James ❤️ A Card Trick - Numberphile

[@trdi]

Matt Parker came with a very similar trick few years ago. The only difference was that his version sometimes works and sometimes doesn't.

[@funbiscuit]

Parker jokes will never get old. Except that sometimes they are.

[@mynewaccount2361]

29 1 47 41 37 1 23 41 29

[@pinaz993]

That one took me a bit to get. Nice subtlety. 10/10, would chuckle again.

[@xyz.ijk.]

Poor Matt ...

[@ESL1984]

@@funbiscuit Once they do, they become a Parker joke of a joke.

[@Abdega]

Cartoon James looks like he just rearranged something in my house in ascending order and is waiting for me to notice what it was

[@rucker69]

I'm going to have nightmares of cartoon James pulling an infinite number of rabbits from a hat.

[@3ckitani]

When people are counting sheeps, you'd be counting rabbits.

[@soleilvermeil]

But have the rabbits to guess what the color of the hat is before he pulls them out ?

[@nathancoulombe6313]

but it's only a countable infinity!

[@GrandMoffTarkinsTeaDispenser]

Sweet dreams*

[@ericbell7]

Someone will do a endless .gif :-)

[@nikanj]

2:59 Did James do something to upset the animator?

[@Adam-pv4qn]

Lol

[@courtney-ray]

😳 my thoughts exactly! 👀

[@GeodesicBruh]

Lol

[@Triantalex]

??

[@jibbiddy]

James: *Goes on Penn and Teller "Fool Us." does trick fools Penn and Teller but immediately explains the math behind it.*

[@toom-zm4bc]

I think that the beginning and the middle are great, but you could improve the "prediction". For example, you could put the James of hearts in the 25th position in the deck and you leave all red aces until nines beneath it. So when you are subtracting the values of the 5 pairs, you just take cards that are below the James of hearts. He will then be in the 25th position after taking the red cards out and you can say that James of hearts knew it and chose that place. If you really want to improve it you could also learn some false shuffles, so you leave James of hearts in the 25th position from the top and do some false shuffles, so that the spectator thinks its totally random where James is. Don't forget that before showing the 25th card, you should remember the spectator that the deck and the spades were shuffled and that he chose which cards would be his and which cards would be yours. You could also let the spectator choose what suit you will use for the trick, so they believe even more that they chose everything. The only tricky thing would be to arrange the deck not knowing what suit they were going to choose, so you would have to arrange it after they said it. Hope you could understand everything and if you have any questions,feel free to ask

[@SVNBob]

Easier method would be to put down 2 cards as the prediction under the James of Hearts: the 2 and 5 of clubs.

[@toom-zm4bc]

@@SVNBob it's not a bad idea, but in my opinion the effect is better if it is in the 25th position

[@karlgiese6100]

Or you could let the spectator put the card where they want, shuffle the deck, and then you do a deck switch.

[@toom-zm4bc]

@@karlgiese6100 that's nice, but a very hard thing for a non magician

[@kirlian5399]

@@SVNBob if you do it with a 2 and a five you could do a riffle force and let the spectator pick the top 2 cards from where you cut the packet (or any other kind of force) This also has another advantage: you can let the spectator pick the predictions from the beginning and put them in plain sight, so the spectator doen't think you did something tricky before showing them.

[@fakjbf3129]

I love the framed section of brown paper from the Graham's Number episode!

[@roryburch861]

So the video goes into this at length, but letting the the participant choose 10 cards, show you and then letting them divide into piles of 5 each. You write down your prediction then, turn them over and order them, find and sum the differences. Letting them have complete control over the cards is usually really impressive for a 1 on 1 trick.

[@brettonjohansen1619]

I appreciate the penrose tiled card backing.

[@georgew.9663]

Whoever drew the cartoons did my boy James dirty, they did him dirty they did

[@scottanderson8167]

Grimy is the best Numberphile by far.

[@sunnzboz9482]

Reminds me of the story that Gauß as a child was ordered by the teacher to summarize all numbers from 1 to 100 to keep him busy. After a minute little Gauß came back with 5050. He arranged the numbers from 1 to 100 like 1+100+2+99+3+98+...49+52+50+51 = 101*50 = 5050.

[@DrJackaloupe]

Wait, so this just boils down to the associative and communicative properties? If the smalls are always negative (subtracted) and the bigs are always positive, then it doesn't matter what order they're arranged in, of course they'd come out to the same value. The way they're dealt only affects one meaningful thing in the whole problem, and it's the sign applied to each number, but because the way they're dealt and then ordered, you're guaranteeing the big-small pairing and therefore guaranteeing the signs of the numbers. Crazy how simple the math is once you strip it down to the basics. Great presentation here.

