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WilburrUK Veteran user 389 Posts |
Hi Mike,
Quote:
On Apr 1, 2014, Mike Powers wrote: You are, I wasn't. I was merely trying to point out that the order of the cards is not the only factor by which they can be rearranged. I don't want to start a fight about it though! Quote:
If so, how do you get a greater number of orders than 52!?? Please show an ordering of the deck that's not among the 52! I think that what lcwright1964 was referring to when talking about subsets was the idea that, in any given order, you can also arrange the deck by splitting it into several piles of varying sizes, so, he also was referring to other types of rearrangement. 52! just isn't a big enough number for some of us (not that size is important, it's what you do with it). Wil. |
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Vlad_77 Inner circle The Netherlands 5829 Posts |
We need Neil deGrasse Tyson, Stephen Hawking, and the lasses and lads at Numberphile.com to visit here!
To WilburrUK, Would Graham's Number be large enough for you? Or, if you really want to get number ****ed, there is always the googolplex. A number so unimaginably large that if the universe measured a googolplex metres across and you could somehow travel it, you would eventually meet a carbon copy of yourself because the possible number of quantum states in our universe is a measly 10^97 and a googolplex chortles at 10^97's puniness. ) |
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R.E. Byrnes Inner circle 1206 Posts |
I'll try to locate and summarize the relevant portion of the Diaconis book. I found it notable because the answer seemed obvious: 52! And he noted that is the obvious response, before going into the counter-intuitive explanation. Perhaps it was in the context of shuffling, where not all 52! permutations can realistically arise from a single shuffle. I only have the book in electronic form, which makes searching for something harder without having the recall of roughly how many pages had passed at a particular point, where it was in the page, etc.
The most vivid way of demonstrating the hugeness of 52! is to compare it with the number of atoms in the universe, which is around 10 to the 81st. Because the exponent that attaches to the total number of deck combinations is something like 37, the two numbers seem closer than they in fact are, but both are plainly massive. "Possible number of quantum states" is a really cool idea, and seems to be about as big as a number gets, short of infinity. |
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martydoesmagic Inner circle Essex, UK 1665 Posts |
Those following this thread might also find this clip from the BBC show QI of interest:
https://www.youtube.com/watch?v=SLIvwtIuC3Y Marty |
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LobowolfXXX Inner circle La Famiglia 1196 Posts |
Quote:
On Mar 31, 2014, Mike Powers wrote: It depends what one means by "arrangements." Considering the deck as an eternal circle, the "arrangement" is the cards' relative order to each other. For instance, one could consider a particular stack as an "arrangement." Whichever card happens to be on top, if they're still in stack order, wherever you cut the deck, you know where all the other cards are. This is not a mathematical error, but a definitional choice, and it would result (as Jamie says in the post before the excerpted post) in 51! "arrangements."
"Torture doesn't work" lol
Guess they forgot to tell Bill Buckley. "...as we reason and love, we are able to hope. And hope enables us to resist those things that would enslave us." |
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LobowolfXXX Inner circle La Famiglia 1196 Posts |
Quote:
On Mar 31, 2014, Mike Powers wrote: You don't get the cards in a row, but it doesn't matter from a probability standpoint; you're dead-on (other than the rounding) - it's one in 221. Alternatively, this can be computed by figuring the number of possible hands: (52)(51)(1/2). (Fifty-two possible first cards times 51 remaining second cards, but divided by two, since it doesn't matter what order you get the cards in, e.g. Ace of Spades followed by Ten of Diamonds is identical to Ten of Diamonds divided by Ace of Spades). That gives you 1326 total hands, and six of them are ace pairs - C+D, C+H, C+S, D+H, D+S, and H+S. 1326/6 = 221.
"Torture doesn't work" lol
Guess they forgot to tell Bill Buckley. "...as we reason and love, we are able to hope. And hope enables us to resist those things that would enslave us." |
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LobowolfXXX Inner circle La Famiglia 1196 Posts |
Quote:
On Mar 31, 2014, Poof-Daddy wrote: 16-1 odds against, actually, or 1 in 17. No matter what the first card is, out of the remaining 51 cards, 3 will give you a pair. 3/51 = 1/17.
