Beautiful trick. The explanation is easier than it seems. You start off with +4 red cards. You then split these to red or black piles. The third pile always removes both 1 back and 1 red card, so doesn't change the imbalance of +4 cards. Simply glorious.
@philbreaden2445 it is impossible to have no red pairs with a +4 imbalance in the deck. Even if it's as perfectly sorted as possible (red, black, red, black) you will end with 4 extra red cards with no blacks between. This also means you can take it a step further. After the first set, you can pocket 2 more blacks to make it +6, and then do a 3rd for even to make the trick even more convincing. You can do this easily by having the spectator pocket the prediction initially and then swipe the 2 extra cards while they are focused on retrieving and unfolding the paper.
@@philbreaden2445You’re right, but, when you say “small chance”, you really mean “next to impossible chance” that out of 48 cards, for all practicable purposes, there’s no real chance that all 48 cards would alternate color pairs if removed sequentially. I’m no math guru to calculate the probability, but it must be in the gazillions. 😊 P.S. … I replied just for fun!
I did this trick with my four year grandson last night. He loved it! He kept asking how I knew how many cards would be different. He went and got more paper for my predictions because he wanted to see it again and again and again. He loved being able to turn over the cards, put them into their piles and then count the number of cards. Thanks for the trick!
Awesome trick, fooled me and my wife 😆 Thank you very much Matt! During the counting process I palmed the 4 cards from my pocket and added them to the red/black pile and got rid of them 😊 Also I suggest (or according to my wife xD) to show the prediction after the counting, so the spectator cannot expect it 😀
This trick fooled me bad also when I first saw it. I forgot about this 💎 A performance tip...is put the deck back in the box rather than your pocket. That would cause some red flags. Love your taste in card magic 🎩
Man I love this trick so much… I’ve always wanted to get serious into magic and even got a busted Spiderpen that never got replaced… I never comment but I love a good card trick. I can still pull a card force from an old method after much practice but it’s not really a good one but use to practice it on strangers…. This is brilliant, used it on a 10, 13, 18 year old and myself when I saw it without understanding and the to the pocket and walk away was brilliant…. Just a random moment with kids whom got to participate and actually enjoyed it even if it took a lot to get them around the table… Simply brilliant, elegant, hats off to you and the magician that came up with it… Now I have to learn more… I did the trick twice but… 😅
Very strong! I did suspect that it was not a 52 card deck with an even number off reds and blacks. But the repeat fooled me, as I did not really notice that the cards were pocketed before beginning that phase, and that the missing black cards were cleverly (and sneakily) added. I really like that when you repeat the trick, you can write the prediction prior to asking them if they want red or black. In fact, they will probably not even remember that in the first phase, you asked them which color they wanted BEFORE writing the prediction
Excellent, thank you, I'll try to practice it with two packs same back not comupulsory but that's what i have at home :D, to see whether i can't start with more different outcomes before the end kicker. I'm not a magician at all but I like observing and it always amazes me how "constant" talk from the magician, spectator brief participation, event if it's just thinking about his chosen card, can have the spectator focus on "the wrong" things :)
Awesome trick, saved to my playlist so I can work on it. but still my favorite one from you is your version of shuffleboard. It's equal to one I learned from Juan Tamariz 30-35 years ago, but I never knew the name of it, or I'd tell you. It's my favorite card trick of all time. I tell people, "in this world we have streets, avenues, boulevards and ways. This trick has nothing to do with streets, avenues, and boulevards, but I'm gonna drive you right down the middle and drop you off on "no friggin way"".
Miraskill by Stewart James. One of the all time best. John Bannon has a version where you start with a clean deck and don't have to remove any cards. I was on a cruise a few years ago grom Miami to South Hampton. Bill Cook was the magician on board. I fooled him badly with this and he made me teach it to him. Absolutely killer.
I changed it a bit because putting the deck in my pocket was a bit too suspicious. I have the four extra cards in the card box facing away from them. And I tell them "any red / black mix goes into the discard box". So as the first trick is happening I'm instantly covering my tracks by putting cards into the box, therefore also resetting getting ready for the second part.
The text in the prediction could be just one, regardless of who choose red or black cards. Just depends on who reads the text. The prediction can be “I’ll have 4 more cards” from the beginning and the, depending what color the spectator chooses, the prediction could be read by the spectator (in case of blacks) or by the magician (in case the spectator chooses reds) Fantastic trick and I love the enthusiasm you putt on it even being a self working trick. Tx
another great fooler Matt !! ❤ 🎩.. i think a better reveal the first time , would be to tell spec. to count his - then count yours , then say OK, the difference is 4 , THEN show the prediction. ( just how id do it )
Nice entertaining card game. Based on how I see it. Math is the reason obviously. It's all done in pairs only of 2 to work this way. And carda can then be revived in 2-4-6-8,ect. The cards can only work that way in pairs ,so their can never be a one pair of a color left as( red, red or black black ) with a one pair of ( a red and black card ) . It will always have to end with at least a double paired colors or with all the cards mixed as black and red as pairs. To break it you have to just remove one card ( in 3's). When doing in pairs it can only end always as a pair of colors equal to each pile or else you get no pairs of same color. Only works as pairs though. Elegant playing card . Thanks for sharing. Great stuff.
Thank you for the trick Mr Matt I really enjoyed it it is a super cool effect if you don't mind could you tell me the name of those cards you are using I would certainly appreciate it thank you very much my friend and I'll catch you on your next video take care your friend Billy Simmons
Knowing what to ignore (the prediction) and that odd pairs always come in pairs made it easy to understand. But nevertheless a beautiful trick. You can go one step further. You can place the prediction before your viewer chooses the color. You simply need a two way out.
Explanation is kind of simple. No matter how much cards are pairs at the end. You do always split the cards into red and blacks. So the amount of color pairs is always the same on both sides. The diverted cards are also the sames 😉😉
@@MattMcGurk I love the simplicity of Miraskill... Stewart James was a genius to have discovered this. You can do it without removing the cards, and let the spectator choose either colour... I usually use a difference of 2 though, not 4. You can then have the cards re-shuffled, and do the second phase without the need to put the cards in your pocket!
