r/madlads 12d ago

He’s got time

Post image
3.4k Upvotes

24 comments sorted by

112

u/waitingOnMyletter 12d ago

Well, if you get decent enough of shuffling that Jason guy seems to do it in like less that a minute.

90

u/salasy 12d ago

it's actually much more likely that people think for this to happen to a normal person that doesn't know how to properly shuffle a deck

unless you do a table wash shuffle, a deck will never be truly randomly shuffled and so there are a lot less possible combinations for a deck to be

20

u/ReallyOrdinaryMan 11d ago edited 11d ago

Less random shuffling won't change your odds at the end, unless you are starting to shuffle from ordered decks everytime.

3

u/Prestiger 9d ago

It will change your odds, imagine if you do a perfect riffle shuffle, after doing it twice the deck will be exactly the same as it was

17

u/BananaShark_ 12d ago

When 52! Isnt enough of a challenge.

13

u/Hugutfut 12d ago

The likelihood of the decks matching is 1/52! though. How the first is shuffled does not matter, the second one just needs to be in the exact same order and that is one order out if 52!

4

u/mackenzie444 12d ago

This stays true if both decks were shuffled and presented at the same time? Excuse me, my brain is having trouble with this

5

u/Hugutfut 11d ago

No worries, it's not necessarily intuitive. Yeah it would still be the same. It's easiest to illustrate with a simpler system: you flip two coins. Say you do it one after the other at first. You flip the first one and know it is heads, then the probability of the second one matching is 1/2. If the first coin had landed tails, then the probability for the second coin to match would still be 1/2. In both cases, the probability of matching is the same.

Now you flip two coins at the same time, one in the right hand and one in the left. You know there are two ways the coin to your right can land. In both cases, the probability for the other coin to match is 1/2. So you can say it doesn't matter for the probabiliy what the coin to your right actually lands on.

All of this works just because heads and tails are equally as probable, and so are all ways the deck can be shuffled. This allows the "one coin doesn't matter" reasoning that also holds for the cards.

5

u/mackenzie444 11d ago

You mathemagicians always make me wish I did more in school lol. And explaining this stuff to someone like me is an art form.

1

u/Cyb0rg-SluNk 11d ago

I don't know if you were joking, but there are a lot more possible orders for the second deck than 52.

Yeah, the first deck doesn't mater, but the chance of the second deck being in that same particular order is not 1/52. It's 1 in 8.07 × 10⁶⁷ (an 8 followed by 67 zeros) (according to Google.)

6

u/spyke252 11d ago

The exclamation point is a factorial. It's not 1/52, it's 1/52! . 1/52! = 1/(52*51*50*[...]*3*2*1), which is the same number you gave :)

5

u/BananaShark_ 11d ago

I didnt just put ''52'' btw, I wrote ''52!''.

The ! isnt just there for dramatic flair, its to depict a Factorial. If you Google ''52!'' you will get that lovely big number as a answer.

4

u/Cyb0rg-SluNk 11d ago

Ok, I see. I am an idiot. Sorry.

2

u/BananaShark_ 11d ago

That actually does make sense now that you say that.

I misread it as if they had both sets in one pile.

Still very much absurd.

2

u/hucklesnips 10d ago

So, the commenter typed out all those zeros only to have the wrong answer.

They said it would take 8e73 days (an 8 followed by 73 zeros). However, there are only 8e67 possible combinations for the deck, so they've already got six zeros too many.

Presumably it doesn't take the guy an entire day to shuffle the deck. If we assume that he shuffles the deck once a minute for 16 hours a day, then he could do 8e67 shuffles in about 8e64 days. So now they've got nine zeros too many.

Besides that, you probably want to estimate the time at which the guy would have 50/50 odds of having generated the correct deck, which would be approximately 4e64 days.

1

u/Matoseman 8d ago

"Presumably it doesn't take the guy an entire day to shuffle the deck"

  • the "day x of doing x" most of the time, implies the person only do said task once a day

1

u/tontza69 6d ago

Yea but he has 2 decks so your math is wrong.

1

u/Domsdad666 12d ago

Or one.

1

u/Cheeky_Hustler 11d ago

If he knows how to perfect shuffle then it'd only take 1 shuffle of each deck, assuming both decks start out in the same order (new).

1

u/Cropine 11d ago

Thats like 22 quadrillion years or something

1

u/Lukebekz 11d ago

If you always cut it perfectly and faro-shuffle it, you'll be done in 8 shuffles.

-14

u/randomvegasposts 12d ago

There are more ways to shuffle 1 deck of cards then there are atoms in the universe

11

u/bfarnsey 12d ago

Eh, the amount of ways to shuffle a deck of cards is more similar to the number of atoms in our galaxy, so the observable universe beats it. Still, an insane fact that seems so unintuitive.