The Final 3 - Amazing Math Card Trick
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- เผยแพร่เมื่อ 5 เม.ย. 2009
- WATCH my REVISED video for this trick • The Final 3 Card Trick... One of the most amazing card tricks revealed if you follow this Incredible Self Working Mathematical Card Trick. The best card tricks tutorial videos revealed at www.thecardtrickteacher.com
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The number of cards in each stack is the key to the trick.
Stack #1=10
Stack #2=15
Stack #3=15
Dealers stack=9
The marks first card is placed on top of stack #1 which makes it the 11th card in the deck from the bottom card#42, the marks second card is placed after the
first stack= 10
+
marks first card 1
+
stack#2 15
+
marks second card 1
= 27th card from the back which is card #26
then the marks third card is put on top of stack #3 and the dealers stack is put on top of that making it card #10. So the marks card ends up being 10, 26,42.
Then after you take the first 4 cards and put them in the back the marks cards become # 6,22,38.
Then when you start turning over the cards alternating face up and face down they come out as so.
First deal is all the odds face up
1,3,5,7.......51
all evens face down
2,4,6,8...…..52
Second deal
Face up 52,48,44,40,36....4
Face down 50,46,42,38.....2
Third deal
Face up 2,10,18,26,34,42,50
Face down 6,14,22,30,38,46
Fourth deal
Face up 46,30,14
Face down 38,22,6
Last 3 cards left are the marks cards 6,22,38
Nice
Respect
By the way, it also works with stacks of 14,15 &15 (instead of 10-15-15) that way you don't need to move 4 cards to the bottom
I didn't put the 4 at the bottom with the 10 15 15 and it worked
@@Krizwar16 really? Nice
@@Krizwar16 Did not work for me
really? I will try
445678?
This is really a good trick. As mentioned by you, there is a mathematical expression / explanation.
As we add the cards,
Bottom of Stack -> (10)+(Card A)+(n1)+(n2)+(Card B)+(n3)+(n4)+(Card C)+ (9) -> Top of Stack
where n1 is the number of first random pick cards, n2 = 15 - n1
n3 is the number of second random pick cards, n4 = 15 - n3
A, B, C are the cards in question
Also we removed 4 cards from top of deck and add to the bottom of stack
Applying these,
Bottom of Stack ->(4)+ (10)+(Card A)+(n1)+(15-n1)+(Card B)+(n3)+(15-n3)+(Card C)+ (9)-(4) -> Top of Stack
which yields:
Bottom of Stack -> (14)+(Card A)+(15)+(Card B)+(15)+(Card C)+ (5) -> Top of Stack
We can now take position of the Cards A, B and C as seen from Top of Deck / Stack
C = 6th
B = 22nd
A = 38th
All are in "Even" positions.
In each iteration of elimination, we remove the ODD positions, finally after 4th iteration, we will have the Selected Cards Only - A,B,C
Hope you like it!!
Yeah, I was just thinking the same thing 😂
@@MP-tf7cc Hmmm...me too! LOL
A slightly slicker variation is to get the spectator to pick four matching cards. Count the piles into 11, 15, and 15, and the remaining pile should have 7 cards. Perform the trick exactly as indicated in the video, getting them to place one card on a pile, cutting, adding, and so on. As shown in the video, pick the piles up in the order 3, 2, 1. However, in this version the spectator will still have a fourth card which you ask them to keep safe. You then tell them we're going to play "snap". When they see any of their cards they are to reveal their matching card, but, of course, they'll never see any of them until only three remain. You can tell them it's like playing Russian Roulette were the odds are increasingly stacking against you as we continue. When only three cards remain, ask them to reveal their card, flip over the last three cards, and watch the reaction of your audience.
I like this version better. But not exactly as in the video, as you don’t take the 4 cards off the top. You just take the piles 3-2-1 and start flipping.
Thinking about making a revised version of this video, where you do not need to bury 4 cards, and I'll include an explanation. Should I do it, or is this old stuff?
That's pretty good but I have two ones that is impossible
Two card tricks, sorry
you should revise it, unless you already have
-it would be cool to know how it works because some people are not mathamaticly smart.
-better video quality would be nice
-being diverse with the video would be cool too
I hope this helped
MisMag822 i really love your vids and contests Love and Peace Stijn Weijters
This is very similar to your Jack The Bounty Hunter. I wish you make another video soon...
If you don't put the 4 cards underneath after stacking the piles, you can still get it right, just that the 3rd time you go through the cards, you have to start with a card down instead of the usual "up".
Espyyyyy Distraction ;)
I love how flexible this trick is with its simplicity allowing you to perform it in so many different ways
It is because of the card position that is created. It is in the 6th, 22nd and 38 card position. You will notice that 38-22 is 16 and 22-6 is 16. This is also 2^4. Given their are 4 times that the cards are removed from the deck, these are the positions that will always end up at the end. If you increased the deck size than the next position is 54, however, you would need to increase the number of total cards by a similar increment. Would show you picture from excel...but it will not upload.
The mathematics was easy.
Let the selected cards be a,b,c.
