[mizuchaud]

Noel, I'm just trying to figure out the pattern here.

The structure of the shuffle seems to be in question.

[TheEMan]

Amount of chips in each stack/different colour takes - times to shullfle pefect to get to starting point.

10 chips = 6 Times
9 chips = 9 Times
8 chips = 4 Times
7 chips = 7 Times
6 chips = 6 Times
5 chips = 5 Times


here is a vid of an perfect shuffle of 9 chips in each stack
http://www.21ace.com/poker-chip-tric...%20Shuffle.WMV

[lamonica]

Originally Posted by mizuchaud

Noel, I'm just trying to figure out the pattern here.

The structure of the shuffle seems to be in question.

Right. The problem isn't well-defined, so the answers can't be either. For example, if I have stacks ABCDE and 12345, then I would consider both A1B2C3D4E5 and 1A2B3C4D5E to be just as perfect, and I would consider it reasonable to randomly alternate between which side is dominant, and also would consider "cutting" (top half goes left vs. top half goes right) to be random as well. And of course, odd stacks introduce additional variables.

The point is that the problem doesn't say "perfect and exactly repeated", and the issue is that there are several different perfect cut and shuffle sequences. If you think about it, you could easily produce an infinite number of perfect steps if you wanted to - it wouldn't happen in practice though.

Even without the odd stacks, just the differences that I mentioned are enough to show that there is no concrete answer -other than a probablility distribution (in fact, several of them, depending on your rules).

Now, if you decide to make some concrete choices - say, for example, stacks always cut top-to-the-left-pile, bottom-to-the-right, AND left-side chips are dominant (shuffled chips from the left always go on top of their peers from the right), then it can be reduced to a precise number, and combinatorics would yield a closed formula for any size stack of 2n. If I'm bored this weekend I'll produce the forumulas.

[TheEMan]

i mostly shuffle with my left hand, and that means that the the bottom chip in the left stack ends up at the bottom after each shuffle.

i dont see any logic in these shuffles though

11 chips = 11 Times
10 chips = 6 Times
9 chips = 9 Times
8 chips = 4 Times
7 chips = 7 Times
6 chips = 6 Times
5 chips = 5 Times
4 chips = 3 Times
3 chips = 3 Times

[Wedge Rock]

As mentioned, it depends if you cut the stack to the left or to the right (and the smae way every time). It also depends on which stack goes to the bottom and whether or not its the same stack everytime.

I recall "discovering" this phenomena a month or so ago. I can't recall if I was shufling stacks of three or four, but I recall it took three perfect shuffles (cutting the same way each time and with the same stack ending up being the bottom chip each time) to "un-dirty" the stacks.

[plz]

This is a classic mathematics problem - it's called the "perfect shuffle" problem. It's typically analyzed using cards, not chips, but obviously the analysis applies to chips as well. Here's a solution that I worked out earlier this year:

Start with any deck (or stack of chips) that has an even number of cards. A card’s position in the deck is numbered from bottom to top, with 1 = bottommost card, and n = topmost card (an even number). A “perfect shuffle” is defined as the action of dividing the deck exactly in half and combining the two halves in a manner that perfectly interlaces cards from each half of the deck.

There are two cases (although Case 2 below can be shown to be a subset of Case 1):

Case 1: The bottom card from the deck does NOT remain in the same bottom position during each shuffle.

Case 2: The bottom card (and thus the top card also) DOES remain in the same position during each shuffle.

Case 1: For any card starting at position p, its position after a shuffle can be calculated as:

S(p) = 2 * p modulo(n+1)

So after k consecutive perfect shuffles, a card starting at position p has the following position:

Sk(p) = 2^k * p modulo(n+1)

Note that if the number 2^k modulo(n+1) = 1, then the above equation reduces to:

Sk(p) = p

And that means that every card has returned to its original position in the deck. So to figure out the number of perfect shuffles necessary to return every card to its original position, you find the integer k such that 2^k modulo(n+1) = 1. For example, for a deck of 10 cards (n = 10), k = 10 since 2^10 = 1024 and 1024 modulo(11) = 1. (By the way, it’s always possible to find an integer k that solves 2^k modulo(n+1) = 1 for any given n, so that means every deck will eventually return to its original order in a finite number of shuffles.)


Case 2: For any card starting at position p, its position after a shuffle can be calculated as:

S(p) = (2 * p) - 1 modulo(n-1), for p = 1 … n-1; for the card in position n, S(p) = n

So after k consecutive perfect shuffles, a card starting at position p has the following position:

Sk(p) = (2^k * p) - (2^k - 1) modulo(n-1)
= (2^k * p) - 2^k + 1 modulo(n-1)
= 2^k * (p - 1) + 1 modulo(n-1)

Note that if the number 2^k modulo(n-1) = 1, then the above equation reduces to:

Sk(p) = (p – 1) + 1 = p

And that means that every card has returned to its original position in the deck. So to figure out the number of perfect shuffles necessary to return every card to its original position, you find the integer k such that 2^k modulo(n-1) = 1. For example, for a deck of 10 cards (n = 10), k = 6 since 26 = 64 and 64 modulo(9) = 1.

