No two card shuffles will EVER be the same.

That doesn’t sound right, does it? In all of human history, and all of human history to come, no matter how many times a deck of cards is shuffled, the same combination will never, ever, ever come up. How can that be? Well…

The 52 cards in a deck can be arranged in...
8,065,817,517,094,390,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,
000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,
000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000
...different ways.

What if we were to put every person alive on the Earth, right now, on this shuffle project? According to the U.S. Census Bureau, the total population of the Earth, as of 02/04/08 at 15:40 GMT (EST+5) is 6,648,429,413.

So, each person would have to shuffle...
1,213,191,419,513,742,831,701,174,895,214,300,000,000,000,000,000,000,000,000,000,000,
000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,
000,000,000,000,000,000,000,000,000,000,000,000,000,000,000
...decks of cards before all the possible shuffles were exhausted. What if we gave them a little time to work on that? How about 100 years…

If every person alive on the Earth, right now, shuffled continuously for the next hundred years, they would have to shuffle...
3,847,004,754,926,886,198,950,960,474,423,900,000,000,000,000,000,000,000,000,000,000,
000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,
000,000,000,000,000,000
...decks of cards, every second, before all the possible shuffles were exhausted.

We don’t seem to be getting any closer to achieving all those shuffle combinations. How long would it really take? Let’s put those 6,648,429,413 people on planet Earth to work, on a more realistic goal: shuffling one deck of cards every 10 seconds, 8 hours a day, 5 days a week, and see how long it takes.

360 shuffles per hour
2,880 shuffles per day
14,400 shuffles per week
5,256,000 shuffles per year
times 6,648,429,413 people is 34,944,144,994,728,000 shuffles per year!

We should see all those shuffle combinations coming up in about...
1,534,592,373,876,406,012,176,560,121,765,600,000,000,000,000,000,000,000,000,
000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,
000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000
...years. Don’t hold your breath.

Flint;429634 wrote:

The 52 cards in a deck can be arranged in...
8,065,817,517,094,390,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,
000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,
000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000
...different ways.

How do you derive this number? The number of permutations is 52! (= 1 x 2 x 3 x 4 x . . . . x 51 x 52), which is 8.0658 x 10^67. Your number seems to be 8.0658 x 1000^67, which makes a slight difference.

Does this work for Old Maid?

Flint is crazy.

SteveDallas;429639 wrote:

How do you derive this number?

I took 8.06581751709439 e+67, copy/pasted 6.7 lines of ten groups of ",000" then put commas in the original figure, like this "8,065,817,517,094,39" and then pasted these on top of the zeros so that it was the same length. Oh, I left the "8," on a serarate line, because it was before the decimal.

What did I do wrong? Oh. I did 67 groups of ",000" instead of just zeros.

Flint;429643 wrote:

What did I do wrong?

Nothing. You're crazy.

Yeah, so? I just wanted to show the zeroes, so you could see how obscenely gigantic the number is. 52! is not likely to be exhausted during the course of human history.

Of course, the part I didn't include is how these are not really random deck shuffles. Card are arranged in typical, repeating patterns, due to the fact that we are shuffling from a partially ordered state, based on the card combinations that happen during the course of card games.

Anybody care to take a crack at that?

Flint,
I respectfully disagree, but don't have time now to work through the math.

flint--i've a huge crush on you.


deal.

HungLikeJesus;429658 wrote:

Flint,
I respectfully disagree, but don't have time now to work through the math.

Combinations of card shuffles, 52! is 8.06581751709439e+67.

Population of the Earth (possible number of people that could shuffle a deck of cards) is 6,648,429,413. That’s 1.2131914195137428317011748952143e+58 shuffles per person.

Divide 1.2131914195137428317011748952143e+58 shuffles by number of seconds per year (31536000) and you get 3.8470047549268861989509604744239e+50.

That means: if every person alive on the Earth started shuffling one deck of cards per second, it would take 3.8470047549268861989509604744239e+50 years to get through all the combinations.

Brianna;429660 wrote:

deal.

