Shuffling Card Trick - Numberphile

[@MyRegularNameWasTaken]

I saw this trick in a magic trick book once, but the book's explanation for why it worked was essentially "It's just a mathematical fact that this works." I'm very happy to see an in-depth explanation for this.

[@shruggzdastr8-facedclown]

Mad kudos to Jason for showing off his upside-down and backwards writing skill in this video!!

[@tristanridley1601]

Actually more impressed by that than the maths lol

[@aNytmare]

The most impressive trick I saw is the Upside Down writing :-)

[@snowfloofcathug]

I can write upside down with my wrong hand, it's not super difficult

[@gui1521]

+Nyt Mare Even more impressive when you already struggle to draw a "spade" or a "club" under normal conditions...

[@Marz2695]

+Nyt Mare Or you could just write 'h pow' E pow '2 pow '1 pow. Look at it upside down

[@leppie]

+Nyt Mare I could read upside down at one stage (havent tried in almost 20 years), but riding a packed subway train/tube teaches you that :D

[@Marz2695]

+Zakatos °°°6 pow' 8 pow' L pow '9 pow

[@garrick3727]

Many many years ago I had a Paul Daniels card trick pack of cards. It was a pack of special cards and a little booklet of maybe 25 tricks. The cards were special in that they were slightly narrower at one end, so you could do a lot of tricks involving getting someone to pick a card and then making sure it went back in the pack upside down. So people could shuffle the cards, and as long as they didn't drop them on the floor, you could still find the chosen card. However, my favorite trick did not use the altered cards property, it just used the ordering of the cards presented here. And even though it was 40 years ago, I never forgot that order - a clear advantage of using a saying as a memory aid. What I didn't know, and learned from this video, is that order is "famous". I assume that means Mr Daniels did not invent it, which - if true - is a little disappointing. Edit: I just Googled it, and apparently it comes from a book from 1902 and may even be older. Mind blown.

[@boumbh]

"Consecutive even though they are mixed up." Jason Davison, 2016. This trick is amazing! Thank you so much!

[@Shilag]

The thing I'm most impressed by is him writing upside-down. What!

[@Marz2695]

+Shilag Or you could just write 'h pow' E pow '2 pow '1 pow. Look at it upside down

[@SteveDice21]

This guy looks like he trains dragons

[@omenakookos]

+Steve Dice Yeah, looks like the dragons tails have hit him in the nose more than a few times

[@misaeldavid6593]

wtf? hahaha

[@Triantalex]

false.

[@vaanar843]

😂😂😂

[@BuddyCrotty]

That's amazing! I don't think I could write upside down nearly as well as Jason.

[@Laborejo]

the real magic here is that he writes upside down on the paper

[@barutjeh]

+NoneG3 not to mention the way he holds the sharpie.

[@dakotacx]

For those interested, there is a book called Mnemonica which captures the sequenced deck and allows a magician to do many things like this. It isn't quite as mathy, but it is very interesting and I suggest you check it out.

[@Uejji]

In the comments: people complaining that the so-titled "card trick" is a trick.

[@SathvickSatish]

Iykury haha

[@DCROSS67]

This is one of the better non skilled card trick I've ever seen.

[@martinlarge2614]

Brilliant explanation. I have known this trick for some years but didn't really understand how it worked. I also didn't know that if it was set in a sequence of thirteen there would be one of each value. Before the cards are counted and shuffled the pack can be cut as many times as you like, which doesn't affect the outcome.

[@SoopaPop]

Ah yes, Brady's fabled Speedmaster. What a sight to behold 0:35

[@CaballusKnight]

If only our math high-school teacher came with a pack of cards when he taught us Permutations. All math teachers need some course on how to sell their merchandise. Cause if they're just writing letters and numbers on a blackboard, kids will look for the nearest exit. Thank you Numberphile, bringing cool thematics back in mathematics. (thematics is not a real word, the youtube red line says so, I imagined it might be. Now I just squared it and looks negative)

[@eeple29]

Someone please correct me if I'm wrong, but to my understanding: The trick requires the deck to be prepared such that each pair has R & B; each foursome has D, C, H, & S; and each 13-some has A-K. The cut, count, and riffle shuffle will always rearrange the cards in an order that also conforms to those patterns. So theoretically, can't the trick immediately be performed again without having to re-prepare the deck, since it's still in an acceptable order?