One very subtle thing I do is try to put some distance between their colour choice and the prediction...ie you hold the cards and show its a normal deck, then ask their colour choice and hand them the cards...get them to shuffle...make piles.....do a casino wash....shuffle again.....ask them now if they are happy its a normal deck and that they have fully mixed it...feel free to shuffle or mix more.....then come in with the prediction and carry on. Its a very subtle point but if you ask for their colour choice and then too quickly make a prediction they will easily link the 2 things, maybe Im being ultra picky ...its a great little trick.
no dont even do it the way it was performed....it looks terrible that you make your prediction AFTER they pick no no thats bad do this instead....Pre write the predictions and have them in the same pocket with the "I have 4 more" on the inside and the "you have 4 more" on the outside....this way you can reach into your pocket AFTER they make their prediction...but it looks like you made the prediction BEFORE they picked....then when you reach in your pocket you know the "I have 4 more" is closer to your body....I = predection touching me and YOU = prediction not touching me
@@roccoVAL Excellent idea! But you meant AFTER they make "their choice" (as opposed to "their prediction"), right? Then, when repeating the trick, I would say, "OK, I'm going to ask you to once again to select red or black, but before you tell me which color you'd like, I'm going to make another prediction." Then show a blank piece of paper and proceed to write and fold it up. This will convey the utmost fairness and will be the last thing they remember seeing you do and thus most likely obscure any memory they may have (if any) that the first prediction was pre-written
@@dowaliby1 on the second one i wouldnt do the pocket trick i would just say...."well you know a magician never performs a trick twice, but for you"....i would write this one in front of them.....looks really sus when if both were pre written
Explanation is pretty simple, if you take a deck and seperate the black cards from the red in two piles they will have the same number of cards in them. When you discard you discard one red and one black so nothing changes.
The unintentionally smooth transition into 2nd time actually comes from the spectators themselves lol. There’s always someone out there who’ll say wait wait wait. Do that again. And bingo, you’ve hooked another one
Great trick! Presentation ideas: 1. Option #1: Say: "Remember the Rain Man movie, where Dustin Hoffman counted cards in the Casino? Let's try to recreate that!" and then go into the trick. 2. Option #2: Present it as a competition - whoever gets the most cards, wins some money. You put a folded bill on the table. If the spectator ends up getting more cards, open the bill and read a prediction pre-written on it: You win the money, unless I knew by how many cards you'll win. On the other side of the bill, there is a pre-written prediction: "You lose the money because you won by 4 cards". If the spectator chose the losing color to begin with, just show the other side of the bill. For the 2nd phase, put the bill in your pocket and when retrieving the deck back, take a new bill (maybe a higher denomination?) with a new prediction on it: "It's a draw, no one wins the money!".
This is wild. I am 61 these days. When I was in jr high, a friend's dad showed me this one cool card trick that I have done my entire life since but for the absolute life of me I cannot figure out how it works. It's definitely in league with this one in the video. I will describe the steps as if the trick is in effect so you can replicate it. If you already know or can figure out how it works please do a video on it. Magician = M Participant = P M shuffles the cards thoroughly in front of P. M has P shuffle the cards thoroughly and cut 3x. M takes the cards face down and flips them face up off the top, setting them on the table before P. M silently counts the cards and makes separate stacks. Example: M turns up a 2 of any suit. M quickly flips over the cards while silently counting up to 10 from two: (three, four, five, six, seven, eight, nine, ten) and stacks it neatly to one side and begins again. M turns up an 8. M flips over the cards silently counting up to 10 (nine, ten) face up, and then starts a new stack. M turns up a 7. M flips over the cards up to 10 (eight, nine, ten) face up, then another stack. M turns up a 3. M flips over the cards up to 10 (four, five, six, seven, eight, nine, ten), face up and another stack. M turns up a Jack and counts it as 10, starting a new stack with the next card. M turns up a 10, counts it as 10, and starts a new stack with the next card. M turns up a 6. M can pick up the single face or 10s cards and use them to count up to 10 (seven, eight, nine, adds Jack for ten or seven, eight, and adds the jack and 10 as the ninth and tenth cards). M will have a few singles and 10s come up so leave them face up by themselves or use them if need be to make a fuller stack. It's optional. This continues through the entire deck. At the end, M tidies up all the stacks face up and then tells P: point to any stack that has 5 or more cards. P points to a stack. M sets it to the side and has P select two more stacks (pointing, not picking up or messing with) and M sets those aside, collects all the remaining stacks and puts them into a single deck and then quickly counts all the cards, face down, through the remaining deck, silently counting up to 19, and then starting again at 1 through the rest of the cards. That final sum after 19 will be important at the end. Let's assume for example, there were 8 cards left over in the deck after counting off the first 19. (total 27 in this stack of remaining cards). M takes the 3 stacks chosen by P, flips the stacks face down and puts them in a line, instructing P: when I turn around, rearrange these stacks "shell game style" keeping them face down, in their respective stacks. Do not shuffle the cards. Just change their position so I don't know which stacks I set aside. M turns around. P rearranges cards as noted and informs M when they've done it. M turns back to P and has them point to any stack. P chooses, in the example, the center stack. M flips the top card face up. It's an Ace. P chooses the right most card. M flips the top card face up. It's a 4. M predicts for P the left card will be a 3. And without fail it will be. Ace=1 so (1+4+3=8) I have spent my entire life wondering how this trick works. It always works. If you can do math and the P doesn't shuffle anything after the initial time so they know you're not cheating of course. How does the final sum of the remaining cards in the stacks have any possible bearing on the painted on numbers in the stacks? I truly hope you or somebody out there has heard of this trick and can kindly explain how it works. Not how it's done, I showed ya how it's done, but how the painted on numbers of the top card in 3 stacks is always adding up to the quantity of cards after 19th one in the other stacks. lol
I showed it to my friend, and he immediately repeated the steps over and over and figured out that the default was that it was going to be an equal amount of cards at the end. He later figured out the trick by randomly removing cards. This is a nice trick. But it's something the spectator can easily replicate.
I know how it works. If you could go through the deck and never get a red and black pair, you would each have twenty six cards. But since you take the same amount of red as you do black when you discard the red and black pairs you end up with the same amount if you don't remove four black cards.
There are an equal number of black and red cards cards in the discards pile, so four black cards must have been missing from the deck (or four extra red cards) when the trick starts.
The simplest (but not the shortest!!) explanation is perhaps this… Imagine that the cards get so well shuffled that the entire pack (minus the 4 black cards) is arranged in a perfect “red, white, red white…” fashion, but with the 4 extra red cards all together at the end. Naturally, the first 44 cards will get discarded (all pairs being red/white) and only the last 4 will get allocated - in this case to the red pile. So clearly the red pile has 4 more cards than black, which has zero!! Now, going back to the start, imagine that the perfectly shuffled pack is altered by having just 1 of cards removed from the middle section of the pack (containing the perfectly arranged “red/white” pairs), and this card gets reinserted somewhere else the front 44-card portion of the pack. At the position where this card was removed, we will now see two cards of the same colour meet (the opposite colour to the card removed). And at the position the card gets reinserted (between a red & black pair) it will naturally create an reciprocal colour-matched pair (the card itself and one of cards it is now next to). The pack now obviously has an additional 2 pairs of matching-coloured cards, one red, one black. The total number of red pairs is now 3 (6 cards) and the total of black pairs is now 1 (2 cards). So there are still 4 now red cards than black. And this difference will remain, no matter how much more the pack is re-shuffled. Each time another card “breaks” the red/white arrangement, an additional colour-matched pair will be created (one red, one white). And, of course, it makes no difference if the 4 red cards at the end are actually shuffled into the pack. They will either create 2 more red pairs or will displace the other cards down the order of dealt pairs, creating pairs further down the line.