[@Lord_Skeptic]

It works the same with any even number of card. Difference between them. 2 = 1 4 = 4 6 = 9 8 = 16 10 = 25 12 (J = 11, Q = 12) = 36 14 (K = 13, Joker = 14) = 49

[@John_259]

At this very instant, Penn and Teller are shivering in their boots!

[@laurent221000]

Found a different, more graphical, but more complex way to prove it: Imagine the cards laying on the table in order. Then you mark half of them blue and the other half red(for the two sides). You know that the cards will each find a partner of the other colour and you know that they are going to start matching from the longest distance to the shortest(if you do the counting in the same order as they did in the video). We will count the connections between the cards for the result. So lets start with an example: no matter what colour the Ace has, it´s connection will always go over the middle, because the other 4 spots between the ace and the middle are not enough for the 5 cards of the opposite colour to fill and the ace will connect to the highest of them. So at least one connection going between 5-6 from the ace. The same will happen with the 2: 3 spots left and 4 cards of the opposite colours to fill. So another guaranteed crossing over the middle(between 5 and 6) this works until we reach 5. So we know that 5 connection go over the middle point, resulting in a value of 5. The same game will work for the connection between 4 and 5, except for the last connection(with the 5 involved) resulting in 4 connections between 4 and 5. this goes down until we reach 1 - 2 so it´s 5 + 4 + 3 + 2 + 1 for one half. because the situation is symmetrical the total has the be 1+2+3+4+5+4+3+2+1 = 25 If we spin this further with other numbers than 5, we can explain why raising x in x^2 will raise the result of mentioned formula by 2x + 1

[@mohithraju2629]

This was a question in Indian RMO(Regional math Olympiad)

[@himanshu9559]

Can you elaborate please ?

[@SathvickSatish]

Justin Weaver wdym? There is a competition called RMO and this was a question apparently

[@andresmartinezramos7513]

I was going to ask why would they have this kind of problem in a Regional Olympiad, then I realized that an Indian region probably has more population than my entire country...

[@GhostyOcean]

James showed how you can't have two small/large numbers paired up so you're always left with groups of (large)-(small). Using the associative property of addition you can rearrange the numbers to have all the addition of large numbers on one side and all the subtracted small numbers on the other side (treat subtraction as addition of negative numbers so the property holds). You'll end up with 10+9+8+7+6-5-4-3-2-1 =(10+9+8+7+6)-(5+4+3+2+1) =40-15 =25

[@kyazarshadala8114]

okay cool, but why did you make animated james nightmare fuel?

[@footballsoccer358]

Great! Used this on my friends, they were mind-blown when they saw my prediction card in his backpack (I put it in there beforehand)!! I put the wrong answer under the "J" card and told him to actually check his bag for my REAL prediction :D sneaky sneaky....

[@surfing_youtube]

This is the one of the best mathematical card trick I've ever seen.

[@UltraCboy]

2:58 Here’s your free Cartoon-James-pulling-bunnies-out-of-a-hat button

[@romajimamulo]

Thanks... I guess?

[@Eliza_Yump]

Take that thing away from me

[@vtron9832]

James grime still looks better in real life

[@QuasiELVIS]

Only just.

[@infoverflow5559]

This is how I understand it: By sorting decks in reverse order and by taking the difference of two numbers we undermine their separation in two decks, Those two operations just move all half high card in one deck and all others move to the other deck. The result is going to be sum(sort(arr)[n/2+1:n) - sum(sort(arr)[1:n/2])

[@profmda]

I liked this a lot not because of its magic trick feature but because it had an interesting fact about sets of numbers. As James mentioned the effect works no matter what the colection of 2n numbers is - and you can have repeats and non-integers as well. The answer is always the sum of the n big numbers minus the sum of the n small numbers. If there are repeats these two subcollections may overlap but it doesn't matter.

[@Vikash137]

I propose we make 'A James Heart' the opposite statement to 'the Parker Square'

[@soveu8237]

2:30 Is that Graham's Number on the wall?

[@KalOrtPor]

Written and signed by Ron Graham himself!

[@AlexKing-tg9hl]

I love you James! You’re the best

[@egorchik69]

I think the trick will be more amazing if a spectator could pick up random, let's say, 10 cards and after some mental calculation you can make a prediction based on those cards; and then you perform the trick.

