deck of cards

[eruheru]

how many times would you have to shuffle a deck to wind up with all the cards in the original order?

[Rakdos]

if you do a perfect shuffle(one card between each one), 8 times

[EvoN1020v]

That's the problem: you can't do a perfect shuffle with 52 cards, it'll take like a million million of times to get the cards back in the original order.

[matt grime]

Who says you can't do a perrfect shuffle?

[BigMoosie]

how many times would you have to shuffle a deck to wind up with all the cards in the original order?

 

To answer your question,

 

52!

= 1x2x3x .... x50x51x52

= 80658175170943878571660636856403766975289505440883277824000000000000

[matt grime]

That certainly works if you mean 'repeat an arbitrary shuffle' until you get back to the beginning, but that is a weak bound, and in general the largest order of an element in S_n (shiffles of n objects) is not n! but considerably smaller. S_4 for instance has order 24=4! but the largest order of an element in S_4 is 4. But then there is the interpretation that you mean 52! shuffles will reorder the pack for all shuffles, which is certainly true (not sure if its optimal with this property since in S_4 the universal number ie the lcm of the orders of elements is 12)

 

Would the OP be so kind as to write his question unambiguously?

 

Ie, what do you mean by shuffle, precisely? A perfect riffle shuffle (interleaving), an arbitrary shuffle repeated, or just a series of random shuffles (when there is no deterministic answer, but merely a probablistic one)

[Bettina]

Hmm....I must be missing something.

 

I would think the answer is unattainable. Order to Chaos....No matter how many times you shuffle the deck.

 

Bettina

[5614]

Just as there is a probability of you getting dealt an Ace of any suite there's also a probability of getting an Ace and a 2. There's also a probability of getting an Ace, a 2 and a 3.

 

Continuing the trend and the probabilities getting even less likely there is also a probability of getting a whole suite in order.

 

etc. etc. until we end up with the probability of getting every single card in order. This is what the OP wants.

[BigMoosie]

That certainly works if you mean 'repeat an arbitrary shuffle' until you get back to the beginning' date=' but that is a weak bound, and in general the largest order of an element in S_n (shiffles of n objects) is not n! but considerably smaller. S_4 for instance has order 24=4! but the largest order of an element in S_4 is 4. But then there is the interpretation that you mean 52! shuffles will reorder the pack for all shuffles, which is certainly true (not sure if its optimal with this property since in S_4 the universal number ie the lcm of the orders of elements is 12)

 

Would the OP be so kind as to write his question unambiguously?

 

Ie, what do you mean by shuffle, precisely? A perfect riffle shuffle (interleaving), an arbitrary shuffle repeated, or just a series of random shuffles (when there is no deterministic answer, but merely a probablistic one)[/quote']

 

Wanna know if he's left handed too?

[matt grime]

IF you read the replies here it is evident that people have interpreted it in several different ways. Which one is it? Is it that unreasonable to ask, Bigmoosie, exactly which meaning the question ought to have?

 

If s is a shuffle in S_52, then the number of times it needs to be done to obtain the initial ordering again is (any multiple of) ord(s).

 

Often s is presumed to be the distinguished shuffle of a perfect riffle shuffle, which is eminently practically attainable as any card sharp can tell you. I don't know its order.

 

If we want a number that works for all shuffles then it is lcm(ord(s)) for s in S_52.

 

If we mean: shuffle at random, then another at random, the initial formation occurs once every 52! times on average, this is the probabilistic version.

 

The person who mentioned chaos has introduced a spurious concept, and it should be ignored.

[shmoe]

Often s is presumed to be the distinguished shuffle of a perfect riffle shuffle, which is eminently practically attainable as any card sharp can tell you. I don't know its order.

