9:28 how did you know the exact last card number? When they shuffle it, won't the card orders(10❤&5🖤) change randomly? Also if suppose the 10 ❤was not on either of piles while disclosing the cards with what alternate way you will decode the last card?
*steventaputoro* It can be explained, it's just not being explained by you. Whatever you do explain, you do it badly. Certainly not recommending your channel.
@@chrisbodum3621 That's a bit harsh. I've found a great explanation of this effect here ua-cam.com/video/wLXJ-o5mB1k/v-deo.html, but I don't think you should be trashing Steven the way you did. Steven is helping a lot of people to get interested in card magic and I personally like his cool Australian accent.
A couple of observations based on the video and the comments: DOING IT FACE DOWN - If you do it face down you'll need to change the patter to something like "cut the deck to see if you can LEAVE half" because the bottom half is what you'll need to count to bring the bottom card to the top. WHICH PILE YOU FORCE MATTERS (if you want the kicker) - They're shuffling two piles with a keycard on top of each. Those 2 cards may end up in positions 1 and 2...or they may not. The second card could be in any position and end up in either pack depending on the shuffle. The only guarantee is the ONE of them will be on top after the shuffle, which if why you'd force the pile that's dealt first. CUE FOR THE KICKER - The cue for the kicker can be anywhere. If the keycards end up in positions 1 and 2 it will be the top card of your pile (like the video) but it can be anywhere, even in the spectator's pile. The position of the second keycard is random, depending on the riffle shuffle. Once you see the keycard (anywhere), you'll know the kicker.
Really important, not sure why he didn’t explain it your way. Maybe he figured once we tried it a few times, we would figure it out. I tried it five times and the cards were on the end, as he had twice, but every time they were in the last two piles. Once you see one, you know the other is at the end. Still cool little trick. The key is to distract them from how you pick black and red, and that the cards are perfect red black alternating.
5:08 order of colour is 1,1,3,2,3. Remember the sequence and set cards aside.... then count the cards upside down and reveil them in colour piles after 😊
@@HolyParsivalin this case, 1 black, 1 red. 3 black, 2 red, 3 black. Starting from the ten of clubs. So you just need to peek the first colour and you're set. Practice in front of a mirror or video yourself until the thoughts become natural and so does the peek, tie it in with the story telling and boom
I can explain how it works: Take two piles of cards, each in alternating black/red order, then place them face-up on the table. One pile with red on top and the other with black on top. These are piles 'A' and 'B'. Take one card from either pile and use it to start building a new pile (Pile 'C'). Let's say you take the red card. The top card in piles 'A' and 'B' are now black. Take either of them and place it on pile 'C'. The top card in piles 'A' and 'B are now different. Take one of them, place it in pile 'C', to leave two same-colour cards on the top of piles 'A' and 'B'. Take one and place it on pile 'C'. You can see that each pair of cards in pile 'C' is one of each colour. The 1st and second cards are one black and one red, the 3rd and 4th are one black and one red etc. Now imagine that instead of taking cards from the top of each face up pile, the piles were face down and were being riffle shuffled. The first card to hit the table could be either red or black, but that leaves the bottom card of both of the piles being shuffled the same as each other, and opposite to the first card. When that card is released, the bottom cards are different again. When the third card is released, the bottom card of both piles is the same as each other (and opposite to the third card). In exactly the same way as when the cards were being dealt from the face-up piles, we could see that pairs of cards were being added to column C, the cards in the riffled pile are also in black-red pairs.
@dawiem6310 I take it that the 'Huh?' at the end of your comment implies you don't understand my explanation. Let me try to make it simpler: 1) Try actually getting a deck of cards while you read through my comment, and follow the actions, rather than just reading it. 2) When you are following the instructions with the 'face up' piles, appreciate that the action you're taking is effectively exactly what happens when you riffle shuffle. Riffle shuffling may look effective (and probably is if the cards are already well mixed) but maybe its limited effectiveness can be demonstrated by this method: Put all the hearts in one face-up pile, arranged Ace to King, with the Ace on top. Do the same with the clubs. Now turn both piles upside down and riffle them together. Now start to take the cards from the top of the face-down pack. For EVERY card you take (apart from the last one), there are only two possible cards that it could be - either the 'next' card in the hearts sequence or the next card in the black sequence. So if the first card you take is the King of Clubs (because it can't be anything other than a king), you know that the next card can only possibly be the Queen of Clubs or the King of Hearts etc. Just get a pack of cards, practise riffle shuffling, and you will see how my original comment makes sense - it's got several 'Likes' already, so it must make sense to those people!
@@kwilson5832thank you for taking the time to write out this well thought out explanation. If I’m not mistaken, is the success/failure of the trick contingent upon a proper riffle shuffle, whereas the shuffle alternates the cards from group A and group B one at a time?
3:15 rotating cut does not change card order. 3:33 you'd think they'd see the cards are alternating at that point and know something is up. 6:20 won't work if you don't do a perfect shuffle, yes? Doesn't this assume that the cards are perfectly interlaced left/right after the shuffle? I'm just guessing because else if two cards are put into the main deck from either side, it is going to throw off things? 4:47 then they say no, I want the one I put my hand on! trick dies.... Okay, I now understand the reason for dividing into piles of 10 so that you have a guide for what the other pile is so you DON'T have to shuffle them equally. Then keeping the 10 of hearts isolated lets you name that one. So, I get it now except if someone was set on spoiling the trick they could. 1) the cards are clearly alternation red/black and 2) they may want to choose from the 2 piles instead of being given what they did not put their hand on.
OK, figure this one out. I was shown this one in high school in the middle 60's. Absolutely no prep is needed, you can even have your mark shuffle the deck all day before performing the trick or not at all and they can even chose the cards used if you wish. 1) Take any 21 cards from the deck. Include jokers if you wish. 2) Splay out 3 rows of 7 cards, left to right, so that all cards are face up and can be identified at one corner. 3) Have someone memorize one card from any row but not disclose it. 4) Have the person only disclose what row the card is in but NOT the card itself. 5) Pick up all 3 rows of cards making sure that the identified row is in the middle. 6) Turn the pack over and splay out 3 rows again, left to right. The identified card can be in any row at this point. 7) Again, have the person identify which row the SAME card is in. 8) Pick up all 3 rows again making sure that the identified card row is in the middle. 9) Turn the pack over and splay out 3 rows again, left to right. The identified card can be in any row at this point. 10) Again, have the person identify which row the SAME card is in. 11) Pick up all 3 rows again making sure that the identified card row is in the middle. 12) Turn the pack over but this time, count out 10 cards flipping them as you go. 13) If the trick is done correctly, the 11th card will be the chosen card EVERY TIME regardless of which row or position it had been in on all 3 rounds. I won cigarette money in high school by counting out 12 cards and betting quarters that the "next" card that I flipped was the card. When the sucker took the bet, I would go back and flip the 11th card again. Cigarettes were less than 40c a pack back then.
Thanks for reminding me of this trick. Its the only i really learnt when i was young a long time ago. Keep on meaning to look it up but wouldnt know what to put in search. Thanks for a simple run through.
Great trick. However, the way it’s done the spectator would notice the red black pattern when counting how many cards were cut. It works exactly the same if the deck was cut and counted with the deck face down
You can count it facing down. The first part of the trick is misdirection. Pretending that you’re testing the spectator’s intuition cutting the deck in exactly half. In reality, you’re just setting up the real trick. So it doesn’t matter whether you count the cards facing down.
You can let spectator choose either pile of 10 because the other pile is the exact opposite color. You finish the trick the same but it gets rid of the magician’s choice.” Great trick!!!
And I remembered 10H on top at the beginning and the reveal. So you should also have them cut face down. I don't think it subtracts from the trick to have them face down.
My intuition is telling me to suggest that you should set your camera to manual focus for the over-hand sequence shots, so that the camera isn’t changing focuses each time your hands move across the center of the screen
I’m 25 , went to Vegas and re found my love for magic . That part when you taught us “magicians force “ was actually genius ! I never could of thought to manipulate the situation like that
Why are you expecting the spectator to shuffle like that? I don’t know anyone who can shuffle like that, so the trick won’t work if they just start shuffling overhand ¯\_(ツ)_/¯
you just tell them the way you want them to do it. you can use hand gestures while you tell them. or just cut and push both piles together on the table, there's lots of ways you can do it.
@@tyrgarion2730 During Shin Lim’s latest appearance on Jimmy Fallon, Shin asks Jimmy to shuffle cards and Jimmy tells Shin he doesn’t know how to ripple shuffle. Lucky for Shin he wasn’t doing this trick 😂
The Gilbreath Principle works a follows: You have two stacks of cards, one that goes R-B-R-B-...-R-B and one that goes B-R-B-R-...-B-R-B-R. They are shuffled together, meaning a new stack is created and 52 times, the bottom-most card of either stack becomes the new top-most card of the new stack. The first card has a choice of being either B (left stack) or R (right stack). Whichever we choose, the second card can only be the opposite, no matter what stack it comes from. Adding this second card necessarily makes it so for the third card, we have a choice again. The fourth card will be the opposite of the third. It's now easy to see how, in the new combined stack, every other card is precisely the opposite color of the card below it.
