Mind-boggling Card Trick (you can try at home)

[@boumbh]

*Spoiler* There are four stacks: - face up red stack => let say there are A red cards in it, 0 black cards - face up black stack => B black cards, 0 red cards - face down red side stack => C red cards, D black cards - face down black side stack => E red cards, F black cards Since there are as many red cards as there are black cards, A + C + E = B + D + F. Since the face up red stack has the same number of cards as the face down red side stack, A = C + D Since the face up black stack has the same number of cards as the face down black side stack, B = E + F Replace A and B in the first equation: (C + D) + C + E = (E + F) + D + F Discard D and E: C + C = F + F And so C = F, the number of red cards in the face down red side stack is equal to the number of black cards in the face down black side stack.

[@karlkastor]

+boumbh Exactly how I did it! You are missing the end where some cards of the stack are swapped, though.

[@patrickwienhoft7987]

+Karl Kastor Swapping Red with Red doesn't do anything, neither does swapping black and black. So left over case one: You take a red card from the red pile and a black card from the black pile and swap them. Now the number of red cards in the red stack is reduced by one and the number of black cards is reduced by one aswell. As they were equal before swapping, they are equal after swapping aswell. If you take a black card from the red pile and a red card from the black pile, something similar happens: the number of red cards in red stack and black cards in black stack both increase by one, so they are still equal. Nice deflection there.

[@thelatestartosrs]

+Karl Kastor LOL

[@thelatestartosrs]

+boumbh well done

[@standupmaths]

That is some mighty fine working out. The same as mine except for choice of letters!

[@RobertRussellComposer]

I think you can prove this with some simple algebra. I'm calling the four piles R, B, X and Y, where R = red pile, B = black pile, X corresponds with R and Y corresponds with B. I'll also define Xr, Xb, Yr, Yb to be the number of reds/blacks in the X/Y piles. Start with R + B = 26 And that R = X and B = Y From these, you can easily get the two equations... R = 26 - Y B = 26 - X Secondly, we have that... Y = Yr + Yb X = Xr + Xb If we substitute these into the previous equations, we get... R = 26 - (Yr + Yb) B = 26 - (Xr + Xb) Finally, it must be true that... R + Xr + Yr = 26 B + Xb + Yb = 26 Which rearranges to... R = 26 - (Xr + Yr) B = 26 - (Xb + Yb) Equating these with the previous equations, gives... Yr + Yb = Xr + Yr Xr + Xb = Xb + Yb Which leads to... Yb = Xr To show the other one, start again from... Y = Yr + Yb X = Xr + Xb Substitute in Yb = Xr to get... Y = Yr + Xr X = Xr + Xb Subtract these to get... 0 = Yr - Xb Yr = Xb :)

[@peppybocan]

+Robert Russell Composer exactly!

[@mathtronic]

+Robert Russell Composer Additionally, the swap does not affect the equality. If like colors are exchanged, Yr, Yb, Xr, Xb remain the same. If n Xb are exchanged with n Yr then (Yr-n) + (Yb+n) = (Xr+n) + (Yr-n) (Xr+n) + (Xb-n) = (Xb-n) + (Yb+n)

[@thelatestartosrs]

+Robert Russell Composer good job you managed to prove that Yr = Xb but that's not true. imagine picking 26 red cards facing up and having Xb = 26... how is Yr 26 if there are no Y cards?

[@thelatestartosrs]

+Robert Russell Composer Here: "Y = Yr + Xr X = Xr + Xb Subtract these to get... 0 = Yr - Xb Yr = Xb" the mistake is that the 3rd line isn't true you can deduct that Xr = Y-Yr = X-Xb from the 1st and 2nd. not Xb= Yr also Xr = Y-Yr means that Xr = Yb so it was close but you can't "show the other one" as you said because the other one isn't true.

[@ericvilas]

wait, in the final step, wouldn't it give you Y - X = Yr - Xb? You don't have the same number of cards on each pile, generally.

[@alaslipknot]

Micheal sent me here, and as always, am glad he did :)

[@thatskap]

You sir, are one of those people who make mathematical videos entertaining. Thank you for that :)

[@standupmaths]

My pleasure! Plenty more where these came from.

[@thatskap]

+standupmaths Omg Thank you for replying! I just got your book, "Things To Make And Do In The Fourth Dimension" and its mathematical yet entertaining. Thank you for the effort you put into everything. Feels great seeing your work. -Sincerely, Kacper, a fan of Mathematics :)

[@AnastasisGrammenos]

That is one strange dice!

[@ScaryPurpleAmpersand]

I went and set up the trick like you said and got pretty excited about what the trick was going to be... But then it was something that seemed totally obvious and intuitive to somebody who plays card games. In a two player or two team card game--particularly trick-taking games such as bridge, spades, or pinochle--knowledge that the number of cards you have in a suit is inversely proportional to the number of cards the other player has in said suit is key to playing the game.

[@beefcakeandgravy]

dafuq is up with that die, or is it a weird camera perspective?

[@Ruminations09]

+George Smith mathsgear.co.uk/collections/dice/products/skew-dice

[@beefcakeandgravy]

Laurelindo die is the singular, dice is used for multiple dice. English for you.

[@beefcakeandgravy]

Laurelindo what about mouse and mouses? Mice is the plural form. We say house but we don't say hice. I'll admit the English language is quirky and I'll never understand why some words are used over others.

[@kyzer42]

Really? Because I always thought they were both "mice". And what about multiple moose? Are they meese?

[@DarqIce]

That supposed-to-be-proper-cube-thing (I'm not getting into the die/dice debate here) is trippy AF indeed :D

[@MsRealReem]

who else is here from Vsauce ??

[@rw4877]

Thats soo me

[@mustangandthef4060]

how do you know!

[@MsRealReem]

***** what's the type of your comment ?

[@MsRealReem]

***** the point of my comment is to see if there are others here from Vsause

[@JL-zw7hi]

i

[@Formulka]

why the hell is that die all twisted and crooked?

[@standupmaths]

That is a "Skew Dice". They are completely fair and available here: mathsgear.co.uk/collections/dice/products/skew-dice

[@Formulka]

+standupmaths my wallet is rea... sold out :(

[@standupmaths]

+Formulka Wow, that was fast. There should be some more in stock soon.

