How Many Shuffles for a Deck of Cards?

[daveed]

at least how many times does a deck of 52 cards need to be shuffled for it to be considered well shuffled?

 

[NateTG]

at least how many times does a deck of 52 cards need to be shuffled for it to be considered well shuffled?

What do you mean by well shuffled?
also
What shuffling technique do you suggest?

 

[check]

I once heard that the technique called "The Riffle", where the deck is cut in two piles and the cards are shuffled together on a table, where a few cards from one side fall, then a few from the other etc. needs to be done seven times to ensure randomness.

Edit

Just found this good site for you:
http://www.encyclopedia4u.com/s/shuffling-playing-cards.html

 

[shmoe]

On a related note, if you perform 8 "perfect rifle shuffles" to a deck of 52 cards, it ends up in the exact same order you started with. How to describe a "perfect rifle shuffle"- split the deck perfectly into 2 piles, the 26 top cards and the 26 bottom cards. As you rifle them together, you first put down one card from the bottom pile, then one card from the top pile, then one from the bottom, etc.

This is pretty near impossible to do quickly in practice, you first have to split the deck in half perfectly then have the cards fall alternately one from each hand. The moral is there has to be some "randomness" to your shuffling tecnique, such as clumsy fingers, or maybe greasy fingers from hot wings.

 

[Hurkyl]

If you believe that, I suggest you actually try it.

 

[shmoe]

If you believe that, I suggest you actually try it.

Already have years ago. Do you not believe it at all or is 8 wrong (could be 7 or 9, but I'm remembering 8)? You can think of any shuffle as an element of the group of permutations of order 52. So if you keep performing the exact same shuffle over and over again, you eventually get back to where you started. What seems suprising at first is a perfect riffle shuffle has order only 8. This is less suprising when you right down it's decomposition as a product of disjoint cycles.

It takes the sequence 1, 2, ..., 51, 52 to
1,27,2,28,...,25,51,26,52

So we get

(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)

Hmm, guess it was 8 after all. I didn't intend to write it all out, but it was easy enough. If n is less than or equal to 26, n->2n-1. If n is greater than 26, n->2n-52.

 

[Hurkyl]

Off by one errors suck. :( I placed card #2 winding up somewhere other than spot 2 at the end.