And x, and y be the variables that we are cutting from the 15 cards set.
So, the piles are.
x-a-10;y-b-15-x;9-c-15-y
Now they are piled in the following order-
9,c,15-y,y,b,15-x,x,a,10
And 4 cards are put from the top to the bottom.
So, it becomes-
5,c,15-y,y,b,15-x,x,a,14
As 15-y and y, and 15-x and x are together, the variable is cut, and this becomes simple mathematics.
And the pile becomes-
5,c,15,b,15,a,14
Now start from up, and down.
The down pile will be after first iteration-
(2,c,7,b,7,a,7)^-1 as it is going from top to bottom
which is
7,a,7,b,7,c,2
And after the second iteration-
1,c,3,b,3,a,3
After the third iteration-
1,a,1,b,1,c
After the last iteration-
a,b,c.
There you have it. I don't know if anybody posted this or not, but I liked doing it, so.
Hahaha, amazing xd
chal be school ja aaj fir bunk marke betha h
Yup, 26 lines of explaining. sounds VERY easy
True, true , but you could make it simpler
haha, Easy, totally makes sense...
To understand how it works you should start at the end. The distance between each chosen card is doubled with each "flipped" card that we add between "non-flipped" ones till we run out of cards for an additional pass, which means the distance at the beginning of the trick has to be a power of 2.
With a standard deck of cards, working our way backwards from the end we'll end up with a starting setup where the chosen cards are 16 cards away from each other, (15 non-chosen cards separating them), that's 33 cards including the chosen cards, and there are 19 cards left, which are not enough for another pass.
The "head" and "tail" of the starting setup can be worked out in the same way: the order of the stack is reversed with each pass, and every time we start flipping cards, (going "forwards"), the "head" needs to be an odd number of cards not including a chosen card, so that the chosen card will not be flipped. We also add the last card onto it, because going forwards we always flip that card first, (removing it).
We end up with 5 cards (odd) as the starting head, and 14 as the starting tail.
With the 5||||...|||[14] setup we get 5>[2]>1, and [14]>7>[3]>1, (square brackets denote "tail").
As you can see, the tail can be an even number as long as half of it is odd for the next pass as head.
(Remark: the head could be an even number, but then you have to remember to start the pass with a non-flipped card).
When we start, it doesn't matter where our volunteer cuts the 15 card piles, the way we stack them back later we always have 15 cards separating each two successive chosen cards.
The only reason I can see to start with 10-15-15-9 and then bury 4 out of the 9 cards is to make the piles look more uniform at the beginning.
Hope that helps
I don't speak English very good, but I 'l try to give you some tips to understand the trick.
I'm a math teacher and it is amazing to look at this simple trick.
The mathematical system is dividing by 2. Every round, you lose the half of your cards:
1. From 1:22 till 1:36: no shuffeling of the deck! The three cards were put on the top of the three decks, independant of the cutting of the decks! The three cards are on position 6, 22 and 38 in the new deck of 52 cards, after replacing the 4 cards from up to down. The numbers of these cards are even. Thats important!
2. When you cut the deck by up and down; the new places of the three cards are now 3, 11 and 19 from 26 left cards.
3. But bottom and top of the deck are changed! So, the numbers of the three cards are now from the new top: 8, 16 and 24 from 26. Again even numbers!
4. Again cutting the deck in half by up and down: the three cards have the numbers 4, 8 and 12. from 13 cards left. Again even numbers!
5. But: you changed top and down, so, the numbers of the three cards are 2, 4 and 8 from 13.
6. Again up and down, so, there are 7 cards left. The numbers of the three cards are 2, 4 and 6 from 7. Again even numbers!
7. Again cutting by up and down and you had the three cards left, because the numbers 2, 4 and 6 are the three cards from the beginning of the trick.
Maybe, my explanation is easy, but English is not my standard language. Sorry.
in order to skip the "bring 4 cards down"
start with 14-15-15.
that's what I do and no fails.
Shut your mouth thank you bro you are the best
I understood the trick. The cards in groups of 10,15,15 will stay together, no matter how many cards you pick from the next pile and put on top of the Ace, therefore when you first piled the cards into a full deck, the Aces were (in order from the top of deck) 10th, 26th and 42nd.
After transferring 4 cards from the top to the bottom of the deck, the Aces are in positions 6, 22, 38. Then start with fsce up, hence all odd cards arrangement wise (1,3,5,7....) face up and the even cards(2,4,6,8....) face down.
After the pile is halved, the Aces are now at positions 8,16,24. Split the deck again, odd ones face up, even face down.
Aces now at positions 2,6,10. Split again with odd cards face up, now the positions of Aces are 2,4,6.
Split again and the only 3 cards face down left are the Aces.
For a better overlook, leave all 3 aces face up in the deck and try the trick. Put the 3 aces face up even if the pile they are in is faced down.
This is outrageous. Mind blown that someone figured this out
GREAT trick! I too used 14-15-15 and did not have to move the 4 cards and it worked perfectly. Many thanks.
