Sal the Magician

[bugzpodder]

Sal the Magician asks you to pick any five cards from a standard deck (52 cards), 13 denominations (2,3,...,10, J, Q, K, A) and four suits (diamonds, hearts, clubs, spades). You do so, and then show them to Sal's assitant Pat, who places one of the cards back in the deck and puts the remaining cards into a pile. Meanwhile you shuffle the reduced deck of 48 cards several times. Sal is blindfolded, and does not witness any of this. Then Sal takes off the blindfold, takes the pile of cards that Pat arranged, examines it, and then picks out your fifth card from the deck! The deck of cards is not marked, and there is no verbal or visual communication between Sal and Pat. Without supernatural powers, how does Sal know your fifth card?

[sql_lall]

Sorry, just a loop hole.
I guess, with your explanation, that Pat can put the card face-up in the pack. That way, when you shuffle it, it will be the only face-up card. So when Sal looks at it, all he/she has to do is pick the face-up card.

Oh, and by marked, does that mean the pictures on the back of the pack are symmetrical??
Otherwise, Pat can put your card back face-down, BUT rotating it 180 degrees from all the other cards. This way, when Sal looks at the shuffled pack from behind, all the pictures on the back of the cards will be up one way except for the chosen one. (This would look cooler, cos Sal could pick out the card without looking at the faces of the deck.)

Oh, and i know both of these are non-maths solutions, so i guess you had another in mind.

[bugzpodder]

no loopholes. math only and yes you can do it.

[alkatran]

You mentioned the suits and numbers.. that's either a distraction, or a hint...

They could have the initial deck in a certain pattern, unless of course you could pick from any point in the deck.

[bugzpodder]

well i didn't make the question up. this is still possible to do if you were the person to pick out any five cards, and put one back of your choice, and re-shuffle the deck. of course you let the assistant rearrange the remaining four cards to his/her likings (hint hint)

[alkatran]

I suppose the assisstant could always give the magician the lowest card in number and suit (picking one suit as lowest...) to raise the odds... then he would fiddle with the cards, using the motions to create morse code, telling the magician what to do.

[SteveCRM]

but theres no visual or verbal communication er....gotta be something with the assistants organizing

[bugzpodder]

again don't look for loop holes

[sql_lall]

I am guessing that it is in the arrangement of the 4 other cards (your hint) that tells what the 5th was.

I.e. arranging the 4 to represent the value and suit of the 5th.
Something like...each card can be represented by a number
1->52, so perhaps the face values of the 4 cards are in such a way that displays this number?? However, to do this i am not sure.

[bugzpodder]

thats the idea.

[alkatran]

So let me get this straight, his assisstant arranges the cards to indicate the 5th one? And there's no visual communication? How does that work?

Oh, and he could always press on his foot, that's not verbal/visual ;p

YES IM JOKING

[Evil_Giraffe]

One way it could work...
The pack is ordered in some way (total order)
be that suit 1 in order, followed by suit 2, etc.. or the 2s, 3s, 4s, etc, it doesn't matter as long as there is some order.

The four that are left out can then encode 24 different states on the order they are laid out in (4!).
We need to differentiate between 48 cards, so you could say if they're in a line then ot's in the forst 24, if they're in a square it's the last 24.

Meh, not sure if that's foolproof. Got to handle cards having been taken out of the ordering when you count, so not sure how easy it'll be to do in your head. Maybe Sal is good at mental arithmetic.

[alkatran]

From what I understand you can pick the card from anywhere in the deck and YOU shuffle it afterwards.

bugz (bunny) keeps hinting at the assisstant doing something, even though he said there was no communication.

[bugzpodder]

that is basically my idea. i just have a few more comments:

1. it does say to put them in a pile so i guess all you need to do is say if the tile is placed vertically on the table then it is 1-24 and if its placed horizontally on the table then it is 25-48

also, it is not hard to keep track of the cards taken out, just figure out the identity number, than add the number of cards outside that has less identity value than the fifth card.

the hard part in my opinion is given a permutation, figure out in the ordered permutation which one is it.

Originally posted by Evil_Giraffe
One way it could work...
The pack is ordered in some way (total order)
be that suit 1 in order, followed by suit 2, etc.. or the 2s, 3s, 4s, etc, it doesn't matter as long as there is some order.

The four that are left out can then encode 24 different states on the order they are laid out in (4!).
We need to differentiate between 48 cards, so you could say if they're in a line then ot's in the forst 24, if they're in a square it's the last 24.

Meh, not sure if that's foolproof. Got to handle cards having been taken out of the ordering when you count, so not sure how easy it'll be to do in your head. Maybe Sal is good at mental arithmetic.

