Consider joining Wrath of Math to get early and exclusive videos and music, and to help support what I do: ua-cam.com/channels/yEKvaxi8mt9FMc62MHcliw.htmljoin This presentation is based on these MIT lecture notes; I was perusing them and thought this card trick was really fun: ocw.mit.edu/courses/6-042j-mathematics-for-computer-science-fall-2005/0d1038c54d262f3cbab53c13d9935f8e_ln10.pdf More math chats: ua-cam.com/play/PLztBpqftvzxXQDmPmSOwXSU9vOHgty1RO.html
@@NonExistentSpace Technically that's not wrong, is it? (I am not a native english speaker, but in german it would work) "I guess the murderer is the man with the blue hat, but it also could be the one ("the man") with the green hat." In my example "one" ist not the number, but a substitution for "man". That should work in this case too, where "one" is just a substituion for "ace".
3:59 4! = 24, and the covered card could be put to the left or the right of the open cards, giving 4! x 2 = 48 possibilities, exactly the number we need to pinpoint each of the remaining cards.
It's trippy that one can use one of 24 possibilities in order to encode one of 48 cards. But having said that, in the setup I missed that the assistant gets to choose which card is not revealed, which is where the extra information is
You're right, but there are two good reasons for using the face value as the major key. First, there are more face values so you're less likely to need a tie-breaker. Second, it's mentally easier to recognize 7 > 3 than Spades > Diamonds. Magicians have enough to think about on stage that the simpler approach is always preferred.
This isn’t really a trick, it’s just a rotation and sequence of cards. That’s the same as me shuffling the deck and looking at the first 5 cards before conducting the trick.
What about this: With the assistant, we agree that the card we need to guess will always be in the 5th position. This leaves us with 4 positions. We agree on which position corresponds to which suit. Then we proceed as follows: first, we lift one of the cards in positions 1-4 to reveal the suit (hearts, diamonds, spades, or clubs). This leaves us with 3 cards, which we then lift from left to right in specific combinations to reveal the value?
Fun to see this content. A couple of adjustments. In your initial description of how the cards were chosen, I either missed, or it wasn't said, that the assistant saw (and ordered) all the cards. Even so, here's a way to make a stunt into a fun trick. First, the assistant goes out into the crowd, not explaining what's happening. They get one person to shuffle the cards, then get another to shuffle the cards and pick one (and hold it), do this until there are five cards chosen, ostentatiously throw the cards over everyone. It is important that still no one knows what the trick is. Here's the part I'm not solving yet. But there must be a conspiracy, a huddle, assistant versus magician. Perhaps the assistant confesses he's jealous of the Master. Let's defeat him! YES! (Conspiracy.) The assistant and people with the cards huddle, and they all look at the Cards together. They make a point that the Master could not possibly guess the right one. (Or maybe not; maybe don't give away the trick yet.) Maybe just insist, "We're going to fool him. He'll finally see ... I'M AS GOOD AS HE IS!" THE POINT IS: the assistant must, at this point, figure out which will be the matching Card (the matched suit). With that determined, he schmoozes the card holders with conspiracy, making sure the match Card is on the bottom, making a point to the person who chose it, "Whose card was this? Yours? YES! You have just the face to fool the Master. Guileless. Dull. He'll never see through you BWAHHAHAHAH!" The point is that the Assistant arranges the Cards then and there, making as much fuss as he can with the cardholder, coaching him about how to fool the master. Because he then sends the guileless one up to defeat the Master. After all, the master is so lame, that even some novice from the crowd can fool him. The guileless one is sent up to smack the Cards down on the table, and then the reveal is down in whatever way they want (turning over one card at a time, laying out 4 face up, one down). Make the Master sit on his hands, but with the force of his magic eye, he sees through the guilelss one. There could be a lot of smug dialogue between the Master and Assistant; or not, as people may suspect you're sending messages (that's another way to do the trick). Of course, in the end, the Master