A quicker explanation. An unaltered deck has the same amount of red and black cards. If you remove any number of pairs that contain a red and a black card, you have removed the same amount of red and black cards. Therefore the number of red and black cards in the deck remain equal. And that's what you'll end up within your final piles.
}Fun trick. It's important to consider your audience with this one though. Mathematically minded people will see what is happening instantly. But this would be a real fooler for kids or people who don't use math and their daily life. It does have a couple things going for it. The palming is confusing. The spectator might forget that they chose their color before you make your prediction. Especially because you can make the second prediction before they choose their color.
I think this trick have a mathematical explaination. But I can't explain it because i got C in mathematics 😂😂😂 very cool trick Matt 🎉❤❤❤ nice video and happy New Year 🎉🎉🎉
I would change the number .say your at a party and end up doing it for multiple people. If you remove 6 it will be 6. If u remove 2 it will be 2. If you had the opportunity to change the amount in the bathroom or somewhere your alone it wouldn't be suspicious if the same people see you do the trick on someone else. But remember it always has to be an even number because 52 is even. If you end up with one card they're gonna know you altered the deck. Someone can easily count the pairs of 2's and relize you altered the deck count.
it works because every time you discard a red and black, you're taking one away from each set of red and black, so the difference of the cards will remain the same.
Not a fan of self working tricks but this is cool how u palm away then steal them back. I would say for the set up, if you’re going to do this in the middle of other tricks, don’t worry about people seeing u remove cards. Nobody knows what you’re doing, plus u can just make it look like you’re tidying up the deck. If u truly want to do it deceptively, I don’t know why you would, but u can leave the jokers in, move the cards u want to remove to the top or bottom, then steal them out when u go to take the jokers out.
@MattMcGurk he was a huskita I owned his nan mum uncle cousins I had 6 they all gone now will have mote just moving yo the right place the dog you see is Mr Ghost the most amazing dog you would ever meet, the only hard thing about havin a dog I sayin goodbye 😔 love your work bruv
When you discard, you always throw away 1 of each, so in the full deck it will always be equal. I don't think I would show this trick to any mathematicians or engineers :) The 4-more was only ever possible with an incomplete deck
I learned this trick when I was just a kid! The way I did it was even easier: My prediction would say I have 4 more cards than you, but I would not remove 4 cards before hand. When the separation of cards was complete I would say, "Count your cards but don't say how many you have until I ask." You both silently count your respective cards, and when you get your total add 4. You then lie and say, "I have ## cards, how many do you have"? As you are announcing your total you casually add your cards to the discard pile and shuffle.
To set up your 2 or 4 removed cards place them between the t.w. jokers and the other non playing cards from a bicycle deck. The jokers face up your cards face down…you can than take your cards out of the box remove the jokers plus the removed cards… place Thatcher one joker on top of the removed cards… your removed cards are now face down on the bottel then do trick one … then for trick two place the jokers with the removed cards on top of the deck… and ask the spectarors do you think it was a lucky guess?? And do the trick the second time … this time de removed cords stay on the deck and remove only the jokers … good trick it fooled me too
This trick was invented in 1936 by a Canadian magician Stewart James. It was titled Miraskill and it's a testament to a lack of simple critical thinking for many individuals. It was also written up in the Encylopedia of Card Tricks published first in 1937. It didn't fool me one bit when someone first showed it to me and that was when I was about 10 years old. I'll turn 80 this year so one should be able to figure out how long ago that was. If one can't then the odds are good that this trick might fool them too.
Another twist for you. The first time you use your deck with 4 cards in your pocket and you deal the cards out. The second time you have a spectator get another deck and you have them deal the cards.
Sadly as someone familiar with logical analysis i could see what was happening from the very beginning. As soon as you said how the pairs were allocated I realised that it never matters what order the cards are in because when you remove cards it's always 1 and 1 maintaining the relative number of red and black cards
You are discarding an equal number of each of the colors. Using your setup, the deck has 22 black and 26 red cards. Suppose the middle (discard) pile has 10 pairs, that means it contains 10 red and 10 black. The black pile will then contain 22 - 10 = 12 cards and the red pile will contain 26 - 10 = 16 cards -> 4 more red than black cards. The key is that you are discarding an equal number of red and black cards. No matter how many pairs are in the discard pile, there will always be 4 more red than black cards remaining in their respective piles.
If the cards happen to be black red black red, then they all end up all in the middle pile, with four left over. For every one pair that is changed to say black, there will always be a corresponding pair of reds together somewhere in the pile. You have to move one black out of the deck to make a black pair, that leaves two reds together where you took the black out. Try it! No math required. Every possible combination in the deck can be replicated by this method starting with red black red black then taking one card out at a time to make the pair, that is why it always works.
Saw this a while ago but dont ask ne to explain it. As regards the first prediction why not simply have a prepared one saying 'there are 4 more red cards than black'. Can be done BEFORE you start anything and is irrelevant what colour they chose?
I know how the trick is done!!! You ready for it.... My Granchildren actaully came up with the solution as this is our Favourite TH-cam channel to watch.. Ok here's the solution... why it works.. It's Magic!!! yup my grandchildren are never wrong! I agree Magic real Magic :)..
