The Brain Teaser Thread

[CPA]

Originally posted by: sirjonk

Originally posted by: CPA
I'm not good at these, but I have one. Probably too simple, though.

Picture a chocolate cake and a cake knife. You need to cut the cake into eight pieces with only three slices with the knife. How do you do it?

I've got this one.

You bring 3 chefs blindfolded into the room. You tell each one that you're going to put a marble in their left hand and a coin in their right. They have to flip the coin behind their back so that it lands in the hand with the marble, freeing their right hand. The first chef to do this walks up to the cake and cuts it horizontally and vertically and then once sideways through the middle.

bah, told you it was too easy

 

[TuxDave]

Originally posted by: brikis98

Originally posted by: QED
Ok... another oldie:

Given a standard 8-by-8 chessboard with two opposing corner squares removed and a stack
of 2-by-1 domino pieces, is it possible to cover the entire chessboard using the domino pieces
without any overlaps? Each half (a square) of each domino must cover exactly one square on the chessboard,
and each square on the chessboard must be covered. If so, how?

I don't think it can be done.

On an 8x8 chessboard, squares are (or can be) colored black and white, so no two adjacent squares are of the same color. If you do this, you'll notice that the squares on opposite corners must be the same color. So, if you removed two corners, your board has either 32 black squares and 30 white squares or 32 white squares and 30 black squares.

The problem states that each half of a domino piece covers exactly one square on the chessboard. Since domino pieces cover two squares total, and these squares are adjacent, the two squares covered MUST be of different color. Since domino pieces can't overlap or hang off the board, it then follows that there is no way to get them to cover the entire board if the number of black and white squares is not equal.

You are teh smarty. I like that approach.

 

[Chronoshock]

Originally posted by: Kyteland
Say you have a sheet of paper, what is the maximum number of pieces you can cut it into given n cuts. You cannot fold the sheet or rearrange already cut pieces (i.e., even after the nth cut, the pieces remain in the same position as they were before any cuts were made.)

And if you really want to geek out generalize the problem to any dimension. You have a X dimensional shape. You cut it N times in the X-1 dimension without moving or rearranging the pieces. How many pieces do you have?

The general formula is n*(n+1)/2 + 1 because each new cut can intersect all previous cuts. This leads to an additional n pieces. You start with one piece at 0 cuts, so you get 1 + sum(1,n) = n*(n+1)/2 + 1

 

[Kyteland]

Originally posted by: Chronoshock

Originally posted by: Kyteland
Say you have a sheet of paper, what is the maximum number of pieces you can cut it into given n cuts. You cannot fold the sheet or rearrange already cut pieces (i.e., even after the nth cut, the pieces remain in the same position as they were before any cuts were made.)

And if you really want to geek out generalize the problem to any dimension. You have a X dimensional shape. You cut it N times in the X-1 dimension without moving or rearranging the pieces. How many pieces do you have?

The general formula is n*(n+1)/2 + 1 because each new cut can intersect all previous cuts. This leads to an additional n pieces. You start with one piece at 0 cuts, so you get 1 + sum(1,n) = n*(n+1)/2 + 1

That's correct. Now generalize it to any dimension.

 

[QED]

Originally posted by: Kyteland

You have 12 golf balls, 11 of the same weight and one of a deviating weight. You are allowed to use a balance scale 3 times. BUT, you do not know whether the deviating golf ball is heavier or lighter than the rest. Using the scale 3 times, you have to point out the deviating ball AND tell whether its heavier or lighter than the rest.

I think I have this one.

At first glance, it doesn't seem possible at all-- but if you do the math, you realize that there are 3 distinct weighings, each with 3 different outcomes (either the left side weighs more, the right side weighs more, or they are even). If you could make each weighing totally independent from the others (or as close to independent as possible), you have a total of 3*3*3, or 27, different outputs. This is more than the 24 different possible ways for exactly one of the golf balls in particular to be lighter (or heavier) than the other 11. So there is, at least in theory, enough information available to determine exactly which ball is different and whether it is heavier or lighter than the rest.