@@mozes42 Quite. Unless it's built into the trick in the first place. In that case you perform the trick using one method which is potentially guessable, then say " I bet you think I did X". Then do apparently the same trick again but using a different method which is clearly not-X.
@@mozes42 The first time you do the trick it entertains and amazes because everything is a surprise ... the second time, there's no surprise anymore because they know what to expect ... they've shifted into analysis mode while they try to detect the gimmick.
Actually, you don't need to do a magician's force for the 10 packs if the cards are "opposite". That's a needless complexity. In fact, it works better if the spectator has a completely free choice between 10 packs.
That suggests a further elaboration. Your subject might ask you to repeat the trick, and that can be dangerous with a simple force. If you sense that the subject is just a little "wise" and suspects the force, you run it through to that point -- then I see two ways to proceed: 1) Do the force. Subject objects, "You did it the other way last time", or even worse pulls a switch on you and says "No, no. I want you to use THAT pack this time", "Why of course", you say, knowing both cards. Or 2) when you sense his suspicion, step back and insist that he pick up his chosen packlet himself -- "No! I insist! You pick it up yourself this time, look, I'll even turn my back, just to assure there's no deception...."
@@TonusFabri2024 One of the first things I was taught when I began studying magic 60 years ago was NEVER repeat a trick. In this case, how are you going to resort the deck red-black without the person watching you set it up?
Cool trick, thanks. Btw, in the one shuffle, they might turn the 10H to the 2nd in the deck (and it will go to the magician's deck) so you'll need to notice and take a pick at the new top card if that happens.
This is the BIG problem with the trick; The SHUFFLE. If the Ten of Hearts does NOT end up on top you lose the ending surprise. The magician can't control the shuffle.
@@mosheshpinel3108 If you count the cut from the bottom half, you force the 5H to the top of the bottom half of the cut, while the 10H remains on top of the top half of the cut. No matter how they riffle, one or the other will be at the top of the combined deck, and that's your final card in the trick. The other card will be in the top 10 so you can eliminate it from your guess. Plus the last card in your flared out pile of 10 will be red if the 5S is the final card in their pile, and the last card in your flared out pile will be black if the 10H is the final card in their pile.
@@mosheshpinel3108 That isn't a problem at all. The last card in their pile has to be either the 10 of hearts or the 5 of spades. It is impossible for it to be any other card. If you see the 5 of spades anywhere in your pile of 10, you know their final card is the10 of hearts. If you see the 10 of hearts anywhere in your stack, you know their final card is the 5 of spades.
Yes, count out the bottom half of the cut and force the bottom card to the top, ensuring the 10H or the 5S will be the first card dealt to the piles of 10, and becomes your final card.
one problem i thought of is what if when you ask the person to shuffle, and they start doing a hand shuffle instead of a card cut. then that would mess the order up wouldn't it?
I don't know card tricks, and I definitely can't "fan shuffle" or interlace them like you did. So, it seems the trick requires that the spectator know how to shuffle in the manner you show, right? (Google says it is riffle shuffle. I would likely ruin the pattern).
Actually, it can't. Any combination of two halves of the deck after inversing of sequence in one of them would result in 26 consecutive pairs of black and red cards which allows the "guessing".
Riffle shuffle is a pretty common shuffle and by far the easiest way to get a decent mix. You probably don’t play many card games if you don’t know it. Learn it though; it’s pretty simple. As the other commenter said though, it doesn’t matter.
If you riffle shuffle a Deck r->b->r->b... with another b->r->b->r... deck you will insert between 1 or more cards between the r and b card and it will only switch the r->b sequence or do nothing. The deck: r -> b -> r -> b -> ... pair them in your head (r -> b) -> (r -> b) -> ... 1. insert one new black card [b] through shuffle r-> [b] -> b -> r -> b ->... 1. new pairing (r-> [b]) -> (b -> r) -> (b -> ...) (r-> b) -> (b -> r) -> (b -> ...) 2. insert more then one because no one can do perfect riffle shuffle r-> [b] -> [r] -> b -> r -> b ->... 2. new pairing (r-> [b]) -> ([r] -> b) -> (r -> b) ->... (r-> b) -> (r -> b) -> (r -> b) ->... So it will switch the red and black sequence on this part of the deck if the card number is odd or does nothing if its even.
Thanks for that. It makes sense now. But,why doesn't this also hold true for an ordinary hand shuffle?. After all you are still just introducing odd or even pairs into another stack .
Steven,regarding the “trick that can’t be explained “would you please re-explain how you know the last card (the kicker) in the spectators pile. Thanks very much! Wayne
Does it matter how the spectator shuffles the cards? Does it have to be very accurately done like the magician does? I am a complete beginner. Thank you
No it doesn’t have to be perfect. That’s what gives the “three black in a row, two red, one black two red” type of randomness as a result. All the magician really needs to do is pick someone who knows how to riffle shuffle without blowing the cards all over the room.
A better way would be to peek the bottom card (Key card 1 or KC1) just before you start the trick and after the 1st cut take the packet that contains that KC1 and count it onr by one - all the cards being face down during this part Now for the trick to work for the very last card (still the KC1) the shuffle MUST be done so that tah KC1 syas et the top like Stephen did in both presentation and explaination :-) Its easy to realize that if the KC1 is in second place from the top for instance it won't be in the spactator's hand after making choose the 1st pile of 10 cards.
I count the pile with the 5S on the bottom face down, which puts it at the top. The 10H is on top of the other pile, ensuring that both will be on the top of the large piles before the riffle shuffle, and one of them will be the bottom card of one pile when I deal out 2 piles of 10. Much better presentation. Doesn't give away the pattern.
Just be authoritative and shuffle for them and then tell them to hold their hand over cut card group while you hold their other hand like you're sensing the color through the person then have them flip the card after you call the color while you close your eyes and peek through your lashes at the spread.
Why do you need to force the pile? Could they not just pic your pile and you can read off the opposites with the pile containing the 10 of hearts, that way you'll predict their 5 of spades?
Suppose the 10 of hearts is on top after the shuffle as happens in the demonstration. The second card from the top does not need to be the 5 of spades. (It is not in the demonstration.). If you don’t force and they pick the “wrong” pile, the bottom card will be the second card from the top after the shuffle. However you don’t need to force. If they pick the “wrong” pile the bottom card of that pile will be one of the top two or the bottom two from the start of the trick. So now you just need to memorize 4 cards ahead of time.
@@zyklos229 At the drawing of the final card all others are already exposed. The lying down card is the one that's missing. While it's impressive to look over them quickly, you don't even need a trick to figure that one out.
Here is the explanation: By starting with alternating colors, you are guaranteed that the two piles are in opposite colors if there is no shuffling . So the question is, will shuffle (once) change that? To change that, the shuffle would have to create the same color in (odd, even) pair such as (1,2), (3,4) and so on, but a (even,odd) pair, such as (2,3), (6,7) would not matter (since 2 will be the first card and 3 will be the second card, so they will not be matched against each oether). Now since the two piles prior to shuffling are arranged in opposite order. Let's say if the first card in pile A is black, then black can only appear as the even number cards in pile B. So for black to appear in connective order, it would have to come from an odd number card from pile A and a even number card from Pile B. But an odd number card from pile A is proceeded by even number of cards while an even number card in pile B is proceeded by odd number of cards. This means a same color pair resulted from shuffling is always proceeded by odd number of cards. In other words, it will always be a (even, odd) pair, and never a (odd, even) pair!
I get it how you are telling the colours of the cards but how did you know the exact Card number of the last revealing card? Did you peek at the card before dealing it down? 6:40
I have 2 questions. What happens if the first cut gives an odd number of cards? And why do you need to force the 10 cards packet as both packets should be the exact opposite of the other?
The key is that you want the number of cards selected to be less than either of the halves in the 'intuition' cut. Most people will do a reasonable job of getting within a card or two of half, so by choosing 10 (20 total), you have some margin relative to 26. But in the event the cut is way lopsided, you could still make it work by using half of the smaller size (if it was 19, you'd use 9 cards in each pile instead of 10, for example...for a total of 18 - must be less than or equal to the smaller pile).
@@vincentv.9729 ..the direct answer is, it doesn't matter if the first cut gives an odd number. The only thing that's important, is that the cut be larger than 19, and less than 33.
I was wondering too about if the intuition cut was odd. But thanks for explaining I get it now. Excellent illusion can't wait to practice it. Great video!!! TY
It was the card that was at the very beginning of the trick. But the wow factor is that you not only knew what color it would be, but the actual card. So a little icing on the cake.
I can explain it. It's not that difficult to explain. I'll give you a clue: when you arrange them red & black all the way through - that's an artificial pattern that never changes it just repeats red/back over & over. Then you do 1 cut & 1 shuffle. That's it. that puts the inverse of the new shuffled pattern with almost perfect reciprocal r/b r/b pattern except that has a slight chance of throwing this off but it's mitigated by the fact that you deal off the top 20 cards - which eliminates the differences at the bottom of the shuffle - making a perfect mirror image of the cards in the left & right hands. All you have to do is read the image on one side or the opposite on the other side. This is easy to discern. It took me much longer to type it out than the 30 seconds it took to realize how it works. I left out details but I told you more than I was going to. The details I'd add would further clarify things that are obvious to me but I've told 90 percent of the story.