[@TofranBohk]

You could have a card selected. Look through the deck to find all of a suit (not the same suit as the selected card) taking them out of the deck. As you look, you count down to the 24th card and catch a break. Continue removing the cards until you get all ten cards. Keep the break as you hold the deck in one hand. Cut the deck at the break and have your spectator place the card in the 25th place. Do a few false cuts/shuffles. Place the deck aside and do your mathematical bit like in this video. Then, miraculously, find the card in the position indicated by the sum you found!

[@leobirtwhistle]

Great video as always. The idea that a lot of magic tricks are dressed up mathematical effects is really intriguing, it would be interesting to see more videos exploring this idea.

[@Pietrac]

To make it more impresive you can handle cards to spectator at the begining and tell them to make 5 pairs. Also you don't have to only write your prediction, it can be anything from turned card in deck on 25th position to something like 25 cards left in deck (and others just be gone).

[@johncowne1442]

And what arrangement would give you five fives?

[@AshAquamarine]

The main thing you could do to dress it is not TELL them that you're using only the spades, infact I would use as many suits and colour as possible but do a false shuffle, stack the deck so that you get Ace through ten.

[@tomasbeltran04050]

I knew I had seen that singing banana before

[@michagrill9432]

Why was 6 afraid to go camping with 7? Because 7 1ted 2 bring 3 knives 4 sur5al, but 6 knew that 7 secretly h8ted him and did not have be9 in10tions

[@joshandrews8913]

What is this, a mnemonic for remembering the first 10 positive base 10 whole numbers?

[@michagrill9432]

@@joshandrews8913 Nope just a punny gag

[@yusuf-5531]

Oh dear

[@SathvickSatish]

Micha Grill that’s not bad

[@fernandoarraes8601]

seems like a pun Michael from Vsauce would make

[@Movie.Hammer]

Amazing! I've seen a very similar concept as a proof for some exponential equations using groups as multiplication

[@20DX00]

James and Magic? I think we all know who to call Brian Brushwood

[@haydenhoes12]

this was just after james and brian had an episode together on s̶c̶a̶m̶ ̶s̶c̶h̶o̶o̶l̶ scam nation.

[@neilsamuel5268]

It's easier to understand as a difference of magnitude So if we take big cards as >5 we have 6,7,8,9,10 And small cards would be 1,2,3,4,5 As red cards are difference in magnitude,so we can consider it's big card's minus the small card's number So big cards are always positive and small are always negative Hence 10+9+8+7+6-5-4-3-2-1=25

[@Sylocat]

If the numbers don't have to be consecutive, you can have the audience member grab any even number of cards from a single suit. You'd have to devise some way of knowing which cards were selected (marking them? sneaking a peek at the remaining cards?), and memorize some mnemonic to help you calculate the prediction quickly, then write down the prediction after the cards are selected and cover it somehow. This may be too much sleight-of-hand for a math-based trick, but on the other hand, you'd be able to perform the same trick multiple times in a row with different predictions each time, to mislead the audience into thinking it's not fixed.

[@LaClimSx]

Here’s a version I just thought of : You ask the person the choose a set, and from this set to pick a card that you’ll remove. You are then left with 12 cards, and ask the person to shuffle them while you make a prediction. You then go on with the trick as shown in the video, you just deal six cards to each instead of five, rearrange them and add the differences, and surprise : this was your prediction. Now how to calculate the prediction regarding the card they chose to remove : It is really easy, if the value of the card is smaller or equal to 7, the prediction should be 35 + the value of said card, and if the value is higher or equal to 7, then the prediction should be 49 - the value of the card.

[@l33td00d17]

Ah, cartoon James is even holding his Little Professor.

[@disguisedhell]

Well, I solved this problem quite some days before. It is published in crux mathematicorum from which I think the inspiration is taken from

[@BillyMcCreery]

Since the value is always fixed no matter the cards, how about a variation where you calculate the value as you and the volunteer pick cards at random?

[@Faladrin]

Should be easy to calculate the number for N cards. Since we know the end total is the same for all different ways the cards come out we can solve for the easiest set. Assuming one person gets all the small numbers and the other person gets all the big cards then the smallest small card will be matched with the smallest big card. The difference between them is N/2 (i.e. the smallest small card will be 1 and the smallest big card will be (N/2)+1). The next pair will be the same since on both sides the values increase by 1. So the end result will be N^2/4 (N/2 multiplied by the number of pairs which is also N/2, so N^2/4).

[@yahyasheikhnejad]

I really enjoyed this video and nice explanation. thank you so much.

