Is this deck in a random order?

# Is this deck in a random order?

lol what no? the cards are in order retard

In relation to what? The answer is yes and no, but it depends on the relation.

Define random

Possibly. The odds are one in 52! that's it was random. My guess is that it's not.

Is it random if I hold up a spork?

it's a deck that could have been assembled by a random process

Ummm... I think there's a little more ways to arrange deck than 52...?

Well how many cards do you think there are?

B T F O

T

F

O

Brainlets on suicide watch

maybe

You're trolling r-right ?

Undergrad detected

Thanks for the laugh. I really hope you're baiting.

You must be literally retarded if you think there's more ways than 52!

Factorial! ! ! !

Randomness is not a property of one sample/event.

that's a bad argument. the probability of any given arrangement is also [math]\frac{1}{52!}[/math]. you need more information than that to determine whether something came from a non-random process

Dont kill yourself user its not worth it.

Didn't know Veeky Forums fell for bait this easily, thought you guys were supposed to be smart

maybe if you make some toast

genius here

the answer is yes, it is possible for that permutation of cards to have occurred by chance

Maybe.

The convention says no. Just because it's a possible permutation doesn't mean it's "random".

Not only that, it is as likely as any other random shuffle of the deck.

More or less, yea. It's simply a well known order, but that doesn't make it any less random.

Actually he is right there is more than 52! because YOU DON'T FORGET THE JOKER, ACE OF SPADES ACE OF SPADES COME ON

>keeping the joker in the deck

Veeky Forums is an 18+ website

>not playing Open Face Chinese with jokers

pleb

the fuck, where you get lefty Bikes?

>not getting the obvious dadrock reference

Veeky Forums is an 18+ website

Solve carefully:

2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2

You may not believe it, but the answer is 4!

>le hilarious reference

>>>/reddit/

>genius here

lost

there is literally nothing wrong with that argument. you could not possibly make it more correct

The odds are [number of decks that are randomly shuffled and have this order / (number of decks that are randomly shuffled and have this order + number of decks that are deliberately arranged in exactly this order)]

For an order with no rhyme or reason this would just be 1/1 (random for sure) because there are 0 such decks deliberately arranged in that order, but for this thing it's more like [(number of shuffled decks * 1/52!) / (number of shuffled decks * 1/52! + 10^6)] which is very fucking small, the reason being that there are like a million decks arranged this way on purpose.

tl;dr it depends on how many decks are arranged exactly this way on purpose

that will heal your wounded pride :)

there's everything wrong with that argument, because it doesn't answer the question. OP did not ask for the probability of acquiring any given arrangement of cards, OP asked whether the given arrangement was acquired through a nonrandom process

[math]\frac{1}{52!}[/math] is a necessary expression in any answer to the question, but it is not itself the whole answer

I think 'random order' refers to the way something was organized, not its current state of organization, so the answer is unknown.