I noticed that you alternated the sequence when counting and that there were a lot of black/red alternating cards, but I couldn’t tell how you predicted the final sequence. At first, I thought it might have been an odd/even red/black clue with the cards.
This is all super smart, great video! I see how it works but I don’t understand how even after a random shuffle the two packs of 10 end up being symmetrical in color???
Would also need to make sure the kicker card is last card to go on top when performing the riffle shuffle. So when you count out the piles of ten it is the first card placed on the table ensuring it to be the final card drawn.
For unexplainable trick, you explained it relatively easily. If you split the deck in half you create two identical packs of alternating cards, due to the number being even in the deck. Both have the same alternating order. Now if you reverse the other pack, when you shuffle them you not only have an alternating order in both of the packs, but also alternating between the hands. So if you shuffle perfectly, they'll be in alternating order, but if you shuffle multiple from the same hand, they are also in an alternating order. As such when you deal them in pairs, they'll always keep the alternating order. When you see one person get multiple of the same color in a row, that's where your shuffling was imperfect, but due to the alternating order in both packs the pairs are still alternating.
Thanks for this explanation...I am going to have to read it again, when I'm not tired. But, I think you are right. HIS explanation alone, didn't make sense to me.
Very cool trick but how do you handle situation where most people start scuffling the deck normally and they don't necessarily know how to do riffle suffle or get it done first try?
3:46 You memorize both cards at the top (bottom of pile) which in this case it's the 5 of spades and the 10 of hearts. 10:26 The card that is face up to the right, which here is the 5 of spades, meaning he now knows that the last card is the 10 of hearts. =)
I found and error on the start of the routine. When the deck is cut for the first time. Make sure the count of cards is odd number. Even count screws up the out come of the routine.
no it doesn't. if the cut is odd, you can completely skip the unneeded reversal. the reversal is only needed if the cut is even, but is harmless if the cut is odd.
I was wondering, when setting up the deck, do you put them in order of number and color? How is it the 10 and 5 come up? Couldnt it be just a random number, but you still know the color?
So it _has_ to be shuffled in that particular way? What if they start shuffling it another way (like the majority of people will do), or want to shuffle it more than once?
Cool trick. I was thinking that if you had the spectator flip the cards over that you forced on them (after you've told them the color of course) then not only are they more interactive in the performance but that distracts them from seeing YOU glance at your cards for color reference.
@@Houjixful It's explained in the video: the card on top will either be the ten of hearts or the five of spades. You see which one is visible so you know the hidden one is the other.
@@vnen how would you know what deck to force if you don't know what pile has the ten? both stacks are face down when you force the pile and video didnt explain that. My only guess is to look at how the spectator shuffles and see if he shuffles the 10 to the top
Found this video by accident, I love it. I'm not a fan of set ups, so I had to tweak my performance. Grab a "random deck" lying around that's set up, do some false shuffles, riffles, whatever you're good at then suggest a trick (perhaps when someone offers to play a game of cards so it seems more spontaneous). Using comments in this video, adapt a face down version and bam. The only problem is if people keep asking you to do more or ask to repeat. Never repeat, and if you can do more, do. If not, go back to the original suggestion of playing whichever card game was suggested and after a few games, people will ask again to do more tricks. Because performing one is never enough. That's my humble take anyway
Why force the pile to the spectator? It works with any pile they pick, does not it? (if you don't want to get the last card I mean, but it is without force whatsoever)
Yea to be fair, I would alter this trick and just do the shuffle myself.. This trick is 100% based on a single rifle shuffle and some a hole will always be a smart ass and mess the whole trick up.
I'm not one to put in negative comments, and I don't think this is negative, just fact. The top and bottom cards (5S and 10H in this case) can very well end up in the same pile of 10 cards, it's pretty much a 50/50 chance. I would say, if you happen to see one of those cards in the 'discarded ten', then you have the bonus of being able to predict the last revealed card's suit and value. With that said, I like the concept and have thought of a nice application, thanks!
Explanation: After counting the cards of the top pile, the two piles contain alternating colors, with different colored cards on the bottom. The riffle shuffle is equivalent to taking cards, one at a time, from the bottom of the left or right pile, in a non-predictable way, and placing them in the resultant pile. Suppose the first card to fall is red. Then both piles have black cards, so a black card must come next. After choosing one of the black cards, we are in the same position as we started (two piles containing alternating colors with different colored cards on the bottom). So, every pair of cards in the resultant pile contain one red and one black, at least until either left or right pile is exhausted. That's all that is needed for dealing the two 10-card hands.
You can shuffle it sloppy and it will still work. The whole deck still retains the pattern of the initial Red/Black alternations but is stirred once and there is still a pattern. They don't have to be one to one shuffle.
Riffle shuffle makes 2 types of sequences in the deck. One is type: 1red, 1 black, 1 red, and so on ... alternating of color by just 1 card. The second type is 1red, 1red or 1black, 1black. Maximum is 2 cards together with the same color. Both types of sequences can be combined together. Bud one can mathematically prove(it is not so easy to explain it here corrrectly) , that first card in every sequence type: red, red or black, black must be always placed on an even position. And therefore at the end you get the desired result.
Hey, quick question: Why must I Magician's Choice the pile? As they are opposite, whichever I have to find, I can do it by peeking at the other. And I can also find the last card in the same way. Thanks! Cool trick.
Because that pile will definitely have one or both of your known cards, which you need for the final reveal. Depending on how the shuffle goes you may have one of your known cards (5S 10H) in each pile, or both in the magician's choice pile but they will never both be in the other pile. Try it a few times and you'll see. I suggest turning those two cards face up to make it easier to follow.
I've had 8 Maker's and cokes and tried to follow along watching this. It seems VERY impressive, even if I have not been drinking... but alas, I've failed even the 1st prompt, as a drunken magician!!! What I've learned: don't drink and mage!!!
What if your friends ask you to do it again because they wish to figure it out? You're screwed as you won't be able to alternate colors as in the initial setup, right?
Doesn’t need to be a perfect riffle shuffle. In fact it’s better if it’s not perfect. Just ask them to push the two piles together. Rosetta shuffle even 🤙🏽
Great and very clever. Having said that, I can't think that there is any "spectator" on this planet who would let you get away with telling them that: a: They can only do a ripple shuffle b: they can only do it once!
Hi! Amazing trick, and it works with just Math principles! I have one question: when you "guess" the last card ("its a 10 of hearths" for example), do you need to shuffle the deck in such a way that the 10 will be on top of it? What if the 10 gets more in the middle? I think thats the only confusing part for me. Thanks!
Either the 10 of hearts or the 5 of spades will be on top. You need to see which one is in the visible deck, then you know the one actually on top after the shuffle.
This relies on a very specific method of shuffling - what if the spectator decides to shuffle it a different method? One that will mess up the alternating order? In my mind, you still need an "inside spectator" for this trick - one who will always shuffle the deck using this very specific (albeit most common) method.
If you put something in a sequence of 50/50 when the cards mix two of the same color fall at once before the first opposite color does then allowing and having no choice for the same opposite color to fall with it. There's only two options . So when you split it in half and force the cards to the player it's opposite. Once again if it's not red then it's black. Do it with less cards and shuffle very slow and you will see how the cards fall.
Pretty simple to explain actually. The only thing is it doesn't work if the audience member doesn't shuffle the top two cards correctly for guessing the number and the suit. You can only do the suit colors at that point.
@@mmvalex6903 Yes! Even an inaccurate rifle shuffle will work. To make it NOT work, you'd have to allow more than ten cards from one side before even one card comes from the other side. Or is it five consecutively from one side?
4:40 the magician's force is not needed here. If they take the other pile, ask them to use their intuition to guess the color of their top card, and flip it over, repeating this for the rest of the ten cards. They'll get some right, some wrong. Tell them, hey, that's pretty good. Now you do the same with your pile, and get them all right with the info you got from their cards.
so they have to shuffle a specific way and only once? wouldn't that tip them off when u tell them this minor detail? lol what if they want to shuffle the other way where u smash the 2 halves together?
I like it . I like doing mostly Impromptu cards tricks, however If I didn't I would snatch this one up. Great job ty for the tutorial. I am subscribing . =)
I know a trick much like this one, where the entire deck is stacked. If I am just doing card tricks for a small group of friends or something, I have another trick I like to do beforehand where a section of the deck is shuffled face-up into the rest of the deck. The great thing about this is that when I gather up all the cards I can stack the entire deck, right in front of them, even if they are paying attention, and it just looks like I am getting all the cards to be facing the same way again. Never had anyone call me on it yet. So, if I want to do the trick with the stacked deck, I do the other trick first (which is impromptu), gather up the whole deck and stack it right in front of them, and go right into my stacked deck trick.
I wonder how much of this depends on them being a decently skilled shuffler. Like... What happens if they make a small mistake, and the 5 of spades and 10 of hearts end up in the same stack? I know you can just abandon that final prediction, but does the rest still work if it's not a perfect shuffle?