@@MattMcGurk Thank you Matt... you know the great thing is it makes them think and collaborate between each other, the side benefit the parents get a little respite.. Now remember you did say you didn't care how the trick works, but you also did say if anyone knows to "feel" free and post the answer :) So you asked for it... please be seated... lol - so due to the math it works if you remove 1,2,3,4,5 or 6, I guess any number of cards the result will be there.. This trick works due to a clever mathematical arithmetic invariance principle rooted in the parity of the number of black and red cards after you remove the 4 black cards. Let’s break it down step by step to understand why the outcome is always 4 more red cards than black cards. For the boffins, the math and formula is at the end... I stand to be corrected if I have made any mistakes. --- **Initial Setup:** - A standard deck has **26 red cards** and **26 black cards**. - You remove **4 black cards**, leaving: - **26 red cards**. - **22 black cards**. - Total cards: \( 26 + 22 = 48 \). **Key Observation:** - When cards are paired during the trick: - **Two reds** are set aside in one pile. - **Two blacks** are set aside in another pile. - **One red and one black** are discarded. - Importantly, discarding a red and black card does not change the difference between the total number of red and black cards in the deck. **The Constant Difference:** - Initially, there are \( 26 - 22 = 4 \) **more red cards than black cards**. - Every operation (pairing or discarding) respects this difference: 1. **Two reds or two blacks:** These don't affect the difference between red and black cards, as only cards of the same color are moved to piles. 2. **One red and one black discarded:** This also doesn't affect the difference because it removes one red and one black simultaneously. **Final Result:** - At the end, all cards are either sorted into piles or discarded. - Because the difference of \( +4 \) (red minus black) is constant throughout, the pile with red cards will always have **4 more cards than the pile with black cards**. **Why Does It Always Work?** The principle hinges on the **constant difference in parity** between the two colors of cards: 1. Removing 4 black cards establishes an initial imbalance of 4 red cards. 2. Pairing or discarding cards never alters the imbalance. 3. Therefore, the trick is mathematically guaranteed to end with exactly 4 more red cards than black cards in the final piles. This is a beautiful example of how simple arithmetic invariance underpins the illusion of a seemingly magical outcome! What Is Arithmetic Invariance? Arithmetic invariance means that a particular property of numbers (like the difference between two totals) remains unchanged, no matter how you manipulate or rearrange the numbers-as long as the operations you perform are balanced. In the card trick, the difference between the number of red and black cards is set at 4 when you remove 4 black cards at the start. This difference is "invariant" (unchanging), no matter what you do during the trick, as long as you follow the rules. Why This Feels "Magical" Even though you shuffle the deck and discard cards seemingly at random, the math ensures that the starting difference (4 more red cards) remains constant throughout. By the time you finish, the red pile will always have exactly 4 more cards than the black pile. It's like starting a race with a 4-step lead-you might shuffle your steps along the way, but you’ll always finish 4 steps ahead! Certainly! The mathematical formula for the trick is based on the principle of **invariance**, where the difference between the number of red cards (\( R \)) and black cards (\( B \)) is maintained throughout the process. Here's how to express it formally: **Key Definitions** - \( R_0 \): Initial number of red cards. - \( B_0 \): Initial number of black cards. - \( n \): Number of black cards removed at the start. **Step 1: Initial Difference** The initial difference between red and black cards after removing \( n \) black cards is: \[ D = R_0 - (B_0 - n) \] For a standard deck: \[ D = 26 - (26 - n) = n \] So, the difference \( D \) is always equal to the number of black cards removed, \( n \). **Step 2: Invariance Through Operations** Each operation during the trick does not alter \( D \): 1. **Two red cards (\(+2R\))**: Adds 2 red cards to the red pile, but \( D \) remains unchanged. 2. **Two black cards (\(+2B\))**: Adds 2 black cards to the black pile, but \( D \) remains unchanged. 3. **One red and one black (\(-1R, -1B\))**: Discards one red and one black card, keeping \( D \) unchanged. Thus, the difference \( D \) between red and black cards remains \( n \) throughout the trick. **Step 3: Final Result** At the end of the trick: - Let \( R_p \) be the total number of red cards in the red pile. - Let \( B_p \) be the total number of black cards in the black pile. The relationship between \( R_p \) and \( B_p \) is: \[ R_p - B_p = n \] **General Formula** If \( n \) black cards are removed initially: \[ R_p = B_p + n \] This means the number of red cards in the red pile will always exceed the number of black cards in the black pile by exactly \( n \), no matter the order or pairing of cards during the trick. **Example: Removing 5 Black Cards** 1. Initial setup: \( R_0 = 26 \), \( B_0 = 26 \), \( n = 5 \). 2. Difference: \( D = R_0 - (B_0 - n) = 26 - 21 = 5 \). 3. Final piles: \( R_p = B_p + 5 \). So, the red pile will always have **5 more cards than the black pile**.
![The BEST Card Trick in the World | Revealed [Self Working]](http://i.ytimg.com/vi/g4pBe6buuXg/mqdefault.jpg)
Beautiful trick. The explanation is easier than it seems. You start off with +4 red cards. You then split these to red or black piles. The third pile always removes both 1 back and 1 red card, so doesn't change the imbalance of +4 cards. Simply glorious.
You got it bro 👊
HOWEVER< there is a small chance that there are no red pairs, e.g. if all the cards in the pack are in odd pairs - then it won't work.....
@philbreaden2445 it is impossible to have no red pairs with a +4 imbalance in the deck. Even if it's as perfectly sorted as possible (red, black, red, black) you will end with 4 extra red cards with no blacks between.
This also means you can take it a step further. After the first set, you can pocket 2 more blacks to make it +6, and then do a 3rd for even to make the trick even more convincing.
You can do this easily by having the spectator pocket the prediction initially and then swipe the 2 extra cards while they are focused on retrieving and unfolding the paper.
@@philbreaden2445 you may missed that "there is 4 more red than black" part. The cards CANNOT pair up evenly.
@@philbreaden2445You’re right, but, when you say “small chance”, you really mean “next to impossible chance” that out of 48 cards, for all practicable purposes, there’s no real chance that all 48 cards would alternate color pairs if removed sequentially. I’m no math guru to calculate the probability, but it must be in the gazillions. 😊 P.S. … I replied just for fun!
I did this trick with my four year grandson last night. He loved it! He kept asking how I knew how many cards would be different. He went and got more paper for my predictions because he wanted to see it again and again and again. He loved being able to turn over the cards, put them into their piles and then count the number of cards. Thanks for the trick!
That’s awesome man! So glad you had such a good time with it
Awesome trick, fooled me and my wife 😆
Thank you very much Matt!
During the counting process I palmed the 4 cards from my pocket and added them to the red/black pile and got rid of them 😊
Also I suggest (or according to my wife xD) to show the prediction after the counting, so the spectator cannot expect it 😀
Love this - Why didn't I think of that! Awesome idea!!
Brilliant trick thank you! very clever...
Brilliant effect and easily explained. Thank you!!!😃
Man ur tricks are really cool!!
Low skill requirements but great effects.
So different to those that came first on TH-cam. 👌👌👌
Thanks and more power
Amazing, beautiful trick. Thanks Mr. Mcgurk, this one is really magic. It blow my mind, love it, its perfect. Thank you sir
Glad you liked it!
This trick fooled me bad also when I first saw it. I forgot about this 💎 A performance tip...is put the deck back in the box rather than your pocket. That would cause some red flags. Love your taste in card magic 🎩
yes either will work, the card box was upstairs, that would have been too much effort 😂
@@MattMcGurk wait , stay here while i get my card box upstairs and my magic wand 😅
@MattMcGurk 🤣 🤣 🤣
@@WaldoWizard lmao 🤣 😂 😆
Man I love this trick so much… I’ve always wanted to get serious into magic and even got a busted Spiderpen that never got replaced…
I never comment but I love a good card trick. I can still pull a card force from an old method after much practice but it’s not really a good one but use to practice it on strangers….
