I am wondering if any one knew what, or where one could find or determine, the formula to calculate how a card's position is affected after 5 out faros? This would allow for a mathematical formula that could be used to obtain the position of any card in a 5 faro stack.

In other words, eliminating the need for any actual memorization, or at least providing a back-up to help.

Thanks
Glenn

Alright, well, through some tough work I manages to calculate a weird formula for a FOUR faro stack that changes every 3-4 cards, lol.

You do not want to know how ridiculously complicated the 5 faro one was. I gave up after the first five cards.


So, for determining where a card is after 4 out Faro's from it's original position;

Basically it is the original position of the card (for this situation we will assume from NDO) minus 1 times 15 plus the cards original position again. Then, if the answer is greater than 52, you subtract 51 until it is less than 52.

So, if we consider the NDO position to be N, and the final stack position to be S we get;

( N - 1 ) 15 + N = S

Or

N + 15 ( N - 1 ) = S

This is further simplified;

N + 15 ( N - 1 ) = S
N + 15N - 15 = S
16N - 15 = S

So, 16 times it's NDO position -15 equals the stack position. Unless S is > 52, in which case you subtract 51 until it is < 52

It is subtraction of multiples of 51 that really throws the wrench into the idea here, thereby changing the formula rather than it being a constant single formula that is applicable to all cards.

Simple enough

It is subtraction of multiples of 51 that really throws the wrench into the idea here, thereby changing the formula rather than it being a constant single formula that is applicable to all cards.

For ease of determining the NDO value, I actually start the faro's with the deck in straight numerical order, A-K and the suits in CHaSD (actually, I do SDCH just because I always organized my cards like that when playing games when younger and is thus ingrained into me as second nature.)

Any who, this makes determining the "NDO" value much more straight forward. Each suit is assigned a value based on it's order. For CHaSD, Clubs would be 0, Hearts would be 13, Spades would be 26 and Diamonds would be 39. Each suit is 13 more than the previous, as each suit contains 13 cards.

So, from here you just take the numeric value of the card, 1-13, and add the suit value of the card, and you have the card's NDO position.

Ace of Clubs would be Ace=1 and Clubs=0, so the Ace of Clubs' is NDO 1.
2 of Clubs would be 2=2 and Clubs=0, so the 2 of Clubs' is NDO 2.
Jack of Diamonds, would be Jack=11 and Diamonds = 39, so J of D is NDO 50, Q of D, 51, K of D 52.

The reason I am asking/looking for this is I am trying to find, or develop, a stack where by each card's location can be determined via the same formula.

Obviously this is a little bit more complicated than that. I would like to find one that can be done in my head on the fly, without qualifying for Mensa (even though I do qualify, I am terrible at doing basic math in my head quickly, but working out and solving formulas etc, just give me a calculator for the basic + - * / and square roots and I am golden, but I digress.)

I know that there is a stack where each card is a simple equation based on the preceding card, but this only works for knowing which card follows which, and not for the specific location of each individual card.

I know that Si Stebbins allows for each card's location to be calculated after a glimpse of the bottom card, but that is not quite what I am looking for.

I just want a single formula that allows me to take any card, and using it's value and suit, calculate its position in the stack, then of course shift it based on the bottom card, if the deck has been cut, thus displacing the stack's start and end points.

One last note, I am curious what formula the program 'StackView' uses to simulate a faro, as this would be what I am looking for, and perhaps it is a more simplified equation than what I have come up with and might actually be usable.

It may also be worth mentioning that I am new to stack work, out side of Si, and a brief time with the aforementioned stack where each card is determined by the one preceding it.

Obviously I am looking for a mathematical out to actually memorizing a deck. Remembering one equation is easier than a whole deck, lol.

Why don't you PM Nick Pudar? He is a very nice gentleman. He has been very helpful to me through PM's.

edh

P.S. His weblog is also very informative. He has created some nice effects using the Aronson Stack.

Yes, I actually just received a response from him, helped a bit. Did not lead to the exact solution I was looking for, lol, but I am beginning to think the 4-5 faro stack is not the solution, but it is nice in how it can be gotten in an out of.

Frankly, gdw, I don't see the connection between the formula or equation you are seeking and the process of subjecting a deck in quasi-new deck order to faro shuffles. Possibly just an exercise to promote understanding of ways to express the permutations of orders of cards?

As I've stated in other threads, you could take the set of "rules" used in certain algorithmic stacks like Doug Dyment's Quick Stack and render it into a formula or mathematical equation. However, the resulting equation would be complex and full of conditional terms making it harder to apply and use than the rules themselves. My experience is that the learning of very complex equations is best done through mnemonics. But that's what you want to avoid.

Best wishes, but I think you are on a quest for a Holy Grail which does not exist. I certainly would be pleased if I turn out to be wrong. And, there is certainly no harm in the search.