I don't get it, wouldn't you just use a marked deck for an effect like that, you'd just pull out a deck and tell them what each card is, all the other stuff seems unnecessary. It's a cool method but the spectator doesn't know or care about method.
@@A.D.Gah yes, a random deck from a friend. perfect for when you need a prepared deck. no one will mind you stacking the cards in alternating colors for 2 mins before hand. never mind you can't do the trick a second time.
@@dfborbaif you can’t spread cull and faro, just say so. seems like an issue with the magician if you can’t color separate and shuffle without raising suspicion
Just for everyone who thinks they understand it but don't really, the trick is the alternating b and red. Now think of a stack of 51 and 1 shuffled. That one card pushes the alternation by 1, but it doesn't break the alternation because it just means the alternation starts again on the second stack, one card later. 2 or any even number of alternating cards added into the sequence doesn't break the alternation. Any odd number of cards is just like adding 1 with and even number after the 1 card, so it effects the sequence as only placing 1. Those facts mean a single shuffle does some number of all of the rules tandomly, but each instance doesnt break the black red alteration. Meaning every card when seperate is still its opposite in the other stack. Every discontinuity in the seperated stacks was an odd number shuffled in. I've seen a lot of answers here that are just wrong, but feel right.
Is this a similar trick as in a video called "Shuffling Card Trick" on the "Numberphile" channel? They were trying to explain the mathematical reason why the trick works but I only understood maybe 30%-40% of video.
If we count the cards, we may get an observant viewer who suspects why the cards alternate in colors, first red, then black..... How to get out of this situation?
Why should you force audience to choose the one pile, if its just the opposite and 10H vs 5S, then you can just work with any of the pile? Just curious.. Thank
Good trick, but from what I saw, the one part that can get dicey is the last card/10 of hearts reveal. I noticed the 5 of spades and 10 of hearts were the top cards of the two piles, then the spectator shuffled. The ending would need those two cards to remain on top, so a bad shuffle could easily have buried those 2 cards past the top 2 positions. I guess you can watch the shuffle and see if that's what happened and if it did, then you can eliminate the last card prediction...the rest of the trick is still very good. I'd be curious if this observation was wrong.
That's why you need to force one particular pile. Either the 10 of hearts or the 5 of spades will be on top after the shuffle. The other one will be visible in deck, it might not be the last card, but it does not matter because the spectator does not know about it.
I don't get the shuffle part, can they pick them up to shuffle from one hand to the outer or must they do it the way you did, what ever that is called?
Similar to one I used to know, but after the riffle shuffle, fan to show the mix, and make a break inbetween a pair of the colour with a cut, I would fan a bit, "You can see they're in no particular order" (make the break, then got through showing them the rest and making the cut) "See... red black, red black, black, red, no particular order, that shuffle is the best for really mixing them up like this..." Then deal the lot into two piles. They have one you have the other, they will deal the cards face up separating them into red and black piles, at the same time, I take a card and match it in front of me, so there's 4 piles on the table, 2 piles that are the face up reds and blacks they dealt, and the 2 I dealt face down. Then when you turn them over, incredibly they are all in order all mine are red and black piles too. I still don't know quite how it works, but it does and is a great trick, like this one.
If I understand correctly, if you're willing to lose the magicians force you can actually let them take either pile then the trick still works. but you simply don't name the final card?
Pretty obvious that the black-red mix results in even numbers and placement of cards. Any inconsistency in one hand will show up as the opposite in the other hand. You don't even need to force a hand because if you know one, you know the other, so it doesn't matter which one they pick.
Here is an explanation of why it works. When you have a setup in half of the deck, a riffle shuffle does not change the order of the cards in your setup. Here we have two halves of the deck with mirror image setups (alternating red and black) after the spectator's cards are counted. The riffle shuffle produces a deck with the two series intermingled-- the deck is composed of "blocks" of one or more cards, alternating from the two shuffled halves. Suppose for example that the top block contains 4 cards, in the order RBRB. These are dealt to the respective piles, producing the pattern of opposite colors. Now suppose the second block, which must be from the other half, contains an odd number of cards such as three. You then know that their order must be BRB because of the mirror image property of the halves. The first two cards are dealt onto the two piles. Now the important point: the third card (B) must be the opposite color of the next card, which will be from the first half. But this is guaranteed since the even number (4) of the first block ensures that an odd number card (5) is next, and this matches the odd number (3) from the second half; due to the mirror image property, this must be a red card, since all odd-numbered cards in this half are red. Similar arguments apply to odd followed by even and odd followed by odd blocks. Ironically, you DO NOT want the spectator to make a perfect shuffle, because then his whole pile will consist of only red or black cards.
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9:28 how did you know the exact last card number? When they shuffle it, won't the card orders(10❤&5🖤) change randomly? Also if suppose the 10 ❤was not on either of piles while disclosing the cards with what alternate way you will decode the last card?
@@akashnarayanan4199He’s not explaining it right
The word is "Inexplicable" - "unexplainable" is a lazy American perversion.
*steventaputoro* It can be explained, it's just not being explained by you. Whatever you do explain, you do it badly.
Certainly not recommending your channel.
@@chrisbodum3621 That's a bit harsh. I've found a great explanation of this effect here ua-cam.com/video/wLXJ-o5mB1k/v-deo.html, but I don't think you should be trashing Steven the way you did. Steven is helping a lot of people to get interested in card magic and I personally like his cool Australian accent.
A couple of observations based on the video and the comments:
DOING IT FACE DOWN - If you do it face down you'll need to change the patter to something like "cut the deck to see if you can LEAVE half" because the bottom half is what you'll need to count to bring the bottom card to the top.
WHICH PILE YOU FORCE MATTERS (if you want the kicker) - They're shuffling two piles with a keycard on top of each. Those 2 cards may end up in positions 1 and 2...or they may not. The second card could be in any position and end up in either pack depending on the shuffle. The only guarantee is the ONE of them will be on top after the shuffle, which if why you'd force the pile that's dealt first.
CUE FOR THE KICKER - The cue for the kicker can be anywhere. If the keycards end up in positions 1 and 2 it will be the top card of your pile (like the video) but it can be anywhere, even in the spectator's pile. The position of the second keycard is random, depending on the riffle shuffle. Once you see the keycard (anywhere), you'll know the kicker.
Really important, not sure why he didn’t explain it your way. Maybe he figured once we tried it a few times, we would figure it out. I tried it five times and the cards were on the end, as he had twice, but every time they were in the last two piles. Once you see one, you know the other is at the end.
Still cool little trick. The key is to distract them from how you pick black and red, and that the cards are perfect red black alternating.
Thank you. That helped more than you think!
“The method can’t be explained…I’ll explain in a second”
What he did was to change "can't" to "could'nt"
😂😂😂😂😂
You definitely fuck
He never did explain how it works, though.
Why does it take a second to explain?. You know it 6 hours before you even see that.
5:08 order of colour is 1,1,3,2,3. Remember the sequence and set cards aside.... then count the cards upside down and reveil them in colour piles after 😊
Huh?
nice! thats a more magical way without them realising you are looking at the cards
@@HolyParsivalin this case, 1 black, 1 red. 3 black, 2 red, 3 black. Starting from the ten of clubs. So you just need to peek the first colour and you're set. Practice in front of a mirror or video yourself until the thoughts become natural and so does the peek, tie it in with the story telling and boom
Wow 🙌 but I still don’t get why it is the 10 of hearts the last one
I can explain how it works: Take two piles of cards, each in alternating black/red order, then place them face-up on the table. One pile with red on top and the other with black on top. These are piles 'A' and 'B'. Take one card from either pile and use it to start building a new pile (Pile 'C'). Let's say you take the red card. The top card in piles 'A' and 'B' are now black. Take either of them and place it on pile 'C'. The top card in piles 'A' and 'B are now different. Take one of them, place it in pile 'C', to leave two same-colour cards on the top of piles 'A' and 'B'. Take one and place it on pile 'C'. You can see that each pair of cards in pile 'C' is one of each colour. The 1st and second cards are one black and one red, the 3rd and 4th are one black and one red etc.
Now imagine that instead of taking cards from the top of each face up pile, the piles were face down and were being riffle shuffled. The first card to hit the table could be either red or black, but that leaves the bottom card of both of the piles being shuffled the same as each other, and opposite to the first card. When that card is released, the bottom cards are different again. When the third card is released, the bottom card of both piles is the same as each other (and opposite to the third card). In exactly the same way as when the cards were being dealt from the face-up piles, we could see that pairs of cards were being added to column C, the cards in the riffled pile are also in black-red pairs.
TL;DR: smtg smtg about learning ABC when you're young to counter illiteracy in Botswana
I think I speak for many t say thank you for this explanation, and 'Huh?'
@dawiem6310 I take it that the 'Huh?' at the end of your comment implies you don't understand my explanation. Let me try to make it simpler:
1) Try actually getting a deck of cards while you read through my comment, and follow the actions, rather than just reading it.
2) When you are following the instructions with the 'face up' piles, appreciate that the action you're taking is effectively exactly what happens when you riffle shuffle.