Deal? There's 8.06581751709439e+67 different ways I could take that.

Flint, here's what I'm thinking:
There are 52! (52 factorial) ways that the first shuffle can come out.

The probability of the second shuffle exactly matching the first is 1/52!.

The third shuffle has a 1/52! chance of matching the first shuffle, and a 1/52! chance of matching the second shuffle, so by the third shuffle, there's a 3/52! chance of two shuffles being the same.

The 4th shuffle has a 1/52! chance of matching the first shuffle, and a 1/52! chance of matching the second shuffle, 1/52! chance of matching the 3rd shuffle, so by the 4th shuffle, there's a 7/52! chance of two shuffles being the same.

Etc...

The question should be, "how many shuffles would be required to have an X % chance of two shuffles matching?"

That's interesting.. I'd never really thought about it. (And believe me, the question of card arrangements is not so esoteric to blackjack and poker players, among others.)

EDIT: HLJ, Flint was being sensationalistic in his thread title. I think his point (though with Flint it can be hard to tell) is that there are so many combinations it would take "forever" to go through them all, even guaranteed that there would be no repeats.

Flint being sensationalistic ?!

eee gadds!

say it isn't so!

Heavens to Betsy!

[/great big globs of sarcasm]

Of course it's sensationalistic. I want someone to prove me wrong.

What are the odds of one particular permutation, of 52! possible permutations, repeating? It's been a few years since I took statistics, but I think this should be pretty easily answerable. With the right (TI-83/84) calculator.

What I'm saying, based on what I can see (and this is very counter-intuitive) is that in the history of this planet, we can be almost %100 certain that the same card shuffle has never come up, and never will. Of course, it could happen five minutes later, but the probability of this would be so vanishlingly small that it would be considered impossible for all practical purposes.

Isn't that the Gambler's fallacy? That one event affects future events, event being the shuffle. From the beginning of one shuffle, to the end of the same shuffle, has no bearing on future shuffles.

(Like the Homeless Guy is sure the Pick 3 numbers are predictable based on what has or has not come up in a while.)

Well, in reality, one shuffle does affect the next one, because it isn't really starting from a random order. But I'm assuming the shuffles are random for the purposes of this discussion. Which I feel okay with, because we're talking about all card decks that exist, have ever existed, and will ever exist. They don't have any effect on each other. Of course, I'm calculationg the factorial as if we're talking about one single deck. That's okay because they're all the same, if you assume we're talking about standard 52-card decks. Which we are.

I guess I'm really just avoiding your question.

There are 52! (52 factorial) ways that the first shuffle can come out.

The probability of the second shuffle exactly matching the first is 1/52!.

I don't think this is right. Every one of the 52 cards has 52 possible starting points, so it's actually 1 in 52x52 at least.

I could be wrong though.

Remember, "1 in 52!" is a short-hand way of saying "1 in 52*51*50*49*48*47*46*45*44*43*42 etc." all the way down.

Flint, I calculate that in 9x10^33 shuffles of the deck, the probably of two shuffles resulting in the exact same ordering is 50% (assuming complete randomness in all shuffles).

Is that right? I would take the words at face value. Maths is supposed to be precise is it not?

Anyway, I'll leave you fellas to this (quite boring) discussion. :)

flint's suppostitions are only accurate if the assumption is that you will experience all of the possible combinations before you get the same one twice.....which is quite ludicrous.

How fast is the treadmill moving?

Er, doesn't it also depend on what game you're playing? In certain games, like bridge, the order of cards dealt to a hand is irrelevant, all that is relevant is that 13 cards are dealt to him, and order of a particular hand is rearranged once dealt. There, one could find certain duplicate hands, if dealt in a different order.

You've assumed that shuffles produce random distributions. They do not. Picture a deck with the 7 of Spades as the top card. You split the deck in half, and shuffle the cards together.

There are not 52 possible locations for the 7S to appear, post shuffle. Given a standard shuffle (small groups of cards falling together at a time), the 7S will always appear in the top 5 or 10 cards. This last number is a conjecture, but it will certainly not appear much deeper than that, and will not appear in the bottom half of the deck at all, unless the shuffler simply "cuts" instead of shuffles.