[@Schindlabua]

+standupmaths How did I not know about mathsgear until now? I _need_ that rhombic dodecahedron die.

[@anatine_banana_69]

@@standupmaths how is that a 'fair' dice?

[@adamjdorfman]

This is so easy. As long as the number you swap at the end is the same number, the black cards that end up in the black deck will just be swapped for red cards and vice versa. There's just a whole bunch of symmetry going on that as long as the numbers are the same, that is the number of cards in each deck are the same, it's always gonna work out, reds for reds and blacks for blacks. Love the channel by the way big fan!

[@Dixavd]

I don't own a pack of cards so I'm just going to go back to lying in bed eating Mcdonald's (because I'm a monster of a human being)

[@DaveScottAggie]

I actually paused the video and got a deck of cards, when you first said to try it at home. The first deck I got was short a few cards, then the next deck was short 1 card so I went through and found which one, and marked one of the Jokers as the missing card. I did the trick as you were doing it, then at the end, you tell us that it works as long as there is the same quantity of red and black. Of course, playing with a small deck makes it a little easier to grasp what is going on. I now have worked it out how the cards get "sorted" so that the number of red cards on left always equals the same quantity of black cards on the right.

[@NeatNit]

I proved it to myself, entirely in my head, but I have NO IDEA how to put it down in words or any form of mathematical proof. Can't wait for your video to see it done!

[@lawrencecalablaster568]

:D I love your math magic tricks, Matt! I have consistently shown my family & friends your 27-card trick at gatherings. Half the time they're impressed, & half the time I forget how to do it. Thank you for this new gift!

[@DemonXeron]

Wow. This was simple to figure out. I won't give spoilers. I had it literally as you said "there should be an equal number of red cards in this deck as black ones in the other deck." I thought something like "Yeah, that's really obvious, next puzzle please?"

[@KeithPeters]

most of this is misdirection. Cut the deck into two equal stacks of 26. Say there are 10 red cards in one stack. Then there are 16 black cards in that stack. Which means there must be 10 black cards and 16 red cards in the other stack. The first step just removes an equal number of reds and blacks from the deck and leaves you with two piles of 13 each and the same thing applies to those piles. Swap as many as you want., shuffle them, still holds true.

[@martind2520]

+Keith Peters Except the two piles aren't both 13. The number of cards in them are random based on the original indicator card.

[@KeithPeters]

+Martin D Good point. I'll have to revise my logic!

[@ubermensch826]

matt parker gives me life

[@MaiFiinta]

i love the fact that you are going to make regular videos! can you convince James Grime to do the same?

[@standupmaths]

No one can control the Grime.

[@Maninawig]

This is amazing. Although it applies the same "oil and water" physics as other card tricks (which I can't explain without looping my explination due to how I understand it) the fact that you broke it up into an additional 2 decks and used the season of random piles as a premise was an absolutely inspired stroke of genius. As revolutionary as tsking a bicycle and inventing a car. In truth, having seen the basis of this trick done in so many ways, it was only at the end that I honestly figgured out what you made me do...

[@DrDherp]

Just a variant of an old trick, but a fun one none the less. One of the first one I learned back in 2010 I think. I did a video proving the maths on an old channel. Thanks for posting!

[@nightmare9566]

I go to school...I get good grades....I get tested.....I get told I have a very high IQ....then I go on and watch your videos and feel dumb as a bread. I love it!

[@Boomshicleafaunda]

I followed along, not knowing what the trick was. After I finished (same time as you), I was like "Of course they're the same count!", and didn't really feel the 'magic' in this trick. I guess mentally proved this while the trick was going on. I know quite a few mathy card tricks.

[@fudgesauce]

I haven't thought this through, but I suspect the logic is similar to this old puzzle: you have a cup of milk and a cup of coffee. You take a spoon of some arbitrary size (smaller than the milk & coffee containers) and you lift some milk and put it in the coffee, then you use the same spoon to move some of the coffee+milk mixture back into the milk cup. Mind you, whether or not you stir the coffee or if you don't and predominantly get more milk or coffee in that second spoon isn't specified. After this, do you have more coffee in the milk cup or more milk in the coffee cup?

[@NoriMori1992]

I suspect you're onto something; the preview of the explanation video shows him doing something with two glasses of what looks like milk (or one glass of milk and one glass of water), so I'm betting he's using that same principle to demonstrate how this trick works. It personally reminded me of the riddle where you have 20 coins, and you can't tell which are heads and which are tails, and you have to somehow split them into two groups of 10, where each group has the same number of heads and tails as the other. The solution is to split them in any way, and then turn over every coin in one group.

[@robertmachado4727]

Awesome video and trick. I love that people are trying to explain this by a certain number of cards in the deck or a certain number taken out from the deck. Thanks for the hint, Matt, that it always works as long as the number of cards of each color are the same. That was the clue I needed to figure out why this actually works.

[@Sashafomin95]

a = A + B - (A + (B - b)) = b. The beauty of associativity.

[@aleks.j.]

Okay I think I got it :D Take the total number of cards to be 2x, the number of reds placed face-up as R and blacks face-up as B. Since there is a pile of random cards corresponding to both R and B, we conclude that 2X=2R+2B, so X=R+B, where X is the total number of cards in the random piles. Take rR, rB, bR and bB to be the number of reds in the red random pile, reds in the black random pile, blacks in the red random pile and blacks in the black random pile respectively. Following logic, rR+bR=R and rB+bB=B. The total number of red cards in the random piles can be written as rR+rB and the total black cards is bR+bB. Since the number of cards of each colour are equal, R+rR+rB=B+bR+bB=X (total cards divided by two), so by rearranging this and the previous formula (X=R+B) we can derive the formulas rR+rB=X-R=B and bR+bB=X-B=R Now comes the fun part :D We now know that rR+bR=R rB+bB=B rR+rB=B bR+bB=R So by substitution, rR+bR=bR+bB therefore rR=bB or alternatively rB+bB=rR+rB therefore bB=rR So the number of red cards in the red random pile is always equal to the black cards in the black random pile, but only if there are equal amounts of each type of card. That's satisfying.... Although my explanation seems a bit incoherent :D

[@aleks.j.]

