here I'm lost ...

what are the other [fewer than 14] possible cards? I don't know how you proceed at this point.

Use any card ordering you like, just list the possibilities: suit and ranks.

[Guest]

I think I'm probably about to prove that it doesn't work, but here goes.....

You say, I comment:, I respond again!

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)

Are you saying 0 clubs, 1 diamond, 1 heart, 3 spades - yes

or does "after" mean next higher in suit rank? My notation always refers to number of cards in that suit, and I'm saying that for each permutation of (0 1 1 3), there is a one-to-one relationship with a permutation of (0 0 1 3) - as long as you predetermine your method of card removal. My method simply chooses a card from the suit which is next (in suit order) after the 'empty' suit. This method can always be reversed because, the missing card is then from the suit before the only remaining 'singled' suit So, to be explicit, I chose:

(0 1 1 3) -> (0 0 1 3)

(0 1 3 1) -> (0 0 3 1)

(0 3 1 1) -> (0 3 0 1)

(1 0 1 3) -> (1 0 0 3)

(1 0 3 1) -> (1 0 3 0)

(3 0 1 1) -> (3 0 0 1)

(1 1 0 3) -> (0 1 0 3)

(1 3 0 1) -> (1 3 0 0)

(3 1 0 1) -> (3 1 0 0)

(1 1 3 0) -> (0 1 3 0)

(1 3 1 0) -> (0 3 1 0)

(3 1 1 0) -> (3 0 1 0)

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) ... does "before" mean next lower in suit rank?As above, the chosen card must be in the suit which is 'before' the '1' suit, ie in this case it would be a diamond, assuming C/D/H/S.

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, ... that part I understand

or is from an [the?] empty suit which we can deduce (by method B), [not clear how] and requires another 13 [26?] Method A permutations. Done!

I think I will understand if you can work this example:

Let's say Amber hands Waldo, in this order: 2 of hearts, 3 of hearts, King of clubs, 4 of hearts. [the 0 0 1 3 scenario]

I will call this (1 0 3 0) by ordering the suits Clubs/Diamonds/Hearts/Spades

Then the missing card is one of the following : 5 6 7 8 9 10 J Q K or A of hearts [10 Method A possibilities](agreed) or

the original suit layout must have been (assuming C/D/H/S) (1 1 3 0) or (1 0 3 1).

But had Amber recieved (1 1 3 0), then she would have given Waldo (0 1 3 0) by removing the club rather than the diamond. Therfore, Amber must have had (1 0 3 1) and the missing card is a spade.

Does this make more sense? - I fear not!

[bonanova]

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You've identified the suit(s) and kept the total possibilities less than 25.

Looks like a win!

the first card identifies the suit, so the encodings are halved to 12 quartered to 6 - given by the order of the other 3 cards.

Care to give that approach a try?

And, yes, a larger deck can be accommodated.

I know how much larger, by another source

124 cards.

[Guest]

the first card identifies the suit, so the encodings are halved to 12 - given by the order of the other 3 cards.

Care to give that approach a try?

Unfortunately, no time to look propoerly now. But on the face of it, there will be some cases where you are forced into which suit is removed, eg. (1 1 1 2). If you identify the suit with the first card, then you are only left with permutating (?) the 3 remaining cards, giving you only 6 possibilities, not the 12 you need? Presumably there needs to be a predetermined straategy again where Amber follows a set routine in choosing which card to remove - eg. in the suit with two cards, if the second card is within six places of the first, she removes the higher one (leaving six possibilities), otherwise she removes a different card accroding to some other method.

Will try properly later - if someone else hasn't got there first.....

[bonanova]

Will try properly later - if someone else hasn't got there first.....

You got it.

And you're right, only 6 encodings manages the task.