This is brilliant, used it on a 10, 13, 18 year old and myself when I saw it without understanding and the to the pocket and walk away was brilliant…. Just a random moment with kids whom got to participate and actually enjoyed it even if it took a lot to get them around the table…
Simply brilliant, elegant, hats off to you and the magician that came up with it…
Now I have to learn more… I did the trick twice but… 😅
Thanks
Thanks so much Joe! 🙌☕
This is another great self working trick. Learn this one years ago. It is such a killer. Great tutorial Matt🎉🪄✨️🥃😊😊👍, Happy New Year 🎉🎉
Thanks Robert! 🤩
I just got a new deck of cards for my birthday and a video from you was the only wish I still had
And Happy new year ! WAS A KILLER BTW
Happy birthday Alex! 🎈
@@MattMcGurk thanks
Very strong! I did suspect that it was not a 52 card deck with an even number off reds and blacks. But the repeat fooled me, as I did not really notice that the cards were pocketed before beginning that phase, and that the missing black cards were cleverly (and sneakily) added. I really like that when you repeat the trick, you can write the prediction prior to asking them if they want red or black. In fact, they will probably not even remember that in the first phase, you asked them which color they wanted BEFORE writing the prediction
Not at all! You got it man, it’s all about that second phase!
Great, and unbelievable, as always your videos! Many thanks, and happy new year!
Happy new year!
Great trick, im becoming quite the hero to my grandchildren 😊. Thanks 👍
Yeah man! That’s awesome 🙌
Excellent, thank you, I'll try to practice it with two packs same back not comupulsory but that's what i have at home :D, to see whether i can't start with more different outcomes before the end kicker.
I'm not a magician at all but I like observing and it always amazes me how "constant" talk from the magician, spectator brief participation, event if it's just thinking about his chosen card, can have the spectator focus on "the wrong" things :)
Nice vid you help me impress my mom thank you❤❤❤❤❤❤❤❤❤❤❤❤❤
Great !!! Thank you 😊👏👏👏
sure is an absolute BANGER of a trick Matt, well done!
Superb mate so clever I can’t wait to show my mates 👍👏🏻👏🏻👏🏻👏🏻
Enjoy Robert!
Nice trick. Love it. I may say that bcoz removing and adding the four cards will resulting the outcome.
A great trick. I will definately perform this.
Very simple and very attractive
You are too kind! But what about the trick? 😂
Absolutely brilliant 👊🏽
Thanks legend.
Awesome trick, saved to my playlist so I can work on it. but still my favorite one from you is your version of shuffleboard. It's equal to one I learned from Juan Tamariz 30-35 years ago, but I never knew the name of it, or I'd tell you. It's my favorite card trick of all time. I tell people, "in this world we have streets, avenues, boulevards and ways. This trick has nothing to do with streets, avenues, and boulevards, but I'm gonna drive you right down the middle and drop you off on "no friggin way"".
Love that punchline 🥊 That’s awesome man
Miraskill by Stewart James. One of the all time best. John Bannon has a version where you start with a clean deck and don't have to remove any cards. I was on a cruise a few years ago grom Miami to South Hampton. Bill Cook was the magician on board. I fooled him badly with this and he made me teach it to him. Absolutely killer.
I love the Banon’s smart idea in this trick. I don’t know if I can say it. Read his greats books
Hey Matt, have you thought about putting these tricks or the course into a book? Would be a great read. Love the channel by the way 👌
I think I may have mentioned a course in the video!!! I have 2 courses so far available at mymagiccourse.store 👍
Brilliant!
I changed it a bit because putting the deck in my pocket was a bit too suspicious. I have the four extra cards in the card box facing away from them. And I tell them "any red / black mix goes into the discard box". So as the first trick is happening I'm instantly covering my tracks by putting cards into the box, therefore also resetting getting ready for the second part.
Great idea!
The text in the prediction could be just one, regardless of who choose red or black cards. Just depends on who reads the text. The prediction can be “I’ll have 4 more cards” from the beginning and the, depending what color the spectator chooses, the prediction could be read by the spectator (in case of blacks) or by the magician (in case the spectator chooses reds)
Fantastic trick and I love the enthusiasm you putt on it even being a self working trick. Tx
Genious!
This is a great trick. Simple, not much of a setup but will get us a great reaction.
By the way, Happy New Year Matt❤
Happy new year! 🥳
Nice and great trick. Excelente truco!
Great SW trick, ty 😊
Any time!
Awesome! Great trick!
another great fooler Matt !! ❤ 🎩..
i think a better reveal the first time , would be to tell spec. to count his - then count yours , then say OK, the difference is 4 , THEN show the prediction. ( just how id do it )
Absolutely 💯
Nice entertaining card game. Based on how I see it. Math is the reason obviously. It's all done in pairs only of 2 to work this way. And carda can then be revived in 2-4-6-8,ect. The cards can only work that way in pairs ,so their can never be a one pair of a color left as( red, red or black black ) with a one pair of ( a red and black card ) . It will always have to end with at least a double paired colors or with all the cards mixed as black and red as pairs. To break it you have to just remove one card ( in 3's). When doing in pairs it can only end always as a pair of colors equal to each pile or else you get no pairs of same color. Only works as pairs though. Elegant playing card . Thanks for sharing. Great stuff.
Thank you for the trick Mr Matt I really enjoyed it it is a super cool effect if you don't mind could you tell me the name of those cards you are using I would certainly appreciate it thank you very much my friend and I'll catch you on your next video take care your friend Billy Simmons
They are called Bicycle Verbana. Such a nice deck. ❤️
Thank you my friend I really appreciate that yeah they look like a nice deck@@MattMcGurk
BRILLIANT
Knowing what to ignore (the prediction) and that odd pairs always come in pairs made it easy to understand. But nevertheless a beautiful trick. You can go one step further. You can place the prediction before your viewer chooses the color. You simply need a two way out.
Cool, have to try it
Another absolutely fantastic trick! More of these videos PLEASE.
But whyyyyy does the trick work even if you shuffle midway through ?
Magic silly!!! 🤪
That's very clever. This will puzzle many laypeople!
Oh it destroys!
Super strong trick
Glad you like it!
Cool man. It was mindblowing
Thank you so much 😀
Explanation is kind of simple. No matter how much cards are pairs at the end. You do always split the cards into red and blacks. So the amount of color pairs is always the same on both sides. The diverted cards are also the sames 😉😉
Does it work with an odd number like 3,5,7 etc?
The numbers have to be even. 👍
You are still removing a red and black cards together in the mix pile . So the 4 in your pocket is the difference.
Such a big pay off for such a simple trick
Cool❤❤❤❤❤❤❤
It's Miraskill by Stewart James.
It is indeed 🙌
@@MattMcGurk I love the simplicity of Miraskill... Stewart James was a genius to have discovered this. You can do it without removing the cards, and let the spectator choose either colour... I usually use a difference of 2 though, not 4. You can then have the cards re-shuffled, and do the second phase without the need to put the cards in your pocket!
