[Guest]

Fun with Perfect Shuffles:

We define the Perfect Shuffle to be an operation that, from a stack of 2n cards, does the following:

- Divide the deck into the "top n cards" (aka Top Deck) and the "bottom n cards" (Bottom Deck)

- reassemble a new single deck (ND) by taking the bottom card from the TD, then adding the bottom card from the BD on top of the ND... rinse and repeat until both the BD and TD are exhausted (the card on top of the ND will thus be the card that was on top of the BD).

This defines a permutation of our 2n cards. The questions are:

After how many Perfect Shuffles will the deck come back to its initial state, if the deck is composed of:

-4 cards ?

-10 cards ?

-52 cards ?

-20 cards ?

[Guest]

Wow, this seems hard, but...

After 2 shuffles, the deck has essentially reversed its order, meaning after 4 shuffles it will return to its original state. I'm not sure of a general formule though to solve the other problems.

[Guest]

Interesting results, I wish I had the time to try and work out a formula.

4 card = 4 / 6 cards = 6 / 8 cards = 12 / 10 cards = 20 / 12 cards = 24 / Heres an interesting one 14 cards = 3! / 20 cards = 6 / 52 cards = 52

Edited February 6, 2009 by WhiteSoxFan

[HoustonHokie]

Unless I'm mistaken, the trick is in

of any invidual card. With each successive shuffle, the card will occupy a different position in the deck than it's hadbefore. And it will do so by moving 2 times its current position down in the deck until it becomes part of the bottom of the deck. A card in the bottom part of the deck moves in the opposite manner, in that the distance it moves is based on its position with respect to the bottom of the deck, but it's still a multiple of 2 away from where it was - just in the opposite direction. Looking at the 52 card case and keeping track of the first card looks like this:

1. 1

2. 1*2 = 2

3. 2*2 = 4

4. 4*2 = 8

5. 8*2 = 16

6. 16*2 = 32

7. 32 - (53-32) = 11

8. 11*2 = 22

9. 22*2 = 44

10. 44 - (53-44) = 35

11. 35 - (53-35) = 17

12. 17*2 = 34

And so on. This should work for any size deck. It turns out that each card will occupy every position in the deck exactly once, so the number of perfect shuffles required to return the deck to its original state is equal to the number of cards in the deck.

Edited February 6, 2009 by HoustonHokie

[Guest]

After how many Perfect Shuffles will the deck come back to its initial state, if the deck is composed of:

-4 cards ?

-10 cards ?

-52 cards ?

-20 cards ?

I had gotten as far as n = 9 (or 18 cards in a deck) before I got the rest from POTM

Thank you POTM

--------------------------------------

n | 2n (deck) | # of shuffles

--------------------------------------

1 | 2 | 2

2 | 4 | 4 <--

3 | 6 | 3

4 | 8 | 6

5 | 10 | 10 <--

6 | 12 | 12

7 | 14 | 4

8 | 16 | 8

9 | 18 | 18

10 | 20 | 6 <--

26 | 52 | 52 <--

The pattern doesn't seem to have much of a pattern: a 50 card deck only needs 8 shuffles but 52 cards need 52 shuffles?

It's all sorts of messed up. It must be algorythmic since they made a program to generate the solutions.

I think I'm going to loose a lot of sleep over this one...

[Guest]

I was wrong earlier it never seems to take more shuffles than there are cards in the deck. I did figure out one interesting thing,

When n+1 is a power of two i.e. 3+1 = 2^2 (6 six card deck) or 7+1 = 2^3 (14 card deck) 15+1 = 2^4 (30 card deck) it takes that power of 2 +1 shuffles. So 6 cards = 3 shuffles, 14 cards = 4 shuffles, 30 cards = 5 shuffles, 62 cards = 6 shuffles and so on.

[preflop]

Fun with Perfect Shuffles:

We define the Perfect Shuffle to be an operation that, from a stack of 2n cards, does the following:

- Divide the deck into the "top n cards" (aka Top Deck) and the "bottom n cards" (Bottom Deck)

- reassemble a new single deck (ND) by taking the bottom card from the TD, then adding the bottom card from the BD on top of the ND... rinse and repeat until both the BD and TD are exhausted (the card on top of the ND will thus be the card that was on top of the BD).

This defines a permutation of our 2n cards. The questions are:

After how many Perfect Shuffles will the deck come back to its initial state, if the deck is composed of:

-4 cards ?

-10 cards ?

-52 cards ?

-20 cards ?

You would come to the answers of ...

4

10

52

20

unless i missed something here

[Guest]

Ok, WSF and eurick found the right answers indeed.

... since it's not always x shuffles for x cards.

4 cards -> 4 shuffles

10 cards -> 10 shuffles

52 cards -> 52 shuffles

but:

20 cards -> 6 shuffles

8 cards -> 6 shuffles...

It will obviously never take more than x shuffles to return a deck of x cards to its initial state.

So the questions become:

- (easy first) If we say that the bottom card of a deck is n°1 and that the top one is n° 2n, what is the formula to get the position of card n° y after one shuffle?

- Can you find a general formula for the number of shuffles it takes to return a deck of x cards to its initial state ? At least give a systematic way to find a solution (if possible one that doesn't include "putting the numbers in such website and get the answer").

[Prime]

This is an interesting problem.

I'll give a partial formula here and the rest, if I have time to formulate it simply.

If total number of cards in a deck is a power of two, (2ⁿ), then the number of perfect suffles required to return the deck into the initial state is 2*n.

E.g., For the deck of (2¹) = 2 cards -- 1*2 = 2 shuffles

For the deck of (2²) = 4 cards -- 2*2 = 4 shuffles

For the deck of (2³) = 8 cards -- 2*3 = 6 shuffles

for 16 -- 8 shuffles, and so on.

I've met some people who could do a perfect shuffle every time. Some of those spent time in prison and had nothing better to do, but practice.

I also noted, some card dealers in Las Vegas come pretty close to perfect shuffle.