Faro New Deck Restoration

[mrmagik68]

Happy Saturday all!

In the wonderful Skinner Tapes, Roger Klause talks about how Jay Ose used to open a new deck of cards, display the new deck order and then do 8 perfect faros, restoring the deck back to new deck order. Does anyone know if this method is in print somewhere? I've tried it several times, both in/out faros, but can't seem to restore the deck back to new order. Thanks!


Roberto

[katterfelt0]

Happy Saturday all!
In the wonderful Skinner Tapes, Roger Klause talks about how Jay Ose used to open a new deck of cards, display the new deck order and then do 8 perfect faros, restoring the deck back to new deck order. Does anyone know if this method is in print somewhere? I've tried it several times, both in/out faros, but can't seem to restore the deck back to new order. Thanks!
Roberto

Paul Gertner's "Unshuffled" is one good resource that uses the principle.
Maestro Tamariz also has info on it in Mnemonica.

[Sean-Dylan]

8 perfect "Out" faro shuffles (where the top and bottom card remain the same) w/ 52 cards will return the deck to the same order you started with.

It is in print and discussed in several places. Off the top of my head:

- Expert Card Technique by jean Hugard and Frederick Braue

- Collected Works of Alex Elmsley vol. II by Stephen Minch

- Card College vol. 3 by Roberto Giobbi

- Ed Marlo's Revolutionary Card Technique (Faro Notes)

- Paul Gertner's Steel & Silver by Richard Kaufman (Unshuffled)

[Leo Garet]

If you've got all day and the spectators can stay awake, 52 in-faros will do the same job. 26 in-faros will take you half way.

At the half-way stage:

Original top card becomes bottom card, original bottom card becomes top card.

Original second from top card becomes second from bottom, original second from bottom becomes second from top.

And onwards.

In the meanwhile, another 26 in-faros and you're back to the start.

If memory serves, one (at least) of the tables in "Expert Card Technique" is incorrect.

[mrmagik68]

Thank you gentlemen! I can definitely work with the resources mentioned. Much appreciated.


Roberto

[Dave Le Fevre]

In 1968, I wrote software which worked out how many out-faro shuffles it takes to restore a deck's order, the size of the deck (even, of course) being a parameter.

I no longer have the data, but I remember that a 52-card deck requires an extraordinarily low number (8, as is well known). A slightly larger or slightly smaller deck would take a huge number of out-faros. So we're fortunate that we use 52-card decks.

[Yehuda]

There's this, if anyone's interested:

https://www.denisbehr.de/faro/

Yehuda

[Dave Le Fevre]

Thanks Yehuda - I thought that someone must've created something like that

[Leo Garet]

There's this, if anyone's interested:
https://www.denisbehr.de/faro/
Yehuda

Thanks!

I had seen this before, but it was a few years ago. A great reminder.

[Gareth]

18 card deck also takes 8 outs to return to the beginning.

[Steve Hook]

Here's a good article that came out today at a cybersecurity site I follow. It includes animations showing the 8-shuffle in-faro results and 52-shuffle out-faro results. And as often happens, Persi Diaconis' writings are highly quoted.

https://nakedsecurity.sophos.com/2022/10/24/serious-security-you-cant-beat-the-house-at-blackjack-or-can-you/

[Bob Coyne]

One useful formula is that decks with 2^N cards will recycle to original order in N out shuffles. So, for example, 32 cards (2^5) will recycle in 5 shuffles.

[Dave Le Fevre]

Useful if one's playing Bezique

[S. Brent Morris]

I hope it's too late to contribute to this thread. I just joined Genii Forums and this thread immediately caught my eye. I apologize for the length of my reply, but there is a bit of math involved in the general answer, and there's no short, complete answer.

The number of faro shuffles needed to return a deck of size N to its original order requires solving for an exponent in a modular equation. This type of math could be covered in discrete math or number theory, probably a senior/graduate level course, but it could be covered earlier

For an even-sized deck—
The number of out shuffles that returns a deck to its original order is the smallest X that satisfies:
2^X ≡ 1 (mod N-1)
The number of in shuffles that returns a deck to its original order is the smallest X that satisfies:
2^X ≡ 1 (mod N+1)

Example: Let N = 52 (an ordinary deck). Then 2^8 = 256 ≡ 1 (mod 51), so 8 out faro shuffles return a deck to its original order. Then 2^52 = 4,503,599,627,370,496 ≡ 1 (mod 53), so 52 in faro shuffles returns a deck to its original order, but the arithmetic is messy!

Special cases: If we note that 53 is a prime, then we have the special case where X = N-1 or X = 53-1 = 52. But now we need to get a little deeper into number theory. Further, if we have the special case where N is a power of 2, i.e. N = 2^k, then X = k.

For an odd-sized deck—
The number of out or in shuffles needed to return a deck to its original order is the smallest X that satisfies:
2^X ≡ 1 (mod N)

Example: Let N = 53 (an ordinary deck plus a joker). Then, as with the above example, 52 out or in shuffles return a deck to its original order. If N = 51 (an ordinary deck with one card missing), 8 out or in shuffles return a deck to its original order.

Bonus: For an odd-sized deck—
If X is the number of shuffles needed to return an odd-sized deck to its original order, then any sequence of X out or in shuffles returns a deck to a cyclical shift of its original order. (This is how you can do ACAAN with faros.)

There is table in both of my faro shuffle books, Magic Tricks, Card Shuffling, and Dynamic Computer Memories and Everything* You Wanted to Know about the Faro Shuffle *But Were Afraid to Ask, the give the number or out or in shuffles needed to restore decks of size 3-197. I can post a PDF of the table, if that's possible.