Nice one, Thanx ! Let's fool my friends once again :-)
One very subtle thing I do is try to put some distance between their colour choice and the prediction...ie you hold the cards and show its a normal deck, then ask their colour choice and hand them the cards...get them to shuffle...make piles.....do a casino wash....shuffle again.....ask them now if they are happy its a normal deck and that they have fully mixed it...feel free to shuffle or mix more.....then come in with the prediction and carry on. Its a very subtle point but if you ask for their colour choice and then too quickly make a prediction they will easily link the 2 things, maybe Im being ultra picky ...its a great little trick.
no dont even do it the way it was performed....it looks terrible that you make your prediction AFTER they pick no no thats bad do this instead....Pre write the predictions and have them in the same pocket with the "I have 4 more" on the inside and the "you have 4 more" on the outside....this way you can reach into your pocket AFTER they make their prediction...but it looks like you made the prediction BEFORE they picked....then when you reach in your pocket you know the "I have 4 more" is closer to your body....I = predection touching me and YOU = prediction not touching me
@@roccoVAL That sounds really good, thats a much better way of getting that distance between choice and prediction...great idea.
@@roccoVAL Excellent idea! But you meant AFTER they make "their choice" (as opposed to "their prediction"), right? Then, when repeating the trick, I would say, "OK, I'm going to ask you to once again to select red or black, but before you tell me which color you'd like, I'm going to make another prediction." Then show a blank piece of paper and proceed to write and fold it up. This will convey the utmost fairness and will be the last thing they remember seeing you do and thus most likely obscure any memory they may have (if any) that the first prediction was pre-written
@@dowaliby1 on the second one i wouldnt do the pocket trick i would just say...."well you know a magician never performs a trick twice, but for you"....i would write this one in front of them.....looks really sus when if both were pre written
This ☝️
incredible
Explanation is pretty simple, if you take a deck and seperate the black cards from the red in two piles they will have the same number of cards in them.
When you discard you discard one red and one black so nothing changes.
your videos are so good i fool my parents all the time😁
Great video ❤ I've been watching since you had 11k subs
Legend! Thanks for the support!
Love it.😊
Thank you! 😊
simply awesome and clear explanation of the steps involved
The unintentionally smooth transition into 2nd time actually comes from the spectators themselves lol. There’s always someone out there who’ll say wait wait wait. Do that again. And bingo, you’ve hooked another one
💥 Yeah baby! Love that 😂
Great trick! Presentation ideas:
1. Option #1: Say: "Remember the Rain Man movie, where Dustin Hoffman counted cards in the Casino? Let's try to recreate that!" and then go into the trick.
2. Option #2: Present it as a competition - whoever gets the most cards, wins some money. You put a folded bill on the table. If the spectator ends up getting more cards, open the bill and read a prediction pre-written on it: You win the money, unless I knew by how many cards you'll win. On the other side of the bill, there is a pre-written prediction: "You lose the money because you won by 4 cards". If the spectator chose the losing color to begin with, just show the other side of the bill. For the 2nd phase, put the bill in your pocket and when retrieving the deck back, take a new bill (maybe a higher denomination?) with a new prediction on it: "It's a draw, no one wins the money!".
Awesome fooler Matt, thanks for all your hard work.
Unbelievable!
This is wild. I am 61 these days. When I was in jr high, a friend's dad showed me this one cool card trick that I have done my entire life since but for the absolute life of me I cannot figure out how it works. It's definitely in league with this one in the video. I will describe the steps as if the trick is in effect so you can replicate it. If you already know or can figure out how it works please do a video on it.
Magician = M Participant = P
M shuffles the cards thoroughly in front of P.
M has P shuffle the cards thoroughly and cut 3x.
M takes the cards face down and flips them face up off the top, setting them on the table before P.
M silently counts the cards and makes separate stacks.
Example:
M turns up a 2 of any suit. M quickly flips over the cards while silently counting up to 10 from two: (three, four, five, six, seven, eight, nine, ten) and stacks it neatly to one side and begins again.
M turns up an 8. M flips over the cards silently counting up to 10 (nine, ten) face up, and then starts a new stack.
M turns up a 7. M flips over the cards up to 10 (eight, nine, ten) face up, then another stack.
M turns up a 3. M flips over the cards up to 10 (four, five, six, seven, eight, nine, ten), face up and another stack.
M turns up a Jack and counts it as 10, starting a new stack with the next card.
M turns up a 10, counts it as 10, and starts a new stack with the next card.
M turns up a 6. M can pick up the single face or 10s cards and use them to count up to 10 (seven, eight, nine, adds Jack for ten or seven, eight, and adds the jack and 10 as the ninth and tenth cards).
M will have a few singles and 10s come up so leave them face up by themselves or use them if need be to make a fuller stack. It's optional.
This continues through the entire deck. At the end, M tidies up all the stacks face up and then tells P: point to any stack that has 5 or more cards.
P points to a stack. M sets it to the side and has P select two more stacks (pointing, not picking up or messing with) and M sets those aside, collects all the remaining stacks and puts them into a single deck and then quickly counts all the cards, face down, through the remaining deck, silently counting up to 19, and then starting again at 1 through the rest of the cards.
That final sum after 19 will be important at the end.
Let's assume for example, there were 8 cards left over in the deck after counting off the first 19. (total 27 in this stack of remaining cards).
M takes the 3 stacks chosen by P, flips the stacks face down and puts them in a line, instructing P: when I turn around, rearrange these stacks "shell game style" keeping them face down, in their respective stacks.
Do not shuffle the cards. Just change their position so I don't know which stacks I set aside.
M turns around. P rearranges cards as noted and informs M when they've done it.
M turns back to P and has them point to any stack.
P chooses, in the example, the center stack.
M flips the top card face up. It's an Ace.
P chooses the right most card.
M flips the top card face up. It's a 4.
M predicts for P the left card will be a 3.
And without fail it will be. Ace=1 so (1+4+3=8)
I have spent my entire life wondering how this trick works. It always works. If you can do math and the P doesn't shuffle anything after the initial time so they know you're not cheating of course.
How does the final sum of the remaining cards in the stacks have any possible bearing on the painted on numbers in the stacks?
I truly hope you or somebody out there has heard of this trick and can kindly explain how it works.
Not how it's done, I showed ya how it's done, but how the painted on numbers of the top card in 3 stacks is always adding up to the quantity of cards after 19th one in the other stacks. lol
Thanks again,
GT in Australia 🦘🐨🇦🇺
I showed it to my friend, and he immediately repeated the steps over and over and figured out that the default was that it was going to be an equal amount of cards at the end. He later figured out the trick by randomly removing cards. This is a nice trick. But it's something the spectator can easily replicate.
Cool trick, though the sharp spectator will notice there must be 4 cards missing, since the discarded pile contains just as many blacks as reds.
are jesters included?
Hey Celeste, It's best to take them out so there's no confusion between if they are red or black.