Riffle shuffling may look effective (and probably is if the cards are already well mixed) but maybe its limited effectiveness can be demonstrated by this method:
Put all the hearts in one face-up pile, arranged Ace to King, with the Ace on top. Do the same with the clubs. Now turn both piles upside down and riffle them together.
Now start to take the cards from the top of the face-down pack. For EVERY card you take (apart from the last one), there are only two possible cards that it could be - either the 'next' card in the hearts sequence or the next card in the black sequence. So if the first card you take is the King of Clubs (because it can't be anything other than a king), you know that the next card can only possibly be the Queen of Clubs or the King of Hearts etc.
Just get a pack of cards, practise riffle shuffling, and you will see how my original comment makes sense - it's got several 'Likes' already, so it must make sense to those people!
@@kwilson5832thank you for taking the time to write out this well thought out explanation. If I’m not mistaken, is the success/failure of the trick contingent upon a proper riffle shuffle, whereas the shuffle alternates the cards from group A and group B one at a time?
@@kwilson5832the video suggested that a riffle shuffle is not needed… not many participants know how to do it!
magician: now shuffle
me: does pile shuffling
magician: 😳
Yeah, better you shuffle yourself and perform a riffle shuffle. Here in India, people hardly know how to riffle shuffle
3:15 rotating cut does not change card order. 3:33 you'd think they'd see the cards are alternating at that point and know something is up. 6:20 won't work if you don't do a perfect shuffle, yes? Doesn't this assume that the cards are perfectly interlaced left/right after the shuffle? I'm just guessing because else if two cards are put into the main deck from either side, it is going to throw off things? 4:47 then they say no, I want the one I put my hand on! trick dies.... Okay, I now understand the reason for dividing into piles of 10 so that you have a guide for what the other pile is so you DON'T have to shuffle them equally. Then keeping the 10 of hearts isolated lets you name that one. So, I get it now except if someone was set on spoiling the trick they could. 1) the cards are clearly alternation red/black and 2) they may want to choose from the 2 piles instead of being given what they did not put their hand on.
OK, figure this one out. I was shown this one in high school in the middle 60's. Absolutely no prep is needed, you can even have your mark shuffle the deck all day before performing the trick or not at all and they can even chose the cards used if you wish.
1) Take any 21 cards from the deck. Include jokers if you wish.
2) Splay out 3 rows of 7 cards, left to right, so that all cards are face up and can be identified at one corner.
3) Have someone memorize one card from any row but not disclose it.
4) Have the person only disclose what row the card is in but NOT the card itself.
5) Pick up all 3 rows of cards making sure that the identified row is in the middle.
6) Turn the pack over and splay out 3 rows again, left to right. The identified card can be in any row at this point.
7) Again, have the person identify which row the SAME card is in.
8) Pick up all 3 rows again making sure that the identified card row is in the middle.
9) Turn the pack over and splay out 3 rows again, left to right. The identified card can be in any row at this point.
10) Again, have the person identify which row the SAME card is in.
11) Pick up all 3 rows again making sure that the identified card row is in the middle.
12) Turn the pack over but this time, count out 10 cards flipping them as you go.
13) If the trick is done correctly, the 11th card will be the chosen card EVERY TIME regardless of which row or position it had been in on all 3 rounds.
I won cigarette money in high school by counting out 12 cards and betting quarters that the "next" card that I flipped was the card. When the sucker took the bet, I would go back and flip the 11th card again. Cigarettes were less than 40c a pack back then.
Didn't work. The chosen card was in 7th position 3rd row.
Cheers mate 😂😂🎉
Thanks for reminding me of this trick. Its the only i really learnt when i was young a long time ago. Keep on meaning to look it up but wouldnt know what to put in search. Thanks for a simple run through.
This was the only card trick I learned as a kid and I still stomp people lol😢
Great trick. However, the way it’s done the spectator would notice the red black pattern when counting how many cards were cut. It works exactly the same if the deck was cut and counted with the deck face down
You can count it facing down. The first part of the trick is misdirection. Pretending that you’re testing the spectator’s intuition cutting the deck in exactly half.
In reality, you’re just setting up the real trick. So it doesn’t matter whether you count the cards facing down.
Your point is well taken. It is very obvious that the cards are in a black/red repeating pattern. Turning cards face down is the answer.
You can let spectator choose either pile of 10 because the other pile is the exact opposite color. You finish the trick the same but it gets rid of the magician’s choice.” Great trick!!!
No nobody can notice when he did it first time I'm sure you didn't understand but now you know
And I remembered 10H on top at the beginning and the reveal. So you should also have them cut face down. I don't think it subtracts from the trick to have them face down.
My intuition is telling me to suggest that you should set your camera to manual focus for the over-hand sequence shots, so that the camera isn’t changing focuses each time your hands move across the center of the screen
Yws
The Gilbreath principle is amazing and i like the idea of the spectator trying to cut half the cards as the excuse to reverse them 👏
I’m 25 , went to Vegas and re found my love for magic . That part when you taught us “magicians force “ was actually genius ! I never could of thought to manipulate the situation like that
'magician's choice' is the name of this principle
Why are you expecting the spectator to shuffle like that? I don’t know anyone who can shuffle like that, so the trick won’t work if they just start shuffling overhand ¯\_(ツ)_/¯
you just tell them the way you want them to do it. you can use hand gestures while you tell them. or just cut and push both piles together on the table, there's lots of ways you can do it.
I don’t know ANYONE who DOESN’T shuffle like that. How would you NOT expect the spectator to shuffle like that?
@@tyrgarion2730 During Shin Lim’s latest appearance on Jimmy Fallon, Shin asks Jimmy to shuffle cards and Jimmy tells Shin he doesn’t know how to ripple shuffle. Lucky for Shin he wasn’t doing this trick 😂
@@jc_alpha I don’t know Jimmy Fallon. Seen him on tv never met him
I shuffle like that, always have.
The Gilbreath Principle works a follows:
You have two stacks of cards, one that goes R-B-R-B-...-R-B and one that goes B-R-B-R-...-B-R-B-R.
They are shuffled together, meaning a new stack is created and 52 times, the bottom-most card of either stack becomes the new top-most card of the new stack.
The first card has a choice of being either B (left stack) or R (right stack). Whichever we choose, the second card can only be the opposite, no matter what stack it comes from.
Adding this second card necessarily makes it so for the third card, we have a choice again. The fourth card will be the opposite of the third.
It's now easy to see how, in the new combined stack, every other card is precisely the opposite color of the card below it.
Nice. That's deep,
I'm pretty good with math. By the time he finished running through the explanation and saying he didn't understand it... I understood it.
Not with the clumsy way that I shuffle cards.
Still trying to wrap my head around this.
But is this guaranteed to work every time when done properly?
One problem that could crop up. If the participant asks you to repeat the trick, you can’t do it without bring a new ordered deck out.
You never repeat a trick
@@mozes42 Quite. Unless it's built into the trick in the first place. In that case you perform the trick using one method which is potentially guessable, then say " I bet you think I did X". Then do apparently the same trick again but using a different method which is clearly not-X.
@@johnnhoj6749 true
@@mozes42 The first time you do the trick it entertains and amazes because everything is a surprise ... the second time, there's no surprise anymore because they know what to expect ... they've shifted into analysis mode while they try to detect the gimmick.
In which case they’ll know you’re not using actual magic
Actually, you don't need to do a magician's force for the 10 packs if the cards are "opposite". That's a needless complexity. In fact, it works better if the spectator has a completely free choice between 10 packs.
you do need the force for the zinger at the end. You would be keeping track of the TOP card which was the ten of hearts.
He said the 5S was known too, so no force is needed.
That suggests a further elaboration. Your subject might ask you to repeat the trick, and that can be dangerous with a simple force. If you sense that the subject is just a little "wise" and suspects the force, you run it through to that point -- then I see two ways to proceed: 1) Do the force. Subject objects, "You did it the other way last time", or even worse pulls a switch on you and says "No, no. I want you to use THAT pack this time", "Why of course", you say, knowing both cards. Or 2) when you sense his suspicion, step back and insist that he pick up his chosen packlet himself -- "No! I insist! You pick it up yourself this time, look, I'll even turn my back, just to assure there's no deception...."
@@TonusFabri2024 Bro.. nobody ever does the same trick twice.. you should know that.
@@TonusFabri2024 One of the first things I was taught when I began studying magic 60 years ago was NEVER repeat a trick.
In this case, how are you going to resort the deck red-black without the person watching you set it up?
Cool trick, thanks. Btw, in the one shuffle, they might turn the 10H to the 2nd in the deck (and it will go to the magician's deck) so you'll need to notice and take a pick at the new top card if that happens.
This is the BIG problem with the trick; The SHUFFLE. If the Ten of Hearts does NOT end up on top you lose the ending surprise. The magician can't control the shuffle.
@@mosheshpinel3108 If you count the cut from the bottom half, you force the 5H to the top of the bottom half of the cut, while the 10H remains on top of the top half of the cut. No matter how they riffle, one or the other will be at the top of the combined deck, and that's your final card in the trick. The other card will be in the top 10 so you can eliminate it from your guess. Plus the last card in your flared out pile of 10 will be red if the 5S is the final card in their pile, and the last card in your flared out pile will be black if the 10H is the final card in their pile.