Also, swapping equal numbers of cards won't make a difference. If S cards are taken from the red pile, and x of them are red, S-x are black, and y black cards are taken from the black pile, S-y reds are taken. So rR is the red cards in the red pile and bB is the black cards in the black pile (with a ' meaning after the swap), rR'=rR-x+(S-y)=rR+S-(x+y) therefore rR=rR'-S+x+y and bB'=bB-y+(S-x)=bB+S-(x+y) therefore bB=bB'-S+x+y knowing that rR=bB, rR'-S+x+y=bB'-S+x+y so rR'=bB'

[@marias6572]

+Alex J. Oh you mathematical genius! xx

[@aleks.j.]

ahahahahaha just a bit of a fanatic ;)

[@Threedog1963]

I watched a couple of your videos first and had a deck of cards ready.

[@Formulka]

the red pile indicates how many black cards are left over face down, so if there is x in the pile in front of red there can only be red-x in the other which is the same as the amount of red in front of the red pile

[@harmidis]

We have two piles of open cards, the red (R) pile and the black (B) pile. Above them we have: on the red pile r1 and b1 (red and black) cards, on the black pile r2 and b2 (red and black) cards. We must prove r1=b2 We have: r1+r2+R=b1+b2+B (1) (all red cards are equal to black cards) We also have: R=r1+b1 (2) and B=r2+b2 (3) (equal cards above to every open pile) We substitute (2) and (3) to (1) (1)=> r1+r2+r1+b1=b1+b2+r2+b2 (r2 and b1 canceled on both sides) => 2r1=2b2 => r1=b2 That was, actually, very easy to prove!

[@kwinvdv]

I initially did not believed it. I immediately thought I had a counter example, namely if all the cards you see are the same color, such that one pile contains all the cards that are facing down. This however would mean that you can't exchange any cards and that all the cards facing down have to be of opposite color, thus still satisfying the criteria of number of same color cards in each pile in front of the up-facing "piles" (in this case the other color pile has size zero). After this I checked the maths myself I can confirm it to be true.

[@petermenzies9193]

Vsauce just stole your thunder :(

[@jacknikolov7478]

Man, i really love the part when you don't give the answer. It makes people think and this is awesome. There should be more people like you. Nice job and keep going. Can't wait to see more!

[@bolerie]

This actually makes a lot of sense. SPOILER ALERT! This is pretty easy to prove yourself, so if you want to try go ahead before you red on Proof: We flip the first card and it is red. The following things can happen: The downwards facing card is black. The balance of red and black cards in the deck stays the same. The downwards facing card is red. This means that there are now 2 more black than red cards left in the deck. This can only be balanced by a later black flip, downwards black move. No amount of flipping red cards can balance it. This cycle is repeated 26 times. It doesn't matter if the flipped card is red or black, since a flip and a downwards facing card of the same colour always balance out. Now we have as many red cards in the red pile as we have black cards in the black pile. When we swap cards the following things can happen: A red card from the red pile for a black card from the black pile. Both piles' "values" decrease by one and the balance is maintained. A red card from the red pile for a red card from the black pile. Nothing happens. A black card from the red pile for a black card from the black pile. Nothing happens. A black card from the red pile for a red card from the black pile. Both piles' "values" increase by one and the balance is maintained. Thus we see that there will always be the same number of red cards in the red pile as black cards in the black pile

[@itaialter]

I tried it, and I got an edge-case scenario that I think clued me in to the logic. My edge-case was that when I looked at the card color, most of the cards were black (by chance), so most of the cards went into one pile. That made it a bit more clear to see what is going on. I didn't do a mathematical proof, but I have a hunch.

[@standupmaths]

I actually use the edge-case in the explanation video. Good work!

[@kwinvdv]

+standupmaths You do not need edge-cases to prove this, or do you mainly use it give viewers more insight as to why it is true? Because you only need a handful of relations to deduce it.

[@itaialter]

Kwin van der Veen I didn't say you _need_ it to prove it, but it helps with the understanding of it, that's all.

[@lednerg]

It has something to do with glasses of milk.

[@ObiWanCannoli]

This guy...

[@audreyoldaccount9233]

+lednerg How dumb are you?

[@lednerg]

justAnotherfanb0y My bad, it was one glass of milk and one glass of water. Still a pretty good guess though.

[@audreyoldaccount9233]

lednerg Are you serious? Dumbass.

[@lednerg]

justAnotherfanb0y I made the original comment before the second video came out. At 5:50 in this video, you can see a little window of it, and he's got what looks like two glasses of milk on the table. You had to be there, I guess.

[@DaveScottAggie]

Pretty neat trick. If you make a smaller deck of cards (for example 4 cards of each color) and run the trick, it is easier to see what is going on: for both the "deal" portion, and the "swap" portion. Then it is not hard to show that it will work for any size deck, as long as there are equal quantity of red and black cards.

[@DaveScottAggie]

I actually tackled the swapping stage first, making the assumption that the number of red cards on the left equaled the number of black cards on right, at beginning of swap, then showing if you swapped only 1 card, what are the possible outcomes, then what if you swap 2, what are the possible outcomes. And then does this expand out for any number of swapped cards. After that was tackled, then looked at the deal portion to see how it would 'happen' to turn out that would make my assumption at the beginning of the deal be true or false.

[@natewright1197]

If the deck is arranged RBRBRB... (I know this is probability nearly zero), all of the cards will be in the red side and all of the cards in the facedown "red" pile will be black. I think this and the reverse (BRBRBR...) will be the only two arrangements that would give a result that could not end with the two piles containing the same red and black, because there would only be one pile.

[@standupmaths]

It still works because the red-side pile contains zero reds and the black-side zero blacks. 0 = 0

[@Quintingent]

+Nate Wright Ah, but in those cases there are zero red cards in the 'red' pile and zero black cards in the 'black' pile. It just so happens that one of the piles has zero cards in total.

[@natewright1197]

Ah true. Didn't think about that!