Great trick. It's easy enough maths of you sit and think about it for a while. I tinkered with this general design and came up with my own version. Fools everyone. Brilliant!!
Another cool variation to the ending would be not to place any cards face up but all face down until the final 3. You’ll know that the final 3 in the pile that was supposed to be the only facedown pile will be the spectators’ cards. Only turn those 3 up at the very end and be like “Wow, you guys must’ve cut them perfectly”
yes teacher finally after 2 hour of math permutation and combination i made its mathematical solution.
its all about the even odd game. all the customer card at even places from top 6 ,24 and 32.
request every one to place any card in that position and remember that.
and then play the game of even odd as per the video first up== second down and so on.....
the three card left are always the magic 3 one always.
So the only thing left to do is to explain that the cuts you make, don't really matter. These cuts does not influence the amount of cards in between the cards.
this is not a complicated problem its really simple
Brilliant! I don't care how it works but I can't wait to try it. Really well explained. Thank you
Thankyou so much for your videos. I am new to your channel. I learnt this trick and sweet 16 today.... waiting for my husband to get back to do them with him... so cool. I keep getting this one wrong because I keep forgetting to pass 4 cards under.... I am also new to magic tricks... I am most grateful and cannot wait to see what else you have on your channel xx
Nice trick. If you set it up slightly differently you don't need to move the four cards to the bottom after all those cuts. Instead of 10 - 15 - 15 - 9. Use 14 - 15 - 15 - 5. (you don't cut the first and last packets anyways)
That way the trick runs more smoothly in the middle. And you can go straight from the cuts to the up and down bit. :)
got to be the best self working card trick ever.. down the pub i never need to buy a drink i win so many doing this amazing trick
First and foremost, the third card gets put 6th place from top when you take the 4 cards to the bottom (otherwise known as the third deck). The first card gets placed as the 15th from the bottom (and 11th before moving the 4 cards). The first and second deck gain cards from the third and second deck respectively, but that does not change the amount of cards on the bottom of the previous deck because it rejoins the next deck when placed on top and so offsets each card as 6th (first), 22ndth (second) and 38th (third).
It might be different for you but it will follow a similar pattern and so be placed as an even card right to the end, leaving you with your the spectators three chosen cards.
Nailed it on the first try, nice! Let's try it again!
I tried it myself.. Worked 3 times.
Then I went to show it to someone. I made a fool of myself and made a mistake and it didn't work. FML.
lol the FML killed me
It's work 💯%
FML... I'm still laughing
LOL 😱 Thats exactly what will happen to me I'm sure.
Ive had spectators forget what cards they chose and stopped me wrongly... Making ME forget if I did "up or down" last and relied on them telling me... Which led to the wrong succession, in turn, fuxked it up... So trust the trick... And know that they wont see their card til the Reveal...
Hey dude this a great trick
I am a huge fan of card trick your my favourite card trick TH-camr
Fun trick. Works with 52 cards. Start up each time. This trick works because (there is always the 15 cards between each chosen card) as you end up stacking the remaining cards from each "cut" stack back in order. Foot note the moving of 4 off top to bottom is not needed if your first stack is 14 instead of 10 as that is where they end up anyway.
Cheers!
Its satisfying to see him flip the cards up lol
you dont really need to put those 4 cards on the bottom of the deck after you did that 4 piles thing, you just need to start with 14 cards then 15 then 15 then 5
Lucia Anita Zancanella that's so much better lol
My didnt work i did it after dvd. Why😀????
@@sren1446 I also do card tricks and I am also giving away a deck of cards to a lucky person so pls go checkout my latest video there you will see my announcement if card deck giveaway. So pls subscribe to my channel.
Your tricks are mind blowing and I love them!
Even cards dealt down and odd up. That's why it only works with the odd number of 1 started with face up in the deal down. Absolutely amazing trick that I still use to this day, but I have a great opening statement that I'll reveal at a later time that keeps this trick as fresh as a delicious steak dinner.
After the spectator places their cards, the piles are collected in specific order, then 4 are taken off the top and put on the bottom. At this point before we deal, no matter how the spectator cut the piles, the position of the 3 cards will always be in positions 6-22-38 out of 52 cards.
Keep in mind that for the selected cards to always be in the down column, they must always be in an EVEN numbered position in the pile before dealing. Also keep in mind that as we deal them into into up/down piles, they are also getting REVERSED in order.
After the first deal, the 3 cards will be in positions 8-16-24 out of 26 remaining cards.
After the second deal, they will be in positions 2-6-10 out of 13 remaining cards.
After the third deal, they will be in positions 2-4-6 out of 6 remaining cards.
After the fourth and final deal, they will be the only 3 cards in the down pile.
The only way this worked for me is I turned my phone upside down and was facing exact same way as in vid..lol
Who on Earth worked this trick put.
I did this at a store. Manager made me buy the deck afterwards.
What do you expect
Wildly Films it’s a joke 😀
Sweet trick. It boils down to getting a card in the 6th, 22nd and 38th position and they will work out like this.. pretty cool.