[@jones1618]

Right. Those properties are preserved but the initial countdown & cut would break those properties if the stacks were just put together. Many tricks that use this principle use a sneaky step to reverse the top stack again so the properties are preserved.

[@original1up]

just did it at home and it worked. I am pretty impressed :D

[@Betacak3]

I'm at 0:06, I don't know what the trick is about, but I can already see that the cards are not in a random order. They're RBRBRBRB all the way through. I think I've seen too many of these by now =p

[@HotDog-yf2je]

+Buttercak3 yeah, every card trick is set up , yeah they came out as you said, but i could only take and shuffle the cards as you said, i dont find that impressive , in math it is, but as a trick

[@jonas1015119]

+Buttercak3 Hallo :D

[@niklasgransjen684]

+Buttercak3 I don't even think about it being to trick the person you show it to, I just find it impressive that the card doesn't "mix"!

[@georgeking7438]

+Buttercak3 that is literally the trick. That is what the goal of the trick is, he was just being sarcastic

[@veeryahoo5215]

To the people who are wondering about the Upside down writing.... Jason is left handed...Few of them write everything upside down...bending their hand other way unlike right handed..so, he can easily write upside down with straight hand..

[@axellno1759]

Thubs up for upside down writing! ..and for the matching hair color to the sweater and for the whole video of course. :)

[@igorvieira344]

love this math with cards and shuffling videos!

[@laserbam]

Great video!

[@Musabre]

Busting this one out next time i'm down the pub. What could possibly go wrong.

[@ThisNameIsBanned]

Writing upside down seems quite a feat , probably valuable for a magician !

[@barnyardhouse]

I've been listening to Jonathan Strange and Mr. Norrell! It's so great!!

[@blazejecar]

numberphile....I have 2 questions maybe you could make a video about... 1: if pi is infinite and can contain any number imaginable....can it contain another pi? Because if it can, it's not non-periodical, as pi would be repeating itself. But if it doesn't contain another pi, it means there's a number combination into found in pi, which means pi isn't infinite. Either way you choose, one half of definition of pi doesn't work... 2: say you have a graph x^∞. What happens? ok for f(0)=0 and for f(1)=1 but what would happen in the negative quadrants? is it treated as an odd or even exponential function? is the limx->-∞ + or -∞?

[@beeble2003]

+blazejecar 1. It's not known whether pi contains every _finite_ sequence of digits. However, it certainly doesn't contain every infinite sequence of digits and, in particular, it can't contain a copy of itself. If, for example, the sequence of digits from the k-th digit onwards was the same as the sequence of all digits, then the sequence of digits would have to repeat every k digits. But that would mean that pi was rational and we know it's not. 2. x^∞ isn't well-defined so you can't graph it.

[@3Ppaatt]

+blazejecar Pi has an infinite number of digits, which means an infinite number of places a sequence of digits can start. The problem is that there are a "countable infinity" of digits in Pi, that is, you can line them up. They go on forever, but they can be put in an order without missing any. The reason you can't find every real number in the digits of Pi, is because there are an "uncountable infinity" of real numbers. That is, you can't even put them in any order before you skip most of them. This is proven by Cantor's diagonalization. So there are too many real numbers for each of them to appear in the too few digits of Pi. The part of your argument which falls through is that Pi _can_ be infinite _and_ still not be periodical _and_ still not contain every number combination.

[@Ddub1083]

+Ppaatt note... just reread your comment and realized you were saying pi was countable... misread. carry on :) lol no... any PARTICULAR decimal number (like for instance pi or 22/7) ALWAYS has a countable set of numbers after the decimal. To be a countable infinite set one must only be able to map each number in the set to an injective function. Since there is an nth number for every n in the set of numbers after the decimal in pi, even though its infinite, it is a countable infinite set. Compare with the numbers between 0 and 1... if we start at n1 0.00001 and go to n2 0.00002, there will always be MORE NUMBERS between those two n's and therefore the set is infinite and uncountable... while pi has PARTICULAR numbers, it is necessarily countable, and infinite.