[@EvanED]

After thinking about it for a while (and wanting a deeper explanation than the equations others are posting), here is my favored explanation. I am going to describe a variant of this game where you get to deliberately decide where every single card in the deck goes. I'll call the face-up cards that determine the column 'key cards,' and the other cards 'target cards.' The way this works is there will be a sequence of 26 draws. (I say "draw" but you have full choice of what cards you draw.) In each draw, you pick up one key card and one target card and put them into their place. There are then four possibilities: (i) red key card, red target card; (ii) red key card, black target card; (iii) black key card, red target card, (iv) black key card, black target card. Really these reduce to two cases: the target and key cards are the same, and the target and key cards are different. You start out with the same number of (red target cards in the red column) as (black target cards in the black column), both zero. Our goal is to set down cards to make those numbers unequal. If I pick up two cards of a different color, that doesn't affect the balance, because I'll be adding a black card in the red column or a red card in the black column. So that can't affect the balance and get us to our counterexample goal. If you pick up two cards of the *same* color, hey, it looks like we've made progress. For example, if I pick up two reds, I've now increased the number of red target cards in the red column by 1 and thrown off the balance. However, what does this mean? Because there are 26 draws and only 26 red cards in the deck, by picking two red cards in a draw we have guaranteed that, at some later point, we will have to pick up a black key card with a black target card. At the point we do that, we'll be adding a black target card to the black column, and that balances out the red/red card we added before. Boom, we're back to even.

[@0xBADFECE5]

I can see how swapping cards between the mystery piles doesn't affect the outcome of the trick because swapping a black for a black or a red for a red maintains the status quo, whereas swapping a black for a red or vice versa will either add one qualified card to each pile or subtract one from each pile.

[@JohnDlugosz]

This reminds me of the puzzle of wine and water. Two pitchers have the same volume of liquid initially. Some quantity of A is poured into B. B is mixed. The same quantity of mix is poured back to pitcher A so they once again have the same volume. Is it true that the amount of wine in the water matches the amount of water in the wine?

[@sirjohnnyg]

Finally came up with an easier way to think about it without all the equations, you just need to accept that when you split the 52 cards into two equal stacks the number of reds in one stack will be equal to the number of blacks in the other (not hard to prove). Since half the cards are in the top two stacks and half the cards are in the bottom two, the above would hold true thinking about the groups of 26 as being top vs bottom, but that doesn't really help. What does help is going diagonally. There are also 26 cards total in the top left and bottom right combined, but the only black cards in that 26 are in the bottom right pile. There are also 26 cards total in the top right and bottom left combined, but the only red cards in that 26 are in the bottom left. Now we have 2 stacks of 26 and the two things we care about must be equal. The swapping doesn't affect the pile sizes and so doesn't change the relationships.

[@rodbotic]

Great Trick! I showed it to the guys at work. And we worked it out. Good thing I work in a physics lab.

[@pietroberardigili562]

I'm happy that you need people like me

[@EmanuelsWorkbench]

Half the deck is removed (the face up red and black cards). This is exactly 13 red and 13 black cards in the deck. What is left are the remaining cards: again 13 red and 13 black cards. These are randomly distributed into 2 piles of 13 cards each. Looking at the first pile, lets say there are 4 red cards in it, the remaining 9 being black. How may cards are left to populate the 2nd pile? Exactly 9 red cards and 4 black cards. Let's say that the two piles end up being identical: All of the red cards are in one pile and all of the black cards are in the second. If you want to mix it up, take a red card from the first pile and add it to the second pile. Now to make the piles even, you have to take a black card and put it in the red pile (the moved red card "forced" a black card to change piles). Red cards not in the first pile force the same number of black cards to be displaced into the first pile. Therefore the total number of a given coloured cards in one pile always equal the number of the other coloured cards in the second pile.

+Emanuel de Matos The face-up cards aren't necessarily 13 reds and 13 blacks. They could be e.g. 26 reds.

[@maximp520]

Hahahaha mate! I though the same at beginning, but it turns out that you dont have an *equal* amount of cards in each pile (13)! Did u even try the trick? You may end up with f.e., 14 red cards in the two left piles, and 12 in the other two! So yeah....

I did try the trick. If, for example, the initial state of the deck is alternating red-black, all face-up cards will be e.g. red, and all face-down cards black. The trick will still work, because the face-down stack will have 0 red cards, and the "other one" will be empty, so also have 0 red cards.

[@LionelSimple]

+Emanuel de Matos while half the deck is removed, they need not be 13 red and 13 black.

[@maximp520]

Ya , there may be 13 cards in each pile , but only if every second card takes turns to be red-black (I mean, if they are alternating), but still, itisnt necesscary for the teick to work, so yeah

[@abcrtzyn]

For some reason I had an odd number of cards and it still worked

[@teachersoule8163]

So I solved the problem using algebra (which wasn't as bad as I thought), but couldn't solve it using just logic at first. After thinking about it for awhile, I realized that it should be possible to take the face down cards and shuffle them and redistribute them into the two piles so that each face down pile has the same number of cards in it as the face up pile and it should still work. I've tested this theory several times and it does work. Still having trouble wrapping my mind around the logic of why. So the swapping is just a "parlor trick" because it seems to work no matter what you do.

[@tggt00]

Anyone else is here not because of Vsauce?

[@tj12711]

What a witty and insightful comment

[@tggt00]

+tj12711 I'm genuinely subscribed to this channel for over a year.

[@acediamond5399]

+tggt00 I am a subscriber as well, I am not here from Vsauce. But I watch Vsauce too.

[@victorvalerio3059]

I saw this video the DAy It went out, and I laughed at the spoiler Vsauce made.

[@unvergebeneid]

+tggt00 Why don't we all pat each other on the back for watching this channel before it was cool? ;)

[@mattlm64]

The left hand decks have 26 cards and there are 26 of each colour. Using basic algebra you can figure it out: R = Number of red cards in top left. B = Number of black cards in top right. Lb = Number of black cards in bottom left Lr = Number of red cards in bottom left Rb = Number of black cards in bottom right Rr = Number of red cards in bottom right All reds = R + Lr + Rr One side = R + Lr + Lb All reds and one side both have 26 cards each, so: R + Lr + Rr = R + Lr + Lb Cancelling out gives: Rr = Lb So the righthand reds are the same number as the lefthand blacks, and this works the other way around of-course. Neat trick for sure.

[@mattlm64]

+Matthew Mitchell No I'm wrong as both sides do not necessarily have 21 cards so... R = Number of red cards in top left. B = Number of black cards in top right. Lr = Number of red cards in bottom left Lb = Number of black cards in bottom left Rr = Number of red cards in bottom right Rb = Number of black cards in bottom right Try to prove Rb = Lr R + B = 21 : Top cards are half of deck Lb + Rb + B = 21 : All blacks make half of deck ∴ R = Lb + Rb R = Lr + Lb : Same number of cards on top left as there is on bottom left. Lb + Rb = Lr + Lb : Both equal to R Rb = Lr : Cancel out Lb QED?