I know this is an old video but it’s still one of my favourite tricks. I do this with putting the extra 4 on the first pile as well but I do it with 3 different people choosing one card each. The reveal feels a little slow to me so I tell the first person I’m going to split the deck with them. I deal out 1 card at a time to a pile in front of them and then a pile in front of me. After doing this with each person, I’m left with 6 cards. I tell the first I’ll give them an extra card if they tell me what their card was. Give them a card and then turn over the next which is theirs. Same again with 2nd and 3rd person, I feel like this makes them more a part of the trick than just observing. Great channel, lots of great tricks here!
That's a terrific adaptation.
So if I'm not mistaken, you deal everything face down to everyone until the final reveal? If so, I think that's gold.
@@jaerivus sorry for the late reply, that's right!
@@Mezillious I had to re-watch this trick, but I still stand by what I said... Your effect sounds really cool.
Question though : do the cards always stay in the same order of Person 1, Person 2, Person 3? If so, that's fickin nifty.
@@jaerivus Yes they do! after you've share with person 3, you back to person 1 then 2 then 3
Nifty, but quite easy to figure out.
THe cards that are being put down face up, are the odd numbered cards. THe three cards chosen by your subject are always even, simply becase you put 9 cards on top of the 3rd one, and 15 on top of 2nd and 3rd. And that keeps going.
Pretty cool though
Saeid Momtaha
I still don't get it. Because the spectator can pick how many ever cards they want to place on top of the other piles
You have to remember that afterwards, all the cards are getting stacked together, so the amount they choose isn't significant.
Here's how it works:
you have your nr. 1 pile of 14 cards, your nr. 2 pile of 15 cards,and your nr. 3 pile also of 15 cards.
Now you take the your first card, that was chosen as key-card, and put it on top of the number 1 pile with 14 cards, making it number 15.
You then till your subject to cut the next pile as explained in the video, and that's where the whole trick takes place:
If he/she takes, let's say 10 cards and cut them of to the first stack, that leaves you with only 5 cards left to put the second key-card on top of, but since those 5 cards are gonna go on top of the first pile anyways, that cut didn't actually chance the order of anything other that the 15 cards that you wheren't going to use anyways. Key-card 1 is still card number 15 in the, and key-card 2 is still number 31 in the stack, and so will key-card 3 still be number 47 in the stack, no matter how they cut. :)
Nice essay I'll tell my grandkids
I concur with your analysis. (even though I didn't read it all lol)
For those of you who say it doesn't work, don't be so sarcastic! This gentle man is trying to show us a secret that we may not know yet! Maybe it's best we go and try another time and watch him more closely! It's meant to work if you do this right!
Hats off to whoever came up with this trick!!
Great video as always!
Beautiful! Here's another one like it: make 3 piles by putting a card faceup on table & then count off cards faceup on it, from the # on the first card to 13 (put 3 cards on a 10, 2 on a Jack, etc.) Do 3 times, turn the 3 piles over, turn up top card on 2 piles. To find the # on the remaining top card, count the cards in your hand: count off 10 extra cards, plus the two #s face up, and the # of cards remaining in hand will be the # on the face down top card! THERE'S MORE: you can do this with any # of cards in deck (D), any # of piles (P), made by counting to any # (C). Do as before, using this formula to find # of extra cards (E) to count off: E = D - CP - P. Wow!
The cards will always b the 6th, the 22nd, and the 38th (I will explain y in a minute), since they r all even, they will end up on the face down side cuz all odd numbers r face up. The next round they will always b the 8th, the 16th, and the 24th, also all being evens.Then they will b the 2nd, the 6th, and the 10th, and then the 2nd, the 4th, and 6th (every other one for the last round). That is why it works, because it will always b in an even numbered spot. The cards will always start as the 6th, 22nd, and 38th, because after it is one deck again you place 4 cards on bottom making only 5 cards above the first card chosen since there were 9 cards placed above it, then the next card is the 22nd cuz there r 15 between the 6th and 22nd card, and the last card is 38th cuz again there r 15 cards between the 22nd and 38th (I know if u subtract u get 16, but there is 15 cards between the 2 not counting either of them). Cutting the piles doesn't change anything since if you place the decks together the right way they touch again.
Exactly what I was about to say (not) most easiest mathever
If it helps, the cuts for the cards are an illusion since the rearrangement of each sub-deck really does not matter. By stacking the cards back at the end, it's the same as if the cuts never happened.
+Heaven Lindenstruth Exactly...i have this similar magic trick with him but i use 7 cards horizontal and vertical a long time ago...what i remember with my magic trick is that you need to spell out a word...a trick will actually work if you know the magic of mathematics even though you are wearing blindfolded too it will work..but if you if you ask me how on earth did i do it, i have a book of magic when i was a kiddo :)
I'm using the 14, 15 and 15 set up too. My padder or is patter, is Queen of Hearts life long love King of Hearts have a son Jack of hearts. The world (of cards) will attempt to seperate them but watch. Great payoff of them sticking together through thick and thin as a family. Young and old can or should be able to relate to this hackneyed love story.