is smug ("You thought you could defeat me. Amateur. ONE OF DIAMONDS!") Oh, I forgot to mention. This is super-important. The Master must smugly repeat the steps the assistant took to fool him. "You let all of them shuffle, as if that would fool me! They truly selected a card at random, just to fool me! And you didn't even come up here yourself, you coward. This fool you've sent me ... is telling me what his card is. I see it in his eyes. That little curl of his lip." This "rewrites" the audiences memory. When they were picking the cards, they didn't know why. ANd now that you are about to reveal the Card after laying a sequence in front of you, you have to remind them that it's quite impossible you could know the fifth Card. The whole thing is a misdirection away from the fact that the assistant was deliberately ordering the very cards they watched him ordering. Etc. This could be a very entertaining trick. It's fun, because the defeated assistant could immediately demand another try. This might be even more startling because if people think you must have somehow memorized the deck, that can't possibly be the case. Also, a nice detail, even if a freak circumstance resulted in five of the same suit coming up (i.e., 3 or more of a suit), the trick would still work. Or, to put it another way, is there a possible group of 5 Cards that could not be arranged as necessary. I'm thinking no. If I had the A through 5 of clubs, I can arrange the 2, 3, and 4 in the necessary sequence to indicate that the 5 is the last Card. In fact, I think I can arrange those five Cards with any arrangement, so long as the first card is smaller than the last card. I could have the A and 3, and then arrange the 2, 4, 5 in the necessary s, m, l order to indicate the last card is 2 away from the A. I don't know if this is the "worst case scenario". If I draw four of a kind, the odd man out will be higher or lower; I get the first and last in place, and then use suits to indicate s, m, l. I don't think I'd ever be in trouble.
Funnily enough, the (very clever) way you encode the 5th card also totally works as a constructive proof that the trick is possible, and one that is very accessible for people that you would otherwise lose once terms like factorials and n choose k are involved. Not that there is anything wrong with the matching approach.
For the remaining three cards, you actually need two conditions. As you described it, you can use the face value of each card to order them into (s, m, l). But you also need to rank-order the suits as a tie-breaker in case two cards have the same value (e.g. Hearts > Clubs).
The math is really interesting but I wonder depending on the rules of the setup if it can be done without thinking about complicated math at all. For example, flipping from the top, bottom, left, or right of the card, the first flip could indicate the suit. The second could indicate 1-3, 4-6, 7-9, 10-13. And the third could indicate which of the 3 or 4 choices it is. Or if you just consider horizontal vs vertical flips, you could incorporate going right to left, or left to right, to encode more information. The suit could be encoded in the order flipped, for example going straight left to right or right to left, or going out-out-in-in (1 4 2 3 for example) or in-in-out-out (like 2 3 1 4) or zigzagging (1 3 2 4) or any other kind of defined pattern since there's more than enough for that. Then you could use the orientation of the flips (vertical and horizontal) to narrow down the values, which would let you have more than enough possible ways to encode 13 numbers since it could accommodate 16 in total. In this sort of way, it wouldn't even matter what the four cards being revealed first are, but in the setting of a magic trick, it could act as a red herring to distract from the method used. Of course in the setting of doing a magic trick, you'd probably put on some level of theatrics, even if it's just a bar trick. And you shouldn't perform it for the same audience twice anyway, so there shouldn't be a way to catch on based on patterns.
A thought I have not even halfway through the video: It's probably relevant that, in the sets of 5 cards, the order of the cards doesn't matter, but the 4 revealed cards can be revealed in a specific sequence.
pretty sure you could have 52 * 51 * 50 * 49 .... you could also convey any card with only 2 other cards one representing the value and the other one representing the suit. but none of this really matters in your trick anyways.