@@mosheshpinel3108 That isn't a problem at all. The last card in their pile has to be either the 10 of hearts or the 5 of spades. It is impossible for it to be any other card. If you see the 5 of spades anywhere in your pile of 10, you know their final card is the10 of hearts. If you see the 10 of hearts anywhere in your stack, you know their final card is the 5 of spades.
I would do the first count into 2 halves with a face down pile so the alternating black and red won't be seen.
Yes, count out the bottom half of the cut and force the bottom card to the top, ensuring the 10H or the 5S will be the first card dealt to the piles of 10, and becomes your final card.
The Gilbreath Principle. "A thing of terrifying beauty." -Max Maven.
Well said 💪🏽
@@steventaputoro That was a well made video. Thank you for sharing.
Awesome. Does the shuffle have to be a perfect riffle shuffle?
No, it doesn't.
Oh man, I saw Max Maven at my school in 1993. My mom and I met him afterward.
I like this trick but calling the last card is hit and miss for me. Could you explain again how to call the last card?
Memorize the top and bottom card when you're gimmicking the deck.
one problem i thought of is what if when you ask the person to shuffle, and they start doing a hand shuffle instead of a card cut. then that would mess the order up wouldn't it?
I don't know card tricks, and I definitely can't "fan shuffle" or interlace them like you did.
So, it seems the trick requires that the spectator know how to shuffle in the manner you show, right? (Google says it is riffle shuffle. I would likely ruin the pattern).
Actually, it can't. Any combination of two halves of the deck after inversing of sequence in one of them would result in 26 consecutive pairs of black and red cards which allows the "guessing".
Riffle shuffle is a pretty common shuffle and by far the easiest way to get a decent mix. You probably don’t play many card games if you don’t know it. Learn it though; it’s pretty simple. As the other commenter said though, it doesn’t matter.
For this trick it's probably even better to do an "imperfect" riffle shuffle, so that the resulting order seems more random.
If you riffle shuffle a Deck r->b->r->b... with another b->r->b->r... deck you will insert between 1 or more cards between the r and b card and it will only switch the r->b sequence or do nothing.
The deck:
r -> b -> r -> b -> ...
pair them in your head
(r -> b) -> (r -> b) -> ...
1. insert one new black card [b] through shuffle
r-> [b] -> b -> r -> b ->...
1. new pairing
(r-> [b]) -> (b -> r) -> (b -> ...)
(r-> b) -> (b -> r) -> (b -> ...)
2. insert more then one because no one can do perfect riffle shuffle
r-> [b] -> [r] -> b -> r -> b ->...
2. new pairing
(r-> [b]) -> ([r] -> b) -> (r -> b) ->...
(r-> b) -> (r -> b) -> (r -> b) ->...
So it will switch the red and black sequence on this part of the deck if the card number is odd or does nothing if its even.
Thanks for that. It makes sense now.
But,why doesn't this also hold true for an ordinary hand shuffle?. After all you are still just introducing odd or even pairs into another stack .
What if they suck at shuffling?
Steven,regarding the “trick that can’t be explained “would you please re-explain how you know the last card (the kicker) in the spectators pile.
Thanks very much!
Wayne
Does it matter how the spectator shuffles the cards? Does it have to be very accurately done like the magician does? I am a complete beginner. Thank you
He explained that he looks at the card as he lifts it.
No it doesn’t have to be perfect. That’s what gives the “three black in a row, two red, one black two red” type of randomness as a result.
All the magician really needs to do is pick someone who knows how to riffle shuffle without blowing the cards all over the room.
does it still work if they do the shuffle to mix them together as an overhand shuffle?
Should you be counting the cards faceup? The obvious alternating red and black pattern seems a dead giveaway.
A better way would be to peek the bottom card (Key card 1 or KC1) just before you start the trick and after the 1st cut take the packet that contains that KC1 and count it onr by one - all the cards being face down during this part
Now for the trick to work for the very last card (still the KC1) the shuffle MUST be done so that tah KC1 syas et the top like Stephen did in both presentation and explaination :-)
Its easy to realize that if the KC1 is in second place from the top for instance it won't be in the spactator's hand after making choose the 1st pile of 10 cards.
... shuffle MUST be done so that the KC1 stays et the top...
I count the pile with the 5S on the bottom face down, which puts it at the top. The 10H is on top of the other pile, ensuring that both will be on the top of the large piles before the riffle shuffle, and one of them will be the bottom card of one pile when I deal out 2 piles of 10. Much better presentation. Doesn't give away the pattern.
If I was a engaging spectator, I would shuffle the cards randomly. How would that work for you? How would you stop me from doing that?
You tell them to do the riffle shuffle and don’t once.
If you don’t then you say the Spectator messed it up.
I can't shuffle like that
Just be authoritative and shuffle for them and then tell them to hold their hand over cut card group while you hold their other hand like you're sensing the color through the person then have them flip the card after you call the color while you close your eyes and peek through your lashes at the spread.
then you, the engaging spectator ignored ur instructions and you should expect for the trick not to work
Why do you need to force the pile? Could they not just pic your pile and you can read off the opposites with the pile containing the 10 of hearts, that way you'll predict their 5 of spades?
Thats for the final card only maybe
Suppose the 10 of hearts is on top after the shuffle as happens in the demonstration. The second card from the top does not need to be the 5 of spades. (It is not in the demonstration.). If you don’t force and they pick the “wrong” pile, the bottom card will be the second card from the top after the shuffle. However you don’t need to force. If they pick the “wrong” pile the bottom card of that pile will be one of the top two or the bottom two from the start of the trick. So now you just need to memorize 4 cards ahead of time.
@@zyklos229 At the drawing of the final card all others are already exposed. The lying down card is the one that's missing. While it's impressive to look over them quickly, you don't even need a trick to figure that one out.
Here is the explanation:
By starting with alternating colors, you are guaranteed that the two piles are in opposite colors if there is no shuffling . So the question is, will shuffle (once) change that? To change that, the shuffle would have to create the same color in (odd, even) pair such as (1,2), (3,4) and so on, but a (even,odd) pair, such as (2,3), (6,7) would not matter (since 2 will be the first card and 3 will be the second card, so they will not be matched against each oether). Now since the two piles prior to shuffling are arranged in opposite order. Let's say if the first card in pile A is black, then black can only appear as the even number cards in pile B. So for black to appear in connective order, it would have to come from an odd number card from pile A and a even number card from Pile B. But an odd number card from pile A is proceeded by even number of cards while an even number card in pile B is proceeded by odd number of cards. This means a same color pair resulted from shuffling is always proceeded by odd number of cards. In other words, it will always be a (even, odd) pair, and never a (odd, even) pair!
As soon as you let the magician touch the cards, it becomes explainable.
I get it how you are telling the colours of the cards but how did you know the exact Card number of the last revealing card? Did you peek at the card before dealing it down? 6:40
Initial setup had 5 of spades top and 10 of hearts bottom. If the magicians pile is the 5 then the other pile has the 10.
I have 2 questions. What happens if the first cut gives an odd number of cards? And why do you need to force the 10 cards packet as both packets should be the exact opposite of the other?
The key is that you want the number of cards selected to be less than either of the halves in the 'intuition' cut. Most people will do a reasonable job of getting within a card or two of half, so by choosing 10 (20 total), you have some margin relative to 26. But in the event the cut is way lopsided, you could still make it work by using half of the smaller size (if it was 19, you'd use 9 cards in each pile instead of 10, for example...for a total of 18 - must be less than or equal to the smaller pile).
@@eventhisidistaken that was not my question but your answer was helpful :)
@@vincentv.9729 ..the direct answer is, it doesn't matter if the first cut gives an odd number. The only thing that's important, is that the cut be larger than 19, and less than 33.
The first cut is not really needed but gives an illusion that the participant has done something random to the deck..I think
I was wondering too about if the intuition cut was odd. But thanks for explaining I get it now. Excellent illusion can't wait to practice it. Great video!!! TY
Cool trick! QW: What does the 10H (that you emphasize at the end of the trick) have to do with the trick?
It was the card that was at the very beginning of the trick. But the wow factor is that you not only knew what color it would be, but the actual card. So a little icing on the cake.
@@hansangb First thing I thought when he predicted the last card was sharp card counting. There were 42 cards face up for long enough.
I can explain it. It's not that difficult to explain. I'll give you a clue: when you arrange them red & black all the way through - that's an artificial pattern that never changes it just repeats red/back over & over. Then you do 1 cut & 1 shuffle. That's it. that puts the inverse of the new shuffled pattern with almost perfect reciprocal r/b r/b pattern except that has a slight chance of throwing this off but it's mitigated by the fact that you deal off the top 20 cards - which eliminates the differences at the bottom of the shuffle - making a perfect mirror image of the cards in the left & right hands. All you have to do is read the image on one side or the opposite on the other side. This is easy to discern. It took me much longer to type it out than the 30 seconds it took to realize how it works. I left out details but I told you more than I was going to. The details I'd add would further clarify things that are obvious to me but I've told 90 percent of the story.
Huh. That's a great explanation. So simple when you know the trick. Thanks.