Cutting the cards creates an illusion of randomness. If you notice how the cards are picked up, each of the three cards are in the exact order they need to be in for the trick to work. First chosen card, Card 1, will always be the 11th card off the bottom. ... the 2nd seems tricky but it isn’t. The stack of 15 is cut in half- let’s say the first 7 cards are placed on top of Card 1. Card 2 is placed on top of the remaining 8 cards and this stack is placed on top of the first pile. You’ll always have 10 cards, Card 1, 7 of the 15 cards, the remaining 8 of the 15 cards and then Card 2, the first cut of the 2nd stack of 15, the second half of the 2nd stack on 15, Card 3 followed by the final 9 cards. Before splitting the decks, you move 4 of the 9 from the top to the bottom.
If you do this methodically, the selected cards will always be the 6th, 22nd & 38th cards before splitting.
Bullshit
Hm it always seems to mess up for me on the 3rd set of flips. Idk what I'm doing wrong. I would love for someone to help me!
Ok i know im too late to help you
On the third round you need to switch it you will go with a down up down up and so on but in the same place
Anyone here from 2019?
Austin Rode ayee
Kaiden Paolino yoooop that’s dope
Me
😁
JefftheWizard Ducker yessah
It's because it doesn't matter how many cards the spectator cuts, the 3 cards which the spectator cuts will always remain in the 6th, 22th, n 38th position after u put the 4 cards from the top to bottom. Which will end up the last three card remaining after doing up down. It's just maths but awesome.
This is an amazing trick. I could make it the second time. Wow! I am amazed.
so say up cards are odd numbers and down cards are even numbers . turning up and down and up and down will leave the "picked" cards every 8 th card. . next up down up down picked cards will be 4th down . then next time picked will be 2nd down . then finally 1st down
I don't understand your explanation
instead of moving four cards to the bottom after picking up the piles just make the original piles 14 15 15 this will create the same effect while also making the trick much smoother
Nice tactic!
That is freakin' amazing! I can't wait to try it!!
This is a simple trick to figure out. Let's start half way through, just before we start alternately dealing the pack face up and face down, and then discarding the face up cards and repeating until we have only three face down cards. At this point it is simple to see that the cards at the positions 6, 22 and 38 (Position 1 being the top card and position 52 being the bottom card) in the pack will become the remaining three face down cards as you repeat the process. The whole trick up until this point is just an elaborate way to get the three chosen cards into those three positions. The only thing that may confuse is that you cut the stack of fifteen cards twice in the process, giving a seemingly random nature to the final configuration of the deck. This is an allusion. There is always 15 cards between the first chosen card and the second, and 15 between the second chosen card and the third, no matter where you cut. Finally moving four cards from the top of the stack to the bottom, leaves the three chosen cards at positions 6, 22 and 38. No trick and no real mathematics either.
This is such a cool trick! I'm gunna try it on my boyfriend XD
You still dating him?
NeosMil lol I was wondering that too
How’d it go?
I think an interesting affect might be to ditch the 3 cards, so their cards are no longer in the deck. Thoughts?
It works because the fist card goes on top of the 10 pile, with 4 more below it, the 15th card. You take “as many as you want from the next pile” then put the next card on top … with the rest of that pile being put on top (underneath that card), so they’re actually spaced out by 16 no matter how you cut the piles. Then the top card is always the 6th one, so it’ll be face down, and the sequence works its magic from there.
It's quite easy actually. When the 2. and 3. pile is putted on the top of the one to the right, in the end the cards will always be on the top of "their" pile. Meaning when you put them together they will end up as; number 11, number 27 and number 43 from the bottom. Then after removing 4 to the bottom...15, 31 and 47. And by going up down, up down. Number 15, 31 and 47 will be the remaining 3. :)
Works everytime, dummy proof, a great trick anyone can do.
If your first pile has 14 cards then when you gather them up you don't need to shift four from top to bottom. lot less sloppy.
+Jerel Mercurio Thanks man
You can do it with 15 in each that way the spectator won't suspect what it is you're doing - you're just creating 3 even piles with 4 cards left to place on top. 1st run has to be in reverse but this will make it seem more random. Also when I do it I keep both piles face down so they havent a clue until the end when you just surprise them with their 3 cards.
The 3 cards you have after doing the up-down thing obviously depend only on their places at the start. So you need to start having them in the right places which in the trick is done before going to the up-down part. This explains the trick completely because you don't have to know the places where the cards should be at the beginning or how they stay at the end but you only need to know that there are such 3 places.
my friend showed this trick to me and also explained it! it has simple mathematics! and if you concentrate on how the cards are placed in the pile and also the pile placing, you can figure it out easily
jesus i remember finding this a long time ago when i was 8 and i was so obsessed and thats what got me into a lot of card tricks, then i forgot them all and when i tried to find this video again i couldnt
2:37 Lol, made me laugh. DOWN
Lol
I know how to do it but how it works every time is a mystery. I know most of my card tricks from you they are great Thanks keep them coming.
You can make this into a much cooler trick than what it is.