[@Ddub1083]

+Jared Thomas Perhaps it is you who needs to reread... first line: "note... just reread your comment and realized you were saying pi was countable... misread. carry on :)"

[@ASOUE]

For number 1: pi can contain any digit, aka 0-9 For number 2: infinity is a concept, not a number, therefore that equation is impossible.

[@AGrayPhantom]

The explanation goes over my head very quickly.

[@K0ggy]

Am I right that this isn't possible with a 32-card set (7,8,9,10,J,Q,K,A)? Because every 8th card had to be the same number and also the same color (because 8 is a multible of 2) which is not possible. Or in general: It isn't possible if the different kind of cards (numbers) are a multible of 2 (or 4, which is included in 2). Or is there any other permutation for these cases to make the trick work?

[@leohalb]

The most impressive part in this video was when he began to write upside down.

[@Mixa_Lv]

A hip looking guy on Numberphile? What's this? :p

[@ikasu00]

He's not a professor?

[@black_platypus]

+Mixa As long as there are enough people around using words like "hip", it all cancels out, don't worry ;P

[@awsomebot1]

+Mixa Dr. James Grime (aka singingbanana) is pretty hip tbh.

[@oldcowbb]

so many card trick this month

[@atorrance]

Does anything get messed up if they initially deal down an odd number of cards? If they dealed so they had to shuffle 1 pack of 25 with 1 pack of 27, could that mess up the ordering?

[@pantouffle]

Shuffling cards will never be the same again.

[@jvnsfunny]

This made me smile :)

[@RichardDominguezTheMagicIsReal]

I have already worked out a presentation and this is going to be a featured effect in my close-up act, ty

[@maxonlinesuccess3934]

Great video. You make maths great!

[@jnewb1]

love numberphile and hello internet!

[@logicalfallacies3352]

Could you make a video explaining why the digit sums of cubes follow a pattern of 1, 8, 9?

[@rasowa2958]

+Jacob Peacock Got my vote. That would be interesting. I wasn't aware of the pattern.

[@mvpsarah]

That upside-down writing tho! Like a Baws! he doesnt even flaunt it either... that blew my mind...

[@igorvieira344]

but it depends on the riffle shuffle, doesn't it? shouldn't 7 riffle shuffles make the deck almost random?

[@avskrap]

Nice Speedy Pro there, Brady!

[@tennison-chan]

And ... Every 52 cards, there are 4 aces! Strange!

[@thefauvel7558]

Indeed.

[@funny_monke6]

+Tennison Chan Not only that... There are 4 of *every card!*

[@thefauvel7558]

funny_monke6 Coincidence? I think not.

[@raymondfabi1140]

Tennison Chan 5o

[@komami1006]

this is amazing👏👏👏

[@venkatbabu186]

Numbers are like three planes geometry. 8 in each plane vertical to one another. Shuffling is the varieties. Split at half the deck.

[@papergamesproductions]

1:00 those cards have NO runs of 3 or more, Witch isn’t expected from randomness

[@logster5670]

Can you post how to make an elliptical pool table/loop table or post the dimensions, please?

[@jasond9304]

+Numberphile, I have a question which I have wondered for a while now, and I was wondering if you could calculate the total number of possible key combinations that are possible on a piano if a human had infinite hands. (So of the 88 keys how many ways are there to play the notes together (and singularly) in any combinations up to all 88 keys at once. (e.g all 88 keys, all keys excluding g7 and b3, no keys, and so forth for all the other combinations which are possible.))

[@Forchune]

Your piano keys have two states they can be in, pressed or not pressed. 1 or 0. You can simplify your question with binary! With binary counting you can get every possible combination. So let's look at just 2 keys, the combinations are 01, 10 and 11. Since you mentioned that no keys being pressed should also be counted we also add 00 as a combination. So for just 2 keys the max combinations is 2^2=4. Now we scale it up. 88 keys gives us 2^88 possible combinations.

[@WrightMaths]

It is a great channel...