Alphabetic ordering is probably too hard to remember, especially with international assistants. In card games, usually cards are ordered Hearts, Diamonds, Spades, Clubs, in decreasing order. Since the colours lign up, this is the easiest to remember
Excellent explanation. Way more in depth than most mathematical explanations for this trick. The only thing I would add is the name, Fitch Cheney's 5 card trick.
a less mathematically interesting solution: you can get an extra bit of information by the order in which the 4 known cards are revealed. so you have 24 permutations that can correspond to cards 1-24 in a normal 52-card deck ordering (minus the 4 known cards). then, the magician walks in, and if the assistant lays the cards out left-to-right (from the magicians perspective), the 5th card is the corresponding card from 1-24. if the assistant reveals them from right-to-left, just add 24 so the card is somewhere from 25-48. problems: i cant remember if the trick requires that the 4 known cards are already laid out before the magician walks in, and im already done with the comment. the keen observer would also be able to notice that the order in which they're laid out changes, upon viewing the trick several times. i also don't have any mathemagically-inclined friends to do this trick anyway
Fascinating demonstration, thank-you! But in the original example before 1m23s you didn't seem to follow the the rules. The first revealed card shown is 4H, but that isn't paired (in the same suit) as the hidden card 2C. Or am I missing something? Thanks again :-)
Thanks for watching! Yeah, that was a quick cheat to just demonstrate the spirit of the trick, that four cards are revealed to predict the fifth - I should have actually followed the rules to avoid retrospective confusion!
@@WrathofMath Actually it is as if you have followed a slightly more cryptic rule; very similar to the 'elementary rule': but for two details (1) the card that reveals the color and starting point is revealed second (=> 6C), and (2) the rotation goes counterclockwise, equivalent to a substraction => [4H, 5H, 6D]=[m,L,s]=4 => counterclockwise, 6C - 4 = 2C. Voilà :)
Nope, it's the order the cards are shown that tells the magician the exact card value and suit. And the magician's assistant isn't necessarily doing this silently. I was taught in CHaSeD order.
@RoderickEtheria Mate, if your whole complaint is that he decided to use the same method with a different order, you do not understand the nature of magic
Ooh, love it. Let's consider the possibilities: AC is revealed card, 2C is hidden card. The assistant needs to communicate "1", which in the video's encoding is (s, m, l). This is pretty straightforward, AD, 2D, 3D in that order. So Ace of Clubs + 1 = 2 of Clubs. 2C is revealed card can't happen because AC is more than 6 steps away clockwise. Going to skip similar examples. AD is revealed card, 2D is hidden card. Again the assistant needs to communicate "1" (s, m, l). Since C < D, we get AC, 2C, 3D. So Ace of Diamonds plus 1 is 2 of Diamonds. AD is revealed card, 3D is hidden card. Now the assistant needs to communicate "2" (s, l, m). Again by the video's logic, this ordering is AC, 2D, 2C. So Ace of Diamonds + 2 = 3 of Diamonds. 2D is revealed card, 3D is hidden card. The assistant needs to communicate "1" (s, m, l). This is clearly AC, 2C, AD by the video's logic. So here there are multiple ways to get to the solution for the given set of cards, any one of which would work. But there doesn't seem to be a set of cards for which there is no way to get to the solution given consistent rules.
@@TooTallForPony My question would be, how is ace related to 2 and to K? Imagine the set of cards 6H 9H KC AD 2S We take 6H as the first card and 9H is the hidden card. We need to communicate (msl) to the magician to signal a 3. But which of K, A and 2 is the largest? 2 is obviously smaller than K. But A can be larger than K and smaller than 2... Where is the beginning and the end of this scale relative to A?
@@WrathofMath Is there a legit reason to call something TONCAS? I feel like when you say something is TONCAS, you already knew it was at that moment and there is nothing to learn from this. Is it only used to name the concept or no? 🤔
This feels overly complicated and requires the assistant to pick the hidden card. Let's simplify things and let a spectator choose which card remains hidden. The last card revealed will tell us the suit by using CHaSeD ordering. The revealed cards each represent a suit based on the order they lay on the table. How the cards are turned over will tell us the value in binary. If the assistant grabs a card by the corner to turn it over, that's a 1. If the card is grabbed in the middle, that's a 0.
Not if you go clockwise, which is the whole point of writing the cards in a circle: there are 7 steps to go from 5 to Q, but only 6 steps to go from Q to 5.