Tell your spectator to pick three random cards and write them down on a piece of paper. Then you could secretly do this entire trick and write down what you think we're the cards chosen and then at the count of three, you both reveal what the predictions were and if they match, it's dope!
what if i told you that when i tried this for the first time my ace of spades came out at the same spot yours did in the video, well thats magic
Me 2!
Sameee
this worked great but I found I was having to go "down up" on the last 2 for it to work for me
watch the video, he said you have to go down up the 3rd time around
Vor Pike no he didn't idiot
just get rid of jokers and it should work
Jyuusankaimon "just get rid of jokers and it should work" Good comment! Works in many situations besides card tricks! :)
+Vor Pike
Spastic!
Cool card trick on my first attempt it was wrong but my second attempt was a success thank you for this :) I really appreciated this vid thanks a lot !
This is one of my favorite card tricks
Oh my
ive been going over it a million times and it never works!
i like this card trick it easy and induces great reactions, thanks!
Alternatively to burning the 4 cards to the bottom of the deck... i.e. to the bottom of the first pile. You can simply deal out pile no 1. 14 cards (10 +4 burned) , pile no. 2 15 and pile no. 3 15... that way the piles are almost identical and the spectator won't realise anything... to make it more convincing when you count the first pile of 14 cards, count 1, 3,4,5,6,7,8,9,10,11,12,13,14,15 (14 numbers, but the spectator think 15) and then count 15 cards to the other two piles. IT WORKS and it doesn't involve burning 4 cards (which means the spectator can't question why you do it). Hope this helped out some of you who were asking if you can use 48 cards etc..
It works it didn't work the first time because I forgot to take four card to the bottom but the second time I followed all the way through and it works!!!!
Never works for me
Best trick EVER works every single TIME!!
You can also deal the cards starting with the card face up on the first 2 passes then start with the card face down on the 3rd pass and you'll end up with the 3 cards left. This replaces the need to count 4 cards at the start. Hope this makes sense.
Fixed places at first deck setup: 10-26-42
The rest is repeated
How do we know 2nd card is at position 26?
Never mind. Figured it out. Duh!
@@dennismartelli3415 If you pay close attention to the movements you'll see that the bottom 10 remain there, the 11th card is one of the aces, then on top of this comes the entire second group of cards (15 of them) which bring the position of the second ace-card to position 10+1+15+1 = 27 (not 26, I did this in my mind and probably made the error). Then on top of this comes the entire second group of 15 cards and hence the position of the 3rd ace is 27+15+1= 43. The rest is automatic.
@@odisseaspanopoulos5627 Thx. I figured it out similarly but working from the top.
@@dennismartelli3415 The whole trick is based on the movements being designed to hide the fact that the groups of cards of 10-15-15-9 are placed in this exact order on top of each other with the 3 aces in between floors. The confusing movement is that the two groups of 15 are split first, for each split the top goes under what was the bottom and then the entire group of 15 is placed on top of the group of 10. This is repeated for the 2nd group of 15.
Anyone here from Christmas of 2019
Christmas of 2020
Yep
This was explained to me about a year and a half ago by a math magician who knew the innovator of magician math. Which was his professor at one point. But how it works is very amazing. But it easier to look it up. But it is calculated by multiples because you put the 4 cards on bottom it rises the aces up and makes the order different. But honestly it doesn't really matter if you did put the 4 cards on bottom even though the cards might come up in a different fashion. But the math too it is basically this You pick up all but three decks, and there's decks SS, TT, and UU on the table. If the top cards are s,t,us,t,u, then you have (14∗3−s−t−u)(14∗3−s−t−u) cards on the table and 52−(42−s−t−u)=10+s+t+u52−(42−s−t−u)=10+s+t+u cards in your hand. You have to put (10+s+t)(10+s+t) cards on the table, leaving you with (10+s+t+u)−(10+s+t)=u(10+s+t+u)−(10+s+t)=u cards in your hand. So the number of cards you have will match the top of the turned-over deck.
I'm sure many of you have figured it out. I was just bored and figured why not try to make some sort of try at a proof of this card trick. I'm not quite sure how to prove this rigorously mathematically. However, it's fairly easy to convince yourself that it's always true by considering the sequence
{ (4 + 10) , card1 , a1 , 15 - a1 , card2 , a2 , 15 - a2 , card3, (9 - 4) }
Where card1-3 is the three cards in question. a1 and a2 are the piles taken by the spectator to put onto the pile next to it (the pile on the right). The trick lies basicly in "tricking" you into believing that you have a say in where the cards are put. But it's completely irrelevant, since you put the piles on top like that. In the sequence we have a1 cards in a row. Then 15 minus that exact amount, so we still have 15 cards there (none of which is card 1, 2 or 3). So we have the sequence of cards like this
{ 14 , card1 , 15 , card2 , 15 , card3 , 5 } Now we start removing cards (remember to do it from the right [the top] ). Up, down, up, down, up. The fifth card - just before card3, is up, so this means that if the number of cards before card123 are odd the cards will be faced down - how convenient. The sequence is now.