"Torture doesn't work" lol
Guess they forgot to tell Bill Buckley. "...as we reason and love, we are able to hope. And hope enables us to resist those things that would enslave us." |
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lcwright1964 Special user Toronto 569 Posts |
Quote:
Please elucidate on the concept of more than 52! orderings to a deck. There are 52! permutations of all 52 cards. There are 52!/1! permutations of 51 cards sampled from a 52 card deck. There are 52!/2! permutations of 50 cards sampled from a 52 card deck. etc. all the way down to 52!/39! permutations of 13 cards from a deck (bridge hands, but in this case card order matters) etc. all the way down to 52!/47! permutations of 5 cards from a deck (poker hands, but taking account of order, so that AKQJ-10 in spades is not the same as AKQ-10-J in spades) etc. all the way down to 52!/51! permutations of 1 card taken from a deck. Add up all of the subset possibilities, and there you go. That's what I mean. Les |
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Mike Powers Inner circle Midwest 2983 Posts |
There still seems to be some confusion flying around in regard to this issue. Non math folks may end up with the wrong idea. Can we all agree on this:
1. We're talking about a normal deck of 52 cards all face down like when you play cards. The backs are identical etc. This is just a regular Bicycle deck all face down with no jokers or add cards etc. Just ace through king of each suit. 2. If we create a list numbered from 1 to 52 and write in the 1 position the name of the top card of the deck and so on down to 52 for the bottom card of the deck. We have recorded the order of the 52 cards from top to bottom. If it were in Aronson Stack, 1 would be the JS and 52 would be the 9D. 3. Given this very clear picture of what's going on i.e. normal deck, all face down sitting on the table waiting for someone to write down the order of the cards, then there are exactly 52! lists from 1 to 52 which show the possible orderings of our nice, normal, all face down deck. There is not even one more 1 to 52 list needed. Every possible arrangement is in that 52! stack of lists. This is what the question "How many ways can a deck of cards be arranged" is trying to get at when a magician or gambler asks. People who ask that question should be told that the answer is 52!. Right?? Mike
Mike Powers
http://www.mallofmagic.com |
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WilburrUK Veteran user 389 Posts |
Quote:
On Apr 5, 2014, Mike Powers wrote: ok, sorry! I wasn't sent the memo that said that you make the rules round here. |
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vindar New user Paris, France 62 Posts |
I have to agree with Mike here. Although one can give many interpretations of what "the number of way to arrange a deck of cards" means, I believe that the canonical answer is 52!. It is the number I would give any layman asking me this question.
However, I also realize that it is interesting (at least from a math. point of view) to remark that other results may be obtained with different assumptions. For instance when taking into account the asymmetry of some cards... |
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Mike Powers Inner circle Midwest 2983 Posts |
I'm not sure where the "making rules" notion comes in. If you don't define the conditions under which you're working, the answers are all over the place. I merely defined the conditions that I'm working under. Under those conditions, which I think are what people generally assume we mean when we say "order of the cards", the answer is 52!. If someone wants to ask the question about multiple decks, decks with jokers, decks with some cards reversed, decks with cards containing different colored backs etc. That's fine. It just doesn't seem to fit the context of magic, gambling etc.
If someone shuffles a deck and places it on the table and asks "how many ways can these cards be ordered" I doubt that they want an answer based on reversed cards, different colored backs and so on. The assumption, unless otherwise specified, would likely be what I described above. I'm not "making rules", I'm defining the conditions so that the answer is unambiguous. Seems fair to me..... Mike
Mike Powers
http://www.mallofmagic.com |
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owen.daniel Inner circle England 1048 Posts |
I agree that the original question was interested in the number of orders of 52 distinct cards, all face down, which is indeed 52!