So, the distribution is not random. Each card will only move by a few positions in a well-ordered shuffle (relative to other shuffles). Assume that a large number of people regularly open brand new decks, which have the same starting order, and then give one shuffle. There is a much more limited (relatively) set of possible distributions for that first shuffle, that first permutation.

....unless you shuffle, cut, shuffle, cut, shuffle, cut..... :p Just thought I'd try and make things more difficult. This thread is hurting my poker game.

smoothmoniker;429781 wrote:

You've assumed that shuffles produce random distributions.

No, I haven't actually:

Flint;429652 wrote:

Of course, the part I didn't include is how these are not really random deck shuffles. Card are arranged in typical, repeating patterns, due to the fact that we are shuffling from a partially ordered state, based on the card combinations that happen during the course of card games.

Flint;429718 wrote:

Well, in reality, one shuffle does affect the next one, because it isn't really starting from a random order. But I'm assuming the shuffles are random for the purposes of this discussion.

Okay I guess I did say that, but with a qualifier. And it's partially because I don't know how you could calculate in that variable, and, if it would even make a difference, considering that all 52 cards have to be in position; doesn't this quickly become just more random data?

smoothmoniker;429781 wrote:

Picture a deck with the 7 of Spades as the top card. You split the deck in half, and shuffle the cards together.

There are not 52 possible locations for the 7S to appear, post shuffle. Given a standard shuffle (small groups of cards falling together at a time), the 7S will always appear in the top 5 or 10 cards. This last number is a conjecture, but it will certainly not appear much deeper than that, and will not appear in the bottom half of the deck at all, unless the shuffler simply "cuts" instead of shuffles.

So, you're cutting down the variables a bit for that one card, or for any cards that were used in the play of the last hand. I'll even further this to say that the same cards might appear in similar places if the same players were playing the same games, but... with the size of the numbers we're looking at here, I don't see these little possibilities having much impact. And, as always, it's all 52 cards in the exact same position we're talking about, not just a few cards, in a somewhat similar position.

smoothmoniker;429781 wrote:

Assume that a large number of people regularly open brand new decks, which have the same starting order, and then give one shuffle. There is a much more limited (relatively) set of possible distributions for that first shuffle, that first permutation.

That's another good point. Of course, that deck is only new once. I would say that not enough new decks of cards have ever been produced to make much impact on the unimaginably huge number of permutations we're dealing with. Remember, I had every person on the Earth shuffling cards as a full time job, from now until an impossible number of years after our sun has burnt out and no trace remains that planet Earth ever existed. So, new decks of cards...I'm not so concerned with that.

deadbeater;429763 wrote:

There, one could find certain duplicate hands, if dealt in a different order.

I'm talking about the whole deck, not just one hand. And order matters.

lumberjim;429732 wrote:

flint's suppostitions are only accurate if the assumption is that you will experience all of the possible combinations before you get the same one twice.....which is quite ludicrous.

No, I just don't know how to calculate the probability that and I want somebody to do it for me. What I have shown, though is how the number of permutations is much larger than our brain can even comprehend.

HungLikeJesus;429726 wrote:

Flint, I calculate that in 9x10^33 shuffles of the deck, the probably of two shuffles resulting in the exact same ordering is 50% (assuming complete randomness in all shuffles).

Now we're getting somewhere. Thanks, HLJ. I'll take you on your word (and these number are so big, I don't think it matters how far off we are from being exactly right).
__________________

Now, to get through 9.e+33 permutations, everyone on the Earth (6,648,429,413 people) shuffling decks of cards as a full time job (one shuffle per 10 seconds, 8 hour days, five day weeks) it would take 257,553,876,374,935,601.83683591322334 years before you had a 50% chance of having the same shuffle come up. Hey, we’re finally out of exponential notation! Too bad, though, I don’t think human civilization has been, or will be, around for that long. Anyone have something concrete to bring this number down into the realm of might-ever-actually-happen?

