Friday Puzzler -- a nice card trick

Re: Friday Puzzler -- a nice card trick

bill earl

Does the assistant see the card? Is he/she allowed to 'assist'?

Re: Friday Puzzler -- a nice card trick

Alex Y

The assistant sees all five cards. He chooses one to give to the audience member, and stacks the other four face down on the table. After that, the assistant's job is done.

Re: Friday Puzzler -- a nice card trick

bill earl

I can think of one way. You would not need to actually turn over any cards to deduce the selected one.

Re: Friday Puzzler -- a nice card trick

Alex Y

Interesting. I can't imagine what that would be, unless you are examining the other 47 cards.

Not the way the trick I am describing works, but you may have a better one!

Re: Friday Puzzler -- a nice card trick

Larry Barrett

I can't imagine Bill's solution either. My guess for the original trick is that since the assistant sees the five cards, he can arrange the remaining 4 cards in such a way that they provide a key to the selected card, the key being something like a pre-arranged order by number or suit. This seems too complicated, since there is presumably no way to ensure that certain cards are in the selected five, but is the best I can come up with.

On the right track

Re: Friday Puzzler -- a nice card trick

bill earl

My thinking was similar to Larry. Since the assistant knows the card, he can arrange the remaining 4 cards in such a way to identify the card.

Since there are 4 cards to work with and 52 cards in the deck, you need a code that can represent at least 52 unique values. That could be done via the orientation of the cards in the stack. If you assign values to the card as follows:

0 = horizontal

1 = 45 degrees to the left

2 = vertical

3 = 45 degrees to the right

Now you can represent 4^4 or 256 different values with a stack of 4 cards. (You really only need 3, since 4^3 = 64).

And you don't need to turn them over to decode the value.

Very clever!

Alex Y

Full credit here. Not the solution I was looking for, but absolutely works. As a magic trick though, it seems like this would telegraph the methodology.

Adding a condition

Alex Y

Bill came up with a very clever solution to the problem as originally stated. But let's add another condition to keep the puzzle solving going.

The four cards are put in a stack in front of another audience member, but out of my sight. That audience member then turns over the cards one at a time and reveals the value of each as they are turned over.

I.e., the information that I need to determine the fifth card is contained solely in the values of the remaining four cards and the order in which they are revealed.

Obviously, a magician would add lots of misdirection. I think I would be tempted to riffle the deck of remaining cards, claiming to be able to see and remember all the cards there, so I only need to see the remaining four before determining the one in your pocket. Or I might do some hocus-pocus with the cards pressed to my forehead.

Hints

Alex Y

I just "dealt" myself five random sets of five cards. I will start giving examples of what my assistant will do with each of these selections.

In each of the hands I was "dealt", my assistant has 2 or more choices of what do do with the cards. (In some hands, she may only have one choice, but not in the five I was dealt.)

Hand 1: KD, 6H, 10D, 4C, 10C

If my assistant leaves me a stack of cards that I turn over in the order 4C, KD, 10D, 6H, I will be able to determine that you hold the ten of clubs.

Re: Adding a condition

bill earl

the information that I need to determine the fifth card is contained solely in the values of the remaining four cards and the order in which they are revealed

Hmmm. There are 24 (4!) ways to order the four cards. Once subtracting those 4 cards from the deck, there are 48 possibilities for the 5th card.

Q.E.D. Can't be done (although I am sure you will show us how )

We'll see

Alex Y

the beauty of this trick is that the better you understand combinatorics, the more likely you are to conclude that it is impossible. It certainly appears that I have only about 1/2 of the information I need. In fact, an article I will link to once this is solved, by a MIT professor, includes this line:

I am forever indebted to a graduate student in one

audience who blurted out “No way!” just before I named the hidden card.

Re: Hints

Larry Barrett

Would another option for your assistant be to leave you a stack of cards that you turn over in the order KD, 6H, 10C, 4C, you will be able to determine that I hold the ten of diamonds?

Re: Hints

Alex Y

With that stack, I would conclude that the hidden card was the 3 of Diamonds. But the convention I and my assistant are using is not unique. There are many ways of using the same principles, and you may have hit on one.

The only other alternative with that set of cards, using our convention, would be for me to turn over the 10D, 6H, 4C, 10C and correctly guess the KD.

What is the system you envisioned?

No 'system' yet

Larry Barrett

I am trying to infer one from your hints. Need more hints.

Next set of cards

Alex Y

9H

AS

9D

AH

You are holding the 2 of Hearts.