[@Tracy_AC]

*SPOILER* The deck is initially divided into two halves: 26 face-up cards and 26 face-down cards. Of the face-down cards, X are on the red side and Y are on the black side, where X+Y=26. Also, because the total number of red cards in the deck is 26, and because X of them have been placed in the face-up cards, there must be 26-X=Y red cards remaining in the face-down cards. Following the same logic, there must be X black cards remaining in the face-down cards. Further, on the red side there are A red cards and B black cards, where A+B=X, and on the black side there are C red cards and D black cards, where C+D=Y. A and C are all of the red cards in the face-down cards, so they are equal to Y, and B and D are all of the black cards in the face-down cards, so they equal X. We now have several equations: X+Y=26 (the two face-down piles) A+B=X (the red and black cards in the face-down red-side pile) C+D=Y (the red and black cards in the face-down black-side pile) A+C=Y (the red cards from both face-down piles) B+D=X (the black cards from both face-down piles) We want to show that A=D (red cards on red side equals black cards on black side). Simply see that if A+B=X and B+D=X that A+B=B+D, and therefore A must be equal to D. We can also look at C+D=Y and A+C=Y to determine that C+D=A+C, and therefore A and D are equal.

[@Quicksilver_Cookie]

I thought this trick was too obvious. In fact, why you even needed this whole splitting into 4? If you just take full deck and randomly split it into two piles amount of reds in pile 1 will ALWAYS be equal to amount of black cards in pile 2. No matter how you shuffle and what you do. If it doesn't work it means you have more reds or black cards in your deck :D

[@SpySappingMyKeyboard]

I put all cards in the first pile, none in the second. The first pile has 26 red cards, the second has 0 black cards (or vice versa).

[@Quicksilver_Cookie]

SpySappingMyKeyboard 0 cards is not a pile.

[@SpySappingMyKeyboard]

I guess you could argue that. I just went for the most extreme example. If you insist on that rule, then 51 cards in the red pile and 1 in the black. Now you either have 26 red 0 black or 25 red 1 black. Either way, the rule is still broken.

[@Quicksilver_Cookie]

SpySappingMyKeyboard Stop breaking my brilliant rule with your extremes please :D I understand that it doesn't work like this. However, I thought it was obvious I'm talking about equally sized piles - just like in a video.

[@SpySappingMyKeyboard]

But the piles in the video aren't necessarily even. If every card you reveal is red, every other card will go in the 'red' pile, with none in the black (or you can reveal 1 black card I guess).

[@ibolakersfan]

More math card tricks please mr.parker!

[@c0rv377]

I did this with a much smaller deck, and it's super easy to intuit. I took 8 of each suit, so I've got 16 cards. I divied out the cards - ended up with a stack of 6 red face-up, and 2 black face-up, leaving 6 cards in the "red" pile, and 2 cards in the "black" pile. Also, there are only 2 red cards face down anywhere, and 6 black cards face-down anywhere. IF (and that's a big if) the "black" pile has the two red cards, the "red" pile must have the remaining 6 black cards. Each pile adds up to zero. There is only one way to swap 2 cards: If you swap a card from each stack, the "black" pile contains one red and one black, and the "red" pile contains one red and five blacks. The black cards in the left pile match the red cards in the right pile. If I swap another card, I can swap a matching suit for a matching suit, which changes nothing, or I can swap two opposite suits. This either takes me backwards to 0 of each in each pile (swapping the black out of the 2-card pile leaves 2 reds, and 6 blacks in the larger pile), OR it can take me forward - two black cards in the small pile, and two reds in the larger pile. If the piles were more even (3 and 5, or 4 and 4), the same could be constructed to more than those 2 cards. Because the stack sizes are opposite to the number of cards left in the deck that you are checking for, you can actually go through every possible combination of cards in either stack through this process. You could take the extra black cards, (or whatever suit you have in excess) and rule them out. They're forced into the larger pile; the pile in which they're not being tallied, because the smaller pile is ONLY big enough to hold the cards of the smaller remaining suit, and no more. I wish I were more eloquent in describing this, but if you try it out for yourself with a smaller hand (try them face-up, perhaps), you'll be able to see what I'm talking about.

[@RizviRab]

Vsauce sent me here... after ruining the trick.

[@whiz8569]

When he pointed and said "We need people like you," I was picking my nose.

[@dino130395]

Its quite simple really... You know that there are equal number of red and black cards in the deck. if you remove the same number of black and red cards from your deck there will still be an equal number of them in the un-turned deck. Now if there are 7 red cards in the "red" deck, then we know for certain that there will be 7 missing red cards from the other deck, and since we know that there are equally many cards this means that there must be 7 black cards in the "black" deck.

[@TheBigBigBlues]

Just realised that the colour palette you use for this channel is the same as that used for Tom Baker as Dr Who's scarf. I hope this is not a coincidence.

[@R0nBurgundy]

The edge cases make it easier to understand. There are 26 red and black cards. If by chance you deal 26 red cards you will have 26 cards in the red pile of which all will be black, and therefore 0 red in red pile and 0 black in black pile. However if you deal 25 red cards and 1 black you will have 25 in the red pile and 1 in the black pile. The red pile will either have 1 red card and 24 black or 25 black cards while the black pile can have either 1 black or 1 red. It doesn't matter how you swap the cards the two piles will have the same number of reds and blacks and if you remove a red from the red pile and gain a black the black pile loses a black and gains a red, canceling the swapping. so if you have a red in the red pile you must have a black in the black pile. The maths behind this. The red pile will contain Nr cards where Nr = number of red cards turned over. The black pile will contain (T/2)-Nr cards. Where T is the total number of cards (standard decks have 52 but you can do this with any number of cards as long as number of red cards = number of black cards) The two piles between them will have (T/2)-Nr red cards and Nr black cards. (they have the inverse of what was turned over) Say there are n red cards in the red pile, this means there is Nr-n black cards in the red pile. Which in turn means their must be Nr(number of black cards between the two piles) - (Nr-n) (black cards in red pile) black cards in the black pile. Nr- Nr + n = n black cards in black pile. Therefore the number of red cards must equal the number of black cards. The shuffling at the end doesn't affect the number of red being equal to the number of black cards because the same number of cards are changed between each pile. It many however change 'n' itself but this doesn't matter in terms of the trick. Ken

[@NoriMori1992]

2:16 - Nice, you're using those "weird but fair dice". XD

[@betaslash-ix1ls]

This is awesome, I've seen your live maths stuff for teenagers. It was great and this is really great too!