So how can you make each weighing as independent from the others as possible? From Information Theory (specifically Bayes Theorem) we know in general that the more independent two events are, the less effect one event will have on the other's probabilities of possible outcomes. We also know that, in general, we gain more information from an event when each possible outcome is equally likely than we do from an event where one outcome is heavily favored.

With those ideas in mind, our first weighing should be 4 against 4. Why? Each of the 3 possible outcomes (left side weighs more, right side weighs more, both sides even) are equally likely. Thus, we maximize the amount of information gained with this first weighing. Say we weigh ABCD against EFGH.

Case 1: ABCD < EFGH
If ABCD is lighter than EFGH, then we know the heavier/lighter ball is not I,J,K, or L so we can effectively ignore them the rest of the way. Our second weighing should again maximize the probability of each possible outcomes, with the knowledge that one of A,B,C, or D is light or that one of E,F,G, and H is heavy. We do not want to weigh 4 against 4 again using all of A,B,C,D,E,F,G, and H because no matter how we arrange it we can only end up with 2 possibilities instead of 3 as we now know for certain one side will have to contain the lighter/heavier ball. So we choose to weigh 3 against 3, with two sitting out. We could choose to remove one ball from each side for our second weighing, but it is not possible for the lighter side to become the heavy side this way-leaving just two possible outcomes instead of the 3 we need. Hence, the two balls that sit out the second weighing must come from the same side of the first weighing-- say G and H. Weighing ABC vs DEF isn't much better though, as the odds are pretty high that ABC will weigh less than DEF ( 5/8 probability). If instead we weight ABE vs CDF, the probabilities of each of the possible outcomes are fairly even (3/8 odds that ABE is lighter, 3/8 odds that ABE is heavier, and 2/8 odds that the two sides are even). If ABE is lighter, then we know either A or B is light or F is heavy-- and we simply weigh A against B for our third and final weighing to determine which is which. If ABE is heavier, then we know either E is heavy or C or D is light and we weigh C against D to determine which is which. If ABE and CDF are equal, then we know G or H is heavy and we weigh G against H to determine which is which.

Case 2: ABCD > EFGH
The logic for when ABCD weighs more than EFGH is identical to Case 1. We again use ABE vs CDF for our second weighing. If ABE is heavier, then we know either A or B is heavy or F is light so we weigh A against B for our third weighing. If ABE is lighter, then we know either E is light or C or D are heavy so we weigh C against D for our third weighing. If ABE weighs the same as CDF, than we know G or H is light and we weigh G against H for our third weighing.

Case 3: ABCD = EFGH
If ABCD and EFGH are even, then we know one of I,J,K or L is heavy or light, and all of A,B,C,D,E,F,G,H are normal so they can (mostly) be ignored the rest of the way. For our second weighing, we could weight IJ against KL, but this would leave only two outcomes (they can't possibly be even!) and thus will not give us enough information to find out which of the 8 remaining possibilities is true. We could weight I versus J for our second weighing, but if they are even then we would have four remaining possibilities (K is light, K is heavy, L is light, and L is heavy) with only one weighing left (which has 3 possible outcomes), so we won't have enough information. Hence, for our second weighing we bring back one of the normal golf balls and try IJ vs AK. If IJ is less than AK then we know either I or J is light or K is heavy, and we weigh I versus J for our third weighing to decide. If IJ is more than AK we know either I or J is heavy or K is light, and we again weigh I versus for our third weighing to decide. If IJ and AK are even, then we know I,J, and K are normal so L must be either heavy or light-- and we weigh L against any of the other golf balls for our final weighing to determince whether L is heavy or light.

 

[chuckywang]

Originally posted by: brikis98
I'm bored as hell today, so lets post some brain teasers to keep my mind busy.

I'll start:

You are sitting at a table that has a whole bunch of coins scattered on it. You notice that exactly ten of these coins are tails up. All of a sudden, all the lights in the room go out and it's pitch black.

The question: how can you make two stacks of coins so that you can guarantee that each stack has the same number of coins tails up? Note that, in the dark, you cannot tell one side of a coin from the other - ie, they feel/taste/smell/etc the same on each side.