{ 7 , card1 , 7 , card2 , 7 , card3 , 2 } Remember that the sequence now has been inverted (The first card is in the bottom and last at the top - You get my point), so instead of doing it from the right, we now do it from the left and obtain
{ 3 , card1 , 3 , card2 , 3 , card3 , 1} From the right again
{ 1 , card1 , 1 , card2 , 1, card3, 0} From the left again
{ card1, card2, card3 }
And there we have it. As I said, not a real proof - but - meh. For youtube this should suffice ;D
Cheers
I tried this trick over and over but it would just not work
U r doing it wrong....go back and study video....trust me the first 8 times or so I got it right once...now I'm always 100 percent...u r doing something wrong....
Jake Yang put the first selected card on the pile of "10"... cut, add.. next...
I am sure by now you have it mastered after a month ...
I've been slacking on watching videos and commenting
Moron
Raghav Raghuvansh theatre y r u so mean? I just didn’t get. U don’t have to be this mean.
Jake Yang some people are just not happy with their own lives and have to try and throw some weight off their own shoulders..
Believe me... everyone who read that knows who the real moron is.. just saying.
I was confident this wouldn't work, this video had to be a joke. There was no way it could work every time with such random placements. Still for whatever reason I tried it out. Sir, you just earned a new subscriber
Jake Wojtanowski obviously you have never heard math.
no bitch this will work because its maths
3 cards. 10, 15, 15, and 9 cards left. cut pile 2 put on 1, cut pile 3...on 2.. Good stuff Mismag. God bless.
Awesome!!! I love this trick.....
hello again everybody ! ^.^
Let A be the card placed in the left
pile, B the card in the middle pile and C the card in the right pile.
After set up cards are in these positions A: 6 B: 22 C: 38
After first iteration A go to 24 (3rd from the bottom of a 26 deck) B go to 16 (11th from the bottom of a 26 deck) C go to 8 (11th from the bottom of a 26 deck)
After second iteration C go to 10 (4rd from the bottom of a 13 deck) B go to 6 (8th from the bottom of a 13 deck) and A go to 2 (12th from the bottom of a 13 deck)
After third iteration A go to 6 (last of a 6 card deck) B go to 4 (3rd from the bottom of a 6 card deck) C goes to 2 (5th from the bottom of a 6 card deck)
After forth iteration C go to the bottom (last of 3 card deck) B go to 2 (2nd from the bottom of a 3 card deck) A go to 1 (3rd from the bottom of a 3 card deck)
These means that A B and C are always last and always in the same order.
So this gave me an idea about a possible change in the setup. Let Alex choose card A, Bob choose card B and Carol choose card C. You do all trick like explained in the video and when 3 cards are left you give first card to Alex second to Bob and third to Carol and ask them to reveal it. Everyone will get exactly the card they choose.
3 is Key, 3 squared is 12, 12 is 1plus 2 that's 3...why haven't you stared he's not using 52 cards!!! That's a Rap...
This works because in the end you are just putting the 10 on the bottom, one of their cards, 15 on top of that, than one of their cards, the other 15 on top of their 2nd card, and then you are placing their last card on top of the 2nd 15, and lastly placing nine on top. With the 4 from 9 on the bottom it all works out that the person's cards will all be on a face down pile when starting face up. You cut onto the pile next to the pile their card is on, so the original piles still join up. Thanks!
Very impressed thanks for sharing this trick !
you guys need to take out the jokers
+Eden Portal I use the jokers and it works... Maybe I'm lucky
+Miloustique most of the times it doesn't works
Actually you just need to remove 2 cards...it doesn't matter which ones cuz you need 52 cards in total.
Yours isn't, though.
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OMG yes it worked and by the way please can you do a face reveal when you hit a million subscribers !!!!
bahat accha
Really amazing, my first ever try works.
The way they're arranged is the reason. All 3 cards are in even positions in the deck(6th, 22nd, and 38th from the top) and since the first card you put is face up, all the face up cards are the ones in odd positions and the face downs are the ones in even positions therefore they're not put where the face up cards are.
when he says pick up the cards in this order "1,2,3" he actually means "pile 3, then 2 then 1
See my above long comment about how he shouldve shown it from the "magicians" POV
pile1(10 + 1(spectator card)+ X(that is cut from pile two))
pile2(15-x +1(spectator 2nd card) + y)
pile3(15-y +1(spectator 3rd card) + 9)
now order of deck is
9+1*+(15-y)+y+1*+(15-x)+x+1*+10
then after he puts 4 cards below then order change and becomes
5+1*+(15-y)+y+1*+(15-x)+x+1*+14
then up down
2+1*+7+1*+7+1*+7 from down to top(these are down cards)
so from top to down become
7+1*+7+1*+7+1*+2
then up down
3+1*+3+1*+3+1*+1 from down to top(these are down cards)
so from up to down
1+1*+3+1*+3+1*+3
then up down
1*+1+1*+1+1*+1 from down to top(these are down cards)
so from up to down
1+1*+1+1*+1+1*
then up down
1*+1*+1*(we left with three cards only that were the spectators cards) :)
2+1*+7+1*+7+1*+7 from down to top(these are down cards)
so from top to down become
7+1*+7+1*+7+1*+2
then up down
3+1*+3+1*+3+1*+1 from down to top(these are down cards)
so from up to down
1+1*+3+1*+3+1*+3
then up down
1*+1+1*+1+1*+1 from down to top(these are down cards)
so from up to down
1+1*+1+1*+1+1*
then up down
1*+1*+1*(we left with three cards only that were the spectators cards) :)
From this point can you explain how you got these answers
ya
u understand the deck up to 5+1*+(15-y)+y+1*+(15-x)+x+1*+14
then he goes for up down so half cards will open and half cards will remain unexposed so the 3 main cards at 6th,22nd &38th position
then all odd number cards exposed and even number cards unexposed then new deck form of 26 cards in which 6th card of original deck is now at 3rd position from bottom&24th position from top, 22nd at 16th from top and 38th at 8th position from top.