Certainly possible. I did a quick search for card trick videos just to see what the titles and thumbnails looked like, and I didn't see any covering this trick - at least not at a glance.
Consider joining Wrath of Math to get early and exclusive videos and music, and to help support what I do: ua-cam.com/channels/yEKvaxi8mt9FMc62MHcliw.htmljoin
This presentation is based on these MIT lecture notes; I was perusing them and thought this card trick was really fun: ocw.mit.edu/courses/6-042j-mathematics-for-computer-science-fall-2005/0d1038c54d262f3cbab53c13d9935f8e_ln10.pdf
More math chats: ua-cam.com/play/PLztBpqftvzxXQDmPmSOwXSU9vOHgty1RO.html
sorting problem i seen and actually made a variant of this once.
Ah yes, the 1 of Diamonds
Did I say that? What do you expect from a guy who doesn't have a single non-prop deck of cards 😂
Favourite card, whenever someone asks me to think of a card I always pick that and so far nobody has managed to guess it
@@WrathofMath it's even better that you used it as an example of a card other than the Ace of Diamonds
12:50 "Ace of diamonds."
13:00 "It could also be the one of diamonds."
@@NonExistentSpace Technically that's not wrong, is it? (I am not a native english speaker, but in german it would work)
"I guess the murderer is the man with the blue hat, but it also could be the one ("the man") with the green hat."
In my example "one" ist not the number, but a substitution for "man". That should work in this case too, where "one" is just a substituion for "ace".
3:59 4! = 24, and the covered card could be put to the left or the right of the open cards, giving 4! x 2 = 48 possibilities, exactly the number we need to pinpoint each of the remaining cards.
It's trippy that one can use one of 24 possibilities in order to encode one of 48 cards. But having said that, in the setup I missed that the assistant gets to choose which card is not revealed, which is where the extra information is
13:01 Babe, new cards dropped for standard playing card decks
Really glad I didn't have to memorize 2.5 million combinations
bro was ready though, respect 👑
0:46 That’s not called shuffling. That’s rotating the deck.
You can also just use the reveal order of the remaining three cards to encode the number, regardless of the card values.
Whether or not you made a mistake at 22:00 depends on which subkey is first. You are correct if the number is the major key. You must have forgotten.
You're right, but there are two good reasons for using the face value as the major key. First, there are more face values so you're less likely to need a tie-breaker. Second, it's mentally easier to recognize 7 > 3 than Spades > Diamonds. Magicians have enough to think about on stage that the simpler approach is always preferred.
17:46 Why is there a name to such a natural intuitive thing
13:29 what an unfortunate notation
i think it's great notation, but i see what you mean 😂
This would have been easier to follow if you did e.g. 4♦️ instead of 4D. That being said, this is a great trick, at least from a math perspective!
As if hes going to draw a whole club or spade everytime a club or spade needs to be represented
Yeah if I was a decent doodler I would have done that, but using the letters seemed safest for consistent readability.
Coincidentally, I predicted the 4 of♦from the thumbnail before I watched the video based on the card that is most unlike the previous four cards.
@@WrathofMath You could buy yourself a set of four stamps and just stamp the symbols
Bro will use anything other than a standard pack of cards you can find in a gas station
This isn’t really a trick, it’s just a rotation and sequence of cards. That’s the same as me shuffling the deck and looking at the first 5 cards before conducting the trick.
What about this: With the assistant, we agree that the card we need to guess will always be in the 5th position. This leaves us with 4 positions. We agree on which position corresponds to which suit. Then we proceed as follows: first, we lift one of the cards in positions 1-4 to reveal the suit (hearts, diamonds, spades, or clubs). This leaves us with 3 cards, which we then lift from left to right in specific combinations to reveal the value?
Very smart way simpler
When I saw this, I just had to write a program to try this out with since I don't have any friends irl... This is such a cool concept.
