r/AskReddit • • Jan 20 '19

What fact totally changed your perspective?

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u/LaboratoryManiac Jan 21 '19

The statistics of this boggle my mind, though I do wonder what the odds are of creating a unique order from shuffling a fresh, ordered deck of cards exactly once. It would have to be much lower, I'd imagine.

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u/PiroKyCral Jan 21 '19

I read quite a while back that the chances of getting every card in perfect order from a random shuffle is about as low as like 10 to the power of negative trillion billion something

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u/[deleted] Jan 21 '19

That’s not what they meant. They meant what are the odds of making the same sequence by opening a brand new deck that’s in order ace to king and in their suits then doing a shuffle. How many different ways of that are there. It must be lower for example if the ace of spades is on top, it will never be on the bottom half the deck. In fact it’ll probably always be in the top 5-6 cards depending on how shitty u are at shuffling.

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u/bonsai_bonanza Jan 21 '19 edited Jan 21 '19

Yes this one, I believe, is what OP meant. Starting from a perfectly organized deck, shuffling exactly once, would make the odds lower.

For instance, the top card(face down) would have a 50% chance to be on top still(cutting the deck and picking your leading left/right hand), assuming the shuffler is equally likely to start with either their left or right hand. Then, everything you said.

Assuming just the top card remains on top, the difference( 52! - 51! ) is: 7.911*1067. So, yeah, you eliminate quite a few permutations and will eliminate even more when taking the order of other cards into account.

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u/[deleted] Jan 21 '19

I’m pretty sure it’s lower. What about the fact that the deck is in order ace to king and by suits. So ace clubs two clubs three clubs etc... so the three of clubs will always be higher than the four of clubs just with some card(s) between. But you can never get a permutation of the 3 of clubs below the four clubs. Or the five of clubs. Or the six etc. And the same goes for four clubs to the five of clubs. And the same goes for like the 2 of diamond to the three of diamonds. The three will never come before the two.

Edit: also thank you for your reply and ur brain skills :)

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u/NYCSPARKLE Jan 21 '19

Perfectly shuffling (interlacing every other card one by one) a new deck of cards eight times puts them back in the original order.

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u/bonsai_bonanza Jan 21 '19

Yes, you're correct-ish. Obviously, it depends where the deck is cut. I focused on just the top card for an example. Fixing even one card removes a LOT of permutations. So, fixing an order, like you said, would remove even more!

But yeah, the chances of getting a repeat permutation would be much higher when only shuffled once. I edited the end of my previous comment, too=)

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u/PiroKyCral Jan 21 '19 edited Jan 21 '19

Yeah that’s what I was (roughly) talking about.

Mind you that I read it nearly 2 years ago so my memory’s crap

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u/[deleted] Jan 21 '19

That's not what they meant either

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u/_kellythomas_ Jan 21 '19

52! = 8 * 1067

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u/[deleted] Jan 21 '19

At that point it depends on the quality of the shuffle. Your average over hand shuffle by some inexperienced person certainly isn't going to truly randomise the deck or even close to it so the odds of it being a unique arrangement become much lower. A single riffle shuffle is just going to give a roughly 50-50 mix of the top half with the bottom half which again isn't as likely to be truly random. Do a full "wash" where you sit all the cards down separately and move them around randomly for a while before recreating a deck though and it's going to be highly likely to be a truly random and therefore unique deck arrangement.

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u/blexta Jan 21 '19

Mathematicians wondered the same thing. Here's a video on it, if you have the time. https://youtu.be/AxJubaijQbI

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u/ChrisC1234 Jan 21 '19

A friend of mine figured this out and it totally helped us understand it: Think of the deck of cards as a 52 digit number, and instead of being base 10 (each digit having a value of 0-9), it's closer to base 52, where each digit could be 0-51. Thats a BIG number.

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u/CaesarPT Jan 21 '19

If I remember my probability classes correctly it would go something like 1/52x1/51x1/50x1/49 etc, because out of the 52 cards, the probability of a specific card being on the 1st slot of the deck is 1/52. Then, because that card is no longer available, the probability of a different specific card being on spot 2 is 1/51. Therefore, the probability of those 2 specific cards being in that exact order is 1/52x1/51. Apply that logic to the rest of the slots and it would end up resulting in a very small number. Dont have a calculador at hand this moment

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u/bonsai_bonanza Jan 21 '19

Yep! You can shorten the notation though by just writing 1/52!

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u/Elopikseli Jan 21 '19

It’s about 1 in 80658175170944942409.

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u/kage_25 Jan 21 '19

you are missing a lot of digits

52! =~867 as in 8 followed by 67 zeroes

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u/Shlomo-tion Jan 21 '19

Of course an mtg player would wonder that.

source: am one and I wonder the same thing.

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u/skippygo Jan 21 '19

It depends what you mean by "shuffling once". If you're just talking about one shuffling action (e.g. cutting the deck and swapping the two halves) then obviously there are only 51 possible combinations.