then again up down then again odd position cards get exposed and even number cards left forming the deck of 13 cards that contain 6th position card of original deck to 2nd position from top,22nd at6th position and 38th at 10th position now he again goes for up and down this time again odd position exposed and even position remains unexposed then new deck of 6 cards form in which again these there cards at 2nd, 4th and 6th position so at last these three cards remain unexposed to through this experiment.
This is a very simple trick: basically, when you do the up/down/up/down thing to a deck repeatedly, there will be 3 cards that come out last: cards (from the top) numbers 6, 22, and 38. This trick simply sets up the deck so that those cards always come then. And as for the "cutting" of the deck, the spectator is not cutting the deck. Rather, they are shifting the positions of other cards. If you watch very closely, you'll find that the three important cards' positions ALWAYS stay the same no matter how the deck is cut; it is only the positions of the other cards that change during the cut.
cause you have an uneven amount of card in each pile after you cut
and since there is an even amount of card in the deck, only the even or the uneven ones are upside or down\
The Final 3:
So, basically all you have to do is:
Let the spectator choose 3 of their favourite cards. The tell them to write the cards down on a piece of paper. Until then make 4 piles of cards consisting 10,15,15,9 cards in each pile respectively. Tell the spectator to put 1 card on top of the first pile (the pile with 10 cards) and then tell them to take as many cards as they want from the second pile and put them on top of the first pile. Then tell the spectator to place their second card on top of the second pile. Then tell them to take as many cards as they want from the third pile and place them on top of the second pile. Finally place the third card on top of the third pile and take all the cards from the last pile and put them on top of the third pile. Then pick the cards up from a right to left order (15,15,10) and make a deck of cards. 𝐓𝐚𝐤𝐞 𝐭𝐡𝐞 𝐭𝐨𝐩 𝟒 𝐜𝐚𝐫𝐝𝐬 𝐟𝐫𝐨𝐦 𝐭𝐡𝐞 𝐝𝐞𝐜𝐤 𝐚𝐧𝐝 𝐩𝐮𝐭 𝐭𝐡𝐞𝐦 𝐚𝐭 𝐭𝐡𝐞 𝐛𝐨𝐭𝐭𝐨𝐦 𝐨𝐟 𝐭𝐡𝐞 𝐝𝐞𝐜𝐤. Then put the cards down in face up, face down manner and continue that until you have only 3 cards remaining. If everything goes well, this 3 cards will be the cards the spectator chose at the start.
Sorry for the mistakes I may have made while writing that.
It did not work sometimes 😶😶
The cards are mathematically arranged so that this trick works perfect every time. The first card placed on the first stack does not change the number of the ten cards below it, nor the four others that get placed there later. Cutting stack 2 then placing the second card onto it, then cutting stack 3 and placing the third card on that stack then stacking them all up in the order shown, stack 3 on top of stack 2 on top of stack 1 creates an illusion of chaos, but really does not change the original spacing of the 15 card piles. There is still 15 cards between each of your mark's chosen three. Put the other 9 on top; fix the pile by placing four of those on the bottom and you have your set up for up down: Marks cards are the 6th add 16 for the 22nd and 16 more to the 38th card, and 14 more for your 52 card deck. Up downs divide those numbers by two, reverse the order creating a stack with Aces in the 8th, 16th, and 24th places with two more cards left over. another set of updowns: divide by two again and reverse the order too, and your Aces are in the 2nd, 6th, and 10th positions with three left over. Another set of updowns leaves the stack with Aces at 2nd, 4th, and 6th positions. Your last updowns save your marks three cards for last,and as we all know, good math never fails. If anyone fails at this trick, they did something wrong.
i know how it done from the first time,it about how talented in maths you are ;)
::::SPOILER:::
Took me about four runs to figure it out. No matter how the cards are cut, you end up with 5 top cards, the first ace, 15 cards, the second ace, 15 cards, the third ace and then 14 cards remaining.
On the second round, you get
7~A~7~A~2
Third round you get
1~A~3~A~3~A~3
Forth and fifth rounds
1~A~1~A~1~A , then A~A~A
Regardless of when you cut it. The way you stack them again will always have 10, 15 15 spacing between them. The spacing remains the same and that's why the math works