Fun to see this content. A couple of adjustments. In your initial description of how the cards were chosen, I either missed, or it wasn't said, that the assistant saw (and ordered) all the cards. Even so, here's a way to make a stunt into a fun trick.
First, the assistant goes out into the crowd, not explaining what's happening. They get one person to shuffle the cards, then get another to shuffle the cards and pick one (and hold it), do this until there are five cards chosen, ostentatiously throw the cards over everyone. It is important that still no one knows what the trick is.
Here's the part I'm not solving yet. But there must be a conspiracy, a huddle, assistant versus magician. Perhaps the assistant confesses he's jealous of the Master. Let's defeat him! YES! (Conspiracy.) The assistant and people with the cards huddle, and they all look at the Cards together. They make a point that the Master could not possibly guess the right one. (Or maybe not; maybe don't give away the trick yet.) Maybe just insist, "We're going to fool him. He'll finally see ... I'M AS GOOD AS HE IS!"
THE POINT IS: the assistant must, at this point, figure out which will be the matching Card (the matched suit). With that determined, he schmoozes the card holders with conspiracy, making sure the match Card is on the bottom, making a point to the person who chose it, "Whose card was this? Yours? YES! You have just the face to fool the Master. Guileless. Dull. He'll never see through you BWAHHAHAHAH!" The point is that the Assistant arranges the Cards then and there, making as much fuss as he can with the cardholder, coaching him about how to fool the master.
Because he then sends the guileless one up to defeat the Master. After all, the master is so lame, that even some novice from the crowd can fool him.
The guileless one is sent up to smack the Cards down on the table, and then the reveal is down in whatever way they want (turning over one card at a time, laying out 4 face up, one down). Make the Master sit on his hands, but with the force of his magic eye, he sees through the guilelss one. There could be a lot of smug dialogue between the Master and Assistant; or not, as people may suspect you're sending messages (that's another way to do the trick).
Of course, in the end, the Master is smug ("You thought you could defeat me. Amateur. ONE OF DIAMONDS!")
Oh, I forgot to mention. This is super-important. The Master must smugly repeat the steps the assistant took to fool him. "You let all of them shuffle, as if that would fool me! They truly selected a card at random, just to fool me! And you didn't even come up here yourself, you coward. This fool you've sent me ... is telling me what his card is. I see it in his eyes. That little curl of his lip." This "rewrites" the audiences memory. When they were picking the cards, they didn't know why. ANd now that you are about to reveal the Card after laying a sequence in front of you, you have to remind them that it's quite impossible you could know the fifth Card. The whole thing is a misdirection away from the fact that the assistant was deliberately ordering the very cards they watched him ordering. Etc.
This could be a very entertaining trick. It's fun, because the defeated assistant could immediately demand another try. This might be even more startling because if people think you must have somehow memorized the deck, that can't possibly be the case.
Also, a nice detail, even if a freak circumstance resulted in five of the same suit coming up (i.e., 3 or more of a suit), the trick would still work. Or, to put it another way, is there a possible group of 5 Cards that could not be arranged as necessary. I'm thinking no. If I had the A through 5 of clubs, I can arrange the 2, 3, and 4 in the necessary sequence to indicate that the 5 is the last Card. In fact, I think I can arrange those five Cards with any arrangement, so long as the first card is smaller than the last card. I could have the A and 3, and then arrange the 2, 4, 5 in the necessary s, m, l order to indicate the last card is 2 away from the A. I don't know if this is the "worst case scenario". If I draw four of a kind, the odd man out will be higher or lower; I get the first and last in place, and then use suits to indicate s, m, l. I don't think I'd ever be in trouble.
Funnily enough, the (very clever) way you encode the 5th card also totally works as a constructive proof that the trick is possible, and one that is very accessible for people that you would otherwise lose once terms like factorials and n choose k are involved. Not that there is anything wrong with the matching approach.
1:25 The sequence does not folow the trick, I'm disappointed.
For the remaining three cards, you actually need two conditions. As you described it, you can use the face value of each card to order them into (s, m, l). But you also need to rank-order the suits as a tie-breaker in case two cards have the same value (e.g. Hearts > Clubs).