Also note that Case 2 is really a subset of Case 1, because you can ignore the cards in positions 1 and n (they stay fixed) and you’re shuffling only cards in position 2…n-1, which is a deck with n-2 cards and thus in this case the solution is the same as Case 1 but with two fewer cards.

I created a spreadsheet that brute force calculates these numbers. Note that for a standard deck of 52 cards, with a Case 2 shuffle, the deck returns to its starting order in only 8 shuffles. For a Case 1 shuffle, it takes 52 shuffles to return to its original order!

[Wedge Rock]

Isn't this what I said?!

If not, its what I meant...

Actually, I remember Chris Ferguson doing a bit on this in the Ask the Pros show on Fox Sports. He could do a perfect shuffle rather quickly and the trick was that he would take a perfectly aligned deck, shuffle it so many times (5?) -- his perfect shuffles looked like ordinary random shuffles -- and he would retun you a perfect deck...

[mizuchaud]

Man, my brain hurts.
I only want to know why there seems to be a pattern...
4 in three shuffles and 8 in four shuffles seem to make sense as the shuffle seems to have in physical space a kind of four quadrant sorting thing going on.
the 10 chips per stack sorting out in 6 shuffles was a surprise to me.

If someone has the full numbers crunched I'd love to see it.

PS e man you rock. I do it rightie and it's the mirror image of your motion. (But not nearly so high as 9) (seven maybe)

[TheEMan]

Originally Posted by mizuchaud

Man, my brain hurts.

hahaha i agree with you.

[jdunford]

Originally Posted by plz

This is a classic mathematics problem - it's called the "perfect shuffle" problem. It's typically analyzed using cards, not chips, but obviously the analysis applies to chips as well. Here's a solution that I worked out earlier this year:

Start with any deck (or stack of chips) that has an even number of cards. A card’s position in the deck is numbered from bottom to top, with 1 = bottommost card, and n = topmost card (an even number). A “perfect shuffle” is defined as the action of dividing the deck exactly in half and combining the two halves in a manner that perfectly interlaces cards from each half of the deck.

There are two cases (although Case 2 below can be shown to be a subset of Case 1):

Case 1: The bottom card from the deck does NOT remain in the same bottom position during each shuffle.

Case 2: The bottom card (and thus the top card also) DOES remain in the same position during each shuffle.

Case 1: For any card starting at position p, its position after a shuffle can be calculated as:

S(p) = 2 * p modulo(n+1)

So after k consecutive perfect shuffles, a card starting at position p has the following position:

Sk(p) = 2^k * p modulo(n+1)

Note that if the number 2^k modulo(n+1) = 1, then the above equation reduces to:

Sk(p) = p

And that means that every card has returned to its original position in the deck. So to figure out the number of perfect shuffles necessary to return every card to its original position, you find the integer k such that 2^k modulo(n+1) = 1. For example, for a deck of 10 cards (n = 10), k = 10 since 2^10 = 1024 and 1024 modulo(11) = 1. (By the way, it’s always possible to find an integer k that solves 2^k modulo(n+1) = 1 for any given n, so that means every deck will eventually return to its original order in a finite number of shuffles.)


Case 2: For any card starting at position p, its position after a shuffle can be calculated as:

S(p) = (2 * p) - 1 modulo(n-1), for p = 1 … n-1; for the card in position n, S(p) = n

So after k consecutive perfect shuffles, a card starting at position p has the following position:

Sk(p) = (2^k * p) - (2^k - 1) modulo(n-1)
= (2^k * p) - 2^k + 1 modulo(n-1)
= 2^k * (p - 1) + 1 modulo(n-1)

Note that if the number 2^k modulo(n-1) = 1, then the above equation reduces to:

Sk(p) = (p – 1) + 1 = p

And that means that every card has returned to its original position in the deck. So to figure out the number of perfect shuffles necessary to return every card to its original position, you find the integer k such that 2^k modulo(n-1) = 1. For example, for a deck of 10 cards (n = 10), k = 6 since 26 = 64 and 64 modulo(9) = 1.

Also note that Case 2 is really a subset of Case 1, because you can ignore the cards in positions 1 and n (they stay fixed) and you’re shuffling only cards in position 2…n-1, which is a deck with n-2 cards and thus in this case the solution is the same as Case 1 but with two fewer cards.

I created a spreadsheet that brute force calculates these numbers. Note that for a standard deck of 52 cards, with a Case 2 shuffle, the deck returns to its starting order in only 8 shuffles. For a Case 1 shuffle, it takes 52 shuffles to return to its original order!

Do you mind posting that spreadsheet? Thanks.