Stack 1: 10 coins
Stack 2: All the other coins

Flip every coin in Stack 1 over.

/teaser

 

[JTsyo]

Originally posted by: Kyteland

You have 12 golf balls, 11 of the same weight and one of a deviating weight. You are allowed to use a balance scale 3 times. BUT, you do not know whether the deviating golf ball is heavier or lighter than the rest. Using the scale 3 times, you have to point out the deviating ball AND tell whether its heavier or lighter than the rest.

How about break them into 4 groups?
ABC vs DEF
Equal - replace DEF with GHI
equal - you know JKL contains the unlike one
not equal - GHI contains the unlike one
Not equal - replace DEF with GHI
equal - you know GHI contains the unlike one
not equal - ABC contains the unlike one

From the group of three you weigh 2 and if they are equal then the one you didn't weigh is the different one.
...
Just realized if it's JKL, you can figure out which one but not if it's heavy or light.

 

[Kyteland]

Originally posted by: Kyteland

Originally posted by: Chronoshock

Originally posted by: Kyteland
Say you have a sheet of paper, what is the maximum number of pieces you can cut it into given n cuts. You cannot fold the sheet or rearrange already cut pieces (i.e., even after the nth cut, the pieces remain in the same position as they were before any cuts were made.)

And if you really want to geek out generalize the problem to any dimension. You have a X dimensional shape. You cut it N times in the X-1 dimension without moving or rearranging the pieces. How many pieces do you have?

The general formula is n*(n+1)/2 + 1 because each new cut can intersect all previous cuts. This leads to an additional n pieces. You start with one piece at 0 cuts, so you get 1 + sum(1,n) = n*(n+1)/2 + 1

That's correct. Now generalize it to any dimension.

Here's the answer to the above in case anybody is interested.

• F(d,n) = F(d,n-1) + F(d-1,n-1)

for any dimension d with n cuts. You can also state it as

• f(d,n) = A0*n^0 + A1*n^1 + ... + Ad*n^d

• f(d,n) = 2^n when n<=d

which yields a solvable system of equations.

 

[QED]

Originally posted by: TuxDave

Originally posted by: brikis98

Originally posted by: QED
Ok... another oldie:

Given a standard 8-by-8 chessboard with two opposing corner squares removed and a stack
of 2-by-1 domino pieces, is it possible to cover the entire chessboard using the domino pieces
without any overlaps? Each half (a square) of each domino must cover exactly one square on the chessboard,
and each square on the chessboard must be covered. If so, how?

I don't think it can be done.

On an 8x8 chessboard, squares are (or can be) colored black and white, so no two adjacent squares are of the same color. If you do this, you'll notice that the squares on opposite corners must be the same color. So, if you removed two corners, your board has either 32 black squares and 30 white squares or 32 white squares and 30 black squares.

The problem states that each half of a domino piece covers exactly one square on the chessboard. Since domino pieces cover two squares total, and these squares are adjacent, the two squares covered MUST be of different color. Since domino pieces can't overlap or hang off the board, it then follows that there is no way to get them to cover the entire board if the number of black and white squares is not equal.

You are teh smarty. I like that approach.

Yep, that is the answer.

It is somewhat convenient for the purposes of this problem that a chessboard is colored black and white that way it is, as it makes the answer obvious. If you don't mention that it is a chessboard or checkerboard and instead just use a generic 8-by-8, or 10-by-10, or 100-by-100 grid of squares using that approach doesn't seem obvious even though the answer is still the same. I know when I first approached this problem years ago I tried every method out there for hours, only to find out the answer was right there all along if only I had looked at the problem in a slightly different way.

 

[chuckywang]

Given 5 points on the surface of a sphere, prove that 4 of them must lie in the same hemisphere.

 

[QED]

Originally posted by: chuckywang
Given 5 points on the surface of a sphere, prove that 4 of them must lie in the same hemisphere.

Oh.. I like this one too.

 

[MysticLlama]

I'm missing something in my reading of the hats thing.

Either way, without knowing if there are an even number of hats, the last guy is 50/50 to die right?