I noticed that you alternated the sequence when counting and that there were a lot of black/red alternating cards, but I couldn’t tell how you predicted the final sequence. At first, I thought it might have been an odd/even red/black clue with the cards.
This is all super smart, great video! I see how it works but I don’t understand how even after a random shuffle the two packs of 10 end up being symmetrical in color???
Unexplainable 🪄
Get a deck of cards and do the whole trick, face up, instead of face down.
Yes I agree ,
I still can't understand how this is possible
Would also need to make sure the kicker card is last card to go on top when performing the riffle shuffle. So when you count out the piles of ten it is the first card placed on the table ensuring it to be the final card drawn.
For unexplainable trick, you explained it relatively easily. If you split the deck in half you create two identical packs of alternating cards, due to the number being even in the deck. Both have the same alternating order. Now if you reverse the other pack, when you shuffle them you not only have an alternating order in both of the packs, but also alternating between the hands. So if you shuffle perfectly, they'll be in alternating order, but if you shuffle multiple from the same hand, they are also in an alternating order. As such when you deal them in pairs, they'll always keep the alternating order. When you see one person get multiple of the same color in a row, that's where your shuffling was imperfect, but due to the alternating order in both packs the pairs are still alternating.
LOL. Embarrassed TO SAY. THE LEAST AS EXPECTED!!!!! YOU HAVE BEEN. HAD !!!!!!!!LOL
Thanks for this explanation...I am going to have to read it again, when I'm not tired. But, I think you are right. HIS explanation alone, didn't make sense to me.
Very cool trick but how do you handle situation where most people start scuffling the deck normally and they don't necessarily know how to do riffle suffle or get it done first try?
It doesn’t matter even if you do a wrong shuffle it still works
I get the color part but I don't understand how you know what the last card is.
3:46 You memorize both cards at the top (bottom of pile) which in this case it's the 5 of spades and the 10 of hearts. 10:26 The card that is face up to the right, which here is the 5 of spades, meaning he now knows that the last card is the 10 of hearts. =)
And in the performance - where exactly do you get to take a peek at the bottom card without spreading them as in the explanation?
I found and error on the start of the routine. When the deck is cut for the first time. Make sure the count of cards is odd number.
Even count screws up the out come of the routine.
no it doesn't. if the cut is odd, you can completely skip the unneeded reversal. the reversal is only needed if the cut is even, but is harmless if the cut is odd.
I was wondering, when setting up the deck, do you put them in order of number and color? How is it the 10 and 5 come up? Couldnt it be just a random number, but you still know the color?
The 10 and 5 was just an example. You just have to remember whatever the top and bottom cards are
So it _has_ to be shuffled in that particular way? What if they start shuffling it another way (like the majority of people will do), or want to shuffle it more than once?
why do you force the first pack if they are opposite?
To guess the last card. You should force the pile with the first card, the one that was on top of the deck.
Cool trick. I was thinking that if you had the spectator flip the cards over that you forced on them (after you've told them the color of course) then not only are they more interactive in the performance but that distracts them from seeing YOU glance at your cards for color reference.
As long as you remember to tell the spectator to do a riffle shuffle, most would just go into an overhand.
Some people can't do a riffle, and it's awkward to specify. But having them do the shuffle isn't essential; you could do it yourself.
what happens if they riffle shuffle the ten of hearts to another position thats not the top?
@@Houjixful It's explained in the video: the card on top will either be the ten of hearts or the five of spades. You see which one is visible so you know the hidden one is the other.
@@vnen how would you know what deck to force if you don't know what pile has the ten? both stacks are face down when you force the pile and video didnt explain that. My only guess is to look at how the spectator shuffles and see if he shuffles the 10 to the top
easier use a marked deck
Found this video by accident, I love it. I'm not a fan of set ups, so I had to tweak my performance. Grab a "random deck" lying around that's set up, do some false shuffles, riffles, whatever you're good at then suggest a trick (perhaps when someone offers to play a game of cards so it seems more spontaneous). Using comments in this video, adapt a face down version and bam. The only problem is if people keep asking you to do more or ask to repeat. Never repeat, and if you can do more, do. If not, go back to the original suggestion of playing whichever card game was suggested and after a few games, people will ask again to do more tricks. Because performing one is never enough. That's my humble take anyway
It seems to not matter which pile they pick, the colors are always opposite, no magician's force needed
Why force the pile to the spectator? It works with any pile they pick, does not it? (if you don't want to get the last card I mean, but it is without force whatsoever)
My friends don’t even know haw to riffle shuffle!
Rifle?
Do a riffle for them, most people don't know how
You don't even have to riffle shuffle the cards. Just an overhand shuffle.
Yea to be fair, I would alter this trick and just do the shuffle myself.. This trick is 100% based on a single rifle shuffle and some a hole will always be a smart ass and mess the whole trick up.
@@WizardPaul An overhand shuffle wouldn't work, because in that case you're actually rearranging the order of cards. Riffle shuffle doesn't do that.
I'm not one to put in negative comments, and I don't think this is negative, just fact. The top and bottom cards (5S and 10H in this case) can very well end up in the same pile of 10 cards, it's pretty much a 50/50 chance. I would say, if you happen to see one of those cards in the 'discarded ten', then you have the bonus of being able to predict the last revealed card's suit and value.
With that said, I like the concept and have thought of a nice application, thanks!
Explanation: After counting the cards of the top pile, the two piles contain alternating colors, with different colored cards on the bottom. The riffle shuffle is equivalent to taking cards, one at a time, from the bottom of the left or right pile, in a non-predictable way, and placing them in the resultant pile.
Suppose the first card to fall is red. Then both piles have black cards, so a black card must come next. After choosing one of the black cards, we are in the same position as we started (two piles containing alternating colors with different colored cards on the bottom). So, every pair of cards in the resultant pile contain one red and one black, at least until either left or right pile is exhausted.
That's all that is needed for dealing the two 10-card hands.
if the specs shuffle is messy this trick wont work
Never mind which pile of 10 cards you choose. Both are the in same oposition.
It's to be able to know their final card.
You explained it very well. Let's just hope the spectator knows how to shuffle properly...🤣🤣🤣
You can shuffle it sloppy and it will still work. The whole deck still retains the pattern of the initial Red/Black alternations but is stirred once and there is still a pattern. They don't have to be one to one shuffle.
@jeffguarino2097 Thanks, Jeff. Good to know. I'll have to try it on the fam...😅😅😅
Does it have to be a perfect ripple shuffle before splitting the last 10? I know everyone won't be able to.
I'm still confused, but if the deck is then shuffled twice it's truly mixed up. Then the magician is screwed.
Riffle shuffle makes 2 types of sequences in the deck. One is type: 1red, 1 black, 1 red, and so on ... alternating of color by just 1 card. The second type is 1red, 1red or 1black, 1black.
Maximum is 2 cards together with the same color. Both types of sequences can be combined together. Bud one can mathematically prove(it is not so easy to explain it here corrrectly) , that first card in every sequence type: red, red or black, black must be always placed on an even position. And therefore at the end you get the desired result.
Hey, quick question:
Why must I Magician's Choice the pile? As they are opposite, whichever I have to find, I can do it by peeking at the other. And I can also find the last card in the same way.
Thanks! Cool trick.
Because that pile will definitely have one or both of your known cards, which you need for the final reveal. Depending on how the shuffle goes you may have one of your known cards (5S 10H) in each pile, or both in the magician's choice pile but they will never both be in the other pile.
Try it a few times and you'll see. I suggest turning those two cards face up to make it easier to follow.
Yes. If the last card in your pile is red, then their last card is a 5S, and if the last card in your pile is black, then their last card is the 10H.
I've had 8 Maker's and cokes and tried to follow along watching this.
It seems VERY impressive, even if I have not been drinking... but alas, I've failed even the 1st prompt, as a drunken magician!!!
What I've learned: don't drink and mage!!!
I failed every time and never touched a drop of alcohol.
What if your friends ask you to do it again because they wish to figure it out? You're screwed as you won't be able to alternate colors as in the initial setup, right?
Hey thanks, very cool ! First card trick that actually works when I do it.
so what happens if they suck at shuffling the cards and don't do a perfect riffle shuffle? wouldnt that make it not work?
Doesn’t need to be a perfect riffle shuffle. In fact it’s better if it’s not perfect. Just ask them to push the two piles together. Rosetta shuffle even 🤙🏽
Great and very clever. Having said that, I can't think that there is any "spectator" on this planet who would let you get away with telling them that:
a: They can only do a ripple shuffle
b: they can only do it once!
Also assuming they will do a PERFECT ripple schuffle
Doesn’t need to be perfect at all…
@@martynsmith5794 does, an imperfect shuffle will often have the important card one or more places away from where you are looking.
Hi! Amazing trick, and it works with just Math principles! I have one question: when you "guess" the last card ("its a 10 of hearths" for example), do you need to shuffle the deck in such a way that the 10 will be on top of it? What if the 10 gets more in the middle? I think thats the only confusing part for me. Thanks!
Agreed...the 10 would have to remain on top for the guessed card part to work.