Flint wrote:

Anyone have something concrete to bring this number down into the realm of might-ever-actually-happen?

No one in the third world works mere 40-hour weeks. You should put most people at 80-90 hours a week, easy.

Clodfobble;429805 wrote:

No one in the third world works mere 40-hour weeks. You should put most people at 80-90 hours a week, easy.

I'm also assuming that infants and people in irreversible comas can shuffle a deck of cards, every ten seconds for eight hours in a row...

...so I think it all works out. I guarantee you'll be working some weekends to take up that slack. The shuffle-per-10-seconds quota is set in stone.

A five year old can shuffle a deck of cards. Since half of the population is not in a coma or under the age of five, doubling the work week will have a bigger positive effect than the negative effect of counting out those unable to shuffle.

But if you don't like that one, then how about you add in population growth, since people will have all this spare time to boink each other? Hey, you're the one who wants to make this happen, I'm just trying to think outside the box for you.

5?

Clodfobble;429809 wrote:

A five year old can shuffle a deck of cards. Since half of the population is not in a coma or under the age of five, doubling the work week will have a bigger positive effect than the negative effect of counting out those unable to shuffle.

But if you don't like that one, then how about you add in population growth, since people will have all this spare time to boink each other? Hey, you're the one who wants to make this happen, I'm just trying to think outside the box for you.

Basically, everyone on the Earth is not going to drop everything and shuffle cards all day. We couldn't even survive that for very long. So, in taking the total poulation of the Earth as my number, I'm making an impossibly generous concession, in order to imagine this actually happening. The 40-hour week is something familiar to us; I'm trying to put it in real terms.

If you like, I could go back to having each person shuffle an exponential amount of decks every second...

Flint;429826 wrote:

The 40-hour week is something familiar to us;

that's less than 1/3 of the hours in a week....what do you do with all of that free time? ya know....it sounds appealing....

This site disagrees

My interest with perfect shuffling started with a simple observation that the standard poker deck containing 52 cards will come back to its or original order if riffle shuffled precisely interleaving the two 26 haves 8 times.

So on the ninth time, the shuffle will be the same as the first shuffle.

He expects a "perfect riffle shuffle", though.

lumberjim;429815 wrote:

5?

:lol:

Flint;429826 wrote:

everyone on the Earth is not going to drop everything and shuffle cards all day. We couldn't even survive that for very long.

Why should you start injecting reality and common sense into this discussion at this late date?? You might as well start thinking about whether those monkeys with the typewriters have had touch typing classes or not.

Okay, I hear what people are saying. You want more shuffles getting done, okay. Let’s look at that.

I’m going by HLJ’s calculation (I’m too lazy to do it myself) that in 9.e+33 permutations you would have a 50% chance of repeating the same shuffle.

I’m using the entire population of planet Earth (6,648,429,413), and I’ll have them shuffle one deck of cards per second, year-round. It would take 42,925,646,062,489,266.972805985537223 years to get through 9.e+33 permutations.

How long is that? Well, the Earth is estimated to be 4.54 billion years old. It’s 9,454,988.1194910279675784109112819 times longer than that.

[SIZE="3"]So, if every living person on the Earth shuffled one deck of cards per second, for 9 million times longer than the Earth has existed, there would be a 50% chance of the same shuffle coming up.

THE SAME EXACT SHUFFLE HAS NEVER HAPPENED, AND NEVER WILL HAPPEN. EVER.[/SIZE]

unless it does...

It totally happened to me the other day. Take my word for it.

Would you settle for 1% chance?

Flint, are you OCD?

HungLikeJesus;430069 wrote:

Would you settle for 1% chance?

If you're offering to calculate how many permutations of 52! you would have to go through to have a 1% probability of repeating a permutation...
I would love to know that.

I'd better present my logic to see if anyone disagrees.

(n*(n-1))/(2*52!) = P; solve for P, where P = probability of 2 or more matching shuffled decks in n shuffles.