This is explicitly mentioned in the video at 21:00, he uses C-D-H-S
The math is really interesting but I wonder depending on the rules of the setup if it can be done without thinking about complicated math at all.
For example, flipping from the top, bottom, left, or right of the card, the first flip could indicate the suit. The second could indicate 1-3, 4-6, 7-9, 10-13. And the third could indicate which of the 3 or 4 choices it is.
Or if you just consider horizontal vs vertical flips, you could incorporate going right to left, or left to right, to encode more information. The suit could be encoded in the order flipped, for example going straight left to right or right to left, or going out-out-in-in (1 4 2 3 for example) or in-in-out-out (like 2 3 1 4) or zigzagging (1 3 2 4) or any other kind of defined pattern since there's more than enough for that. Then you could use the orientation of the flips (vertical and horizontal) to narrow down the values, which would let you have more than enough possible ways to encode 13 numbers since it could accommodate 16 in total.
In this sort of way, it wouldn't even matter what the four cards being revealed first are, but in the setting of a magic trick, it could act as a red herring to distract from the method used. Of course in the setting of doing a magic trick, you'd probably put on some level of theatrics, even if it's just a bar trick. And you shouldn't perform it for the same audience twice anyway, so there shouldn't be a way to catch on based on patterns.
A thought I have not even halfway through the video: It's probably relevant that, in the sets of 5 cards, the order of the cards doesn't matter, but the 4 revealed cards can be revealed in a specific sequence.
Absolutely!
This is genius, thanks!!
Gonna use it at tomorrows party, wish me luck!
That party has no idea what force of nature is coming!
pretty sure you could have 52 * 51 * 50 * 49 .... you could also convey any card with only 2 other cards one representing the value and the other one representing the suit. but none of this really matters in your trick anyways.
Hall’s Condition seems like a classic TONCAS.
Definitely!
Alphabetic ordering is probably too hard to remember, especially with international assistants. In card games, usually cards are ordered Hearts, Diamonds, Spades, Clubs, in decreasing order. Since the colours lign up, this is the easiest to remember
Except Clubs < Diamonds < Hearts < Spades also.
xenoblade chronicles music at 10:03 from satorl marsh?
Excellent explanation. Way more in depth than most mathematical explanations for this trick. The only thing I would add is the name, Fitch Cheney's 5 card trick.
Thank you for your comment and for sharing the name, it wasn't known to me!
a less mathematically interesting solution: you can get an extra bit of information by the order in which the 4 known cards are revealed. so you have 24 permutations that can correspond to cards 1-24 in a normal 52-card deck ordering (minus the 4 known cards). then, the magician walks in, and if the assistant lays the cards out left-to-right (from the magicians perspective), the 5th card is the corresponding card from 1-24. if the assistant reveals them from right-to-left, just add 24 so the card is somewhere from 25-48.
problems: i cant remember if the trick requires that the 4 known cards are already laid out before the magician walks in, and im already done with the comment. the keen observer would also be able to notice that the order in which they're laid out changes, upon viewing the trick several times. i also don't have any mathemagically-inclined friends to do this trick anyway
What if we encode using order and also whether it is flipped horizontally or vertically? that's exactly 48 encodings.
Fascinating demonstration, thank-you! But in the original example before 1m23s you didn't seem to follow the the rules. The first revealed card shown is 4H, but that isn't paired (in the same suit) as the hidden card 2C. Or am I missing something? Thanks again :-)
Thanks for watching! Yeah, that was a quick cheat to just demonstrate the spirit of the trick, that four cards are revealed to predict the fifth - I should have actually followed the rules to avoid retrospective confusion!
@@WrathofMath Actually it is as if you have followed a slightly more cryptic rule; very similar to the 'elementary rule': but for two details (1) the card that reveals the color and starting point is revealed second (=> 6C), and (2) the rotation goes counterclockwise, equivalent to a substraction => [4H, 5H, 6D]=[m,L,s]=4 => counterclockwise, 6C - 4 = 2C. Voilà :)
Nope, it's the order the cards are shown that tells the magician the exact card value and suit. And the magician's assistant isn't necessarily doing this silently. I was taught in CHaSeD order.