However, Quote:
On Apr 2, 2014, R.E. Byrnes wrote: So, R.E. Burnes, I imagine that perhaps the calculation you're considering is the number of arrangements that can be reached after a single riffle shuffle of the deck. This is discussed under the topic of Gilbreath shuffles, p.68, where the authors (of Magical Mathematics) claim that the number of shuffles is 2^51 which is roughly 2.25 x 10^15. They justify this on p.80. My justification of this (which differs from theirs) is as follows: * When performing a riffle shuffle you cut off 0 <= k <= 52 cards from the top. * Given that you cut of exactly k cards, there are Binomial(52,k) ways to "weave" them into the remaining cards (whilst maintaining their relative order). * So, summing over all values of k, the number of distinct riffle shuffles is the sum of k Binomial[52,k], from k=0 to 52. Which is exactly 2^51. I'm not certain that's the claim you were thinking of, but thought it could well be. Owen |
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vindar New user Paris, France 62 Posts |
Hi Owen,
Although I agree with most of your argument, I just want to remark that the exact number of arrangements that can be obtained with one riffle shuffle of a deck of n cards is not exactly 2^(n-1). Indeed, during your argument, you tacitly assume that two distinct riffle shuffles (with different cutting values k) yield different orderings of the cards. However, this is not always the case. For example, cutting 0 or 52 cards leave the deck unchanged in both case. Similarly, for any k, there is exactly one configuration in binomial(n,k) which leaves the deck unchanged. Luckily, in all other cases, the resulting arrangement of the deck is indeed unique. Thus, the total number of ordering of a deck of n cards with one riffle shuffle is exactly the sum from k=0 to n of binomial(n,k) minus n which equals 2^n - n. Ok, I admit that I am being pernickety here The exact result does not really matter anyway. The point you made that this number is much smaller that n! remains true... Cheers, |
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Mike Powers Inner circle Midwest 2983 Posts |
It's clear that a single shuffle doesn't even come close to randomizing the deck. I believe that Diaconis' research shows that it requires 7 "good" riffle shuffles. BTW Wikipedia has a great article on shuffling: http://en.wikipedia.org/wiki/Shuffling
The number of possible orders of the 52 card deck (face down etc) is 52!. But the notion of shuffling and ending with an order someone else shuffled into depends on whether the deck has been truly randomized. There are some nuances that can be brought in, but my understanding is that Diaconis' research shows that 7 "good" riffle shuffles randomizes the deck. That's an interesting fact for card guys to know as well as the 52! possible orders. I did a presentation on gambling for a Rotary Club awhile back. I used the 52! bit as well as the 7 shuffles thing. While I was talking about Dianconis' work, I was doing false shuffles - both Zarrows and Push Throughs. After six shuffles, I mentioned that the deck might not be truly random and spread the face up. They were in new deck order which blew everyone away. Then the topic became "card control." Mike
Mike Powers
http://www.mallofmagic.com |
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vindar New user Paris, France 62 Posts |
Yes, 7 shuffles is the number agreed upon to say that a deck of 52 cards is sufficiently shuffled (such that very little information remains from the initial configuration).
This reminds me of a nice little card trick that is usually given as an example why 3 riffle shuffles do not mix the deck sufficiently. Effect: The magician gives a deck of card to a spectator and then turns his back on him. The spectator is invited to perform two riffle shuffles. He then cut the deck and look at the top card and buries it anywhere in the deck. He perform another riffle shuffle and give the deck back to the magician. The magician spreads the card face up and reveals the chosen card ! Ok, the trick is not foolproof but it does work about 84% of the time with a deck of 52 cards. And when it does work, it is pretty close from real magic ! By the way, does anyone know the name of the creator of this trick ? Thanks ! |
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owen.daniel Inner circle England 1048 Posts |
Vindar.
You're quite correct, and in fact my claim that Sum[ k Binomial[n,k] ] is 2^(n-1) is in fact incorrect as well (shouldn't type without thinking)! I attended a talk recently by Diaconis where he spoke at length about many of his shuffling results. It would appear as though he's currently trying to understand how one could model the "smooshing" style shuffle, where cards are spread on the table and then messily regathered in sweeping motions. Owen |
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MuscleMagic Special user 794 Posts |
W0w!
very cool info |
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martydoesmagic Inner circle Essex, UK 1665 Posts |
Quote:
ok, sorry! I wasn't sent the memo that said that you make the rules round here. Mike doesn't make the rules, Hoyle does. Every card player knows that! Got to back Mike's stance here, there's no need to be pedantic about the maths involved, this just serves to confuse the matter. In most card games, the back colour, symmetry, and so on, is not a factor, so shouldn't be considered when calculating this number. So, 52! FTW! Marty |
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itsmagic Inner circle middle earth 1117 Posts |
I appreciate Mike's input - so back off him!
Marty, just can't figure out what is "FTW!" |
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