(n-1)^2 < n*(n-1) < n^2, so

(n-1)^2/(2*52!) < n*(n-1)/(2*52!) < n^2/(2*52!),
so the first equation can be simplified (for large values of n) as

n^2 ~= P*2*52!

n = square_root(P*2*52!)

If P = 0.5 (50%), n = 8.98e33
If P = 0.1 (10%), n = 4.0e33
If P = 0.01 (1%), n = 1.27e33

I just noticed a flaw in this logic. Can you see it?

lol - I can't even follow that!

pull my finger!

I forgot to check one of the boundary values. For n = 52!, P should equal one. Instead, for n = 52!, P = (52!-1)/2, so something is not right.

classicman;430138 wrote:

lol - I can't even follow that!

Yeah I was thinking n=52????????????????????, not 52!

and I thought "n" was a letter not a number - silly me.

HungLikeJesus;430196 wrote:

I forgot to check one of the boundary values. For n = 52!, P should equal one. Instead, for n = 52!, P = (52!-1)/2, so something is not right.

Then, what is this n? Does n=52 or does n=number of shuffles?

HungLikeJesus;430116 wrote:

I'd better present my logic to see if anyone disagrees.

(n*(n-1))/(2*52!) = P; solve for P, where P = probability of 2 or more matching [COLOR="Red"]shuffled decks in n shuffles.[/COLOR]

I think n=n.

lmao

You damn kids and your new math.

n represents the number of shuffles. When the decks have been shuffled 52! times (52x51x50x...x2x1), all possible combinations --- hey, I think you're right! I've been thinking about this wrong. You don't have to encounter every possible combination to get two that match. Maybe my original calculations were right. I will think about it while I shower. I hope this B&B has a lot of hot water.

Can we watch...um, er...I mean watch the mathematical genius as his mind dazzles with brilliance and...

Just give me a number so I can slap you in the face with it. I haven't met my quota for scathing remarks this morning.

Shawnee123;430216 wrote:

Can we watch...um, er...I mean watch the mathematical genius as his mind dazzles with brilliance and...

Sorry, I have to be up to the big house in 10 minutes, to get the second B of this B&B.

Buttsecks and barbecue?

Well, it is Montana. Rules are rules.

After last night, I hope it's not too spicy.

classicman;430138 wrote:

lol - [COLOR="Red"]I can't even follow that[/COLOR]!

I think that's the point

:lol: busting a gut

HungLikeJesus;430196 wrote:

I forgot to check one of the boundary values. For n = 52!, P should equal one.

No, P will never be one. You are never guaranteed to get a match.

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.

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

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

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.

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

Happy Monkey;430289 wrote:

Eight, if you know what you're doing.

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.

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&#8217;ve got four pairs of socks in a box, and you&#8217;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&#8217;ve got a red, green, blue, and orange sock; you could draw:

[COLOR="red"]R [/COLOR][COLOR="SeaGreen"]G[/COLOR] [COLOR="Blue"]B[/COLOR] [COLOR="DarkOrange"]O [/COLOR]
[COLOR="red"]R [/COLOR][COLOR="seagreen"]G[/COLOR] [COLOR="darkorange"]O[/COLOR] [COLOR="Blue"]B[/COLOR]
[COLOR="red"]R [/COLOR][COLOR="blue"]B [/COLOR][COLOR="seagreen"]G[/COLOR] [COLOR="darkorange"]O [/COLOR]
[COLOR="red"]R [/COLOR][COLOR="blue"]B[/COLOR] [COLOR="darkorange"]O[/COLOR] [COLOR="seagreen"]G[/COLOR]
[COLOR="red"]R [/COLOR][COLOR="darkorange"]O [/COLOR][COLOR="seagreen"]G[/COLOR] [COLOR="blue"]B[/COLOR]
[COLOR="red"]R [/COLOR][COLOR="darkorange"]O[/COLOR] [COLOR="blue"]B[/COLOR] [COLOR="seagreen"]G[/COLOR]

And that&#8217;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&#8217;s not only possible that the same shuffle has never happened, it&#8217;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&#8217;t required to go through all of them to get a repeat, but&#8230;

...we&#8217;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.