Why not just have the last guy call out the color of the hat in front of him and hope his is that color too? Seems easier than everyone counting all the time.

 

[ng12345]

Originally posted by: chuckywang
Given 5 points on the surface of a sphere, prove that 4 of them must lie in the same hemisphere.

Hmm -- I don't know how to say it mathematically, but I'll give it a shot

any 2 points on the sphere will lie on a circle that goes through the center of the sphere -- thus dividing the sphere into two hemispheres (A&B), with both points in both hemispheres

the three remaining points can either:
1. lie in hemisphere A
2. lie on the same circle as the first two points (in both A & B)
3. lie in hemisphere B

Given this, at the very least 2 points will lie in one hemisphere and 1 in the other, and thus there will be at least 4 points in a given hemisphere

Now you guys are definitely much more math inclined than I am nowadays, but here is one that took me a little bit of time:

I have 3 integer weights in front of me. Using these three weights, I am able to measure all of the integer weights from 1 to 13 pounds. What are the weights?

show work

 

[brikis98]

Originally posted by: ng12345

Originally posted by: chuckywang
Given 5 points on the surface of a sphere, prove that 4 of them must lie in the same hemisphere.

Hmm -- I don't know how to say it mathematically, but I'll give it a shot

any 2 points on the sphere will lie on a circle that goes through the center of the sphere -- thus dividing the sphere into two hemispheres (A&B), with both points in both hemispheres

the three remaining points can either:
1. lie in hemisphere A
2. lie on the same circle as the first two points (in both A & B)
3. lie in hemisphere B

Given this, at the very least 2 points will lie in one hemisphere and 1 in the other, and thus there will be at least 4 points in a given hemisphere

Now you guys are definitely much more math inclined than I am nowadays, but here is one that took me a little bit of time:

I have 3 integer weights in front of me. Using these three weights, I am able to measure all of the integer weights from 1 to 13 pounds. What are the weights?

show work

I used trial and error for a while and the thing that finally set me on to the answer was the fact that there are 3 possible locations for a weight: left pan, right pan, or not used at all. That implied the answer probably used powers of 3, and not just random numbers. The three weights I would use:

1 3 9

Explanation:

I'll show how you can check whether something weighs 1-13lbs by using the 3 weights above on either the left pan (L), right pan (R) or not at all (I just won't include them). The object you're trying to measure is ALWAYS in the LEFT pan.

1 = 1R
2 = 1L 3R
3 = 3R
4 = 1R 3R
5 = 1L 3L 9R
6 = 3L 9R
7 = 3L 1R 9R
8 = 1L 9R
9 = 9R
10 = 1R 9R
11 = 1L 3R 9R
12 = 3R 9R
13 = 1R 3R 9R

cool problem

 

[Rowboat]

Edit: Heh didn't realize this thread had so many replies making my comment pretty pointless sorry.

 

[chuckywang]

Originally posted by: ng12345

Originally posted by: chuckywang
Given 5 points on the surface of a sphere, prove that 4 of them must lie in the same hemisphere.

Hmm -- I don't know how to say it mathematically, but I'll give it a shot

any 2 points on the sphere will lie on a circle that goes through the center of the sphere -- thus dividing the sphere into two hemispheres (A&B), with both points in both hemispheres

the three remaining points can either:
1. lie in hemisphere A
2. lie on the same circle as the first two points (in both A & B)
3. lie in hemisphere B

Given this, at the very least 2 points will lie in one hemisphere and 1 in the other, and thus there will be at least 4 points in a given hemisphere

Now you guys are definitely much more math inclined than I am nowadays, but here is one that took me a little bit of time:

I have 3 integer weights in front of me. Using these three weights, I am able to measure all of the integer weights from 1 to 13 pounds. What are the weights?

show work

Yep, that's correct. The formal term for that circle that goes through the center of the sphere is the "great circle".

 

[QED]

OK... new puzzle.