[@aayushpatra3823]

Cool trick! Numberphile always has new things to learn

[@johanrichter2695]

If you take all the large numbers in one set and all the small one in another the differences becomes the consecutive odd integers and you get as a corollary the well-known fact that the sum of the first N odd integers is N^2.

[@screwhalunderhill885]

The "average difference" is 5. What I mean by that is with 5 cards you always get 5*5 with 5 cards because when your difference is 7 for example there has to be a difference that is 3 so you get two pairs of 5 and so on until you are left with one difference that is 5.

[@AngrySanta]

I kept your beginning but added at the end to add 2-5 and made my prediction the seven card (controlled). Fun trick.

[@thealienontheinternet]

“Mom can we have a Numberphile mathematician?” “We already have James Grime at home” James Grime at home: 2:59

[@Pier_Py]

You can use the second characteristic of the Staistical Mean to explain this. The sum of n number is equal to the mean of the n number multiplied for n

[@Oscee613]

I was watching a Numberphile card trick video from 2012 while this was uploaded/publicized. Eerie

[@harmidis]

Once again: James is magic! Thanks!

[@dippingbird7533]

I figured out another proof, if you take two cards that are consecutive and on oppisite sides, f.ex. 6 and 7 in the first example and swap them over both sets of cards are still correctly ordered and the sum of the differences is unchanged. You can swap the cards until you have only consecutive numbers on each side, meaning one side has 1,2,3,4,5 and the other 6,7,8,9,10 and thus sum of the differences is always the same. This proof also works for any ordered set of even length, swaping elements which are adjacent according to the order-relation.

[@andrewkelley7062]

You should take a machines learning program and place a tube with a steady single vibration going down it and train the program to separate the output of a fluid going through it into two tubes of separate temperatures by adding structure. Then see how far you can go.

[@pretzelbob1640]

You've intrigued me now. Perhaps I can come up with an effect that uses this? Perhaps controlling a selected card to the 25th position of a deck, and using that sum to find it?

[@blunderr6113]

7:23 gotta love the subtle flex of the million sub plaque

[@GhostLightPhilosophy]

You know how you can do completing the square for quadratics...can you find a way to complete the cube for a cubic?

[@Matthewkyle12]

I would love a follow up video on how to build the formula for the general case - i.e. when you pick any number of cards and/or any values of cards

[@PopeLando]

This is in the same family of tricks as one based on the following : Think of any 2 digit number. Sum the digits and subtract from the original number. The result is always the first digit times 9. Eg, 64, sum of the digits is 10, subtract from 64 gives 54 equals 6 x 9. There was a magician's web page which did a mind reading trick on the viewer, based on this fact.

[@fitzchevalerie6597]

that's some nice math, and it satisfies Poincaré's saying : "A true mathematics concept must be elegant"

[@talastra]

Three minutes in, I'm not working out the details in my head, but it is clear that it does not matter how the cards are shuffled or how they are distributed on the table, because the differences of the cards individually are unaffected by their position on the table, and presumably all of the differences that result (the middle cards) will result in 25. since there are 5 cards from 1 to 10. But instead of adding pairs of numbers in that way that yields the n(n+1)/2 formula, we have the difference of pairs. Clearly, that must always be 25 (using 10 cards).

[@talastra]

Okay, for even numbers of cards, n, that's going to be n^2/4 [or, better, (n/2)^2], I think. So, n = 12 --> 36, n = 14 --> 49.

[@ytHMMG]

Love the framed Graham's Number in the background

[@jzieba0204]

I love how they've got the peper from graham's number episode signed and framed on wall

[@EVENTUCATOR]

The very first time I tried this, my pairs were...10-1=9; 8-2=6; 6-5=A; 7-4=3; 9-3=6: I only have one 6 card. Do I use the 4 &2??? (They still add up to 25.)

[@DaviddeKloet]

Seeing Lulu made it completely worth sitting through the ad!

[@Galva94a]

A problem which seems interesting at the start became absolutly trivial after the explanation. wow!

[@paganmadnessYT]

The effect can be increased using two volunteers I think. Either they choose alternating a card or bring more fake randomness by letting them play rock paper scissors each time.

[@FerielBouchaib]

I came after 3 years to watch this . Thanks to Eddie Woo . This Channel is great ✨

[@yaerius]

This is really incredible. How did James find that out?

[@olerocker3470]

I recognized this right away since I do loads of kakuru. 30 in 4 places is always 6,7,8,9.15 in 5 places is always 1,2,3,4,5. The only factor you added was a 10. So 40-15=25.