[@Zaraken]

Actually, removing a Black card from the deck DOES make it more likely that the next card is Red. Not sure what you were trying to say there at 1:35

[@zivush3d]

it has to do with matter preserve (hope I wrote that right).for example if you take poker chips from two different color. n pieces from each color. then separate with your eyes close to two even piles. you get for example x blue in the red pile and x red in the blue pile.

[@TomasMira28]

Hi Matt! I really enjoyed this video. It was a great detail when you said that you won't be explaining the logic of this trick right away and the top view of the table slade away. However I really dislike the ending: the more you turned up the volume the less I could hear you, there were some of your last words that were difficult to hear (especially if you're watching the video in some crapy-sound quality phone as I am). Besides that I'm looking forward to discover this trick by my self :D Keep it going man!

[@trillman1111]

after I watched this video I was thinking about it and went to the bathroom. visualizing how many cards in each pile and the amount of reds and blacks in the face down piles during edge cases. now I know what it was like to have an apple hit me on the head. felt like a 17century scientist discoverering something on the toilet.

[@Ollervo100]

Simple. The turned cards have an equal amount of blacks and reds, so the ratio of blacks to reds in one pile is the complement of the ratio of blacks to reds in the other pile. When you swap same colour cards nothing obviously changes. When you swap different cards the ratio changes but the other piles ratio is still the complement of the first pile.

[@R0nBurgundy]

1:20 ... ok you got me If I'm procrastinating this much I might as well go get a pack of cards.

[@chrisberry7640]

I'm writing this having not seen any of the other comments, and I'm late to the party, but whatever. -The face up black cards has "W" black cards in it. -The way you're dividing the deck has you take every odd numbered card and use it as a decider for which pile the next card goes in. There are 26 odd numbered cards, which implies that the face up red cards has "26-W" red cards in it. -It follows that the face down black pile must have W cards, and the face down red pile must have 26-W cards. -Let "A" and "B" be the number of black and red cards, respectively, in the face down black pile. Because the face down black pile has W cards in it, this means that A+B=W. -Let "D" be the number of red cards in the face down red pile. There are 26-W red cards in the face up red pile. Since there are 26 total red cards, there are 26-(26-W)=W red cards in both of the face down piles combined, which means that B+D=W -Now, substitution: A+B=W=B+D A+B=B+D -Subtracting B from both sides gives: A=D This shows that "A", the number of black cards in the black pile, is always the same as "D", the number of red cards in the red pile.

[@StiiguX]

A lot of people seem to assume there are 13 red and 13 black cards. Also I see many explanations with just a lot of words so I fixed it (as a mathematician myself): Let's assume that in the whole deck there are 26 reds and 26 blacks. Let's mark: n=amount of reds dealt face up -> there are 26-n reds face down in total m=amount of blacks dealt face up -> there are 26-m blacks face down in total Half the cards were dealt face up (every other one) so n+m=26 Let's mark: a=the amount of red ones faced down under the red pile (->there are n-a blacks under red pile) b=the amount of black ones faced down under the black pile. We want to prove that a=b Now 26-m=b+(n-a) 26-m-n+a=b 26-(n+m)+a=b 26-26+a=a=b

[@dbliss314]

Suppose the stack of face-up red cards contains R cards, and the stack of face up black cards contains B cards. Further suppose that the face-down stack by the reds contains Nr red cards, and the face down stack by the blacks contains Nb black cards. We can see that N+R=26, because every other card went into N or R. We can also add up the total number of black cards in all of the piles: B, zero, Nb, and R-Nr. We already know there is a total of 26 blacks, so B+Nb+R-Nr=26. Since B+R=26, this simplifies to 26+Nb-Nr=26, or Nb=Nr.

[@mathmethman]

The dealing and swapping of cards is simply one way to create the following situation: a) On the left there are two equal (in number) stacks of cards, one face up and one face down. All the face up cards are red. b) On the right there are two equal (in number) stacks of cards, one face up and one face down. All the face up cards are black. c) In total there are the same number of red cards and black cards. Half the total cards (i.e. all the reds) can be expressed as Face up red cards + face down red cards on the left + face down red cards on the right. [*] Half the total cards (i.e. all those face up) can also be expressed as Face up red cards + face up black cards which is numerically equivalent to Face up red cards (on the left) + face down cards on the right which is equal to Face up red cards + (face down black cards on the right + face down red cards on the right). [**] If the expressions at [*] and [**] are equal, then face down red cards on the left = face down black cards on the right.

[@nefastusamator]

Every face up card doesn't just add one to the stack above it, it subtracts one of that color from the total number of cards face down. So if I have 23 red face up, I only have 3 red face down. With the black pile being 3 total face down(23 face up were red, so 3 were black), it always contains just enough room for all the red face down cards. Any number of red NOT in the black pile, has to be black. So it doesn't matter if the red pile has 1,2 or 3 red cards, the other side has to have an equal number of black.

[@marvinabarquez8915]

i understand the concept completely way before watching the explanation. do i get a prize now?

[@pjrkearsley87]

tempted to write 'first', thanks for the videos matt. keep them coming!

[@mihauinfamous]

+pjrkearsley87 +1

[@standupmaths]

Thanks for you thanks! I don't mind the odd "first" but this is much nicer.

[@Big-The-Dave]

You can disregard the swap 'n' cards at the end. If you swap a black card for a red card the number of 'correct' colours in each pile changes by 1 (plus or minus depending on the direction you swap). If you swap two cards of the same colour, the number changes by 0. I'm working it out on paper, with no deck, and it's hard :(

[@Big-The-Dave]

What I know is that: Rt + Bt = 26 (the total in the two upper piles) and that Rb + Rr + Br + Bb = 26 (the total in the two bottom piles, split into Red/Black) And that we need to prove that Rr = Bb. Somehow.