Penn and Teller perform a simple card trick. With Penn out of the room, Teller hands a deck of ordinary cards (without the jokers) to an audience member. He jestures for the audience member to pick any 5 cards from the deck as they see fit and hand them to him. The audience member grabs 5 cards, looks at them, and then hands them to Teller.
Tellers takes one of the 5 cards and gives it back to the audience member to hold, and places the other 4 cards face down in a pile. Penn then comes back in the room, picks up the pile of 4 cards, and then correctly announces what card the audience member has in their hand.

How does he do it? How is it possible Penn is able to tell what card the audience member is holding based soley on the four cards lying in a pile?

 

[QED]

Originally posted by: QED
OK... new puzzle.

Penn and Teller perform a simple card trick. With Penn out of the room, Teller hands a deck of ordinary cards (without the jokers) to an audience member. He jestures for the audience member to pick any 5 cards from the deck as they see fit and hand them to him. The audience member grabs 5 cards, looks at them, and then hands them to Teller.
Tellers takes one of the 5 cards and gives it back to the audience member to hold, and places the other 4 cards face down in a pile. Penn then comes back in the room, picks up the pile of 4 cards, and then correctly announces what card the audience member has in their hand.

How does he do it? How is it possible Penn is able to tell what card the audience member is holding based soley on the four cards lying in a pile?

Bump for the bored at work day crowd.

 

[brikis98]

Originally posted by: QED
OK... new puzzle.

Penn and Teller perform a simple card trick. With Penn out of the room, Teller hands a deck of ordinary cards (without the jokers) to an audience member. He jestures for the audience member to pick any 5 cards from the deck as they see fit and hand them to him. The audience member grabs 5 cards, looks at them, and then hands them to Teller.
Tellers takes one of the 5 cards and gives it back to the audience member to hold, and places the other 4 cards face down in a pile. Penn then comes back in the room, picks up the pile of 4 cards, and then correctly announces what card the audience member has in their hand.

How does he do it? How is it possible Penn is able to tell what card the audience member is holding based soley on the four cards lying in a pile?

i'm guessing teller "encodes" the value of the 5th card using the first 4, although i'm still working out exactly how. are all 4 cards face down, or is just the top card face down? are these cards symmetrical, or can you tell if one is "upside" down?

 

[jonks]

Originally posted by: brikis98

Originally posted by: QED
OK... new puzzle.

Penn and Teller perform a simple card trick. With Penn out of the room, Teller hands a deck of ordinary cards (without the jokers) to an audience member. He jestures for the audience member to pick any 5 cards from the deck as they see fit and hand them to him. The audience member grabs 5 cards, looks at them, and then hands them to Teller.
Tellers takes one of the 5 cards and gives it back to the audience member to hold, and places the other 4 cards face down in a pile. Penn then comes back in the room, picks up the pile of 4 cards, and then correctly announces what card the audience member has in their hand.

How does he do it? How is it possible Penn is able to tell what card the audience member is holding based soley on the four cards lying in a pile?

i'm guessing teller "encodes" the value of the 5th card using the first 4, although i'm still working out exactly how. are all 4 cards face down, or is just the top card face down? are these cards symmetrical, or can you tell if one is "upside" down?

By the placement of the cards on the table? Horizontal/vertical/face up/face down = Hearts/spades/clubs/diamonds. Say the center of the table is the middle of a clock, where he places them corresponds to the face value. 1 o'clock = ace, 2 = 2, etc. 11 = J, 12 = Q, dead center = K?

 

[brikis98]

Originally posted by: brikis98

Originally posted by: QED
OK... new puzzle.

Penn and Teller perform a simple card trick. With Penn out of the room, Teller hands a deck of ordinary cards (without the jokers) to an audience member. He jestures for the audience member to pick any 5 cards from the deck as they see fit and hand them to him. The audience member grabs 5 cards, looks at them, and then hands them to Teller.
Tellers takes one of the 5 cards and gives it back to the audience member to hold, and places the other 4 cards face down in a pile. Penn then comes back in the room, picks up the pile of 4 cards, and then correctly announces what card the audience member has in their hand.