 

it's 8, what the second poster was refering to. It can be represented as

 

(1)(2, 3, 5, 9, 17, 33, 14, 27)(4, 7, 13, 25, 49, 46, 40, 28)(6, 11, 21, 41, 30, 8, 15, 29)(10, 19, 37, 22, 43, 34, 16, 31)(12, 23, 45, 38, 24, 47, 42, 32)(18, 35)(20, 39, 26, 51, 50, 48, 44, 36)(52)

 

where 1 is the bottom card. As you say, a perfect shuffle is possible with practice and a good deck. Getting 8 in a row takes *much* talent (or a heap of luck) but is still possible.

 

If we want a number that works for all shuffles then it is lcm(ord(s)) for s in S_52.

 

I wonder what this turns out to be. Crude upper bound is 52! like you said, but we can improve this without much work. Any permutation can be written as a product of disjoint cycles. In this form it's order is the lcm of the length of these cycles. Thus any s in S_52 has order dividing lcm(52,51,50,...1)=2^5*3^3*5^2*7^2*11*13*...*47=3099044504245996706400

 

On second thought, this appears to give lcm(ord(s)), since we can easily demonstrate a shuffe of each admissible prime ower order. There's no lone element of this giant order though. I would think it's not too difficult to come up with the highest order element, just consider its decomposition into disjoint cycles, they'd have to be prime powers and the sum would have to be less than 52, so not too many options a computer could search quickly.

[matt grime]

Now there's an interesting function:

 

F(n) = sup { lcm(p_1,p_2,..,p_r) : p_1|p_2|...|p_r is a partition of n}

 

No idea if there's a nice formula for it. What can we say about it though? It is increasing but not strictly: F(5)=F(6)=6. Obviosuly F(n)=>n, with possible equality as we've just seen. But beyond that nothing really springs to mind.

[entwined]

If you are talking about random mixing of the cards and what the odds are against the deck comming out exactly as packed, then the formule is 52x51x50x49 etc.

 

Look at it this way: Suppose you had 5 cards- an ace, a duce, a trey, a four and a five.

 

Now you mix them up and turn one over. The odds are 1:5 that the turned over card is the ace.

 

If the card turned over is the ace, then the odds are 1:4 that the next card would be the duce, so one could expect on an average that the ace would have to come up first 4 times before it was followed by the duce. That means that there would be 5x4 (20) mixings on average before the first two cards were ace-duce in that order.

 

when the first two cards are the ace-duce in that order, then the odds are 1:3 that the next card will be the trey therefore the mixing will, on an average have to occure 5x4x3 (60) times to get the cards to come up ace duce trey in that order.

 

In order to get the 4 to come up on scedual, a 1:2 shot, it will take an average of 5x4x3x2 (120 mixings). and then, of course the only card left is the 5 which is a 1:1 shot, so the total mixing for the 5 cards is 120 mixings, which just happens to be the product of 5x4x3x2x1.

 

See how that works?

[shmoe]

Now there's an interesting function:

 

F(n) = sup { lcm(p_1' date='p_2,..,p_r) : p_1|p_2|...|p_r is a partition of n}

 

No idea if there's a nice formula for it. What can we say about it though? It is increasing but not strictly: F(5)=F(6)=6. Obviosuly F(n)=>n, with possible equality as we've just seen. But beyond that nothing really springs to mind.[/quote']

 

You can make some more assumptions on the partitions when looking for a max, the p_i's can be taken to be relatively prime to one another and also as prime powers. I don't know of a nice answer either.

 

I did a rough search for 52. The highest I found has lcm 4*9*5*7*11*13=180180. This is obviously the order of an element, but I won't guarantee it is the highest.

[uncool]

If you assume a perfectly random shuffle, meaning that any of the 52! possibilities are equally likely, then you are describing a geometric random variable. The expected value of a geometric random variable is 1/ the parameter, which in this case is 1/52!. So it would take, on average, 52! times to get back to the original order.

[matt grime]

Is it a Geometric rv? You are after all *composing* shuffles. A proof that it is indeed geometric would not go astray (nor should it be hard to provide, if it exists, I think; it is a group action here after all).

[zaphod]

i am pretty sure the original poster was talking about perfect riffles, but you guys are on a roll, it seems