[@prasetiobudiono9424]

Never knew Harry Kane is a math and cards enthusiast

[@chriskennedy00085]

pls tell me what does it imply when we get 'negative' area as an answer to a question in maths ?

[@bvlampe6801]

So after the person took out some number of cards and shuffled them in that way, they should still cycle through correctly, which means you should be able to repeat that step as many times as you want, right?

[@dexter9313]

+Peter Schmock No because the pattern will be different, even if some properties are conserved. The starting case is in a very particular order.

[@ATRonTheGamer]

Maybe I missed something but what prevents someone from shuffling two hearts/two aces etc together upon that single riffle shuffle?

[@justinknutson1046]

+ATRonTheGamer It has to do with the fact that each card value (ace, 10, 9, etc.) is set exactly 13 positions away from cards with the same value. So even if you rigged the shuffling in such a way as to place two aces next to each other, they would always be in separate "chunks." With suits, it's a similar thing, except each suit is 4 positions away from similar suits. A way of imagining it would be so: Say you've reversed and split the deck, and are about to riffle shuffle the two halves together. You take a peek at the bottom card of the left portion of the deck, and see that it's an ace. So you plan on shuffling in cards from the right portion of the deck until you see another ace, whereupon you will have succeeded in "rigging" the shuffle so that they are next to each other. What you'll find out, though, is that it always takes twelve cards before you find another ace - meaning one ace is in position #13 and the other is in position #14. Even if you shuffled a few cards before choosing one to try to rig it with, the total will always be 13. The best way to understand this is to just take out a physical deck and try it. it might make the pattern easier to see.

[@shaded_scars8223]

Kakegurui has taken me here. (Trying to do the gilbreath shuffle just to be like Runa)

[@HunterJE]

"for _a_ sick knave" (A = Ace) works too

[@OriginalPiMan]

and this is why you should always shuffle multiple times, and preferably with multiple shuffle methods.

[@robin888official]

If those modulo-properties exist before the "Brady-shuffle" and persists after it couldn't Brady have done it repeatedly? And if cutting doesn't disturb the order isn't any sequence of cuts and "Brady-shuffles" doing the trick? Would make it more impressive. (Even more as it already is!)

[@Dreadpirate404]

No, the properties do not entirely persist. I'm no math expert, but I know magic. Shuffle two or more times (7 for perfect randomization) it will mess up the order.

[@robin888official]

+Robin Koch Ok, I tried it. It doesn't work, indeed. It turns out, that the inertial properties are stronger then the one after the "Brday shuffle". Before it the colors, suits and values repeat *in the same order*. After it they don't . Therefore the whole modulo-trick doesn't work a second time. But it's cool anyway. (I had little hope one won't have to sort the deck before every execution. ;-))

[@metallsnubben]

+Robin Koch I guess one thing that would work at least is to cut the deck as many times as you want, since that only means you're starting at another point in the "cycle"

[@alexbabits770]

What is the probability that I shuffle a deck of cards and there are no pairs together? (eg. 99/88/22.... next to each other?) I've tried this a ton and always get a pair!

[@gregorioteran263]

I am not sure if this is a question that has been created (yet), but... "If a road was pi meters long, would it go on indefinitely or would it end?"

[@stevewright8787]

It would end, however we could never know at what point.

[@Ddub1083]

+Slenderman Greg Both... it would end at a particular length... but that length would be indefinite.

[@MMmk1]

Do the cards have to be in the Kings threaten some sick knave something in the beginning?

[@herseIf]

+matekusz1 No, but they look properly shuffled like this, and it's easy to remember. If you do it randomly you might make a mistake (any other sequence would work though, you can even do 4 different sequences if you want).

[@dew9103]

Talk about the gilbreath principle

[@jwso]

What watch is Jason wearing? Missed the entire video because I was trying to find a good shot of it.

[@retepaskab]

I don't get it, how do you ensure that no two same colours are shuffled next to eachother?

[@Kazutoification]

Paused at 0:03, I see a pattern. I'M SUSPICIOUS!

[@RDL15100]

The pattern is the Eight kings threaten thing

[@dominiclynch5618]

I have a question for you, can you actually divide 100 by 3, I have been told there is a way to do it with remainders but I still believe that it's impossible.