[sql_lall]

to figure out the ordered permutation:

give the suits an order (i.e. "hearts" = highest, then spades, diamonds and clubs)

the four cards can then be ranked in order (lowest to highest) 1,2,3,4
then, "x" is first(top of deck), the number is 6x-5->6x
then, if the lowest of the remaining is next, the number is 6x-5 or 6x-4, depending on the arrangement of the last two.
if the middle of the remaining is next, the number is 6x-3 or 6x-2, depending on the arrangement of the last two.
if the highest of the remaining is next, the number is 6x-1 or 6x, depending on the arrangement of the last two.

i.e. if the order was 3142, this would represent #14

1st is "3" => number from 13 to 18
2nd is "1" (lowest remaining) => number 13 or 14
4>2, 14>13 => number is 14

[bugzpodder]

nice. maybe i'll try the trick sometimes. i just need a good assistant that wouldn't screw me up

[kedaman]

Pal places one of the cards into the pack, which has to count for a bit of information. For instance:
odd amount of red cards, gives +24 in the encoding

[sql_lall]

You see, your whole argument comes based on the fact that Pat knows where in the pack the 5th card is. to do this, they would have to stand before the audience and count up to 48 cards. Surely the audience may get bored/catch on the Pat is counting them by then, and ruin the mystery??

oh, and did you think this up, of did you hear it and wonder how it was done??

[bugzpodder]

?? Pat doesn't need to know that. i am sure you can do some calculations to figure it out

[Judd]

...or you've quoted the problem incorrectly. As stated, this trick cannot be performed by any sort of mathematical arrangement which points to a particular card (e.g. fifth card) in the pile.

You do so, and then show them to Sal's assitant Pat, who places one of the cards back in the deck and puts the remaining cards into a pile. Meanwhile you shuffle the reduced deck of 48 cards several times.

The pack is shuffled after Pat has put the single card back and arranged the four cards, thus destroying any numeric 'arrangement'. (I'd feel a bit pedantic pointing out that we are assuming you know which card Pat put back).

So, any 'arrangement' theory has to identify that card by suit and number, not by position in the pile (i.e., you could name the fifth card without looking in the pile).

Also, as the problem is stated, or at least as I read it, you can pick ANY five cards, so it could be Ace, 2, 3, 4 and 5 of spades for example. Not that this matters, we just need to figure out a way to use any four cards to identify any of the remaining 48.

[bugzpodder]

and that is exactly what we did Judd. by putting 4 cards in different order, we can emulate 24 possibilities. if we add another variation (ie deck straight/sideways) we can emulate 2*24=48 possibilities. therefore we can emulate any number from 1-48 using the 4 cards that are outside, which means it can emulate any card in the remaining deck

[Judd]

but I must have missed it....?

Where is the identification of the suit? And the number on the fifth card?

If you are referring to sql_lall's 'ordered permutation', that appears to me to be locating the 14th card?? Or am I misunderstanding it? As I said earlier, it can't rely on the position of the fifth card in the large pile, since it has been shuffled.

You therefore, should be able to name the fifth card, just by looking at the four remaining 'picked' cards and not even referring to the remaining 48.

I do get the feeling I'm not latching on to something...

[bugzpodder]

we number 1-13 for A-K of diamonds, 14-26 for A-K of hearts, etc... we will have all 52 cards numbered (who cares about what position there are in in the deck?) then after the guy picks out 5 cards and one is put back, Pat emulates the identity number minus the number of cards out that has a lesser identity number according to the list above.

kedaman: odd number of red cards outside does not necessary gives you more information, although i may have misunderstood your point.

[Judd]

I knew I was missing something

[kedaman]

bugz: yeah just noticed, i just thought putting them in a certain agnle on the table was a loophole, you could get as much information from there if you could read the angle correctly.

how about this:
Find the first of the smallest gaps, and in case these gaps come in a row take the first in the row, If the gap to the card after the next card exceeds 23 take out the card before the gap, otherways take out the card after the gap. EX:

A 24 B 0 C 12 D 11 E 0

BC is the first smallest gap
CE is 24 which exceeds 23
take out A. its 24 - now we encode this to 23, since we know that 0 is not possible, because that would make it first in a row of smallest gaps. We can now use the order of the 4 cards to encode this number 23, as we have 4!=24 permutations which is just what we need to encode all numbers from 0 to 23.

Sal will recieve

E 25 B 0 C 12 D 11 with the 23'th permutation of the ordered cards. He will find the smallest gap BC and find that CE is 24, pointing to that the card is before the smallest gap. He can now reconstruct
A 24 B 0 C 12 D 11 E 0
and decode A.

Note this example was deliberately the worst case, to make sure that the card before the gap can be encoded if needed.

[bugzpodder]

i don't know what you mean by a gap

[kedaman]

take the five cards and sort them by their encoded order 1-52. The gap between two cards a and b is |code52(a)-code52(b)|
just typing out 52 explicitely not to confuse this with the code48 encoding the chosen card which is put back into the deck