Shuffling Card Trick - Numberphile 

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

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

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

+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

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

Actually more impressed by that than the maths lol

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

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

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

wtf? hahaha

false.

😂😂😂

Iykury haha

+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

+Buttercak3 Hallo :D

+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"!

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

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.

Indeed.

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

funny_monke6 Coincidence? I think not.

Tennison Chan 5o

He's not a professor?

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

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

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

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.

+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.

+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.

+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).

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

+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?

+Jonathan Krillington LOL GET PWNED

+Crazy drummer it is

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.

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

+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.

+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.

+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.

+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 :)"

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.

The pattern is the Eight kings threaten thing

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.

+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. ;-))

+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"

+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.

+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

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

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

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

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.

+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

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

+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)

+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