This Trick CANNOT Be Explained - Revealed

This insane card trick has an unexplainable method. You will look like you have super natural powers.

[brucejohnson]

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.

[xChikyx]

magician: now shuffle
me: does pile shuffling
magician: 😳

[Jbuckets]

“The method can’t be explained…I’ll explain in a second”

[kwilson]

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.

[achildofthelight]

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 😊

[jeffmccrea]

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.

[tylermeyer]

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

[ScottMiller-smrn]

I learned this trick last week and have already done it on about ten people at dinners, etc. But I've made changes: I have them cut the deck face down--I don't want them to see the red/black/red/black pattern. Although before the cut I do fan out the deck face up telling them it's a normal deck of cards, but I don't fan it out such that they can spot the pattern.

Then then cut the deck, and I see how close they came to getting the deck perfectly in half, then shuffle the two half decks back together.

After dealing out two hands of 10 cards, I tell them that whichever deck they pick I will tell them the color of each card before looking at it. I tell them explicitly to put their finger on the pile they want me to correctly predict (so I'm NOT forcing a deck, they legit pick the one I use). This is more impressive that knowing the kicker card, IMO, because they often say: "There no way you've memorized 20 cards after they've been shuffled!"

Then, I predict the first two cards (correctly, of course), then I change it up with the remaining cards saying: "Ok, now let's see if you can predict the next card?" If they guess it right, before flipping it over I tell them they are right! And if they guess it wrong, I'll say something like: "Nope, you are wrong!" then flip it over with absolute confidence. So, I'm still getting each card right, but I'm also getting the other person more involved, while still proving I'm always right and they only get a few right.

This trick always gets lots of praise when it's done.

[Jawnderlust]

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

[davidtyler]

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

[finnkoepke]

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.

[Sudikinoko]

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.

[robertbrandywine]

As soon as you let the magician touch the cards, it becomes explainable.

[TruthSurge]

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.

[Danny]

The Gilbreath Principle. "A thing of terrifying beauty." -Max Maven.

[orga]

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.

[crashmagic]

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 👏

[Longie]

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

[amazingcato]

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.

[taiwanfocus]

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!