Did you even watch the video? That's literally what he ends up doing.
@meltingkeith7046 except his order is clubs, diamonds, hearts, spades, so not CHaSeD order.
@RoderickEtheria Mate, if your whole complaint is that he decided to use the same method with a different order, you do not understand the nature of magic
@meltingkeith7046 And my response of nope was to the thought that it should be impossible. There's plenty of reasons it should not be impossible.
@@RoderickEtheria which means you didn't watch the video, because he literally goes over how it can be done
thank for explaining the trick
AC, 2C, AD, 2D, 3D
Ooh, love it. Let's consider the possibilities:
AC is revealed card, 2C is hidden card. The assistant needs to communicate "1", which in the video's encoding is (s, m, l). This is pretty straightforward, AD, 2D, 3D in that order. So Ace of Clubs + 1 = 2 of Clubs.
2C is revealed card can't happen because AC is more than 6 steps away clockwise. Going to skip similar examples.
AD is revealed card, 2D is hidden card. Again the assistant needs to communicate "1" (s, m, l). Since C < D, we get AC, 2C, 3D. So Ace of Diamonds plus 1 is 2 of Diamonds.
AD is revealed card, 3D is hidden card. Now the assistant needs to communicate "2" (s, l, m). Again by the video's logic, this ordering is AC, 2D, 2C. So Ace of Diamonds + 2 = 3 of Diamonds.
2D is revealed card, 3D is hidden card. The assistant needs to communicate "1" (s, m, l). This is clearly AC, 2C, AD by the video's logic.
So here there are multiple ways to get to the solution for the given set of cards, any one of which would work. But there doesn't seem to be a set of cards for which there is no way to get to the solution given consistent rules.
@ Great reply. But I can’t even remember why I wrote my comment in the first place. 😂🎉
@@TooTallForPony My question would be, how is ace related to 2 and to K? Imagine the set of cards 6H 9H KC AD 2S
We take 6H as the first card and 9H is the hidden card. We need to communicate (msl) to the magician to signal a 3. But which of K, A and 2 is the largest? 2 is obviously smaller than K. But A can be larger than K and smaller than 2... Where is the beginning and the end of this scale relative to A?
I thought this would be a trick thats actually usable irl.
It is, you just need an assistant in on the trick. The last section of the video goes over how it would be done.
Very nice! It is true magic!
10:22 wow TONCAS strikes again
True!
@@WrathofMath Is there a legit reason to call something TONCAS? I feel like when you say something is TONCAS, you already knew it was at that moment and there is nothing to learn from this. Is it only used to name the concept or no? 🤔
Lovely math but a fraught magic trick I think. Too many ways to go wrong.
This feels overly complicated and requires the assistant to pick the hidden card. Let's simplify things and let a spectator choose which card remains hidden.
The last card revealed will tell us the suit by using CHaSeD ordering. The revealed cards each represent a suit based on the order they lay on the table.
How the cards are turned over will tell us the value in binary. If the assistant grabs a card by the corner to turn it over, that's a 1. If the card is grabbed in the middle, that's a 0.
5 and Q are both 6 spots away from each other
Good point! In that case the assistant has a choice. Regardless of which one they choose to expose, the value + 6 will equal the other.
Not if you go clockwise, which is the whole point of writing the cards in a circle: there are 7 steps to go from 5 to Q, but only 6 steps to go from Q to 5.
Ooh, I think I've seen this from Vsauce
Certainly possible. I did a quick search for card trick videos just to see what the titles and thumbnails looked like, and I didn't see any covering this trick - at least not at a glance.
@WrathofMath I actually can't find the video I'm thinking of, maybe I imagined the whole thing
what language is this? 5 minutes in and I understand nothing.
Impossible 😮
❤❤❤❤💙💙💙💙