James ❤️ A Card Trick - Numberphile

James Grime has a card trick (and wants your help improving it).
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[@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.

[@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.

[@julienl.5045]

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 :-)

[@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

[@jibbiddy]

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

[@nikanj]

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

[@Adam-pv4qn]

Lol

[@courtney-ray]

😳 my thoughts exactly! 👀

[@GeodesicBruh]

Lol

[@Triantalex]

??

[@georgew.9663]

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

[@UltraCboy]

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

[@romajimamulo]

Thanks... I guess?

[@Yump0123]

Take that thing away from me

[@brettonjohansen1619]

I appreciate the penrose tiled card backing.

[@fakjbf3129]

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

[@kyazarshadala8114]

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

[@vtron9832]

James grime still looks better in real life

[@QuasiELVIS]

Only just.

[@John_259]

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

[@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.

[@scottanderson8167]

Grimy is the best Numberphile by far.

[@AlexKing-tg9hl]

I love you James! You’re the best

[@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....

[@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.

[@Vikash137]

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

[@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.

[@yahyasheikhnejad1877]

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

[@aayushpatra3823]

Cool trick! Numberphile always has new things to learn

[@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...

[@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.

[@jonnomurray4483]

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

[@surfing_youtube]

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

[@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.

[@philtwoframes3983]

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

[@harmidis]

Once again: James is magic! Thanks!

[@budz2355]

Amazing-- thanks for sharing!

[@simon_patterson]

Clever, and very well explained.

[@DaviddeKloet]

Seeing Lulu made it completely worth sitting through the ad!

[@AngrySanta]

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

[@l33td00d17]

Ah, cartoon James is even holding his Little Professor.

[@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

[@Oscee613]

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

[@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.

[@ytHMMG]

Love the framed Graham's Number in the background

[@muslimali8388]

Really cool card trick!

[@blunderr6113]

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

[@rannov.4707]

You could give them the impression that they have more free will by letting them place their cards in any order they like, and then putting your cards down, so that the smallest card is next to the biggest card etc. Then you could let them move any pair around as they like (this will not affect the pairs themselves).

[@badrunnaimal-faraby309]

The last step doesn't really help. Since you'll end up summing the red cards, shuffling them around is too easily seen as pointless.

[@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.

[@Galva94a]

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

[@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.

[@eliyasne9695]

Wow! So beautiful

[@nubskr]

Beautiful

[@apratimghosh109]

Amazing video!

[@soveu8237]

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

[@KalOrtPor]

Written and signed by Ron Graham himself!

[@jzieba0204]

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

[@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

[@punkkap]

Love me some Penrose card backs

[@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

[@MrMineHeads.]

James Grime is a treasure

[@bhaveshverma8629]

Very informative

[@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.

[@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

[@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.

[@thealienontheinternet]

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

[@yaerius]

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

[@FerielBouchaib]

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

[@khunalhoollah6621]

Went to listen to your talk in Greenwich

[@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?

[@rdxr10]

Interesting trick.

[@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.

[@__Mr.Long__]

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

[@ChandrasegaranNarasimhan]

The trick was simple after the explanation. Congrats for originality.❤❤❤❤❤. Btw I thought you are using 52 cards.

[@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?

[@yashrawat9409]

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

[@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.

[@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

[@nathanderhake839]

I thought I got a heart ❤️ from numberphile but then I realized the video title had a heart in it.

[@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

[@PplsChampion]

these animations going ham

[@greengirl4985]

James❤️

[@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

[@rillloudmother]

oh yeah, it's Grime time!

[@Tatiana-jt9hd]

I see James I click :)

[@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

[@janernn]

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

[@TommiHimberg]

Perhaps the amazing Steven Bridges could give James a hand in improving this trick.

[@ascetic3312]

It's Grime Time!

[@drenz1523]

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

[@nonothebot]

I'm proud because even if I'm not able to explain it mathematicaly, I found out the answer at the begining of the video.

[@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])

[@UpcycleElectronics]

Hail the James of Hearts

[@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.

[@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.

[@xvnbm]

Boy at school: Where do I need math? Teacher: To do magic tricks.

[@erakkarthik]

Nice video

[@elieli336]

this is 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.

[@randomdude9135]

*A Grime video a day keeps the Grim mood away* 😊

[@websterlowe2258]

Could you do a video on the mathematics of card games?

[@johanrichter2695]

That is a massive subject. The mathematics of poker alone could fill many videos.

[@EVENTUCATOR]

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

[@keomarahmat]

Numberphile knows how to make mathematician look cool

[@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?

[@tomasbeltran04050]

I knew I had seen that singing banana before

[@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).

[@sakshamsingh4378]

James with cards is like me dancing in wedding(^o^)(^o^) A rare event

[@crueI]

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

[@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