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-   -   No two card shuffles will EVER be the same. (http://cellar.org/showthread.php?t=16562)

Happy Monkey 02-06-2008 11:49 AM

Doh! You're right. You're never guaranteed to get any particular match, but you are guaranteed to get A match.

classicman 02-06-2008 12:22 PM

OK, how many shuffles does it take to get them back in the same order they were originally in - ie: right outta the box?

Happy Monkey 02-06-2008 12:44 PM

Eight, if you know what you're doing.

If not, then that's the situation where P will never be one, and there are no guarantees.

classicman 02-06-2008 12:49 PM

holy crap! - :::head spins off:::

Flint 02-06-2008 01:18 PM

Quote:

Originally Posted by Happy Monkey (Post 430289)
Eight, if you know what you're doing.

Quote:

A perfect shuffle occurs when the deck is divided exactly in half, and the cards are perfectly interlaced, with one card coming from one hand, then one card coming from the other hand, then one card coming from the first hand, etc. There are two types of perfect shuffles, the "in-shuffle" and the "out-shuffle." Let's assume that the deck is divided in two with the top cards going into the left hand and the bottom cards into the right hand. Then an in-shuffle begins with the first card coming from the left, the second from the right, the third from the left, etc. An out-shuffle begins with the first card coming from the right. If the right hand originally took the top cards, then the definitions are reversed (the in-shuffle begins with the first card coming from the right...). It has been shown that eight perfect out-shuffles returns the 52-card deck to its original order. Apparently, it takes more in-shuffles to do that.
I'm not calculating this scenario based on professional magicians doing elaborate card tricks. I'm talking about people playing a hand of cards, and then shuffling a few times, and then playing another hand. And, maybe they drop them on the floor, or one of the cards get bent, etc.

Saying the possible shuffles is 52! assumes randomness, which isn't entirely accurate, but it's more believable than a series of "perfect" shuffles.

If one card gets out of order in your series of "perfect" shuffles, you've started down the long road of 8.06581751709439 e+67 permutations.

Flint 02-06-2008 02:19 PM

Quote:

Yes you are, because the outcomes are limited--it's like if you have four different colors of socks, and you choose five socks. Your fifth sock must match one of the other four.
Except, in this case, you’ve got four pairs of socks in a box, and you’re pulling them out one at a time and lining them up in that order. You have to get them in the exact same order, not just draw the same ones.

If you’ve got a red, green, blue, and orange sock; you could draw:

R G B O
R G O B
R B G O
R B O G
R O G B
R O B G


And that’s just what could happen if you draw the red sock first. It carries on for a total of 24 friendly, manageable permutations.

But the factorial of 52 is 8.065 817 517 094 39 x10 to the 67th power (roughly 8 with 67 zeroes). That's how many shuffles there are.

Any card of 52 could be in the first position, then for the 51 choices for the second card, there are 50 choices that could be the third card. But they might not be first, second, or third; they could be anywhere in the deck. The number of possible shuffles is so large that the human brain cannot comprehend it directly.

It’s not only possible that the same shuffle has never happened, it’s the most likely outcome; considering the number of permutations, and the number of chances we have had to crunch through them. Certainly you aren’t required to go through all of them to get a repeat, but…

...we’re talking about something like taking all the grains of sand in the world, throwing them up in the air, and having them all fall back down in the exact same place. That isn't going to happen very often, and if you don't have enough time to keep trying, it will never happen. We haven't had enough time to get the same shuffle twice. And before we get the chance, we'll be long gone. It will never happen.

Happy Monkey 02-06-2008 02:27 PM

Quote:

Originally Posted by Flint (Post 430302)
I'm not calculating this scenario based on professional magicians doing elaborate card tricks. I'm talking about people playing a hand of cards, and then shuffling a few times, and then playing another hand. And, maybe they drop them on the floor, or one of the cards get bent, etc.

Right. I believe I addressed that in the second sentence. You are never guaranteed to match any particular shuffle (including the initial order), no matter how many times you shuffle.

The perfect shuffle thing was an amusing aside.
Quote:

If one card gets out of order in your series of "perfect" shuffles, you've started down the long road of 8.06581751709439 e+67 permutations.
Worse than that. If you make one mistake, and then continue a series of perfect shuffles, you will never get the original order back. You'll cycle through a sequence of eight incorrect shuffles until you make another mistake.

edit:
Another aside- out-shuffles have a cycle of eight, and in-shuffles have a cycle of 52.

Happy Monkey 02-06-2008 02:32 PM

Quote:

Originally Posted by Flint (Post 430326)
Except, in this case, you’ve got four pairs of socks in a box, and you’re pulling them out one at a time and lining them up in that order. You have to get them in the exact same order, not just draw the same ones.

In Clodfobble's example, each sock represents a shuffle, not a card, and "four" represents "52!". So you are guaranteed to get a match with five socks, or 52!+1 shuffles. Clodfobble was correcting my misunderstanding of HungLikeJesus' post.

lookout123 02-06-2008 02:40 PM

Wow. I've been visiting the cellar for nearly four years now for threads like this. I honestly couldn't care less about this subject and certainly am incapable of following, let alone creating the calculations you are all doing. In most circles I find myself in I (all arrogance aside) would rank near the top in intelligence and mental ability. Then I come to the cellar and feel like a true simpleton. You guys amaze me. While this subject holds no interest for me, the fact that it has captured your attention enough so that you actually calculate the truthiness of the thread title fascinates me.

Well done, geniuses. Well done.

Flint 02-06-2008 02:46 PM

Quote:

Originally Posted by Happy Monkey (Post 430333)
I believe I addressed that in the second sentence.

Well, I believe I addressed this in post #54:
Quote:

Originally Posted by Flint (Post 430221)
I haven't met my quota for scathing remarks this morning.

Plus, I was addressing the perfect shuffle thing, because it had been mentioned once before. And, 52!+1 ... ha! That's funny.

The interesting thing is that you really must get a match eventually, it's just that we don't have that kind of time. It reminds me of the idea that, given the universe is infinite in size, you can calculate how far you would have to go before you encounter an identical Earth, down to the last atom. Think about that.

By the way, what happened here is that my dad mentioned that the same card shuffle has never happened, and it bothered me. A few weeks later I asked him, did he mean one person has never had the same shuffle in their life? And he said, no, nobody. Ever. It's hard to believe, but we got out some scratch paper and a calculator and started pecking away at it. I'll be damned if I'm not completely convinced. I don't think it's possible that any given shuffle has ever repeated. So, lookout, you can thank my dad for this thread.

I really don't want to have to get out my statistics textbook, but...
I'm curious to know how many permutations it would take to have a 1% chance of repeating a shuffle.

Flint 02-06-2008 03:02 PM

Quote:

Originally Posted by lookout123 (Post 430337)
Well done, geniuses. Well done.

And, yes, I too love the Cellar for having five pages worth of interest in this subject, in just two days. You guys are top-notch.

Cicero 02-06-2008 03:54 PM

Actually- Sheldon is all the proof we need K?

:)

Happy Monkey 02-06-2008 04:46 PM

1 Attachment(s)
Here's the formula to calculate the chance, in n shuffles, that there will be NO matches:

Clodfobble 02-06-2008 05:51 PM

So if Flint sets that equation equal to 99, and solves for n, then he'll have his 1%-chance-of-a-duplicate answer, right?


Note I said Flint does it, because I sure as hell don't have the energy. :)

Happy Monkey 02-06-2008 06:29 PM

0.99, but yeah.


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