[@heimegut6133]

if you do it with remainders you'll be left with 33 and 1 remaining

[@robertfontaine3650]

That hurt my head. I'm going to have sit down and think through why the shuffle doesn't place even 2 blacks or 2 reds together.

[@DraganAlves]

+emsaaron if left half of the shuffle has rbrbrb, right half has rbrbrb, and they shuffle perfectly, you'd have rr, bb, rr and the trick fails. So I don't think your explanation works.

[@idokatz6808]

i was riffleshuffling the other day and thought, how many perfect riffleshuffles can you perform on a deck of cards until it returns to the starting point. So i wrote a small program to calculate it, and found some interesting results: first of all, for 52 cards, the answer is 8. next, the numbers had no clear consistency. i tried to find any sort of formula to calculate it but the results seemed pretty random to me, Except for powers of 2. for any number 2^n, the result is n. I've tried for a couple of minutes to figure out why but didn't seem to get anywhere. And so I turn to you Numberphile, oh lords of the mathematics, I have results, yet no conclusions, and it would be amazing if you could get to this topic because I'm really interested.

[@haoli5733]

+ido katz step 1: number your cards with binary step 2: for 2^n cards, a perfect riffleshuffle is a circular right shift operation on all the card numbers, take n=8 for example: 11111111 stays the same, 11111110 turns to 01111111 step 3: n bit number returns to original value after n circular shift

[@WalkIntoMyFace]

Look up how to do a faro shuffle. It's all rather simple but fascinating

[@andrewzuo86]

Any sufficiently advanced technology is indistinguishable from magic.

[@JaySay]

You know guys... this is numberphile... every card trick will be math based and meant to just "trick" you, not amaze with smoke and juggling blades... I mean, for christ's sake, the man doing the trick is a mathematician, not a magician! He's no mathemagician! He's a MATHEMATICIAN!!! (Sorry, Just realized that mathematician and magician rhyme)

[@vitalisacharenko4273]

This trick does not work with decks of 32 cards (7 up to Ace) (very popular in Germany) :-( but you can (kinda) still do it is you leave out the mod4 part of the trick...

[@gsurfer04]

5:31 Fitting music is fitting

[@Ajax-0137]

Nyquist-Shannon's theorem video please

[@cedricmael704]

everyday I´m shuffling

[@harshalkulat7997]

Does this trick work with Si Stebbin's stack?

[@vasodestarbuks]

Gracias johnny rotten!

[@theelephantintheroom1055]

Would this work with the standard CHaSeD deck setup of increasing by 3 each time?

[@vilmarci]

Nice trick :) Just wondering, is it possible to adapt it to a 32 card Hungarian deck? Basically the same, but cards from 2 to 6 are missing.

[@JaneHowitt]

Arghhhh! My brain hurts :-( But it's SO clever, I can't stop watching...

[@DiegoTuzzolo]

Didn't uderstand why (something I dont know the name) of 12 is 2, and of 6 is 1. What is the name of this "function" and how do I represent it? Thanks

[@rosebynoothername]

The function is 'modulo' or often just 'mod.' It gives the remainder after dividing, so 6 modulo 5 is 1 because 6 = 5*1 + 1, and 12 = 5*2 + 2 so 12 mod 5 is 2.

[@AcheronInTheUK]

+deigo tuzzolo Modulo arithmetic is the same as dividing, but we only care about the remainder after all whole number divisions. 12 modulo of is 2, because 5 goes into 12 twice (which we ignore) and leaves 2, which 5 does not go into. Similarly, 5 goes into 6 once with 1 left over, so 6 modulo of 5 is 1.You notate this as, for the 12 modulo of 5 example, as 12 mod(5)=2Here is the wikipedia article on the subject, which no doubt explains it better than me en.wikipedia.org/wiki/Modulo_operation

[@dogeeseseegod5450]

modula of a number is the remainder after you have divided by that number. modula 5 of 12 is 2 because 5 goes into 12 twice but then had a reminder of 10. modula 5 of 11 is 1 because 5 goes into 11 twice but has a remainder of 1. the remainder is what you are concerned about, however I do not know the notation I'm sure it can be found in Google