@@hobie1115 Ok! Thanks! :)
Either the 10 of hearts or the 5 of spades will be on top. You need to see which one is in the visible deck, then you know the one actually on top after the shuffle.
@@vnengood idea!
This relies on a very specific method of shuffling - what if the spectator decides to shuffle it a different method? One that will mess up the alternating order? In my mind, you still need an "inside spectator" for this trick - one who will always shuffle the deck using this very specific (albeit most common) method.
That is why the trick doesn't work....most spectators don't shuffle cards that way. They hold the deck in their hands.
But when you counted out to 28, they can clearly see it's alternating...shouldn't you count facedown?
If you put something in a sequence of 50/50 when the cards mix two of the same color fall at once before the first opposite color does then allowing and having no choice for the same opposite color to fall with it. There's only two options . So when you split it in half and force the cards to the player it's opposite. Once again if it's not red then it's black. Do it with less cards and shuffle very slow and you will see how the cards fall.
Pretty simple to explain actually. The only thing is it doesn't work if the audience member doesn't shuffle the top two cards correctly for guessing the number and the suit. You can only do the suit colors at that point.
What if someonenis a bad shuffler and does it in chunks and not a perfect 1 to 1 shuffle will it still work?
Of course, this requires the spectator to do a good riffle shuffle.
NO, even a poor riffle shuffle works.
No
@@mmvalex6903 Yes! Even an inaccurate rifle shuffle will work. To make it NOT work, you'd have to allow more than ten cards from one side before even one card comes from the other side. Or is it five consecutively from one side?
4:40 the magician's force is not needed here. If they take the other pile, ask them to use their intuition to guess the color of their top card, and flip it over, repeating this for the rest of the ten cards. They'll get some right, some wrong. Tell them, hey, that's pretty good. Now you do the same with your pile, and get them all right with the info you got from their cards.
I followed along every step and it only worked like once outta 5 times...
Keep practicing lol, that's not bad for your first 5 times
so they have to shuffle a specific way and only once? wouldn't that tip them off when u tell them this minor detail? lol
what if they want to shuffle the other way where u smash the 2 halves together?
Wow great trick and very unexplainable. Even if the shuffle isn't perfect it still works. Amazing tutorial again. Great work
0:48 Two even piles of ten. Starts on his left but doesn't start counting until he lays the card down on his right.
I like it . I like doing mostly Impromptu cards tricks, however If I didn't I would snatch this one up. Great job ty for the tutorial. I am subscribing . =)
I know a trick much like this one, where the entire deck is stacked. If I am just doing card tricks for a small group of friends or something, I have another trick I like to do beforehand where a section of the deck is shuffled face-up into the rest of the deck. The great thing about this is that when I gather up all the cards I can stack the entire deck, right in front of them, even if they are paying attention, and it just looks like I am getting all the cards to be facing the same way again. Never had anyone call me on it yet. So, if I want to do the trick with the stacked deck, I do the other trick first (which is impromptu), gather up the whole deck and stack it right in front of them, and go right into my stacked deck trick.
I wonder how much of this depends on them being a decently skilled shuffler. Like... What happens if they make a small mistake, and the 5 of spades and 10 of hearts end up in the same stack?
I know you can just abandon that final prediction, but does the rest still work if it's not a perfect shuffle?
I don't get it, wouldn't you just use a marked deck for an effect like that, you'd just pull out a deck and tell them what each card is, all the other stuff seems unnecessary. It's a cool method but the spectator doesn't know or care about method.
In this case you need a prepared deck.
@@A.D.Gah yes, a random deck from a friend. perfect for when you need a prepared deck. no one will mind you stacking the cards in alternating colors for 2 mins before hand. never mind you can't do the trick a second time.
@@A.D.G Not everyone is as skilled as you. Be nice
@@dfborbaif you can’t spread cull and faro, just say so. seems like an issue with the magician if you can’t color separate and shuffle without raising suspicion
Their guess will be that it's a marked deck, but this way you can have them freely inspect the cards.
The deck was mirror imaged when you counted and shuffled it . how do you not get how that works?
Brilliant thanks man!!!
Does the shuffle need to be this shuffle type. The spectator may shuffle differently
Just for everyone who thinks they understand it but don't really, the trick is the alternating b and red.
Now think of a stack of 51 and 1 shuffled. That one card pushes the alternation by 1, but it doesn't break the alternation because it just means the alternation starts again on the second stack, one card later. 2 or any even number of alternating cards added into the sequence doesn't break the alternation. Any odd number of cards is just like adding 1 with and even number after the 1 card, so it effects the sequence as only placing 1.
Those facts mean a single shuffle does some number of all of the rules tandomly, but each instance doesnt break the black red alteration. Meaning every card when seperate is still its opposite in the other stack. Every discontinuity in the seperated stacks was an odd number shuffled in.
I've seen a lot of answers here that are just wrong, but feel right.
So how come a shuffle doesn't alter the black red alteration?
I just recently learned it: Spidey - Out of my Mind. However, Peter Turner also included it in his How to Read Minds (2?). Good trick!
Does Peter add any good tips or subtleties to his?
@@superdaven I only have the Spidey version, seems to me that he is the original creator. He did not add anything other than here in this video
@@justcameforthecomments It goes back a long time before Spidey :)
@@malcolmstockbridge2569 oh okay, you know better than me :) No matter how we put it, it’s a great trick
Love it. Thankx for the good tutorial
Why do you have to force the pile closest to them if the two packs are opposite?
What happens if they use another shuffling method?
The intrigue died immediately when he said set the cards up black red black red.
Yeah, that's what learning how to do card tricks will do.
Is this a similar trick as in a video called "Shuffling Card Trick" on the "Numberphile" channel? They were trying to explain the mathematical reason why the trick works but I only understood maybe 30%-40% of video.
Card trick .....can be explained....women....cant be explained....
But sexism can be explained lol
If we count the cards, we may get an observant viewer who suspects why the cards alternate in colors, first red, then black..... How to get out of this situation?
Why should you force audience to choose the one pile, if its just the opposite and 10H vs 5S, then you can just work with any of the pile? Just curious.. Thank
Good trick, but from what I saw, the one part that can get dicey is the last card/10 of hearts reveal. I noticed the 5 of spades and 10 of hearts were the top cards of the two piles, then the spectator shuffled. The ending would need those two cards to remain on top, so a bad shuffle could easily have buried those 2 cards past the top 2 positions. I guess you can watch the shuffle and see if that's what happened and if it did, then you can eliminate the last card prediction...the rest of the trick is still very good. I'd be curious if this observation was wrong.
That's why you need to force one particular pile. Either the 10 of hearts or the 5 of spades will be on top after the shuffle. The other one will be visible in deck, it might not be the last card, but it does not matter because the spectator does not know about it.
In my first attempts, the 10H does not come up at the end most of the time. The rest of it works well.
are you from NZ?
I’m from New Zealand 💪🏽
I don't get the shuffle part, can they pick them up to shuffle from one hand to the outer or must they do it the way you did, what ever that is called?
Similar to one I used to know, but after the riffle shuffle, fan to show the mix, and make a break inbetween a pair of the colour with a cut, I would fan a bit, "You can see they're in no particular order" (make the break, then got through showing them the rest and making the cut) "See... red black, red black, black, red, no particular order, that shuffle is the best for really mixing them up like this..."
Then deal the lot into two piles. They have one you have the other, they will deal the cards face up separating them into red and black piles, at the same time, I take a card and match it in front of me, so there's 4 piles on the table, 2 piles that are the face up reds and blacks they dealt, and the 2 I dealt face down. Then when you turn them over, incredibly they are all in order all mine are red and black piles too.
I still don't know quite how it works, but it does and is a great trick, like this one.
If I understand correctly, if you're willing to lose the magicians force you can actually let them take either pile then the trick still works. but you simply don't name the final card?
I missed the part how you knew that last card was 10 of hearts. I’ll go back
Pretty obvious that the black-red mix results in even numbers and placement of cards. Any inconsistency in one hand will show up as the opposite in the other hand. You don't even need to force a hand because if you know one, you know the other, so it doesn't matter which one they pick.
Here is an explanation of why it works. When you have a setup in half of the deck, a riffle shuffle does not change the order of the cards in your setup. Here we have two halves of the deck with mirror image setups (alternating red and black) after the spectator's cards are counted. The riffle shuffle produces a deck with the two series intermingled-- the deck is composed of "blocks" of one or more cards, alternating from the two shuffled halves. Suppose for example that the top block contains 4 cards, in the order RBRB. These are dealt to the respective piles, producing the pattern of opposite colors. Now suppose the second block, which must be from the other half, contains an odd number of cards such as three. You then know that their order must be BRB because of the mirror image property of the halves. The first two cards are dealt onto the two piles. Now the important point: the third card (B) must be the opposite color of the next card, which will be from the first half. But this is guaranteed since the even number (4) of the first block ensures that an odd number card (5) is next, and this matches the odd number (3) from the second half; due to the mirror image property, this must be a red card, since all odd-numbered cards in this half are red. Similar arguments apply to odd followed by even and odd followed by odd blocks. Ironically, you DO NOT want the spectator to make a perfect shuffle, because then his whole pile will consist of only red or black cards.