Third set

Alex Y

KC

4H

3H

6D

Hidden card is 2 of Clubs

Fourth Set

Alex Y

9S

KS

KH

8D

Hidden card is 2 of Spades

I think from these examples at least part of the answer is becoming obvious. Anyone want to speculate on a partial solution?

Re: Hints

bill earl

I went back and re-read the puzzle. I missed an important detail in my first reading. The fact that the 5th card is selected by the assistant changes the nature of the problem considerably.

The first half of your code is pretty obvious from the examples. The rest is not as impossible as I first thought.

Re: Hints

Alex Y

The fact that the 5th card is selected by the assistant

And that is a key that is well-hidden, in both the written puzzle and in the presented trick. After all, the audience member chose all five cards, and after handing them all to the assistant, the assistant becomes flustered and "remembers" that she was supposed to have the audience member keep one, and gives him back a "random" card from the ones he has chosen. Nobody sees that.

Fifth set

Alex Y

Following up on Bill's observation in the way I present this one:

The five cards the audience member draws are

8C

4D

KS

JH

10H

The assistant looks at these cards and hands the Jack of Hearts back to the audience member to hold. She then arranges the other cards so I turn them over in this order:

10H

4D

8C

KS

Seeing those four in that order is enough to tell me taht the audience member is holding the Jack of Hearts.

Possible system

Larry Barrett

As Bill said, part of the system is evident from the hands you dealt, and is also the reason that 5 cards are necessary to guarantee that there are at least two of the same suit. I think the rest of the system depends on the relative value of the remaining 4 cards, and their order. The absolute value can't be guaranteed, but there will always be a relative value.

To use one of your hands as an example, if 4C KD 10D 6H are turned over in this order, the 4C indicates that the held card is a club. Then the relative values and order of K, 10, and 6, which can be coded as H(igh), M(edium), L(ow) indicate that the held card is a 10. The suits of these 3 cards does not matter, only the relative values. I think that any set that can be ordered as H, M, L would indicate a 10, so a QS 8D 3H would work just as well.

A partial system using the hands you dealt would be:

H, M, L -> 10

M, L, H -> K

M, H, L -> 3

L, H, L -> 2

L, M, H -> J

H, H, L -> 2

Since there are 27 sets of relative values, there will be multiple sets for some or all of the 13 cards.

Getting very close!

Alex Y

As Bill said, part of the system is evident from the hands you dealt, and is also the reason that 5 cards are necessary to guarantee that there are at least two of the same suit.

Correct.

I think the rest of the system depends on the relative value of the remaining 4 cards, and their order.

Close, but remember that you won't have discretion in ordering the card that is the same suit as the hidden card.

The absolute value can't be guaranteed, but there will always be a relative value.

To use one of your hands as an example, if 4C KD 10D 6H are turned over in this order, the 4C indicates that the held card is a club. Then the relative values and order of K, 10, and 6, which can be coded as H(igh), M(edium), L(ow) indicate that the held card is a 10.

Yes.

The suits of these 3 cards does not matter, only the relative values.

True in this case. But the system works just as well if instead of a 10D, I had a 6S.

I think that any set that can be ordered as H, M, L would indicate a 10, so a QS 8D 3H would work just as well.

I don't know. Depends on context. True if you are talking about changing out those three cards in the "hand" we are discussing.

A partial system using the hands you dealt would be:

H, M, L -> 10

M, L, H -> K

M, H, L -> 3

L, H, L -> 2

L, M, H -> J

H, H, L -> 2

Since there are 27 sets of relative values, there will be multiple sets for some or all of the 13 cards.

Going a little off-track in this last part.

Re: Getting very close! Correction

Larry Barrett

Correction: Where I said " I think the rest of the system depends on the relative value of the remaining 4 cards, and their order" I meant the remaining 3 cards that are then turned over. I agree that the first card is only used to indicate the suit and can not be part of the rest of the system.

Where I said "a QS 8D 3H would work just as well" I meant in the hand in the example - so where in the example, 4C KD 10D 6H turned over in that order indicated a 10C, I think a 4C QS 8D 3H would also indicate a 10C; so would any other set of three cards that had a relative order of H, M, L, at least in the system I envision.

Since it is quite possible to have two or three cards in the set of the three remaining cards that are of equal value, I should expand my 'system' to include an E(qual). So relative values for a KS KH 8D would be E, E, L.

This may still be getting off track, so will think a little longer.