Dennis Loomis

Lol, yeah, I am sadly coming to that realization Mr Loomis.

Mind you, I am now contemplating working backwards, lol. That is, working out what un-faroing the deck does, so I can look at a five faro as if it were only three, making it, hopeful, a simpler formula then the four faro one.

Who knows, in the mean time I may just work on a straight memorization, but I will continue my search.

Most folks find that memorization and mnemonics are not as difficult as they expected once they jump in and try. In fact, many people find the use of menmonics to be fun. And, you have one big advantage over trying to find that simple formula: if you ever did, the stack would not have any planned built in effects. If you choose to learn the Aronson Stack, (my choice) or the Tamariz Menmonica Stack, there are already many effects worked out for them.

Dennis Loomis

Well, I have sort of done it.

I have discovered a set of "rules" that work with the 4 Faro deck. They are simpler than most algorithmic decks out there, and have only one single exception, which is quite obvious when ever it needs to be applied.

Basically, you only need to remember 13 numbers, and from there you can calculate ANY card's position in the deck.

If you can remember 13 numbers, you can remember 52 cards. If you remember 52 cards, you don't need any calculation. That speed is the essence of Memorised stack.

If you prefer to remember 13 things, how about remember 13 cards in Quadruple Cyclic Stack. A Quadruple Cyclic Stack not in CHASED order looks very random.

Hideo Kato

Kato-san got a very good point. Thanks for that

I've simplified it even more, now you really only need to remember the first three umbers, then the rest can all be derived from there.

Gdw,
I'd love to hear the details.

Dennis Loomis

Basically, you really only need to remember the positions of the first three, and from there you can extrapolate the position of all the other cards.

You can calculate position of any card knowing bottom card of Si Stebbins Stack. Can you caluculate more quickly than Si Stebbins method. If so (and the order looks seemingly random), you have invented a great Stack.

Hideo Kato

Well, the order is very random looking, but I really invented nothing as it is just a four faro stack. I honestly cannot remember the calculations for the Si Stebbins work, lol. I used to know it, but I never ended up using it. I'll have to look them up again before I can answer the relative speed question.

Also, I have discovered that this works for a five faro stack as well. The numbers are different but the basic concept still applies.

I am not 100% sure if this would be completely efficient yet. For me it is great just to know that it can be done, lol. I imagine that, at least for me, it would work more as another step in the process of eventually having the stack memorized. Again, as I mentioned, perhaps in another thread, mnemonics don't click with me, but math and logic does. This is why I am searching (or perhaps WAS searching) for a stack that could be calculated like this.

Although you can start with only the first few positions and calculate from there, I imagine that through continuous use, all thirteen would eventually become second nature and thus cut the calculations down immensely. Luckily it is all just basic addition any ways, so nothing too complicated. As of now, with knowing the first three positions, it is basically two calculations, both addition, to know any card.

The other benefit of this one, over Si, for example, is the in and out of numerical order application. I have a few poker routines that I do now based off of "new" deck order (for me, straight numerical order) so having one deck that can easily serve the purpose of two is a nice benefit.

Also, I am still working on this just to see if it is indeed even worth while.

When I have got things down a bit more I will let everyone know.

Again, in comparison to Si Stebbins, I also feel the far more random appearance is also a huge benefit.

Thanks
Glenn

You need to read Mnemonica. It's all in there.

I'm pretty sure Marlo has formulas worked out for this.

I do not know how I missed this thread earlier. With a five faro stack, the order is basically every eighth card in the original stack: (ace of spades and ace of hearts are exceptions). So it goes AH, 9H, 4S...(multiples of eight).

All of the cards in the stack are eight apart from the NDO (or original deck order), which is why if you give a ND five faros, then the nines are adjacent to the aces, the kings next to the fives, and jacks next to threes, and tens next to deuces. The cycle starts over again by the time you get to the eighth card (six of hearts).

1 - 1 (AH)
2 - 9 (9H)
3 - 17
4 - 25
5 - 33
6 - 41
7 - 49
8 - 6 (6H)
9 - 14
10 - 22
11 - 30
12 - 38
13 - 46
14 - 3H
15 - 11 (JH)
16 - 19
17 - 27
18 - 35
19 - 43
20 - 51

(Obviously 1-13 are the hearts; Off the top of my head, 14 = AC, 27 = KD, 51 = 2S.)

It might be easier to decipher patterns using A-K (repeat) than the mirror in NDO.

Yes, you can find a complete instruction HOW you calculate one or several cards for perfect faros in Edward Marlos booklet FARO NOTES. The Faro Calculator and The Chain Calculator, very useful and extremely powerful tools for any faro worker.

Yo can also learn this useful calculation in the huge book:
Revolutionary Card Technique by Ed Marlo - Book
http://www.murphysmagic.com/Revolutionar......que-Book