[@markkeilys]

you can also do this with even/odd if you take out an odd number of sets of face cards e.g. all kings or all kings, quines and aces

[@ausizzie10]

Cool trick, had fun working out the math :)

[@DensetsuNoShoujoA]

I did this using algebra too! Say there are R cards in red pile and B cards in black pile. We know 1. R+B = 26 (Since every 2nd card is placed into either the red or black pile, assuming a regular deck of cards with no jokers 2. B reds and R blacks remains in both pile (Since R reds and B blacks were already taken out, and there are same no of red and black cards in the deck) 3. If there are A reds in the red pile, then there are B - A reds in the black pile. So the no of blacks in the black pile is given by no of cards in the black pile (B), minus no of cards in the red pile (B-A). This gives B - (B - A) = A

[@jackripper0987]

Just tried this is two packs, works well Odd thing happened was that the random number I picked form the cards (11) was the amount of simotanoius red/blacks

[@Vesker3322]

For any of you interested in the Magic side of things, here's a cool alternative presentation for this trick: Have the spectator to shuffle the cards thoroughly. Once shuffled, ask if he wants red or black. Let's say he chooses black. Instruct him to deal the first card face up, if it is black, then he is to deal the very next card to himself. If it is red, then he is to deal the very next card to you. Have him go through the entire deck dealing cards in this way and talk about how random this process is. Have your back turned the entire time he is dealing so that you can't possibly know anything. Once finished dealing, tell him to take his pile and hide it. Turn around and grab your pile. Explain that you are going to make a prediction. Look through your cards and count the number of red cards, but don't make it obvious that you are counting. Then count the total number of cards. In your head, subtract the total number of cards from 26 to get how many total cards he is holding. You know how many black cards he is holding because it is equal to the number of red cards in your hand, so subtract that number from the total number of cards he is holding to get the number of red cards he is holding. Write the total, the number of blacks, and the number of reds in his hand on a piece of paper and fold it up and put it in plain sight. Reiterate that there is no way you could know anything about the cards he is holding. Ask him to lay his cards on the table and have him count total number of cards. Reveal just that part of your prediction. Now ask him to count the number of black cards and red cards. Reveal that part of the prediction and amaze everybody!

[@arbivator]

There is one instance I can think of that the trick doesn't work, and that is if you manage to shuffle the deck originally such that the deck is in alternating colours (red card, black card, red card...etc). You'd only get one pile.

[@kaitlyn__L]

i tried it three times, and every time one pile had one more than the other pile. which makes me wonder if the pack is missing a card. maybe i'm just losing track of if i've put one in a pile or not.

[@Quintingent]

Okay, I'm going to attempt to prove this in this comment. We start with an equal number of red and black cards. We shall call this number *X*. Then we do the dividing into two piles. Lets look at the 'red' pile. Suppose that *n* cards end up in that pile. Then *X* - *n* cards are in the 'black' pile. The reason for this is that there were 2*X* cards initially. Half of those are the 'deciding' cards, leaving *X* cards in the 'red' and 'black' piles. Thus these piles are size *n* and *X* - *n*. This also means that between the two piles there are *X* - *n* red cards and *n* black cards. Let the number of red cards in the 'red' pile equal *r*. Then there must be *n* - *r* black cards in that pile. Since there are *n* black cards between the two piles there must be *n* - (*n* - *r*) = *r* black cards in the 'black' pile. Thus the number of red cards in the 'red' pile equals the number of black cards in the 'black' pile. However, this is before we randomly select cards to be swapped between piles. Let's suppose each pile swaps *s* pairs of cards with the other. This can be simulated by swapping individual pairs *s* times. Thus, if we can show that swapping a single pair does not change the fact that the number of red cards in 'red' equals the number of black cards in 'black' then swapping any number of pairs won't change it. There are 4 outcomes for swapping a pair. Red gives a red, Black gives a red, or RrBr, as well as RrBb, RbBr, RbBb. If both swap red, or both swap black, then the number of reds and blacks hasn't changed for either pile. If RrBb occurs, then the number of reds in 'red' diminishes by 1, as does the number of blacks in 'black'. Thus the difference remains the same. If RbBr occurs, then the number of reds in 'red' increases by 1, as does the number of blacks in 'black'. Thus the difference remains the same. Therefore after any given pair swaps, the difference between the number of reds in 'red' and the number of blacks in 'black' remains the same. Thus at the end of the trick, the number of red cards in the 'red' pile equals the number of black cards in the 'black' pile.

[@Pommesflasche]

+David Ward Very nicely put into words!

[@igNights77]

+David Ward Your reasoning is flawless, nice post. I ended up doing almost the same. To simplify a bit, this number that you call X, you could just call it 26. You lose a bit of generality, but people that have trouble reasoning with letters for quantities will understand what's going on a lot better IMO.

[@thelatestartosrs]

+igNights77 but it works with 2 cards and 10000000 cards. as long as half are red and half are black.

[@brianrichter8255]

If you have X face-up red cards and Y face-up black cards, X + Y = 26. You also have X face-down cards on the red side and Y face-down cards on the black side. Suppose all the X face-down cards on the red side are black and all the Y face-down cards on the black side are red. There are X face-down black cards (on the red side) and Y face-up black cards, and Y face-down plus X face-up red cards, so everything adds up. This meets the condition, as there are the same number of face-down red cards on the red side as face-down black cards on the black side (namely zero!). Any other arrangement of face-down cards can be obtained from that position by swapping one or more cards between the two sides. Swapping a red-red or black-black pair doesn't affect color distribution. If you swap a red-black pair you either subtract one red from the red side and subtract one black from the black side, or add a red to the red side and a black to the black side, so the condition is still met, QED.

[@thee0581]

Just consider each pile (6 in total) as a variable in a system of linear equations where you use that you know that the total number of black and red cards of all piles together is the same. Also use that each sum of the lower two piles equals the pile on the top. Write this down, solve the equations and you have proven it.

[@Rote2Zwei]

Step 1: Get the solution Step 2: Shake head because its so simple but fascinating cool video!

[@luketimothy]

Are you ever going to explain the near-impossible matching matches card trick?

[@standupmaths]

Probably not. If I do any more of the same style, I'll label them a bit more clearly.