Flint;430302 wrote:

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.

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.

Flint;430326 wrote:

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.

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.

Happy Monkey;430333 wrote:

I believe I addressed that in the second sentence.

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

Flint;430221 wrote:

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.

lookout123;430337 wrote:

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.

Actually- Sheldon is all the proof we need K?

:)

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

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?


[size=1]Note I said Flint does it, because I sure as hell don't have the energy. :)[/size]

0.99, but yeah.

As long as there are a finite number of cards, it is completely possible that two identical shuffles could happen, although highly unlikely.

Why not write a computer program to see if you can get any repeat shuffles in a reasonable time frame?

Or even do it for one suit and then extrapolate how long it would take for the same thing to occur in an entire deck?

I shuffle cards to pass time and break boredom. I know the montony of shuffling can produce boredom, but for some reason it relaxes and helps sooth tension, stress and anxiety.

One thing I have been noticing lately when I shuffle the deck and look through it afterwards to just see the distribution.

sometimes there is only on set of double number (same number, varying suit) and at other times as many a five such pairings in a single after-shuffle. Recently too I have been finding instances of 3 same number of differing suits grouped together and at others straights and the such.

I am a math affecianado too, and probability/Permutations fascinate me, and I am just wondering why this occurs, it seems that shuffling the deck is just as chancy as playing a game of Poker and that everything is determined by the luck of the draw.

I would appreciate some banter on this phenomenon because it bugs me that it happens, but I accept it begrudingly. I mean seeing one, two and as many pairings as 5 from a single overhand shuffle of 8 to 20 passes really annoys me.

And I guess we are all looking for that perfect shuffle, aren't we? Well I guess it'll never happen, not with the number of permutations available for a standard Poker deck. Thanks.

I think the perfect shuffle is merely an entirely random shuffle. And that means there are going to be pairs and whatnot... because it's random. You shouldn't have the expectation of no pairs if it's more likely that pairs will happen.

Anyone expecting a perfect shuffle from a random shuffle is doing it wrong. One should do a stratified random shuffle to break up the patterns. The Wizard of Oz told me so.

Faro shuffle.

How many permutations does the Icky Shuffle have?

So what happens when I shuffle down the road?

Will you be doing the Airborne Shuffle? Sounds like something one would do before playing 52 Pickup; but, noooooooooooo ...

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tw;874990 wrote:

So what happens when I shuffle down the road?

You end up in Buffalo.

Tangentially pertinent. I read somewhere years ago that 7 shuffles between hands was optimal. After that, the deck does not get more random.

[YOUTUBE]SNYBh4ryl5k[/YOUTUBE]

Since we're drifting, kinda...

I don't know how many times the cards were shuffled, but, this one time, at whiskey camp, I got dealt this hand. Deuces were wild. True Story®.

:yelgreedy

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:D

[/drift]

Gravdigr;875400 wrote:

Since we're drifting, kinda...

I don't know how many times the cards were shuffled, but, this one time, at whiskey camp, I got dealt this hand. Deuces were wild. True Story®.

:yelgreedy

[ATTACH]45397[/ATTACH]

:D

[/drift]

How many did you draw? ;)

Hell, I almost didn't give the cards back after the hand!

[YOUTUBE]ohlFjyCE0gA[/YOUTUBE]

Fool in the rain?

I like it, but I don't get the reference

teh beat

eh? still not getting it. is it in reference to spexx's youtube?, cuz i cant view that.

A shuffle is a type of swing drum beat

http://en.wikipedia.org/wiki/Shuffle_note#Rosanna_shuffle

Oh. Um. Ok.

Flint is usually funny.

Your face is usually funny.

His face is always fun - - - - - ny

lumberjim;876754 wrote:

Oh. Um. Ok.

Flint is usually funny.

Flint;981263 wrote:

Your face is usually funny.

~3.5 years, and that's the best comeback ya could come up with?

:lol2:

Yeah, and I still feel that way.

Feel these

bro, get this lump checked out

lol

*compassionate assholeism