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Old 02-04-2008, 03:56 PM   #16
Shawnee123
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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.)
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Old 02-04-2008, 04:10 PM   #17
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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.
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Old 02-04-2008, 04:10 PM   #18
Aliantha
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Quote:
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.
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Old 02-04-2008, 04:14 PM   #19
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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.
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. . . . . . . . . . . . . . . . . . . . . . . . . . Terry Bozzio
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Old 02-04-2008, 04:21 PM   #20
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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).
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Old 02-04-2008, 04:22 PM   #21
Aliantha
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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.
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Old 02-04-2008, 04:44 PM   #22
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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.
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Old 02-04-2008, 04:59 PM   #23
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Old 02-04-2008, 06:51 PM   #24
deadbeater
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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.

Last edited by deadbeater; 02-04-2008 at 06:59 PM.
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Old 02-04-2008, 07:26 PM   #25
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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.
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Old 02-04-2008, 07:35 PM   #26
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....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.
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Old 02-04-2008, 09:55 PM   #27
Flint
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Quote:
Originally Posted by smoothmoniker View Post
You've assumed that shuffles produce random distributions.
No, I haven't actually:
Quote:
Originally Posted by Flint View Post
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.
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Originally Posted by Flint View Post
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?
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Originally Posted by smoothmoniker View Post
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.

Quote:
Originally Posted by smoothmoniker View Post
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.
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Originally Posted by deadbeater View Post
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.
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Originally Posted by lumberjim View Post
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.
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Originally Posted by HungLikeJesus View Post
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).
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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?
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There's a level of facility that everyone needs to accomplish, and from there
it's a matter of deciding for yourself how important ultra-facility is to your
expression. ... I found, like Joseph Campbell said, if you just follow whatever
gives you a little joy or excitement or awe, then you're on the right track.

. . . . . . . . . . . . . . . . . . . . . . . . . . Terry Bozzio
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Old 02-04-2008, 10:07 PM   #28
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Quote:
Originally Posted by Flint
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.
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Old 02-04-2008, 10:09 PM   #29
Flint
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Originally Posted by Clodfobble View Post
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.
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There's a level of facility that everyone needs to accomplish, and from there
it's a matter of deciding for yourself how important ultra-facility is to your
expression. ... I found, like Joseph Campbell said, if you just follow whatever
gives you a little joy or excitement or awe, then you're on the right track.

. . . . . . . . . . . . . . . . . . . . . . . . . . Terry Bozzio
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Old 02-04-2008, 10:14 PM   #30
Clodfobble
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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.
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