[@firstnamelastname307]

But can this be generalised to any N>10 ?

[@Lightning_Lance]

Interesting, but my intuition (uninformed guess) told me the number would be the same regardless of which player got which cards before the trick was over. So a magician would need to dress this up a bit (make it less straightforward).

[@punkkap]

Love me some Penrose card backs

[@oliviermiakinen197]

Hello James, because I'm not very fluent in English I'm not sure whether you noticed the following math property (I didn't hear it, among what I understood of your video). In order to represent the differences, you used red cards (diamonds or hearts). The interesting property is that you will never need to use another deck of cards, because it is impossible to get the same difference more than twice. Maybe you will have to use a 4 of hearts and a 4 of diamonds, but then you can be confident that no other '4' difference will appear.

[@badrunna-im]

He states that 4:12 onwards.

[@oliviermiakinen197]

@@badrunna-im You are right. Thank you for pointing this out.

[@JCResDoc94]

☼ Get 25 scantily clad models, and a saw that is intent on cutting them in half, if the audience member chooses wrong. That's as far as I have got, what do you think so far?

[@EtoileLion]

Feel like there needed to be just a little more emphasis on the "I don't know what the value will be" - important distinction from the consecutive set, the answer is not simply N^2 at that point, and you'll have to do the maths properly to determine your value.

[@crueI]

You are friends with Brian Brushwood, make it happen😁😁😁😁

[@EVENTUCATOR]

Perhaps incorporate a "Magic Square" with the resulting combinations, lines diagonals and sets, adding up to *25*

[@jonnomurray4483]

Love it ❤️ love your channel too 👍keep em coming 👍

[@budz2355]

Amazing-- thanks for sharing!

[@piciaxel]

for this to be used in a magic trick i feel that there are too many fixed things to have to work through, but with at least a force it can be made work a bit better

[@maswal2051]

Gm sir... Even know the answer... Always eager that what and how will you explain... You always explain v very nicely... Thank you..

[@shubhamaswal9712]

Nice

[@drenz1523]

2:59 this will be in my dreams one day and im looking forward of waiting for it.

[@simon_patterson]

Clever, and very well explained.

[@janernn]

I wanna see the queen and king cards! Queen is Hannah Fry maybe?

[@keomarahmat]

Numberphile knows how to make mathematician look cool

[@Furiends]

I kind of picked up on this from discrete mathematics. Your card set has no duplicates so you can rely on the properties of a set.

[@yokmp1]

This is a scaled down version of what Gauß did in School. He was told to add every number from 1 to 100.

[@AkashGupta-th2nm]

3:38 why is it 126 ways or arranging the cards? shouldn't it 10C5 = 252 ways? Edit: I think I get it. Is it cos when u choose 1 combination, ur actually choosing 2 combinations and aligning these. So the cards in a sense have rotational symmetry

[@jomo87]

Thanks for this, I had the same question and having read it, I agree with your edit!

[@MrMineHeads.]

James Grime is a treasure

[@Theraot]

0:14 it is Alfred E. Neuman

[@__Mr.Long__]

i'd love to be able to buy a replica of that deck

[@Lambent_Haze]

... Did he actually get a fully custom deck printed?

[@Juansonos]

Numberphile was offering the Numberphile deck a while back. I just don't remember if it was a special limited run or regular merch.

[@yashrawat9409]

I pretty much remember this was an RMO problem a few years back (It's the mathematical Olympiad in India 2nd level)

[@VivekPandey-gi3ou]

Please explain 'negative times of negative is positive ' without using distributive law

[@nekogod]

Love the framed paper in the background, that's Matts doodles from the Grahams Number vid if I'm not mistaken.

[@KalOrtPor]

It's Ron Graham's own doodles from when he was on, autographed and dated.

[@leonhardeuler9839]

I didn't catch something from this video, James said there are 126 ways of arranging the cards but 5!=120. How?

[@angelmendez-rivera351]

Sobhan Haggı Well it's not 5 factorial. Why? Because you have *two* sets of 5 cards. The maths are different.

[@maswal2051]

Sobhan haggi.... It will like this... 10C5/2! =10!/5!*5!*2! =126

[@eliyasne9695]

Wow! So beautiful

[@khunalhoollah6621]

Went to listen to your talk in Greenwich

[@moanilsson3448]

Where can I get the deck of cards with Grime on it?

[@Tahgtahv]

I believe they printed a limited run, which has long ago sold out. Hopefully they will sell them again at some point.