I know how it works. If you could go through the deck and never get a red and black pair, you would each have twenty six cards. But since you take the same amount of red as you do black when you discard the red and black pairs you end up with the same amount if you don't remove four black cards.
There are an equal number of black and red cards cards in the discards pile, so four black cards must have been missing from the deck (or four extra red cards) when the trick starts.
You got it!
The simplest (but not the shortest!!) explanation is perhaps this… Imagine that the cards get so well shuffled that the entire pack (minus the 4 black cards) is arranged in a perfect “red, white, red white…” fashion, but with the 4 extra red cards all together at the end. Naturally, the first 44 cards will get discarded (all pairs being red/white) and only the last 4 will get allocated - in this case to the red pile. So clearly the red pile has 4 more cards than black, which has zero!! Now, going back to the start, imagine that the perfectly shuffled pack is altered by having just 1 of cards removed from the middle section of the pack (containing the perfectly arranged “red/white” pairs), and this card gets reinserted somewhere else the front 44-card portion of the pack. At the position where this card was removed, we will now see two cards of the same colour meet (the opposite colour to the card removed). And at the position the card gets reinserted (between a red & black pair) it will naturally create an reciprocal colour-matched pair (the card itself and one of cards it is now next to). The pack now obviously has an additional 2 pairs of matching-coloured cards, one red, one black. The total number of red pairs is now 3 (6 cards) and the total of black pairs is now 1 (2 cards). So there are still 4 now red cards than black. And this difference will remain, no matter how much more the pack is re-shuffled. Each time another card “breaks” the red/white arrangement, an additional colour-matched pair will be created (one red, one white). And, of course, it makes no difference if the 4 red cards at the end are actually shuffled into the pack. They will either create 2 more red pairs or will displace the other cards down the order of dealt pairs, creating pairs further down the line.
And by white, I clearly mean black!!😫🫣
A quicker explanation.
An unaltered deck has the same amount of red and black cards. If you remove any number of pairs that contain a red and a black card, you have removed the same amount of red and black cards. Therefore the number of red and black cards in the deck remain equal. And that's what you'll end up within your final piles.
}Fun trick. It's important to consider your audience with this one though. Mathematically minded people will see what is happening instantly. But this would be a real fooler for kids or people who don't use math and their daily life.
It does have a couple things going for it. The palming is confusing. The spectator might forget that they chose their color before you make your prediction. Especially because you can make the second prediction before they choose their color.
I think this trick have a mathematical explaination. But I can't explain it because i got C in mathematics 😂😂😂 very cool trick Matt 🎉❤❤❤ nice video and happy New Year 🎉🎉🎉
Me too 🤣
I would change the number .say your at a party and end up doing it for multiple people. If you remove 6 it will be 6. If u remove 2 it will be 2. If you had the opportunity to change the amount in the bathroom or somewhere your alone it wouldn't be suspicious if the same people see you do the trick on someone else. But remember it always has to be an even number because 52 is even. If you end up with one card they're gonna know you altered the deck. Someone can easily count the pairs of 2's and relize you altered the deck count.
@@billesposito2112 😵😵😵 I am not good with numbers
it works because every time you discard a red and black, you're taking one away from each set of red and black, so the difference of the cards will remain the same.
Not a fan of self working tricks but this is cool how u palm away then steal them back. I would say for the set up, if you’re going to do this in the middle of other tricks, don’t worry about people seeing u remove cards. Nobody knows what you’re doing, plus u can just make it look like you’re tidying up the deck. If u truly want to do it deceptively, I don’t know why you would, but u can leave the jokers in, move the cards u want to remove to the top or bottom, then steal them out when u go to take the jokers out.
Sound advice!
You could have 2 prepared predictions first phase and second phase folded second fold if that makes sence
I love your husky, we have one also. They are the best ❤️
@MattMcGurk he was a huskita I owned his nan mum uncle cousins I had 6 they all gone now will have mote just moving yo the right place the dog you see is Mr Ghost the most amazing dog you would ever meet, the only hard thing about havin a dog I sayin goodbye 😔 love your work bruv
When you discard, you always throw away 1 of each, so in the full deck it will always be equal. I don't think I would show this trick to any mathematicians or engineers :) The 4-more was only ever possible with an incomplete deck
yes, i don't think that I would show it to anyone who masters additions. Excellent for kids though.
I did this years ago.
I learned this trick when I was just a kid! The way I did it was even easier: My prediction would say I have 4 more cards than you, but I would not remove 4 cards before hand. When the separation of cards was complete I would say, "Count your cards but don't say how many you have until I ask." You both silently count your respective cards, and when you get your total add 4. You then lie and say, "I have ## cards, how many do you have"? As you are announcing your total you casually add your cards to the discard pile and shuffle.
Either you're high or i am
@@porkchp6369 What part don't you understand?
Depending on the spectators colour choice in phase 1 either you read out or have the spectator read out the prediction to you 😊
Yes absolutely 💯
Steward James
To set up your 2 or 4 removed cards place them between the t.w. jokers and the other non playing cards from a bicycle deck. The jokers face up your cards face down…you can than take your cards out of the box remove the jokers plus the removed cards… place Thatcher one joker on top of the removed cards… your removed cards are now face down on the bottel then do trick one … then for trick two place the jokers with the removed cards on top of the deck… and ask the spectarors do you think it was a lucky guess?? And do the trick the second time … this time de removed cords stay on the deck and remove only the jokers … good trick it fooled me too
This trick was invented in 1936 by a Canadian magician Stewart James. It was titled Miraskill and it's a testament to a lack of simple critical thinking for many individuals. It was also written up in the Encylopedia of Card Tricks published first in 1937. It didn't fool me one bit when someone first showed it to me and that was when I was about 10 years old. I'll turn 80 this year so one should be able to figure out how long ago that was. If one can't then the odds are good that this trick might fool them too.
😂 Yes Mr James was one of the greatest thinkers!
You fooled me!
What cards are those?
Bicycle Verbana. 🤩
Another twist for you. The first time you use your deck with 4 cards in your pocket and you deal the cards out. The second time you have a spectator get another deck and you have them deal the cards.
Because when They are mixed.You're taking one of each out of the deck. So the difference of four' still remains the same
Wouldn't it work if you took 6 or even eight cards out, say with six, three in one pocket and three in another, you could have three outcomes then?
Yes absolutely! But I think 3 repeats may be a bit much. Try it and see 👍
Great trick, but simple to understand with the 4 cards
removed.
I'm the kind of guy that when you ask, "do you want the red cards or the black cards", I would say, "I want the diamonds".
Sadly as someone familiar with logical analysis i could see what was happening from the very beginning.
As soon as you said how the pairs were allocated I realised that it never matters what order the cards are in because when you remove cards it's always 1 and 1 maintaining the relative number of red and black cards
What is the name of this trick?