[@BlueishBefore]

+deigo tuzzolo It's the modulus function, usually (in programming anyway) represented by %. 12%5=2 since the whole remainder of 12/2 is two (12=5+5+2). 6%5=1 since the whole remainder of 6/1 is one (6=5+1) 12%4=0 since there is no remaining whole number when 12 is divided by 4 (12/4=3 exactly)

[@K0ggy]

+deigo tuzzolo Its called modulo 12 modulo 5 is 2 6 modulo 6 is 1 modulo means the rest of a division, so 12 modulo 5 means the rest of 12 divided by 5 -> 12 / 5 = 2 with rest 2 or 12 = 2*5 + 2 thats why 12 modulo 5 is 2

[@MrRachter]

Does this also work with the Si Stebbins stack?

[@griffcore]

+Leonard Dobre Yes it does. I thought that was the stack he was going to say he used. Much easier to remember in my opinion.

[@DarkAmikari]

So that's how Matt did the perfect separation trick..

[@pbgrafx]

The Gilbreath Principle

[@Wargon2013]

Was about to write a comment about not starting with 0 when numbering something, then I realized its Numberphile not Computerphile^^

[@hendrik7368]

Nice video

[@crazydrummer4827]

+Jonathan Krillington How can you know? You just wanted to post a comment first, but not write first. It was uploaded 2 minutes ago and it is long 13 minutes so how can you know it is nice?

[@dj-up4nc]

+Jonathan Krillington LOL GET PWNED

[@samuelabreu4349]

+Crazy drummer it is

[@crazydrummer4827]

Samuel Abreu I honestly did not know was it good video, because I came 2 minutes after upload and commenter at that moment, but you kinda missed my point.

[@SpitefulAZ]

Do a video about rubiks cube!

[@dizont]

dont get it when u mix up the cards, how its not possible to 2 reds from different decks to be one after another ??

[@SomeLoops]

It is possible, but never three, and you will always get something like: br/bR/Rb/rB/Br So the two red or black cards will always be split up, so they are in a different pair each.

[@dizont]

ahh right right right ! :D u filled the hole in my brains, thnx

[@SomeLoops]

Ergo Proxy haha, you're welcome

[@drsusredfish]

Here after watching some anime about gambling on netflix.

[@wtfiswiththosehandles]

What kind of sorcery is this??

[@xmagistrtyx6064]

Numberphile i need an answer! what is :- Infinity * 0 = ? infinity / 2 = ? infinity / infinity = ? infinity / 0 = ? Hope you can answer it ! - The Magistrist

[@aleksandarprodanov4454]

+X Magistrty X Infinity isn't actually a number, so you can't do any math operations with it!

[@xmagistrtyx6064]

But think it as it for a while so what can it be?

[@daleftuprightatsoldierfield]

0, infinity, 1, undefined

[@RoelfvanderMerwe]

Nice Speedmaster

[@gordonfrohman250]

didnt get what he said at 8:36 "so to do this trick, we're going to (unknown word) the cards.."

[@SanctumZero]

Sounds like "celibate" >_> but that doesn't really make sense

[@nevillestyles]

I believe "setup the cards"

[@SanctumZero]

Ah, yea, that's it 8D

[@FrodeJohannessenSlartibartfast]

ehhhhhh..... what?! I must watch this a few times!

[@Samyrax]

But can he do it on a rainy night at Stoke?

[@B3Band]

This guy looks like he's gonna try to sell me his mixtape

[@DanDart]

I used to do the one where you added 3 so A 4 7 10 K 3 6 9 Q 2 5 8 J

[@DanDart]

(not the same trick just the same stack type)

[@damienw4958]

Brown paper???

[@nikkehautapelto1323]

cool

[@syfqzk7249]

neat!

[@rdoetjes]

gilbreath... Immiediately when I heard deal down and riffle shuffle I knew it was about Gilbreath This one made famous by Leannart Green

[@Sonicgott]

Props for upside-down writing.

[@georgavagyan]

What’s wrong with the English subtitles?