How does he do it? How is it possible Penn is able to tell what card the audience member is holding based soley on the four cards lying in a pile?

i'm guessing teller "encodes" the value of the 5th card using the first 4, although i'm still working out exactly how. are all 4 cards face down, or is just the top card face down? are these cards symmetrical, or can you tell if one is "upside" down?

Ok, assuming each card has an "up" and a "down" - that is, the pictures and numbers on them are not symmetrical vertically - I think you can do the following:

1. Before the show, Penn and Teller agree on a north, south, east, west with respect to the stage, the table, or any arbitrary object. Teller then can point the pile so that "up" faces one of these four predefined directions to identify the suit. North could be spades, south could be hearts, etc.

2. The top card is face down, but the 3 cards below it don't necessarily have to be. These can have at least 3 "orientations": up and down (with respect to the top card) and upside down. Since there are 13 choices for the "type" of card (1-10, jack, queen, king), we can use the same strategy as the "weights" problem I answered earlier.

That is, before the show, they agree that the second card has a value of 1, the 3rd card a value of 3 and the 4th (bottom) card a value of 9. They also agree that for each of these 3 cards, if it is facing "up" (that is, the same direction as the top card) that is "R", if is facing "down" that is "L" and if it is face up, then it is ignored. Penn can then just use the following to figure out what "value" the 5th card had:

1 = 1R
2 = 1L 3R
3 = 3R
4 = 1R 3R
5 = 1L 3L 9R
6 = 3L 9R
7 = 3L 1R 9R
8 = 1L 9R
9 = 9R
10 = 1R 9R
11 = 1L 3R 9R
12 = 3R 9R
13 = 1R 3R 9R


There are probably even cleaner/simpler ways to encode this information, but this is probably the general idea.

 

[QED]

Originally posted by: sirjonk

Originally posted by: brikis98

Originally posted by: QED
OK... new puzzle.

Penn and Teller perform a simple card trick. With Penn out of the room, Teller hands a deck of ordinary cards (without the jokers) to an audience member. He jestures for the audience member to pick any 5 cards from the deck as they see fit and hand them to him. The audience member grabs 5 cards, looks at them, and then hands them to Teller.
Tellers takes one of the 5 cards and gives it back to the audience member to hold, and places the other 4 cards face down in a pile. Penn then comes back in the room, picks up the pile of 4 cards, and then correctly announces what card the audience member has in their hand.

How does he do it? How is it possible Penn is able to tell what card the audience member is holding based soley on the four cards lying in a pile?

i'm guessing teller "encodes" the value of the 5th card using the first 4, although i'm still working out exactly how. are all 4 cards face down, or is just the top card face down? are these cards symmetrical, or can you tell if one is "upside" down?

By the placement of the cards on the table? Horizontal/vertical/face up/face down = Hearts/spades/clubs/diamonds. Say the center of the table is the middle of a clock, where he places them corresponds to the face value. 1 o'clock = ace, 2 = 2, etc. 11 = J, 12 = Q, dead center = K?

The only information used are the 4 cards themselves, and (here's a hint) the order they are in. Nothing else matters-- not their placement or location on the table, not their orientation (upside-down or rightside-up), or anything else.

 

[alkemyst]

these threads fail because everyone just googles without even thinking.

 

[QED]

Originally posted by: brikis98

Originally posted by: brikis98

Originally posted by: QED
OK... new puzzle.

Penn and Teller perform a simple card trick. With Penn out of the room, Teller hands a deck of ordinary cards (without the jokers) to an audience member. He jestures for the audience member to pick any 5 cards from the deck as they see fit and hand them to him. The audience member grabs 5 cards, looks at them, and then hands them to Teller.
Tellers takes one of the 5 cards and gives it back to the audience member to hold, and places the other 4 cards face down in a pile. Penn then comes back in the room, picks up the pile of 4 cards, and then correctly announces what card the audience member has in their hand.

How does he do it? How is it possible Penn is able to tell what card the audience member is holding based soley on the four cards lying in a pile?

i'm guessing teller "encodes" the value of the 5th card using the first 4, although i'm still working out exactly how. are all 4 cards face down, or is just the top card face down? are these cards symmetrical, or can you tell if one is "upside" down?