Miraskill
You are discarding an equal number of each of the colors. Using your setup, the deck has 22 black and 26 red cards. Suppose the middle (discard) pile has 10 pairs, that means it contains 10 red and 10 black. The black pile will then contain 22 - 10 = 12 cards and the red pile will contain 26 - 10 = 16 cards -> 4 more red than black cards. The key is that you are discarding an equal number of red and black cards. No matter how many pairs are in the discard pile, there will always be 4 more red than black cards remaining in their respective piles.
It works because every time you discard the red and black there is still 4 more reds than black.
Matty just send me a Signed card by you please 🥺🙏
Curse you TH-cam for hiding this from me for 10 hours😂
Hit the bell then 🔔 😂
great trick fooled my parents. performance tip: write your prediction before and use the magicians force
I think giving them an absolutely free choice is a better option. Or just have 2 predictions in 2 different pockets, then pull out the correct one.
oh nice, and the best thing is that you can modify it for 2,4,6,8 cards lol
If the cards happen to be black red black red, then they all end up all in the middle pile, with four left over. For every one pair that is changed to say black, there will always be a corresponding pair of reds together somewhere in the pile. You have to move one black out of the deck to make a black pair, that leaves two reds together where you took the black out. Try it! No math required. Every possible combination in the deck can be replicated by this method starting with red black red black then taking one card out at a time to make the pair, that is why it always works.
You got it! ☝️
Saw this a while ago but dont ask ne to explain it. As regards the first prediction why not simply have a prepared one saying 'there are 4 more red cards than black'. Can be done BEFORE you start anything and is irrelevant what colour they chose?
You could absolutely do this 👍
I know how the trick is done!!! You ready for it.... My Granchildren actaully came up with the solution as this is our Favourite TH-cam channel to watch.. Ok here's the solution... why it works..
It's Magic!!! yup my grandchildren are never wrong! I agree Magic real Magic :)..
Your grandkids have impeccable taste 😂
@@MattMcGurk Thank you Matt... you know the great thing is it makes them think and collaborate between each other, the side benefit the parents get a little respite..
Now remember you did say you didn't care how the trick works, but you also did say if anyone knows to "feel" free and post the answer :) So you asked for it... please be seated... lol - so due to the math it works if you remove 1,2,3,4,5 or 6, I guess any number of cards the result will be there..
This trick works due to a clever mathematical arithmetic invariance principle rooted in the parity of the number of black and red cards after you remove the 4 black cards. Let’s break it down step by step to understand why the outcome is always 4 more red cards than black cards.
For the boffins, the math and formula is at the end... I stand to be corrected if I have made any mistakes.
---
**Initial Setup:**
- A standard deck has **26 red cards** and **26 black cards**.
- You remove **4 black cards**, leaving:
- **26 red cards**.
- **22 black cards**.
- Total cards: \( 26 + 22 = 48 \).
**Key Observation:**
- When cards are paired during the trick:
- **Two reds** are set aside in one pile.
- **Two blacks** are set aside in another pile.
- **One red and one black** are discarded.
- Importantly, discarding a red and black card does not change the difference between the total number of red and black cards in the deck.
**The Constant Difference:**
- Initially, there are \( 26 - 22 = 4 \) **more red cards than black cards**.
- Every operation (pairing or discarding) respects this difference:
1. **Two reds or two blacks:** These don't affect the difference between red and black cards, as only cards of the same color are moved to piles.
2. **One red and one black discarded:** This also doesn't affect the difference because it removes one red and one black simultaneously.
**Final Result:**
- At the end, all cards are either sorted into piles or discarded.
- Because the difference of \( +4 \) (red minus black) is constant throughout, the pile with red cards will always have **4 more cards than the pile with black cards**.
**Why Does It Always Work?**
The principle hinges on the **constant difference in parity** between the two colors of cards:
1. Removing 4 black cards establishes an initial imbalance of 4 red cards.
2. Pairing or discarding cards never alters the imbalance.
3. Therefore, the trick is mathematically guaranteed to end with exactly 4 more red cards than black cards in the final piles.
This is a beautiful example of how simple arithmetic invariance underpins the illusion of a seemingly magical outcome!
What Is Arithmetic Invariance?
Arithmetic invariance means that a particular property of numbers (like the difference between two totals) remains unchanged, no matter how you manipulate or rearrange the numbers-as long as the operations you perform are balanced.
In the card trick, the difference between the number of red and black cards is set at 4 when you remove 4 black cards at the start. This difference is "invariant" (unchanging), no matter what you do during the trick, as long as you follow the rules.
Why This Feels "Magical"
Even though you shuffle the deck and discard cards seemingly at random, the math ensures that the starting difference (4 more red cards) remains constant throughout. By the time you finish, the red pile will always have exactly 4 more cards than the black pile.
It's like starting a race with a 4-step lead-you might shuffle your steps along the way, but you’ll always finish 4 steps ahead!
Certainly! The mathematical formula for the trick is based on the principle of **invariance**, where the difference between the number of red cards (\( R \)) and black cards (\( B \)) is maintained throughout the process. Here's how to express it formally:
**Key Definitions**
- \( R_0 \): Initial number of red cards.
- \( B_0 \): Initial number of black cards.
- \( n \): Number of black cards removed at the start.
**Step 1: Initial Difference**
The initial difference between red and black cards after removing \( n \) black cards is:
\[
D = R_0 - (B_0 - n)
\]
For a standard deck:
\[
D = 26 - (26 - n) = n
\]
So, the difference \( D \) is always equal to the number of black cards removed, \( n \).
**Step 2: Invariance Through Operations**
Each operation during the trick does not alter \( D \):
1. **Two red cards (\(+2R\))**: Adds 2 red cards to the red pile, but \( D \) remains unchanged.
2. **Two black cards (\(+2B\))**: Adds 2 black cards to the black pile, but \( D \) remains unchanged.
3. **One red and one black (\(-1R, -1B\))**: Discards one red and one black card, keeping \( D \) unchanged.
Thus, the difference \( D \) between red and black cards remains \( n \) throughout the trick.
**Step 3: Final Result**
At the end of the trick:
- Let \( R_p \) be the total number of red cards in the red pile.
- Let \( B_p \) be the total number of black cards in the black pile.
The relationship between \( R_p \) and \( B_p \) is:
\[
R_p - B_p = n
\]
**General Formula**
If \( n \) black cards are removed initially:
\[
R_p = B_p + n
\]
This means the number of red cards in the red pile will always exceed the number of black cards in the black pile by exactly \( n \), no matter the order or pairing of cards during the trick.
**Example: Removing 5 Black Cards**
1. Initial setup: \( R_0 = 26 \), \( B_0 = 26 \), \( n = 5 \).
2. Difference: \( D = R_0 - (B_0 - n) = 26 - 21 = 5 \).
3. Final piles: \( R_p = B_p + 5 \).
So, the red pile will always have **5 more cards than the black pile**.