[@andreathecat100]

Nice! Call x the number of red cards in the face up pile, and y the number of face up black cards.You have 2x+2y=52. So that x + y = 26. Call rx the number of red cards in the pile below the red face up pile, and bx the number of black cards in the same pile. Same for the face down pile of cards below the black face up pile: ry is the number of red cards and by is the number of black cards.Of course ry + by = y so you get x + ry + by = 26.Also all the red cards are 26 so you get x + rx + ry = 26.Subtracting the two equations you get by - rx = 0, that means rx = by. The number of red cards in the face down pile below the face up red cards pile is always equal to the number of black cards in the other face down cards pile.

[@abdullahfaisal2290]

You're show is so good....you deserve more subscribers:)

[@jumbo6498]

Actually, I will subscribe because I appreciate quality content.

[@tfwmemedumpster]

a question: In Italy we use deck of 40 cards (1-7+jacks, queens, and kings) instead of 52+2 jokers will the trick always work with that deck too?

[@TheEpicEraser]

This one is easy. Since there's en equal number of black and red cards in the open piles, there's an equal number in the closed piles. A card can only be black or red, so the number of black cards in each pile is "pile size" minus the number of red cards in this pile. Combining all that, the claim reduces to the number of red cards in both piles combined being equal to pile size, which, since the number of black and red cards is equal, is trivially true.

[@undergroundmonorail]

"What we perceive as random often isn't truly random." but teh penguin of d00m is still random right

[@standupmaths]

Don't start. Put down spork.

[@undergroundmonorail]

regardless of how random that is, i feel like python's `random` module is pretty random and for lack of a real deck of cards that's what i used and the trick still worked. (obviously, since it's a mathematical thing, but there's some testimonial for you :))

[@sirjohnnyg]

The top two piles must add to 26, but so must the total red and total black cards (and also the diagonals but I didn't end up using that below). Once you "choose" a number for a top pile and a number for a color in either bottom pile, the rest is determined by these relationships. - If there are N red cards face up (top left), then there are (26 - N) black cards face up (top right). N could be anything from 0 to 26. - There could be any number less than or equal to N of red cards in the bottom left, let's call that M (could pick either pile and either color if you like). So the black cards in the bottom left would then be (N - M). - Since the black cards must sum to 26, we can take off the (N - M) from the bottom left and the (26 - N) from the top right. Thus there are 26 - (N - M) - (26 - N) = M of them, which is the same as the number of reds in the bottom left that was randomly chosen.

[@Cyntaxe]

Seems to only work if you use the randomness of the deck itself to determine the two piles. Tried it four times with using the deck to determine the piles, was successful each time. However, just to see if it was something with the deck, I decided to use random.org to determine which pile the card would go in, evens in one pile, odds in another (and also how many cards to swap). I've tried it seven times with this method, and not once has this worked. Really interesting trick. I'll be trying to figure this out all day, thank you! =)

[@aljawad]

I didn't try the trick, but I can follow the logic: first the two revealed stacks of red and black account for half the deck (13 each of a solid color). Next the "random" swap is something of a subterfuge, the two hidden piles account to the remaining half of the deck (13 & 13) of mixed colors: where the number of reds in one (call it X) means there are (13-X) blacks in the same pile. But the remaining number of red cards in the two piles is 13, so in the other pile the number of reds is (13-X), and the number of blacks is X.

[@justinwhite2725]

wow. I didn't have a standard deck of cards but I used the otherworld cards from call of cthulhu (equal number of red, yellow, green and blue cards totalling 48 cards). I made one pile red&yellow and the other pile blue&green and the trick still worked (I wound up with 2 cards of the correct co!our in each deck)

[@pihungliu35]

My attempt to prove this problem, no case-by-case analysis required: Put black index cards to red pile, call it "Black group"; put red index cards to black pile and call it "Red group". It is easy to see that black group and red group have same amount of card, namely half of a deck. The equal numbers in the trick now become # of red cards in black group and # of black cards in red group. Since we begin with equal number of black and red cards and ends with equal number of cards in both group, the number of "outsiders" (red cards in black group / black cards in red group) should be equal.

[@1johniq22]

I don't know the math, I think I have a fair idea of how it works. 26 cards face up, if there are 15 red cards face up, there are 11 reds face down; which means 11 blacks face up and 15 blacks face down. Above the red would be 15 unknown cards, a maximum of 11 red, which would leave only 11 blacks in the 'black unknown' pile. That's as simple as I can explain it, but it doesn't count as a proof so far as mathematicians would be concerned. This'll blow my little cousins minds though, thanks! -New Subscriber

[@davidmurphy563]

Yes, it works every time. I'm going to have a go at explaining it - I'm sure the correct answer is already in the comments but I'm deliberately not going to read them for sport. I'm a layman and simply don't have anything like the mathematical education to provide a proper proof so I'm just going to try to use logic. Matt said that it was equally likely that the face down card would be black or red - but is this true? In the unlikely event the cards were arranged black, red, black, red then the average sequence length (the number of reds or blacks in a row) in the deck would be 1. However, this is the only case once in a ridiculously large number of outcomes where it is so low, in every other case it is higher. Knowing the distribution curve for sequence lengths would need maths which are way better than mine so I won't try. Plus, even though there is a probability curve, the average length each time will be different and so is effectively random - it's just more likely to be some numbers than others. So, let's instead pick an average sequence length of 5 (out of the hat to be high but not inconceivable) and see what happens. Well, this means that even though the cards are random, because on average the sequence length is 5 you have effectively the same overall result as arranging the cards five red, five black, all the way through (even though in reality it's 3 black, 8 red, etc). Well, in this case it follows that it is much more likely that a red will be followed by a red. Again, I'm lacking the maths to describe this but it 'sounds right' to say that with every card there is maybe only a 1 in 5 chance that the next card will be with a black. It's much greater than 1:1 as Matt claimed is the point and the chances of a red being followed by a red are precisely equal to the chances of a black being followed by a black. Here's the rub: whatever the sequence length for red, it follows that black is governed by the same sequence - given they're in pairs - something which exchanging cards randomly at the end doesn't change. That is to say, exchanging cards between the packs doesn't affect the statistical distribution between the packs. Therefore, although different average sequence lengths will yield different numbers of reds and blacks, those numbers will always tally.

[@klopo333]

Love how the video is 50 fps. You're british in all the strangest ways.

[@standupmaths]

50 fps 4 eva.

[@GoldphishAnimation]

+Ccidaho Crafter woah

[@essentialatom2]

+haved Except for being Australian ;)

[@Pendoza84]

Subscribed cause of "Hello, Michael here!"