Ok, assuming each card has an "up" and a "down" - that is, the pictures and numbers on them are not symmetrical vertically - I think you can do the following:

1. Before the show, Penn and Teller agree on a north, south, east, west with respect to the stage, the table, or any arbitrary object. Teller then can point the pile so that "up" faces one of these four predefined directions to identify the suit. North could be spades, south could be hearts, etc.

2. The top card is face down, but the 3 cards below it don't necessarily have to be. These can have at least 3 "orientations": up and down (with respect to the top card) and upside down. Since there are 13 choices for the "type" of card (1-10, jack, queen, king), we can use the same strategy as the "weights" problem I answered earlier.

That is, before the show, they agree that the second card has a value of 1, the 3rd card a value of 3 and the 4th (bottom) card a value of 9. They also agree that for each of these 3 cards, if it is facing "up" (that is, the same direction as the top card) that is "R", if is facing "down" that is "L" and if it is face up, then it is ignored. Penn can then just use the following to figure out what "value" the 5th card had:

1 = 1R
2 = 1L 3R
3 = 3R
4 = 1R 3R
5 = 1L 3L 9R
6 = 3L 9R
7 = 3L 1R 9R
8 = 1L 9R
9 = 9R
10 = 1R 9R
11 = 1L 3R 9R
12 = 3R 9R
13 = 1R 3R 9R


There are probably even cleaner/simpler ways to encode this information, but this is probably the general idea.

Very good, but in this case assume orientation on the table does not matter, and that the cards are symetrical (i.e. there is no way to tell if a card is right-side up or upside-down). In other words, the only information available to Penn are the 4 cards themselves, and perhaps the order they are in the pile.

It seems almost impossible, doesn't it? But you are on the right track.

 

[DrPizza]

Originally posted by: QED

Originally posted by: sirjonk

Originally posted by: brikis98

Originally posted by: QED
OK... new puzzle.

Penn and Teller perform a simple card trick. With Penn out of the room, Teller hands a deck of ordinary cards (without the jokers) to an audience member. He jestures for the audience member to pick any 5 cards from the deck as they see fit and hand them to him. The audience member grabs 5 cards, looks at them, and then hands them to Teller.
Tellers takes one of the 5 cards and gives it back to the audience member to hold, and places the other 4 cards face down in a pile. Penn then comes back in the room, picks up the pile of 4 cards, and then correctly announces what card the audience member has in their hand.

How does he do it? How is it possible Penn is able to tell what card the audience member is holding based soley on the four cards lying in a pile?

i'm guessing teller "encodes" the value of the 5th card using the first 4, although i'm still working out exactly how. are all 4 cards face down, or is just the top card face down? are these cards symmetrical, or can you tell if one is "upside" down?

By the placement of the cards on the table? Horizontal/vertical/face up/face down = Hearts/spades/clubs/diamonds. Say the center of the table is the middle of a clock, where he places them corresponds to the face value. 1 o'clock = ace, 2 = 2, etc. 11 = J, 12 = Q, dead center = K?

The only information used are the 4 cards themselves, and (here's a hint) the order they are in. Nothing else matters-- not their placement or location on the table, not their orientation (upside-down or rightside-up), or anything else.

If the cards were non-symmetrical, this problem would be trivial as there are 4x3x2 (24) different orders possible for the 4 cards, times two orientations for each card = 48 different arrangements. (and 48 remaining cards in the deck) Likewise, with face up/face down instead of symmetry, you get 48 different arrangements possible. This is the direction I was going until QED's post that I just quoted.

Ditto if arrangement on the table meant anything - heck, just using a 2 and a 3, players keep score from 0 to 10 points easily while playing Euchre. That would draw boos rather than awe.

4 remaining cards: 24 possible permutations. However - 48 unknown cards! I'll toss out a 2nd hint: Teller gets to pick which card to hand back to the audience member.

I hope QED's solution is more elegant than mine to this problem; mine was rather lengthy in explanation as to the rules I used for which card to give back to the audience member. I'm sure there's a simpler solution, but I was satisfied with any solution